Paid Quantitative Aptitude Crackers 2019

Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

1

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Chapter

Simplification, Approximation and Number Series

1

BEST APPROACH TO SOLVE THE QUESTIONS Simplification: Simplification simply means simplifying the complex expressions. There are ways available always which are simpler, easier and faster to calculate a particular complex expression. Questions related to simplification can be solved in variety of ways. But in examination where one needs to calculate as fast as possible, applying good simplification techniques may serve a good purpose. The most interesting about numbers is that they can be expressed in an innumerable number of ways and thus comes the concept of breaking the expression in such a way that one reaches the answer in less time. Following are some examples citing the ways to simplifying an equation or expression. 9

1.

2

+

11

+

3

17 6

=? +

⇒

27+22+17

⇒

66

6 6

=? +

12 5

+

16 10

24+16 10

+ 27× 33

+ (30-3)(30+3)

=? +4 + 900 − 9

⇒ ?=11+9-900-4= –884 2.

? × 65 ÷ 72 = 195 × 312 + 260 × 6 ⇒

?×65 72

=65× 6(3× 52 + 4 × 1)

⇒ ? = 72× 6 × 160 = 432 × 160 = 69120 2

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Approximation Questions based on approximation are usually easy to solve but turn out to be a bit difficult when approximate values are not clearly visible. In case where a value like 17.97 is given, it is easy to replace it with 18. But cases where values like

120 13

or

101 11

are given, then

approximate values can be used by making appropriate adjustments to other terms of the expression. Let’s, for example, try to find out the approximate value of expression given below: 66 50 + + 41% of 97 =? 16 17 The value of

66 16

is a bit more than 4 and that of

may take value of

66 16

and

50 17

50 17

is a bit less than 3. So we

to be equal to 4 and 3 respectively.

We know 40% of 100 is equal to 40. As 41 near to and greater than 40 and 97 is near to and smaller than 100, we may take the approximate value of 41% 𝑜𝑓 97 to be equal to 40 itself. This way, the approximate value of the above expression= 4+3+40=47 A few sample problems: 178 217 − + 33 × 41 =? −32% 𝑜𝑓 207 12 18 ⇒ 15 + 12 + 33(40+1) = ? – 33% of 200 ⇒ 27 + 1355 = ? – 66 ⇒ ? = 1450 19.95 4

⇒

+

20 4

68.12

+

17 68 17

− 12.5% 𝑜𝑓 127.9 =? 1

− × 128 =? 8

⇒ 5 + 4 − 16 =? 3

⇒ ? = −7 Adda247 Publications

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Number Series When different numbers are arranged in any mathematical and logical pattern, a number series is said to be formed. There may be infinite such patterns. So it’s really difficult to find out one out of them. But there is one element that gives us the idea of any pattern. That element is the difference of the terms of a number series. By finding the difference of consecutive terms of a series we can make out the pattern on which the number series is based. Some patterns are clearly visible but others can be found out by cracking the pattern of the differences. A few examples follow: 1.

45,

2.

3,

3.

10,

4

44,

9,

48,

23,

12,

39,

97,

15,

55,

479,

22,

38,

30,

2881,

70,

66

20159

127

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Practice Exercise Based on new Pattern

Directions (1-15): What will come in place of the question mark (?) in the following questions ? 1.

(4 × 4)3 ÷ (512 ÷ 8)4 × (32 × 8)4 = (2 × 2)?+4 (a) 8 (b) 12 (c) 6 (d) 14 (e) None of these

2.

(2√392 − 21) + (√8 − 7) = (? )2 (a) 4 (b) –4 (d) 2 (e) 6

2

1

3.

1

1

1

6

8

12

1 + 1 − 1 =? +1 4

(a) (d) 3

4.

19

5 24 7 12

7

(c)

24

5 12

(e) None of these

of 30% of 3420 = (? )2 × 2

(a) (81)2 (d) 81 5.

(b)

(c) 12

(b) 7 (e) 49

(c) 9

65% of √3136 × 5 =? +154 (a) 56 (b) 28 (d) 32 (e) None of these

(c) 35

1

29% of 193 – 20% of 204.85 = (3375)3 − (? )2 (a) 1 (b) 4 (d) 64 (e) None of these

6.

5

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1

3

1

1

7

2

7

1

3 ÷ 6 × 1 × 3 = (? )2

7.

3

(a)

22

(d) 8.

(b)

3 484

484 3

22 9

(e) None of these

81

60% of 68.05 × 15% of 8 × 1.02 = ? (a) 49.97592 (b) 49.73482 (d) 41.84822 (e) None of these (112.6 ×114.4 )

9.

(c)

117.5 ÷115 ×√11

(c) 54.77442

= (? )2

(a)√11 (d) 1331

(b) 11 (e) 14641

(c) 121

1

10. √3481 ÷ 7 × (441)2 = 12.5% of 1200 + ? (a) 3 (b) 9 (d) 81 (e) None of these 11. 3

19 27

(c) 27

1

% of 675 + 25 % of 184 = x 4

(a) 70.64 (d) 70.46

(b) 71.64 (e) None of these

(c) 71.46

2

12. [2.5 ÷ 0.1)3 ÷ (6.25)2 ] × [(1.25)2 ÷ (5)2 ] = (25)3+x (a) 1

(b) 2/3

(d) 2

(e) None of these

161

13.

377

6

×

1595 253

÷

7 13

(c) 1/3

1

= (x)3

(a) 343

(b) 216

(d) 125

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2

14.

3

4

7

5

11

of of

of

2145 1288

(a) 1 (d) 4 √8649

15.

√961

×

3

√15625 3

√1728

(a) 0.5 (d) 3.5

169

÷√

529

=x (b) 0.5 (e) None of these

(c) 2

(b) 1.5 (e) None of these

(c) 2.5

= x2

APPROXIMATION Directions (1-20): What approximate value will come in place of (x or ?) in the following questions ? 1.

294.01 ×

𝑥 8.01

(a) 10 (d) 5 2.

55.2 ×

− 19.99% 𝑜𝑓 119.99𝑥 = 254.9 ÷ 1.9 (b) 16 (e) 12

410 69

(a) 17.5 (d) 15

(c) 8

+ 499.9 – 127.9 =? % of 5600 (b) 12.5 (e) 8

(c) 10

3.

630 × ? + 599.85 – 55% of 12000 = 37 ½ % of 9200 (a) 12 (b) 18 (c) 15 (d) 20 (e) 8

4.

11999.87 ?

7

1

+ 54.9% of 1800 – 389.9 = 11 % of 9900 9

(a) 24

(b) 28

(d) 18

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5.

8099.9 ×

31 27

(a) 22500 (d) 18200

5

1

3

2

+ 2699.8 × – 1799.8 = 62 % of ? (b) 15200 (e) 19200

(c) 16200

6.

119.89% of 2000.23 + (35.96)² – 59.49% of 2999.89 = (49.89)² – ? (a) 559 (b) 569 (c) 589 (d) 549 (e) 539

7.

(11.87)² + 12.493 × 15.89 – √13224.98 – (?)² = (14.96)² (a) 9 (b) 2 (c) 8 (d) 5 (e) 10

8.

359.93 ?

= (8.89)3 – 14.5 × 39.89 + (1.95)2 – 34

(a) 8 (d) 9

(b) 2 (e) 5

(c) 3

√429.87 + 520.23 + √120.97 = (? )2 + √35.98 (a) 2 (b) 3 (c) 8 (d) 5 (e) 9

9.

10. (23.89)² + √3598.97 – 171.93 + (?)² = (27.98)² + 4 (a) 18 (b) 24 (c) 28 (d) 22 (e) 26 11. (13.97)² – 39.87% of 239.97 + (29.87)² + 330.97 = (?)³ (a) 18 (b) 19 (c) 17 (d) 11 (e) 15 ?+134.5

12.

24

+ 209.87 + (69.87)² – 1999.83 = (54.87)² + 99.85

(a) 235.5 (d) 215.5 8

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13. ? × 12.97 + (24.97)² – (19.89)² + 363.83 = (27.93)² (a) 10 (b) 20 (c) 18 (d) 12 (e) 15 593.89

14.

?

+ 14.87 × 35.88 +

(a) 12 (d) 9

1259.81 17.93

= (25.89)2

(b) 15 (e) 11

(c) 13

15. 48.01% of 249.99 + 𝑥% 499.99 = (11.01)2 + (2.11)2 (a) 2 (b) 3 (c) 5 (d) 1 (e) 6 2

16. 49.11 × ÷ 14.09 + √𝑥 = 1025 ÷ 257 (a) 9 (d) 16 3

17.

4

7

(b) 8 (e) 25

(c) 27

× (15.99) + 31.9% 𝑜𝑓 3199 + 294.9 = 𝑥 3

(a) 10 (d) 15

(b) 11 (e) 8

(c) 12

18. ? % of 6519.97 + (21.97)2 – 29.98% 𝑜𝑓 9139.93 = (25.91)2 (a) 25 (b) 15 (c) 30 (d) 45 (e) 40 19.

577.93+? (15.97)

+ 61.87% of 1449.87= –√2024.87 + 13.89 % 𝑜𝑓 7099.97

(a) 222 (d) 200

(b) 282 (e) 308

(c) 262

20. 32.87% of ? – 17.87 × 44.86 – √399.81 = (24.98)2 + 29.89 (a) 4450 (b) 4500 (c) 4350 (d) 4600 (e) 4900 9

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NUMBER SERIES Wrong Series Directions (1-15): Find the wrong number in the given series that does not follow the pattern? 1.

480,

2.

960,

320,

(b) 272

(d) 320

(e) 1536

4, 3, (a) 60 (d) 14

4.

9, 63, (a) 25 (d) 512

5.

6.

10

272 ,

(a) 960

210, 197, (a) 197 (d) 171

3.

1280,

5,

171,

14,

25,

135,

60,

216,

1536 (c) 1280

80, 15 (b) 15 (e) 135

(c) 80

528 (b) 4 (e) 528 49,

(c) 5

512 (b) 216 (e) 49

4498, 4888, (a) 5846 (d) 7070

5336, 5846, 6422, (b) 4498 (e) 6422

122, 125.2, (a) 125.2 (d) 170

131.6,

141.2, 152, (b) 131.6 (e) 152

(c) 63

7070 (c) 5336

170

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7.

3671, 2341, 4537, (a) 6072 (d) 3671

8.

15, 91, (a) 457 (d) 15

457,

10, 17, (a) 220 (d) 108

45,

10. 10, 30, (a) 10 (d) 3150

90,

9.

1831,

108,

450,

11. 325, 546, (a) 936 (d) 1078

754,

12. 192, 202, (a) 210 (d) 192

210,

13. 3, 11, (a) 1135 (d) 49 14. 23, 30, (a) 42 (d) 200 15. 8, 30, (a) 30 (d) 8 11

49,

42,

60,

1163, 6072, (b) 2341 (e) 1163

120,

5497, 10997 (b) 91 (e) 10997

(c) 5497

(c) 10

3150, 34650 (b) 34650 (e) 30

(c) 3150

216,

63,

(c) − 783

220, 390 (b) 390 (e) 45

936,

191,

− 783

1078, 1170 (b) 546 (e) 1170 225, 243, (b) 202 (e) 243 569, 1135, (b) 1134 (e) 11

(c) 325 288 (c) 288 1134 (c) 3

95, 140, (b) 140 (e) 23

200

180, 180, (b) 120 (e) 60

90

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MISSING SERIES Directions (1-15): What should come in place of the question mark (?) in the following series ? 1.

2.

3.

10, 15, 45, (a) 72250 (d) 85280

270,

12, 26, (a) 9981 (d) 9876

328,

81,

1024, 512, (a) 64 (d) 216

4.

193, 97, (a) 15 (d) 13

5.

17, 24, (a) 30 (d) 28

6.

7.

12

?,

49,

13,

3240, ? (b) 77760 (e) 762560

16,

25,

26,

?,

9,

(c) 75260

1645, ? (b) 8281 (e) 9978

(c) 7321

1, 0.03125 (b) 128 (e) 324

(c) 256

7 (b) 12 (e) 11

(c) 17

? (b) 32 (e) 26

(c) 29

18, 6, 2, (a) −36 (d) −32

−2 , − 10, ? (b) – 48 (e) − 52

56, 1144, (a) 4556 (d) 4296

2104,

2944, 3672, (b) 4456 (e) 4552

(c) − 40

?

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8.

9.

21, 1750, (a) 3556 (d) 3544

2752,

9, 15, 23, (a) 57 (d) 51

33,

10. 141, 197, 269, (a) 607 (d) 603 11. 25, 33, (a) 175 (d) 190 12. ? , 5, (a) 2 (d) 5 13. ?, 10, (a) 6 (d) 9

46,

12,

14. 2160, ? , (a) 1080 (d) 810 15. 2,

13

35,

69,

46,

360,

156,

45,

359,

39,

20,

3267, 3487, (b) 3456 (e) 3558

?

? (b) 59 (e) 49

(c) 53

469, ? (b) 609 (e) 605

(c) 601

112, ? (b) 180 (e) 195 160,

96,

180,

477,

?,

(c) 3536

(c) 185

805 (b) 4 (e) 8

(c) 3

178 (b) 7 (e) 5

(c) 8

60, 30 (b) 720 (e) 1440

(c) 900

959

(a) 858

(b) 564

(d) 727

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Solutions

SIMPLIFICATION (c); (4 × 4)3 ÷ (512 ÷ 8)4 × (32 × 8)4 = (2 × 2)?+4

1.

or, or,

3

(42 ) ×(44 )

4

= (4)?+4

(43 )4 6 4 ×416 412 10

= (4)?+4

or, 4 = 4?+4 or, ? = 6 2.

(e); ?2 = 2√14 × 14 × 2 − 21 + 8 + 49 − 28√2 = 28√2 − 21 + 57 − 28√2 = 36 = 62 ∴?=6

3.

(a); 1 + + 1 + − 1 − =? +1

1

1

4

1

6

1

1

8

1

1 12

1

⇒? = 1 + + 1 + − 1 − − 1 − 4 6 8 12 1 1 1 1 = + − − 4 6 8 12 6+4−3−2 5 = = 24

4.

(c); 3420 ×

24

30 100

×

3

19 2 )

= (? )2 × 2

⇒ 162 = (? × 2 162 ⇒ (? )2 = = 81 2

∴ ? = √81 = 9 65

5.

(b); √3136 × × 5 = ? +154 100 ⇒ 56 ×

65

100

× 5 =? +154

⇒ 182 = ? +154 ⇒ ? = 182 − 154 = 28 14

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6.

(e); (55.97 – 40.97) = 15 – (?)2 (?)2 = 0 ?=0

7.

(d);

10 3

7

×

45

1 2

(? ) = 484

?=

3

22

2

7

× ×

1

= (? )2

22 9

81

8.

(a); 40.83 × 1.02 ×1.2 = ? ? = 49.97592

9.

(c);

117

= (? )2 (? )2 = 14641 ? = 121 113

10. (c); 59 × 3 = 150 + ? ? = 27 11. (c);

100 27 675 27

% of 675 +

101

of 184

100×4

+ 46.46

25 + 46.46 = 71.46 12. (c); [(25 [ [

253 254 1 25

)3

÷ (6.25

)2 ]

× 1002 ] × [

×

100×100

×

100×100 2 +𝑥 3

2

× ((1.25) ÷ 5

1252 1002

×

125×125 5×5

1 52

2)

2 +𝑥 3

= (25)

2 +𝑥 3

] = (25) 2

] = (25)3+𝑥

25 = (25) 2

⇒ +𝑥 = 1 ⇒𝑥 = 1− 3

𝑥= 15

1

2 3

3

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

13. (d);

161 377

×

1595 253

×

13 7

1

= (𝑥 )3

Explanation 161 = 23 × 7 377 = 29 × 13 1595 = 29 × 55 253 = 23 × 11 23×7 29×13

29×55

×

23×11

×

13

1 3

7

1

= (𝑥 )3

5 = (𝑥 ) ⇒ 𝑥 = 125 14. (a);

2

3 13 23

15. (c);

4

7

× ×

5 11 23

×

13

√8649

31

×

12

2145 1288

529

×√

169

=1 3

√15625

×

√961 93 25

×

3

√1728

= 𝑥2

= 6.25 = 𝑥 2

𝑥 = 2.5 APPROXIMATION 1.

𝑥

1

255

8

5

2

(a); ≈ 294 × − × 120𝑥 =

≈ 36.75𝑥 − 24𝑥 = 127.5 ≈ 12.75𝑥 = 127.5 ≈ 𝑥 = 10 2.

(b);

16

X

× 5600 = 8 × 41 + 500 – 128 700 X= 56 X = 12.5 100

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3.

(c); 630X + 600 = 37.5 ×

9200 100

+

55 100

× 12000

630X + 600 = 3450 + 6600 630X = 10050 – 600 9450 X= 360 X = 15 4.

(a); =

5.

(e);

12000

55

+ × 1800 = X 100 12000 = = 1490 – 990 X 12000 X= 500 X = 24 62.5X 100

= 8100 ×

31 27

100 9

(c);

120 100

1 100

× 9900 + 390

5

+ 2700 × – 1800

5x = 9300 + 4500 – 1800 8 12000 × 8 X= 5 X = 19200 6.

×

× 2000 + (36)2 –

59.5 100

3

× 3000 = (50)2 – ?

2400 + 1296 – 1785 = 2500 – ? 2500 – 1911 = ? ? = 589 7.

(b); (12)² + 12.5 × 16 – √13225 – (?)² = (15)² 144 + 200 – 115 – (?)² = 225 (?)² = 229 – 225 ?=2 17

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8.

(e);

360 ? 360 ?

?=

= (9)3 – 14.5 × 40 + (2)2 – 81 = 729 – 580 + 4 – 81 360 72

?=5 (d); √430 + 520 + √121 = (? )2 + √36 √430 + 520 + 11 = (? )2 + 6 31 – 6 = (?)² ?=5

9.

10. (a); (23.89)² + √3600 – 172 + (?)² = (28)² + 4 576 + 60 – 172 + (?)² = 784 + 4 464 + (?)² = 788 (?)² = 788 – 464 ? = 18 11. (d); (14)² – 40% of 240 + (30)² + 331 = (?)³ 196 –

40×240 100

+ 900 + 331 = (? )3

1000 + 331 = (?)³ ? = 11 12. (c);

?+134.5 24 ?+134.5 24

+ 210 + (70)2 – 2000 = (55)2 + 100 + 3110 = 3125

? + 134.5 = 15 × 24 ? = 360 – 134.5 ? = 225.5 18

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13. (e); ? × 13 + (25)² – (20)² + 364 = (28)² ? × 13 + 225 + 364 = 784 ? × 13 = 784 – 589 ?=

195 13

? = 15

14. (d);

594 ? 594 ?

+ 15 × 36 +

1260 18

= (26)2

+ 540 + 70 = 676

594 = ? (676 – 610) ?=

594 66

?=9 15. (d);

48 100

× 250 +

𝑥 100

× 500 ≈ 112 + 22

120 + 5𝑥 ≈ 121 + 4 5𝑥 ≈ 5 𝑥≈1 2

1

7

14

16. (a); 49 × ×

+ √𝑥 ≈

1024 256

1 + √𝑥 ≈ 4 √𝑥 ≈ 3 𝑥≈9 17. (b);

3 4

× 16 +

32 100

× 3200 + 295 ≈ 𝑥 3

12 + 1024 + 295 ≈ 𝑥 3 𝑥 3 ≈ 1331 𝑥 ≈ 11 19

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18. (d);

? 100 ? 100

?=

× 6520 + (22)2 –

30 100

× 9140 = (26)2

× 6520 = 676 + 2741 – 484 2934×100 6520

? = 45 19. (a);

578+? 16 578+? 16 578+? 16

+

62 100

× 1450 =– √2025 +

14 100

× 7100

+ 899 =– 45 + 994 = 949 – 899

578 + ? = 50 × 16 ? = 800 – 578 ? = 222 20. (b);

33 100

?=

×? = 625 + 30 + 810 + 20 1485×100 33

? = 4500 WRONG SERIES 1. (b);

So wrong number is 272 2.

(e);

So wrong number is 135 20

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3.

(a);

So wrong number is 60 4.

(c);

So wrong number is 63 5.

(d); Wrong number = 7070

5.

(e); Wrong number = 152

6.

(a); Wrong number = 6072 Pattern of series —

21

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8.

(e);

So wrong number is 10997 9.

(b);

So wrong number is 390 10. (a);

So wrong number is 10 11. (d);

So wrong number is 1078 12. (b);

So wrong number is 202 22

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13. (b);

So wrong number is 1134 14. (e);

So wrong number is 23 15. (a);

So wrong number is 30 MISSING SERIES 1.

(b); Series is ×1.5, ×3, ×6, ×12, ×24 So, 3240 × 24 = 77760

2.

(d); Series is ×2+2, ×3+3, ×4+4, ×5+5, ×6+6 So, 1645 × 6 + 6 = 9876

3.

(b); Series is ÷2, ÷4, ÷8, ÷16, ÷32 So, ? = 512 ÷ 4 = 128

4.

(d); −96, −48, − 24, − 12, So, 25 – 12 = 13 23

−6

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5.

(d); Series +7, −11, +13, −17, +19 So, 9 + 19 = 28

6.

(d); Pattern of series – 18 × 0.5 − 3 = 6 6×1−4=2 2 × 1.5 − 5 = −2 − 2 × 2 − 6 = − 10 ? = − 10 × 2.5 −7 = − 32

7.

(d);

8.

(a); 21+ (123 + 1) = 1750 1750 + (103 + 2) = 2752 2752 + (83 + 3)= 3267 3267 + (63 + 4) = 3487 ? = 3487 + (43 + 5) = 3556

9.

(b);

24

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10. (c); 141 + 7 × 8 = 197 197 + 8 × 9 = 269 269 + 9 × 10 = 359 359 + 10 × 11 = 469 ? = 469 + 11 × 12 = 601 11. (e);

12. (b);

13. (c);

14. (a);

15. (c); (2 + 5) × 5 = 35 (35 + 4) × 4 = 156 (156 + 3) × 3 = 477 (477 + 2) × 2 = 958 (958 + 1) × 1 = 959

25

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1

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Chapter

2

Ratio Proportion and Mixture & Alligation

BEST APPROACH TO SOLVE THE QUESTIONS Alligation is a method used to solve problems concerned with weightage average. Or, in other words, alligation is just a way of writing equations of weightage average for the easy calculation. Problems involving mixtures or any form of weightage average can be solved easily using the method of alligation. Questions related to replacement, addition or omission of components of mixture may be easily solved using fractions. Be it in form of ratio or by taking volume into consideration. A very useful formula for problems based on mixtures is given below: Final ratio= Initial ratio (fraction of component remained) Final volume= initial volume (fraction of component remained) A few sample problems explaining the concept of alligation are given below. Example 1: ‘An alloy containing 50% gold is formed by melting two metals 2

A and B. Metal A contains 66 % gold while metal B contains 40% gold. Find 3

the ratio in which metal A and B are mixed to form the alloy.’ Solution: Let’s take the fractional equivalents of given percentages and calculate using alligation. 2

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Hence, required ratio is 3:5. Example 2: ‘Average weight of boys in a class is 55kg while average age of girls in the same class is 35. If the average of all the students in the class is 40 kg, find the ratio of boys and girls in the class.’ Solutions:

Required ratio is 1:3. Example 3: ‘A can contains mixture of two liquids A and B in the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?’ Solution: Since B is filled by replacing some amount of mixture, we should take A into consideration Initial ratio of A to the whole mixture = 7/12 7 Final ratio of A to the whole mixture = Using formula, 3

7 16

=

7 12

×𝑦

Or, 𝑦 =

16 12 16

=

3 4

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This shows the mixture remained

3 4

th of itself which means

1th 4

part was

drawn off. 1

ATQ, of total = 9 4

Or, total mixture = 9 × 4 = 36 litres Initial amount of liquid A =

7 12

× 36 = 21 litres

Practice Exercise Based on new Pattern

1.

150 kg of alloy containing copper and zinc in the ratio 3 : 2 mixed with ‘X’ kg of alloy containing copper and zinc in the ratio 2 : 3. If the overall alloy should contain copper between 45% to 55%, what can be minimum value of X? (a) 450 kg (b) 100 kg (c) 50 kg (d) Cannot be determined (e) None of these

2.

Three different liquids which have 10% water, 20% water and x% of water are mixed in the ratio of their quantity 2 : 3 : 4 respectively. If 12% of water is present in final mixture. Calculate value of x. (a) 9% (b) 20% (c) 7% (d) 15% (e) 17%

3.

‘x’ liters of a 30% alcohol solution is mixed with 40 liters of 60% alcohol solution & a resultant of 50% alcohol solution is formed. Now ‘3x’ liters of y% alcohol solution is added to 30 liters of 50% alcohol solution which resulted in 45% alcohol solution. The ratio of y : x is

4

(a) 17 : 6

(b) 16 : 15

(d) 14 : 5

(e) 17 : 8 Adda247 Publications

(c) 7 : 15

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4.

There are three varities of sugar with their quantity in the ratio of 3 : 4 : 5. If 9 kg of first variety and 4 kg of second variety are added to their respective quantity and x kg of 3rd type is removed from it, then final ratio becomes 9 : 10 : 10. Find the sum of initial quantities of these varieties. (a) 120 kg (b) 96 kg (c) 84 kg (d) 108 kg (e) None of these

5.

A jeweler mixed gold and copper in 2 proportion. In type ‘A’ alloy, 6 gm gold is mixed with 5 gm copper and in type ‘B’ alloy, 5 gm gold in mixed with 3 gm copper. If jeweler have 122 gm gold and 90 gm copper, then find the weight of type ‘B’ alloy. (a) 60 gm (b) 80 gm (c) 70 gm (d) 100 gm (e) 90 gm

6.

Two liquid containing (X and Y) mixture mixed, in the ratio 2 : 3. X contain 40% sugar in water, while Y contain alcohol in water. After adding 20 gram sugar in the final mixture ratio of Sugar, alcohol and water becomes 8 : 5 : 16. Given that all ratio is calculate on weight. Find the weight alcohol in mixture ‘Y’. (a) 25 grams (b) 50 grams (c) 60 grams (d) 80 grams (e) 75 grams

7.

Two vessels A and B of equal capacities contain mixtures of milk and water in the ratio 4:1 and 3:1, respectively. 25% of the mixture from A is taken out and added to B. After mixing it thoroughly, an equal amount is taken out from B and added back to A. The ratio of milk to water in vessel A after the second operation is

5

(a) 79: 21

(b) 83: 17

(d) 81: 19

(e) None of these Adda247 Publications

(c) 77: 23

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8.

A 280 ml mix real juice pack contains mixture of Mango juice and orange juice in the ratio of 9 : 5. If x ml juice taken out from pack and 20 ml of orange juice mixed in pack so the new ratio of Mango juice and orange juice become 7 : 5, then find what quantity of mango juice taken out from pack initially? (a) 48 ml (b) 42 ml (c) 54 ml (d) 60 ml (e) 64 ml

9.

There are two vessels A and B which contains mixture of sulphuric acid and nitrous oxide in the ratio of 7 : 2 and 3 : 4 respectively. Mixture of both vessels are mixed to obtain a mixture of 390 ml, in which quantity of nitrous oxide is 160 ml. Find ratio of quantity of mixture in vessel A quantity of mixture in vessel B? (a) 7 : 6 (b) 6 : 7 (c) 5 : 7 (d) 7 : 9 (e) 4 : 7

10. There is two vessels A and B. Vessel A & B contains the mixture of milk, mango juice and water in the ratio of 8 : 5 : 3 & 6 : 5 : 2 respectively. If both vessels are mixed respectively in the ratio of 16 : 13 into another vessel C then the total quantities of mango juice in vessel C will be 20 liters. Find the difference between the quantity of milk and water in vessel C? (a) 15 liters (b) 10 liters (c) 18 liters (d) 25 liters (e) 20 liters 11. Two container P and Q have mixture. In P three liquid A, B and C are mixed in a ratio of 2 : 1 : 1 and in Q two liquid B and C are mixed in ratio of 3 : 2. Mixture of both container P and Q poured into third container X which show the quantity ratio of A, B and C is 8 : 13 : 10. Find the quantity in Q is what % less or more than the quantity in P. (a) 3.25% (b) 12.5% (c) 6.25% (d) 11% (e) 25% 6

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12. In vessel A 80% is milk and rest is water. Vessel B also contain same 1

mixture in which milk is 62 % of total volume. 25ℓ of mixture taken 2

out from vessel A and poured in vessel B so that milk becomes 100% more than water in vessel B. If new quantity of mixture in vessel B is 16% less than initial quantity of mixture in vessel A then find initial quantity of milk in vessel A? (a) 100l (b) 125l (c) 115l (d) 105l (e) 112l Directions (13-14): Three milkman P, Q and R have mixture of milk and water in the quantity of (X + 24) liters, (X+54) liters and (X + 84) liters and milk and water in the ratio of 2 : 3, 8 : 7 and 13 : 5 respectively. 13.

If P sold 50 liters of his mixture and 8 liters of milk added in remaining mixture, new ratio of milk and water becomes 6 : 7 . What quantity of water should be added by Q and R, so new ratio of their mixture becomes 5 : 7 & 5 : 4 respectively ?(1 marks) (a) 56 liters& 40 liters (b) 54 liters& 42 liters (c) 50 liters& 30 liters (d) 60 liters& 50 liters (e) 64 liters& 30 liters

14. Q sold 40% of his mixture and R sold 50% of his mixture, remaining mixture of both became equal. If all three charge Rs. 2 for one liter of water and production cost of milk for P, Q and R in the ratio of 5 : 6 : 7 and total mixture, which all three milkman have cost Rs. 3644. Find production cost of milk per liter for P, Q and R respectively? (1 marks) (a) 5 Rs. 6 Rs & 7 Rs.

(b) 25 Rs. 30 Rs & 36 Rs.

(c) 10 Rs. 12 Rs & 16 Rs.

(d) 10 Rs. 12 Rs & 14 Rs.

(e)10 Rs. 14 Rs& 15s. 7

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15. Vessel A & B contains mixture of orange juice and mango juice in the ratio of 5 : 4 and 5 : 2 respectively. Some quantity of mixture taken out from vessel A and B and mixed in another vessel C, if vessel C contains mixture of orange juice and mango juice in the ratio of 5 : 3 and total quantity of mixture in vessel C is 96 l. then find the difference between mixture taken out from vessel A & B? (a) 16 l (b) 12 l (c) 8 l (d) 10 l (e) 18 l 16. Mixture of petrol and water in vessel A are in the ratio of 4:1 and in vessel B are in the ratio of 5:3. If 20 l of mixture taken out from vessel A and proud in vessel B so new ratio of petrol to water in vessel B becomes 19:10, then find initially quantity of petrol in vessel A, if new quantity of mixture in vessel B is 26 l more than initial quantity of mixture in vessel A? (a) 64l (b) 72l (c) 68l (d) 56l (e) 48L 17. Two container P and Q have mixture of different liquids. In P three liquid A, B and C is mixed in a ratio of 2 : 1 : 1 and in Q two liquid B and C is mixed in ratio of 3 : 2. Mixture of both container P and Q poured into third container X which show the quantity ratio of A, B and C is 8 : 13 : 10. Find the quantity in Q is what % less or more than the quantity in P. (a) 3.25% (b) 12.5% (c) 6.25% (d) 11% (e) 25% 18. Ratio of Vodka and Wine in vessel A is 5 : 3 and same mixture in vessel B in the ratio of 3 : 2, 16ℓ of mixture from vessel A taken out and poured in vessel B new ratio of Vodka to wine becomes 29 : 19. If new quantity of mixture in vessel B is equal to initial quantity of mixture in vessel A, then find quantity of Vodka after 16 ℓ of mixture has been taken out from vessel A ? (a) 50𝑙 (b) 48𝑙 (c) 54𝑙 (d) 80𝑙 (e) 84𝑙 8

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19. Sameer and Veer entered into partnership business with the capital of Rs. x and Rs. (x + 12000) respectively. After One year Divyaraj joined them with capital of Rs. (x + 8000). At the end of 2 nd year Sameer and Divyaraj withdraw their capital and Veer invested for one more year. If Sameer, Divyaraj and Veer gets profit in the ratio of 8 : 6 : 21 respectively. Find sum of capital invested by all three ? (a) 66000 Rs. (b) 54000 Rs. (c) 64000 Rs. (d) 68000 Rs. (e) 70000 Rs. 20. Wheat of x Rs./kg is mixed with wheat of y Rs./kg in the ratio 2 : 3. On 2

selling the mixture at 49 Rs./kg there is a gain of 16 %. If the quantity 3

in which they were mixed is reversed and selling price remains same, then gain percent becomes (a) 25 Rs./kg (d) 32 Rs./kg

550 19

%. What is the volume of ‘x’.

(b) 30 Rs./kg (e) 20 Rs./kg

(c) 28 Rs./kg

21. Three vessels A, B and C of same capacity are all full. Vessel A and B contain mixture of milk and water in ratios 4 : 1 and 7 : 3 while vessel C contains only pure milk. The contents of all the vessels are poured into a larger vessel D. After drawing 30 liters of the mixture from vessel D, the final quantity of water in Vessel D is 40 liters. Find the capacity of each vessel (a) 80 (b) 85 (c) 95 (d) 70 (e) 90 22. Two container P and Q have mixture. In P three liquid A, B and C is mixed in a ratio of 2 : 1 : 1 and in Q two liquid B and C is mixed in ratio of 3 : 2. Mixture of both container P and Q poured into third container X which show the quantity ratio of A, B and C is 8 : 13 : 10. Find the quantity in Q is what % less or more than the quantity in P. (a) 3.25% (b) 12.5% (c) 6.25% (d) 11% (e) 25% 9

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23. Vessel A contains (X + 24) 𝑙 of mixture of milk and water in the ratio of 7 : 8, while vessel B contains (X + 54) 𝑙 of same mixture of milk and 1

water in the ratio of 3 : 2. If 37 % of mixture from vessel A and 40% 2

of mixture from vessel B taken out and mixed in vessel C, then the remaining mixture in vessel B is 15𝑙 more than that of in vessel A. find quantity of milk in vessel C? (a) 49 litre (b) 57 litre (c) 55 litre (d) 53 litre (e) 51 litre 24. Three container (A, B and C) have capacity in the ratio of 6 : 8 : 7 respectively. These three container are pour with three type of liquid X, Y and Z in the ratio 2 : 3 : 1 (in A), 1 : 2 : 5 (in B) and 3 : 1 : 3 (in C) respectively. Equal quantity of mixture is taken out from these containers and poured in container D. Quantity of liquid Y is 1/8 liter more than quantity of liquid X in container D. Find out the quantity taken out from each container. (a) 24 (b) 35 (c) 21 (d) 20 (e) None of these 25.

A shopkeeper has mixtures of nitrous oxide and water in two vessels A and B. Vessel A and vessel B contains mixtures of nitrous oxide and water in the ratio of 7 : 2 & 5 : 3 respectively. The shopkeeper has taken out the mixtures from vessels A and B in the ratio of 9 : 8 and mixed it in vessel C. If shopkeeper sold 68 gm of mixture from vessel C on the cost price of nitrous oxide, which is Rs. 80 per gm, find profit of shopkeeper? 2

(a) 39 % 3

2

(d) 37 % 3

10

2

(b) 41 % 3

2

(c) 45 % 3

2

(e) 35 % 3

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26. Amount invested in business respectively by Rana, Nikhil and Yash, was in AP respectively, and Yash invested thrice amount as that of Rana. Rana was manager and Nikhil was a programmer, and they both get salary of 8% and 5% of total profit respectively. If Rana invested for a month, Nikhil for 8 months and Yash invested for 4 months and Nikhil got Rs.6000 more than both of them together, find their total profit. (a) Rs. 100000/(b) Rs. 150000/(c) Rs. 174000/(d) Rs. 180000/(e) Rs. 20000/27. There are three vessels A, B & C having Milk to water ratio as 3 : 5, 1 : 3 and 7 : y. After adding mixture of vessel B to A, the milk to water ratio becomes 1 : 2 in A . Now 60% of mixture is taken out from vessel A and mixture of vessel C is mixed with remaining mixture of vessel A which results in final Milk to Water ratio of 3 : 4. Find the value of y given that volume of B is 20 ltrs and volume of C is (7 + y) ltrs. (a) 1 (d) 3

(b) 4 (e) 2

(c) 5

28. There are three mixtures A, B and C. Mixture A contains 10% of water and B contains 15% of water. If A and B are mixed together than resultant mixture contains 12% water. while If B is mixed with C, which contains 22% water so the resultant mixture contains of 18% water. All these mixture are mixed in a jar and then 30 ml water is also added. Find total volume of water preset in jar, if initially mixture C contains 66 ml water.

11

(a) 144 ml

(b) 216 ml

(d) 196 ml

(e) 200 ml Adda247 Publications

(c) 184 ml

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29. There are two containers A and B filled with oil with different prices and their volumes are 140 litres and 60 litres respectively. Equal quantities are drawn from both A and B in such a manner that the oil drawn from A is poured into B and the oil drawn from B is poured into A. The price per litre becomes equal in both A and B. How much oil is drawn from each of A and B: (a) 40 litre (b) 80 litre (c) 30 litre (d) 21 litre (e) 42 litre 30. A shopkeeper has two types Wheat i.e Type A & Type B and cost price per kg of Type A wheat is 10 times of cost price per kg of type B wheat. Shopkeeper cheated a costumer by saying that he will give him 200 kg of Type A wheat but shopkeeper gives 200 kg of mixture of Type A & Type B wheat on 10% above the cost price per kg of Type A wheat. If shopkeeper made a total profit of 71

7 8

% in this transaction,

then find ratio between quantity of Type B wheat to Type A wheat in sold mixture? (a) 2 : 5 (b) 2 : 3 (c) 1 : 3 (d) 2 : 7 (e) 1: 5 31. Two farmers A & B have mixture of urea & Potassium in the quantity 1

of 120 kg and 80 kg respectively. The mixture of farmer A has 37 % 2

potassium and the mixture of farmer B has 40% potassium. If farmer A used 60% of his mixture & B used 50% of his mixture and both gives their remaining mixture to farmer C, who have already an equal quantity of urea & Potassium. If ratio of urea & Potassium in resulting mixture of farmer C is 7 : 5, find total initial quantity of mixture farmer C had? (a) 24 kg (b) 36 kg (c) 32 kg (d) 42 kg (e) 48 kg

12

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32. Vessel A, B and C contains mixture of milk and water. Ratio of the quantity of A, B and C is 25 : 20 : 18 respectively. In vessel A milk to water ratio is 2 : 3 and in vessel B water to milk ratio is 2 : 3. If vessel C contain milk to water ratio is 4 : 5 and all the mixture of 3 vessel poured to another vessel D then in vessel D quantity of water exceeds by 6 litres than the quantity of milk in that vessel. Find out the quantity of mixture contained in vessel A, B and C. (a) 50 Ltr., 40 Ltr., 36 Ltr. (b) 25 Ltr., 20 Ltr., 18 Ltr. (c) 75 Ltr., 60 Ltr., 54 Ltr. (d) 100 Ltr., 80 Ltr., 72 Ltr. (e) None of these 33. In a party a tank is filled with some quantities of juice. Three vessels of different volume are used to serve the juice. Volume of the smallest vessel is 25% less than volume of second largest vessel and the 1 volume of second largest vessel is 33 % less than volume of the 3

largest vessel. If 5 times the juice is served by largest vessel, 8 times 1 juice is served by second largest vessel and finally when 3 times 3

juice is served by the smallest vessel, the tank gets emptied completely. Find the total quantity of juice served by second largest vessel is what percent of volume of the tank? 4 4 4 (a) 42 % (b) 44 % (c) 48 % 9 4

(d) 40 % 9

9 4

9

(e) 36 % 9

34. There are three fruit sellers Ramesh, Suresh and Vikash, all of them have a mixture of two types of apple. Ramesh has (x+18) kg of apple, Suresh has 12 kg of apple more than that of Ramesh while Vikash has 3 kg of apple more than that of Suresh. The two types of Apple are mixed in the ratio of 5 : 3, 2 : 3 and 4 : 3 for them respectively. If Ramesh sold 16 kg of mixed apple and added 12 kg of second type of apple then the new ratio becomes 5 : 6 for him. Vikash added y kg of second type of apple to his mixture and thus the new ratio of both types for him is 1 : 1. Then find the value of y. (a) 8 kg (b) 9 kg (c) 10 kg (d) 6 kg (e) 7 kg 13

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35. There are two vessels, vessel A and vessel B. Vessel A contains milk and water in the ratio 5 : 4 and vessel B contains Y litre of wine. 18 litre of mixture from vessel A is taken out and poured in vessel B. Now, 10 litre of mixture from vessel B is taken out and poured in vessel A. If quantity of wine in vessel A is found to be 4 litre then, find the initial quantity of wine in vessel B. (a) 8 L (b) 9 L (c) 10 L (d) 12 L (e) 6 L Solutions

1.

(c); If overall alloy contain copper as 45% 3

2x

45

5 2x

5

100

(150 × + ) = (90 +

5

)=

9 20

(150 + x)

(150 + x)

1800 + 8x = 1350 + 9x x = 450 kg If overall alloy contain copper as 55% (90 +

2x 5

)=

55 100

(150 + x)

1800 + 8x = 1650 + 11x 3x = 150 ⇒ x = 50 kg Minimum value of X is 50 kg 2.

(c); Let the quantity of three liquids is 200a, 300a and 400a 10% of water in first type means 20a water 20% of water in second type means 60a water x% of water in third type means 4xa water ∴ ATQ, 20a+60a+4xa 900a

=

12 100

⇒ 4xa = 108a – 80a 14

⇒ x=

28a 4a

=7

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3.

(e); From statement I 30

60

x×100+40×100 (x+40)

=

1 2

⇒

(2400+30x) 100(x+40)

=

1 2

⇒ x = 20 Now from statement II y

50

(3x×100+30×100) 3x+30

=

45 100

Here 3X = 3 × 20 = 60 litres ⇒

(

60y )+15 100

90

=

45 100

⇒

60y 100

=

81 2

– 15

⇒ y = 42.5 ∴ y ∶ x = 42.5 : 20 = 17 : 8 4.

(d); Let the initial quantity is 3y, 4y & 5y of these varities According to condition 3y + 9 ∶ 4y + 4 ∶ 5y – x = 9 ∶ 10 ∶ 10 From 1st 2 ratios ⇒

3y+9 4y+4

=

9 10

⇒ y = 9, Hence sum of initial quantities is (3 + 4 + 5) × 9 = 108 kg NOTE: No need calculate Value of x 5.

(b); Let gold and copper in Type A alloy be 6a and 5a Let gold and copper in type B alloy 5b and 3b ⇒ 6a + 5b = 122 … (i) 5a + 3b = 90 … (ii) Solving equation (i) and (ii) we get a = 12 ⇒ b = 10 Weight of type ‘B’ alloy = (5 + 3) × 10 = 80 gm 15

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6.

(a); Let weight of alcohol in Y = a gram Let weight of mixture X = 100x gram So, sugar weight in X = 40x gram Water weight in X = 60x gram So, weight of Y = 150x gram Weight of water in Y = (150x – a) gram ATQ, 40x+20 : a : 150x–a+60x = 8 : 5 : 16… (i) a 5 ⇒ = 210x–a 16 ⇒ a = 50x … (ii) Put value of ‘a’ in eqn. (i) 40x+20 8 40x+20 8 1 = ⇒ = ⇒ x= a

5

50x

5

1

2

weight of alcohol in Y = 50 × = 25 gram. 2

7.

(a); Assume there is 20litres of the mixture in both the vessels. In vessel A, milk = 16litres and water = 4litres If 25% mixture from A added to B, milk in B = 15 + 4 = 19litres water in B = 5 + 1 = 6litres Ratio = 19: 6 Equal amount from vessel B to vessel A 19 79 milk in A = 12 + = 5

6

5 21

= water in A = 3 + = 5 5 Hence, the ratio is 79: 21 8.

(c); Mango juice in mixture 9 = 280 × (9+5) = 180 ml

5

Orange juice in mixture = 280 × (9+5) = 100 ml ATQ— 9x

180–14

5x 100–14+20

16

=

7 5

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5(2520 – 9x) = 7(1400 – 5x + 280) 45x – 35x = 12600 – 11760 10x = 840 ⇒ x = 84 ml Quantity of mango juice taken out from pack initially 9 = 84 × = 54ml 14

9.

(b); Let total mixture of sulphuric acid and nitrous oxide in vessel A and B be P mℓ and Q mℓ respectively ATQ, P + Q = 390 … (i) 2P 4Q + = 160 9 7 14P + 36Q = 10080 … (ii) From 14 × (i) – (ii) Q = 210 mℓ P = 390 – 210 = 180 mℓ 180 Required ratio = =6:7 210

10

(c); Let quantity of mixture mixed in vessel C from vessel A & B be 16x and 13x respectively. ATQ— 5 5 16x × + 13x × (6+5+2) = 20 liters (8+5+3) 5x + 5x = 20 liters x = 2 liter Required difference = 8 6 3 2 [(16 × 2) × + (13 × 2) × ] – [(16 × 2) × + (13 × 2) × ] 16 13 16 13 = (16 + 12) – (6 + 4) = 18 liters

11. (c); Let in P, Quantity of A, B and C is 2x, x and x respectively In Q → B and C is 3y and 2y Now in final mixture → (X) A : B : C 2x : x+3y : x+2y = 8 : 13 : 10 Solving → x = 4 unit y = 3 unit Quantity in P → (2 + 1 + 1) × 4 = 16 unit Quantity in Q → (3 + 2) × 3 = 15unit 1 Required % = × 100 = 6.25% 16

17

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12. (a); Let milk and water in vessel A 4a and a respectively Let Milk and water in vessel B 5b and 3b respectively ATQ— 4

5b+25×5 1 3b+25×5

5b+20

=

2 1

2

= 3b+5 1 6b – 5b = 20 – 10 b = 10 New quantity of mixture in vessel B = (5 × 10 + 20) + (3 × 10 + 5) = 105ℓ Initial quantity of mixture in vessel A 100 = 105 × = 125ℓ 84 Quantity of milk in vessel A 4 = 125 × = 100ℓ 5

13. (b); In 50 liters sold mixture of P 2 Milk = 50 × = 20liters 5

3

Water = 50 × = 30 liters 5 ATQ 2(P+24) −20+8 5 3(P+24) −30 5

=

6

⇒

7

2P+48−60 3P+72−150

=

6 7

7P – 42 = 9P = 234 2P = 234 – 42 192 P= ⇒ P = 96 liters 2 Mixture, which Q have = (96 + 54) = 150 liter Mixture which R have = (96 + 84) = 180 liters For Q – 8(96+54) Milk = = 80 liters Water = 18

15 (96+54) 15

= 70 liters Adda247 Publications

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Lets Q added x liters of water 80 5 = 70+x 7 5x = 560 – 350 5x = 210 ⇒ x = 42 liters for R13(96+84) Milk = = 130 liters 18 5(96+84)

Water = = 50 liters 18 Lets R added y liters of water 130 5 = 50+y

4

5y = 520 – 250 270 y= ⇒ y= 54 liters 5

14. (d); ATQ(100−40) (100−50) (P + 54) × = (P + 84) × 100 100 60P + 3240 = 50P + 4200 10P = 960 ⇒ P = 96 liters Let production cost be Rs 5x, Rs 6x and Rs 7x ATQ_ (48 × 5x + 72 × 2) + (80 × 6x + 70 × 2) + (130 × 7x + 50 × 2) = 3644 240x + 480x + 910x = 3644 – (144 + 140 + 100) 1630x = 3260 x = 3260 ⇒ x = 2 Rs production cost for P = 5 ×2 = 10 Rs production cost for Q = 6 × 2 = 12 Rs production cost for R = 7 × 2 = 14 Rs 15. (b); Part of Orange juice in vessel A = Part of Orange juice in vessel B =

5 9 5 7 5

Part of Orange juice in vessel C = 8 By allegation —

19

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=9:7 (9–7) Required difference = 96 × = 12ℓ 16

16. (b); Lets petrol and water in vessel A be 4x and x Petrol and water in vessel B by 5y and 3y ATQ— 4

5y+20×5 1 3y+20× 5

=

19 10

⇒

5y+16 3y+4

=

19 10

50y + 160 = 57y + 76 7y = 84 ⇒ y = 12 New mixture in vessel B = (5 × 12 + 16) + (3 × 12 + 4) = 76 + 40 = 116 ℓ Initial quantity of vessel A = (116 – 26) = 90ℓ 4 Quantity of petrol in vessel A = 90 × 5 = 72 ℓ 17. (c); Let in P, Quantity of A, B and C is 2x, x and x respectively In Q → B and C is 3y and 2y Now in final mixture → (X) A : B : C 2x : x+3y : x+2y = 8 : 13 : 10 Saving → x = 4 unit ⇒ y = 3 unit Quantity in P → (2 + 1 + 1) × 4 = 16 unit Quantity in Q → (3 + 2) × 3 = 15unit 1 Required % = × 100 = 6.25% 16

20

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18. (a); Let ratio of Vodka and wine in vessel A be 5x and 3x and ratio of Vodka and wine in vessel B be 3y and 2y ATQ— 5 8 3 2y+16× 8

3y+16×

=

29

⇒

19

3y+10 2y+6

=

29 19

58y – 57y = 190 – 174 y = 16 New quantity of mixture in vessel B = (16 × 3 + 10) + (16 × 2 + 6) = 96 ℓ = initial quantity of mixture in vessel A Quantity of Vodka remaining in vessel A 5 16×5 = 96 × – = 60– 10 = 50ℓ 8

8

19. (d); Investment Ratio of Sameer : Divyaraj : Veer = x × 2 : Rs. (x + 8000) × 1 : Rs. (x + 12000) × 3 = 8 : 6 : 21 ATQ— 2x 8 = (x+8000) 6

6x – 4x = 32000 x = 16000 Rs. Required sum of capital (Sameer + Divyaraj + Veer) = 16000 + (16000 + 8000) + (16000 + 12000) = 68000 Rs. 6

20. (b); C.P. of mixture = 49 × = 42Rs/kg 2 3

=

7

y –42 42 –x

84 – 2x = 3y – 126 3y + 2x = 210 … (i) New cost price of mixture after reverting the quantity in which they are mixed. (1 +

11 38

)  new CP = 49

New CP = 49 × 21

38 49

= 38 Rs./kg

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So,

3

y –38

= 2 38 –x 114 – 3x = 2y – 76 190 = 2y + 3x Solving (i) and (ii) x = 30 Rs/kg

... (ii)

21. (e); Let the capacity of each vessel be 10x liters. Then, vessel A contains 8x liters of milk and 2x liters of water. Vessel B contains 7x liters of milk and 3x liters of water. Vessel C contains 10x liters of milk. Hence, Vessel D contains 25x liters of milk and 5x liters of water i.e. in the ratio 5 : 1. According to the question, 30 liters of the mixture is drawn from vessel D. Final quantity of water in Vessel D = 40 liters 1 ⟹ 5x − × 30 = 40 6 ⟹ 5x = 45 ⟹x=9 Capacity of each vessel = 10x = 90 liters 22. (c); Let in P, Quantity of A, B and C is 2x, x and x respectively In Q → B and C is 3y and 2y Now in final mixture → (X) A : B : C 2x : x+3y : x+2y = 8 : 13 : 10 Saving → x = 4 unit y = 3 unit Quantity in P → (2 + 1 + 1) × 4 = 16 unit Quantity in Q → (3 + 2) × 3 = 15unit 1 Required % = × 100 = 6.25% 16

22

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23. (b); ATQ, (100−37.5) (100−40) (x + 24) × + 15 = (x + 54) × 100

5

100

3

(x + 24) × + 15 = (x + 54) × 8 5 5(5x + 120 + 120) = 8(3x + 162) 25x – 24x = 1296 – 1200 x = 96 liters total mixture take out from vessel 3 A = (96 + 24) × = 45 litre 8 Total mixture taken out from vessel B 2 = (96 + 54) × = 96 liter 5

7

3

Milk in vessel C = 45 × + 60 × 15 5 = (21 + 36) = 57 litre 24. (c); Let P ltr. of mixture are taken out from all 3 containers. Quantity of X, Y and Z from A container 2P 3P P → , , 6 6 6 From B container P 2P 5P → , , 8 8 8 From C container 3P P 3P → , , 7

7

7

Quantity of Y → =

84P+42P+24P 168

=

Quantity of X → =

56P+21P+72P

Diff. =

168 150P 168

–

3P

2P

6 8 150P

+

P 7

168 2P P

3P

6 8 149P

7

=

149P 168

+

+ +

168 1

=

8

⇒ P = 21 liter

25. (b); In vessel A Nitrous oxide : Water = 7 : 2 In vessel B Nitrous oxide : Water = 5 : 3 In 68 gm of mixture 23

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Mixture taken from vessel A = 68 ×

9 17

= 36 gm

Mixture taken from vessel B = 68 ×

8 17

= 32 gm Total nitrous oxide in 68gm of mixture 7

5

9

8

= 36 × + 32 ×

= 28 + 20 = 48 gm

Total cost of 68 gm mixture = 68 × 80 = 5440 Rs Cost of nitrous oxide = 48 × 80 = 3840 Required profit =

5440−3840 3840

2

× 100 = 41 % 3

26. (a); Let Rana invested Y Rs. ∴ Yash invested 3Y Rs. And these are in AP. ∴ middle term is

Y+3Y 2

= 2Y.

2Y is amount invested by Nikhil. Let total profit be 100x. Then as a salary Rana got 8X. And Nikhil got 5X. Remaining 87X was divided according to share. Y × 1 : 2Y × 8 : 3Y × 4 Y : 16Y : 12Y 1 : 16 : 12 Out of 87X Nikhil got. 16 29

× 87X = 48X.

Total money Nikhil got is 53X Hence ATQ, 53X – 47X = 6000 ⇒ X = 1000 ∴ 100X = 100000/- Which is total profit. 24

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27. (b); ATQ, Volume of B is 20 liters 1 Milk (vessel B)= × 20 = 5 liters. 4

3

Water (vessel B) = × 20 = 15 liters. 4 When this is added to vessel A. 3x+5 1 = ⇒ 6x + 10 = 5x + 15 5x+15 2 x= 5 liters Hence, milk in vessel A = 3x + 5 = 3 × 5 + 5 = 20 liters Water in vessal A = 5x + 15 = 5 × 5 + 15 = 40 liters 60 60% of mixture of A = × (40 + 20) 100 = 36 liters. Remaining mixture = 24 liters. Milk remaining = 8 liters. Water remaining = 16 liters. After adding the content of Vessel C. 8+7 3 12 = ⇒ 60 = 48 + 3x ⇒ x = = 4 16+x

4

3

28. (b); ATQ –

Therefore, Ratio of mixture, A to B is 3:2. And ratio of mixture B to C is 4:3. Hence, ratio of volume of mixtures A, B & C is A:B:C=6:4:3 It is given that contains 22% water, which is 66 ml. Hence total volume of mixture C is 100 66 × = 300 ml 22 B is 400 ml A is 600 ml 25

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Water in B =

15 100 10

× 400 = 60 ml

Water in A = ×600=60 ml 100 Total water in jar after adding 30 ml water = 60 + 66 + 60 + 30 = 216 ml. 29. (e); Let the price of oil per litre in container 'A' be 'x' and that in 'B' be 'y' and the litres mixed be 'a' Initial price of A is 140x. When 'a' litres are removed,it should be 140x-ax. When 'a' litres from B are added,it is 140x-ax+ay , similarly for B is 60y-ay+ax. Now, the overall litres of both mixtures do not change as ‘a’ litres are taken away and are replaced, So price per litre after mixing for :1. Solution A :- (140x-ax+ay)/140 2. Solution B :- (60y-ay+ax)/60. Equating them, 8400(x − y) = 200a(x − y) Or, a = 8400/200 Or, a = 42 litres 30. (b); Let cost price per kg of Type B wheat = x Rs. And, Cost price per kg of type B wheat = 10x Rs. Selling price of mixture of Type A & Type B wheat = 11x Rs. Cost price per kg of mixture of Type A & Type B wheat = 11x ×

32 55

= 6.4x Rs.

ATQ – Let N kg of type B wheat in mixture of 200 kg of Type A & Type B of wheat 6.4x × 200 = N × x + 10x (200 − N) 1280 = N + 2000 – 10N 9N = 920 ⇒ N = 80 kg Required ratio = 26

80 120

=

2 3

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31. (c); Farmer A has ratio of urea & Potassium in mixture = 5 : 3 Farmer B has ratio of urea & Potassium in mixture = 3 : 2 Let farmer C have x kg of urea & x kg of potassium Farmer C has Urea in resulting mixture = 120 ×

40 100

5

50

8

100

× + 80 ×

3

× +x 5

= (54 + x ) kg Farmer C has Potassium in resulting mixture = 120 ×

40 100

3

50

8

100

× + 80 ×

2

× +x 5

= (34 + x) kg ATQ – (54 + X ) (34 + x)

=

7 5

270 + 5x = 238 + 7x 2x = 32 ⇒ x = 16 kg initial quantity of mixture farmer C had = 16 × 2 = 32 kg 32. (a); Let quantity of vessels A, B and C is 25x, 20x and 18x respectively Now in vessel A → Milk : Water 2 : 3 10x milk and 15x water In vessel B → Milk : Water 3 : 2 12x milk and 8x water In vessel C → Milk : Water 4 : 5 8x milk and 10x water ATQ→ 33x– 30x = 6 ⇒ 3x = 6 ⇒ x = 2 Quantity in A, B and C 50 liter, 40 liter and 36 liter

27

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33. (b); Let volume of largest vessel = 3x liter Volume of second largest vessel = 3x ×

2 3

= 2x liter Volume of smallest vessel = 2x ×

75 100

= 1.5x liter ATQ— Q = 5 × 3x + 8 × 2x +

10 3

× 1.5x

Q = 15x + 16x + 5x Q = 36x liter Required percentage =

16x 36x

4

× 100 = 44 % 9

34. (b); Ramesh has = (x + 18) kg Suresh = (x + 18) + 12 = (x + 30) kg Vikash = (x + 30) + 3 = (x + 33) kg For Ramesh, 5

First type of Apple = (x + 18) × kg 2nd type of Apple = (x + 18) ×

8 3

18

kg

ATQ, 5 8

5 8

(x+18)× −16×

3 3 (x+18)× −16× +12 8 8

⇒ (x + 18) × ⇒ (x + 18) ×

30 8 15 8

=

5

⇒

6

5 8 3 (x+18)× +6 8

(x+18)× −10

− 60 = (x + 18) ×

=

15 8

5 6

+ 30

= 90 ⇒ x = 30 kg

For Vikash, Total quantity = 63 kg 4

First type = 63 × = 36 kg 7

3

Second type = 63 × = 27 kg 7

ATQ, 36 27+y

28

=

1 1

⇒ y = 9 kg Adda247 Publications

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35. (d); Let vessel A has 5x L and 4x L of milk and water After first process ratio of wine, milk and water in vessel B will be 5

4

9

9

= Y ∶ 18 × ∶ 18 ×

= Y : 10 : 8

Now 10 litre of mixture from B is poured in A Then, wine will be = y y+18

y y+10+8

× 10

× 10 = 4

10y = 4y + 72 ⇒ 6y = 72 y = 12 litre

29

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1

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Chapter

Partnership

3

BEST APPROACH TO SOLVE THE QUESTIONS Any partnership in business is based on two components. One is the money or capital that is invested and the other is the time for which the money is invested. Mathematically, Profit ∝ investment Profit ∝ time Profit ∝ investment × time Profit share of an individual is directly proportional to investment and time. Hence, any profit share of a person out of total profit is decided by the net effect of these two components. If ratio of investment by two persons is X:Y and ratio of time for which they invested is A:B respectively, then their profit share is in the ratio of 𝑋 × 𝐴 ∶ 𝑌 × 𝐵= XA : YB Example 1: A, B and C entered into a business by investing their intial sum of Rs. 12600, Rs. 14400 and Rs. 13200 respectively. After 6 months C left the partnership and B also left the partnership after 8 months. If after a year total profit was Rs. 69600, then find profit share of C.’ Solution. Here, instead of taking absolute values of money invested, we may use their simplest values i.e. ratio. Ratio of investment of A, B and C is 21:24:22 respectively. And ratio of time for which they invested is 12: 8 : 6 = 6 : 4 : 3 Ratio of their profit share = (21× 6): (24 × 4): (22 × 3)= 21:16:11 Now, profit share of C= 2

11 48

× 69600 = 15950. Adda247 Publications

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

Example: ‘Aman started a business investing Rs. 70,000. Rakhi Joined him after six months with an amount of Rs. 1,05,000 and Sagar Joined them with Rs. 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Aman, Rakhi and Sagar respectively, 3 years after Aman started the business?’ Solution. Ratio of profit share is given by the ratio of (investment×time) Or ratio of their investment is 70000 : 105000 : 140000 = 2 : 3 : 4 Ratio of the time for which they invested = 36 : 30 : 24 = 6 : 5 : 4. Hence, ratio of profit share among them = (2× 6): (3 × 5) ∶ (4 × 4) = 12:15:16 Practice Exercise Based on New Pattern

1.

A and B started a business with a capital of Rs. 32,000 and Rs. 56,000. C join the business on a condition that they all will share the profit equally. For the loss of A and B, C gives Rs. 19800 to A and B to compensate their loss. How much amount A get out of Rs. 19800? (a) 1200 (b) 1600 (c) 1800 (d) Data inadequate (e) None of these

Direction (2-3): A, B and C start a business. Ratio between investment of A and B is 5 : x while ratio between investment of B and C is 9 : 11. A, B and C invested for 8 month, 6 month and 12 month respectively. 2.

If ratio between A’s share in profit to B’s share in profit is 5 : 6 and difference between C’s share in profit to B’s share in profit is 2340, then find the total profit earned by all three together? (2 Marks) (a) 7700 (b) 5390 (c) 6160 (d) 6930 (e) Cannot be determined

3.

Find the value of ‘x’. If out of total profit of Rs 2460, A’s share in profit is Rs 600. (1 Mark) (a) 8 (b) 7 (c) 6 (d) 5 (e) 4 3

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4.

Rahul and Rohit start a business by investing Rs.4900 and Rs.3600 2 1 respectively. After 4 months, Rahul and Rohit withdraw th and of 7 3 their initial investment and after 6 months of starting of business they invested 60% of what they withdraw. Find the share of Rohit in total profit of Rs. 8490 (a) Rs.3555 (b) Rs.4935 (c) Rs.1855 (d) Rs.4550 (e) Rs.4850

5.

‘P’ and ‘Q’ entered into a partnership with 50,000 and 60,000 for 8 and 6 months respectively. ‘P’ buy a car from its whole profit and ‘Q’ invested his whole profit in a scheme which offer 10% interest on S.I. 1 P’s car value depreciated 11 % every year. If after 2-year difference 9 between their money (excluding initial investment) is 23,480 then find P’s profit. (a) Rs. 72000 (b) Rs. 35000 (c) Rs. 64000 (d) Rs. 54000 (e) Rs. 81000

6.

A and B started a business in partnership by investing in the ratio of 7 : 9 . After 3 months A withdraw 2/3 of its investment and after 4 1 months from the beginning B withdraw 33 % of its investment. If a 3 total earned profit is Rs. 10201 at the end of 9 months, find the share of each in profit. (a) Rs. 3535, Rs.6666 (b) Rs. 3055, Rs. 5555 (c) Rs. 4503, Rs. 1345 (d) Rs. 3545, Rs. 3333 (e) None of these

7.

Satish and Saurbh entered into a partnership business for eight months. Satish invested Rs. 40x and Saurbh invested 40% more than Satish, after four months from starting of business Rituraj joined the both with the equal capital of Satish and Saurbh invested together initially. If they got a total profit of Rs. 7200, then find profit share of Rituraj? (a) 1600 Rs. (b) 1800 Rs. (c) 2000 Rs. (d) 2200 Rs. (e) 2400 Rs. 4

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8.

Veer and Sameer enter into a business by making the investment in the ratio of 3 : 4. After six months Veer added Rs. 2000 more in his initial investment, while Sameer withdraw Rs. 4000 from his initial investment. If at the end of one year Veer and Sameer shared profit in the ratio of 7 : 9, then find initial investment of both? (a) 69000 Rs. 92000 Rs (b) 36000 Rs. 48000 Rs. (c) 27000 Rs. 36000 Rs. (d) 45000 Rs. 60000 Rs. (e) 24000 Rs. 32000 Rs.

9.

P and R entered into partnership business with the capital of Rs. x and Rs. (x + 12000), after One year Q joined them with capital of Rs. (x + 8000) at the end of 2nd year P and Q with draw their capital and R invest for one more year, if P, Q and R gets profit in the ratio of 8 : 6 : 21 respectively. Find sum of capital invested by all three? (a) 66000 Rs. (b) 54000 Rs. (c) 64000 Rs (d) 68000 Rs. (e) 70000 Rs.

10. Three partners A, B and C invested their amounts in ratio of 2 : 5 : 8. At the end of four months, A invests some amount such that, his total investment will be equal to C’s initial investment and C withdraw some amount that its total investment will be equal to B’s initial investment. If 10% of B’s share in profit of one year is Rs 2550. Then share of A and B in total annual profit will be? (a) 81500 (b) 61200 (c) 56100 (d) 75000 (e) None of these 11. A and B enter into a partnership business and decided to distribute 60% of profit as per (investment× 𝑡𝑖𝑚𝑒) ratio and rest in the ratio of 7:10. If the amount invested by A is 25 % more than B and B withdraw half of his total investment after 9 months. Find the profit distribution ratio of A and B according to given condition at the end of a year. (a) 43:41 (b) 44:41 (c) 48:43 (d) 49:47 (e) None of these 5

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Directions (12-13): Three partners Abhishek, Neeraj and Aniket enter into a partnership business with the capitals in the ratio of 3 : 5 : 8. After four month Abhishek, Neeraj and Aniket added Rs. 6000, Rs. 9000 and Rs. 12000 respectively, after next four months Abhishek and Aniket withdraw Rs. 5000 and Rs. 8000 respectively and Neeraj added additional Rs. 6000. 12. Two persons Ankur and Saurabh enter a business. Ankur invested capital 6000 more than what Abhishek invested for first four month and Saurabh invested what Aniket invested for last four months. If Ankur and Saurabh got profit share after one year in the ratio of 15 :28 then find theinvestment of Neeraj for last four months?(1 marks) (a) 24000 Rs. (b) 18000 Rs. (c) 16000 Rs. (d) 30000 Rs. (e) 36000 Rs. 13. Amir and Mayank started a business in partnership. Amir investmentfor six months isRs. 6000 more thanwhat Neeraj invested for first fourmonthsand Mayank invested for eight months, Rs. 2000 lessthan whatAniket invested for first four monthsIf Mayank got profit share of Rs. 15400, out of total profit Rs. 22750 thenfind the investment of Amir is how much less than investment of Mayank? (2 marks) (a) 218000 Rs (b) 216000 Rs. (c) 210000 Rs. (d) 220000 Rs. (e) 232000 Rs. 14. Veer, Sameer and Gopal enter into a business by making investment in the ratio of 5 : 6 : 7 respectively. After six months Veer and Sameer 1 withdraw th and 25% of his initial investment respectively. After 5

1

eight months from the starting of business Gopal added 𝑡ℎ of his 7 initial investment. If at the end of one year profit share of Gopal is Rs. 2000 more than Sameer, then find profit share of Veer? (a) 4420 Rs. (b) 4520 Rs. (c) 4820 Rs. (d) 4320 Rs. (e) 4720 Rs. 6

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15. Manish and Rituraj invested Rs. 12000 and Rs. 16000 in a business. After four months Manish and Rituraj both added Rs. 4000 in their initial investment. At the end of one year the total profit was Rs. 172500, if Manish and Rituraj invested their profit share on compound interest at the rate of 20% and 10% respectively then find difference between interests got by both at the end of two years? (a) Rs.10250 (b) Rs.11520 (c) Rs.12210 (d) Rs.13110 (e) Rs.12660 Direction (16 – 17): Data given below about investment of three partners in a business. Read the data carefully and answer the questions: Swati, Monika and Anshika three business partners enter into a business by making investment Rs. 4000, Rs. 6000 and Rs. 8000 respectively. After first quarter Swati, Monika and Anshika added some amount in the ratio of 2 : 3 : 2 1 5. After second quarter Swati, Monika and Anshika withdrew 𝑡ℎ , 𝑟𝑑 and 5

3

3

𝑡ℎ respectively of capital invested in second quarter. At the end of nine month profit share divided among Swati, Monika and Anshika in the ratio of 24 : 37 : 49 respectively. 7

16. Satish and Veer enter into a business, Satish invested 200% more than Swati capital for second quarter, while Veer invested 300% more than Anshika capital for third quarter. And profit share of Satish at end of two year is Rs. 7250 out of total profit of Rs. 15950. Then find for how many months Veer invested his capital? (a) 14 months (b) 12 months (c) 18 months (d) 10 months (e) 8 months 17. Monika and Anshika enter into a business with another partner Shikha. Monika invested 144% of what she invested in second quarter, while Anshika invested 160% of what she invested in second quarter. If all three invested their capital for a year and Shikha gets a profit of Rs. 9600 out of total profit of Rs. 28000, then find investment of Shikha? (a) 12,800 Rs. (b) 11,600 Rs. (c) 14,400 Rs. (d) 14,200 Rs. (e) 13,600 Rs. 7

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18. Satish, Veer and Neeraj entered into a business by investing in the ratio of 4 : 6 : 3. After two years, Satish and Veer withdrew 25% and 1

33 % of their respective initial investment, while Neeraj added 50% 3

of his initial investment more. If at the end of three years they received total profit of Rs. 93750, out of which 22.5% are used for maintaining business and the remaining profit was divided among them according to their investment ratio. Find difference between profit share of Satish and Neeraj? (a) 978.75 Rs. (b) 976.75 Rs. (c) 974.25 Rs. (d) 968.75 Rs. (e) 966.75 Rs. 19. A, B and C enter into a partnership and invested some amount. After 1

one year A double its investment, B increase its investment by 33 % 3

and C increase its investment by 20%. In the third year A and B withdraw their investments and D joins the partnership with C. After three year they got profit in the ratio of 12 : 14 : 17 : 8 (A : B : C : D). If difference between initial investment of B and C is 1150. Then Find out the total initial investment made by A and D together? (a) 12100 (b) 14400 (c) 13800 (d) 15000 (e) None of these 5

4

6

4

5

5

20. Three partners A, B and C invested in the ratio of ∶ ∶ in a business. After 3 months A increased his capital by 40%, after 3 months more A again increased his capital by 20%. B increased his capital by 25% after every 4 months. C increased his capital by 25% after 3 months and after 4 months more C increased his capital by 20%. If the total profit of Rs. 30,130 is earned at the end of the year, find what was the A’s share of profit? (a) Rs. 12,420 (b) Rs. 14,420 (c) Rs. 13,240 (d) Rs. 15,340 8

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21. A, B and C entered into a partnership business, A invested Rs. x, B invested 25% more than A and C invested 20% more than B for first 6 months. After that A left the business and B withdraw 50% of his investment. B left the business after 3 more months whereas C 2

increased his investment by 16 % . If at the end of year difference 3

between profit share of C and (A + B) together is Rs. 12375, then find profit share of C ? (a) 58500 Rs. (b) 56500 Rs. (c) 55680 Rs (d) 55580 Rs. (e) 52680 Rs. 22. Veer, Sameer and Satish entered into a business by investing equal amount each. After four months Sameer added half of his initial investment, while Veer and Satish withdrew half of their initial investment. At end of one year the total profit is Rs. 8000 and Veer and Sameer invested their profit share at the rate of 10% compounded annually for some time in a scheme. If the difference of the amount received by Veer and Sameer from the scheme at the end is Rs. 2420, find for how much time Veer and Sameer invested in the scheme?

23.

1

(a) 4 yr

(b) 3 yr

(d) 2 yr

(e) 1yr

2

(c) 3 yr

P, Q and R entered into a partnership with their initial capital in the 1

ratio of 3 : 5 : 8 respectively. After third month P withdrew rd of his 3

initial capital and Q withdraw half of his initial capital. After next six months P increased his capital by 100% more, Q increased by 300% more and R increased by 25% more. If at the end of an year, difference between profit share of Q and R was Rs. 7000, then find profit share of P? (a) Rs. 6600 (b) Rs. 7700 (c) Rs. 5500 (d) Rs. 5750 (e) Rs. 5950 9

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24.

Veer, Sameer and Divyaraj enter into a business by making an 1 1 1 investment in the ratio of ∶ ∶ respectively. After six months 129

301

602

1

from starting of business Veer withdrew 33 % of his initial 3 investment, while Sameer added 50% of his initial investment and Divyaraj added Rs. 1800 respectively. If after one year profit share of Veer is Rs. 10500 out of total profit of Rs. 20850, then find difference between initial investment of Veer and Sameer? (a) 6400 Rs. (b) 7200 Rs. (c) 7800 Rs. (d) 5600 Rs. (e) 8400 Rs. 25. A, B & C, three friends started the business in which A invested for 6 months, B initially didn’t invest & started as working partner, while C invested for 4 months. They decided to donate 1% of total profit and to give 22% of total profit to B as salary. Find the ratio of their profit shares (A:B:C) if A and C invested in the ratio of 2 : 3 and B also invested an amount which is 80% of total amount invested by A and C together for a single month. (a) 1 : 3 : 1 (b) 1 : 1 : 3 (c) 3 : 1 : 3 (d) 1 : 1 : 1 (e) None of these

10

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Solutions

A B (c); Ratio of share of profit = 32 : 56 = 4 : 7 = 11 Let, total profit = Rs. 11 Now, A, B and C will distribute Rs. 11 equally i.e. A : B : C

1.

11 3

:

11 3

:

11 3

Loss of A = 4 – Loss of B = 7 –

11 3 11 3

= =

1 3 10 3

Ratio of Loss = 1 : 10 Amount that got to compensate the loss =

1 11

× 19800 = 1800

Solution (2-3): Ratio between A’s, B and C’s profit share = 45 × 8 : 9x × 6 : 11x × 12 = 60 : 9x : 22x 2.

(d); Let profit sharing of A, B and C be 60z, 9xz and 22xz respectively. ATQ, 22xz – 9xz = 2340 ⇒ 13xz =2340 ⇒ xz = 180 B’s and C’s share in profit is 9 × 180 and 22 × 180 respectively. A’s share in profit =

9×180 6

× 5 = 1350

Total profit earned by all three together = 1350 + 1620 + 3960 = 6930 11

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3.

(c); ATQ, 60 60+9𝑥+22𝑥

=

600 2460

⇒ 246 = 60 + 31𝑥 ⇒𝑥 = 4.

186 31

=6

(a); Investment of Rahul For first four month = 4900 For fifth and sixth month 5

= × 4900 = 3500 7

3

2

5

7

For Last 6 month = 3500 + × × 4900 = 4340 Rs. Investment of Rohit For first 4 month = 3600 Rs. 2

Next 2 month = × 3600 = 2400 Rs. 3

3

1

5

3

Last 6 month = 2400 + × × 5600 = 3120 Profit ratio Rahul Rohit 4900 × 4 + 3500 × 2 + 4340 × 6 : 3600 × 4 + 2400 × 2 + 3120 × 6 329 237 : Total profit = 8490 Rs. Share of Rohit out of total profit = 5.

8490 566

× 237 = 3555 Rs.

(e); Profit ratio of P and Q 50,000 × 8 : 60,000 × 6 : 10 9 Let P’s and Q’s profit → 10𝑥 and 9𝑥 Value of P’s money after 2 years = 10𝑥 ×

8 9

8

640𝑥

9

81

× =

Value of Q’s money after 2 years 12

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=9𝑥 +

9𝑥×2×10 100

Difference =

= 9𝑥 ×

108𝑥 10

640𝑥

–

81

120 100

=

108 10

𝑥

= 23,480

𝑥 = 8100 P’s profit = Rs. 81,000 6.

(a);

:

A Capital → New Ratio →

B

7×3 : 9×3 :

A

B

21x : 27x Total capital invested by A in 9 months = 21x × 3 + 7x × 6 = 105x Total capital of B invested in 9 months = 27x × 4 + 18x × 5 = 108x + 90x = 198x A : B Capital → 105x : 198x According to question, (105x + 198x) = Rs. 10201 303x = 10201 x = Rs.

10201 303

Hence, Share of A =

10201 303

× 105 = Rs. 3535

Share of B = 198 × 7.

10201 303

= Rs. 6666

(e); Satish capital = 40x Rs. Saurbh capital = 40x + 40x ×

40 100

= 56x Rs. Rituraj capital = 40x + 56x = 96x Rs. 13

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Ratio of profit (Satish : Saurbh : Rituraj) = 8 × 40x : 8 × 56x : 96x × 4 = 320x : 448x : 384x =5:7:6 6

Rituraj profit share = 7200 × (5+7+6) = 2400 Rs. 8.

(a); Lets Veer and Sameer invested Rs. 3x and Rs. 4x respectively. Ratio of investment of Veer and Sameer = [3x × 6 + (3x + 2000) × 6] : [4x × 6 + (4x – 4000) × 6] = (36x + 12000) : (48x – 24000) ATQ— 36𝑥+12000 48𝑥 –24000

=

7 9

36𝑥 × 9 + 12000 × 9 = 7 × 48𝑥 − 24000 × 7 12000(9 + 14) = 12(7 × 4𝑥 − 3 × 9𝑥) 1000 × 23 = 28𝑥 − 27𝑥 x = 23000 Rs. Investment of Veer = 3 × 23000 = 69000 Rs. Investment of Sameer = 4 × 23000 = 92000 Rs. 9.

(d); P : Q : R Rs. x × 2 : Rs. (x + 8000) × 1 : Rs. (x + 12000) × 2 = 8 : 6 : 21 ATQ— 2𝑥

= (𝑥+8000)

8 6

6x – 4x = 32000 x = 16000 Required sum of capital (P + Q + R) = 16000 + (16000 + 8000) + (16000 + 12000) = 68000 Rs. 14

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10. (c); Let amount invested by A, B and C are 2x, 5x and 8x respectively. ⇒ At the end of 4th month A’s investment 2x + 6x = 8x = c’s initial invested ⇒ C’s investment = 8x – 3x = 5x = B’s initial investment Ratio of investment A : B : C = (2x × 4 + 8x × 8) : (5x × 12) : (8x × 4 + 5x × 8) 6 : 5 : 6 Let B’s profit = 5y Now ATQ =

5𝑦×10 100

= 2550

y = 5100 share of A and B = 5y + 6y = 11y = 11 × 5100 = 56100 11. (b); Let the total investment of B = 100 Investment of A = 100×125/100 Ratio of 60% profit = 125×12 : 100×9+50×3= 10:7 Let the total profit = 170 Profit for A for only 60 % of profit = 170×0.6×(10/17)=60 Profit of B for only 60 % of profit = 102-60 = 42 Total profit of A = 60+ 68×7/17= 88 Total Profit of B = 42+68×10/17 = 82 Required ratio = 88:82 = 44:41 15

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Solutions (12-13); Let investment of Abhishek, Neeraj and Ankit is Rs. 3x, Rs. 5x and Rs. 8x respectively. ATQ—Investment ratio of Abhishek, Neeraj and Aniket =

[3x ×

4

+

(3x +

6000)

×

4

+

(3x

+

6000

–

5000)

× 4] : [5x × 4 + (5x + 9000) × 4 + (5x + 9000 + 6000) × 4] : [8x × 4 + (8x + 12000)

×

4

+

(8x

+

12000

–

8000)

× 4] = (36x + 28000) : (60x + 96000) : (96x + 64000) 12. (d); Investment of Ankur = Investment of Abhishek for first four months and next four months = (3x + 6000) Rs. Investment of Saurabh = Investment of Aniket for last four months = (8x + 12000 – 8000) Rs. ATQ— (3𝑥+6000)×12

= (8𝑥+4000)×12

15 28

120x - 84x = 168000 – 60000 36x = 108000 x = 3000 Rs. Neeraj invested for last four months = (5x + 9000 + 6000) Rs. = (5×3000 + 9000 + 6000) = 30000 Rs 16

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13. (e); Investment of Amir = Investment of Neeraj for first four months +6000 = 5x + 6000 Investment of Mayank = Investment of Aniket for first four months – 2000 = 8x -2000 (5x+6000)×6

= (8x−2000)×8 (5𝑥+6000)

= (8𝑥−2000)

22750 –15400 15400

7 11

56x – 55x = 80000 x = 80000 required difference = 8x-2000-5x-6000 3x – 8000 240000 – 8000= 232000 14. (d); Let Veer, Sameer and Gopal invested Rs. 5x, Rs. 6x and Rs. 7x respectively Ratio between profit share of Veer, Sameer and Gopal 4

3

5

4

= (5𝑥 × 6 + 5𝑥 × × 6) : (6𝑥 × 6 + 6𝑥 × × 6): 8

(7𝑥 × 8 + 7𝑥 × × 4) 7

= 54𝑥 ∶ 63𝑥 ∶ 88𝑥 = 54 : 63 : 88 2000

Profit share of Veer = 54 × (88

–63)

= 4320 Rs. 17

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15. (d); Ratio of profit of Manish and Rituraj = [(12000 × 4)+ (12000 + 4000) × 8] : [(16000 × 4) + (16000 + 4000)× 8] = 17600 : 224000 = 11 : 14 Profit share of manish = 172500 ×

11 25

= Rs.75900 Profit share of Rituraj = 172500 ×

14 25

= 96600 Rs Equivalent CI of two year at the rate of 20% = 20 + 20 +

20×20

= 44%

100

Equivalent CI of two year at 10% = 10 + 10 +

10×10

= 21%

100

Required difference between interest = 75900 ×

44 100

− 96600 ×

21 100

= 13110 Solutions (16-17): Let total amount added by Sawti and Monika after first 3

quarter be Rs. 2x and Rs. 3x

4000 ×3+(4000+2𝑥)×3+(4000+2𝑥)×5×3 2 6000×3+(6000+3𝑥)×3+(6000+3𝑥)×3×3

1

12000+12000+6𝑥+(36000+18𝑥)×5 18000+18000+9𝑥+12000+6𝑥 156000+48𝑥 240000+75𝑥

=

=

=

24 37

24 37

24 37

240500 + 74x = 240000 + 75x x = 500 Rs. 18

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16. (c); Total amount invested by Sawti in second quarter = (4000 + 2× 500) = 5000 Rs. Total amount invested by Anshika in third quarter = (8000 + 500 × 5 − 4500) = 6000 Rs. Satish investment = 5000 ×

300 100

= 15000 𝑅𝑠. Veer investment = 6000 ×

400 100

= 24000 𝑅𝑠.

Let Veer invested for T months ATQ – 15000 ×24 24000×𝑇 120 8×𝑇

=

=

7250 (15950−7250)

145 174

T = 18 months 17. (c); Total amount invested by Monika in second quarter = 6000 + 3 × 500 = 7500 𝑅𝑠. Monika investment = 7500 ×

144 100

= 10800 𝑅𝑠. Total amount invested by Anshika in second quarter = 8000 + 5 × 500 = 10500 𝑅𝑠. Anshika investment = 10500 ×

160 100

= 16800 𝑅𝑠. Let Shikha invested = P Rs. ATQ – 𝑃×12 (10800+16800)×12 𝑃 9600 27600

=

=

9600 (28000−9600)

18400

P = 14,400 Rs. 19

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18. (d); Let the amount invested by Satish, Veer and Neeraj initially be 4x, 6x and 3x respectively. Ratio of profit share of Satish, Veer and Neeraj 3

2

4

3

= 4x × 24 + 4x × × 12) ∶ (6x × 24 + 6x × × 12) : (3x × 24 + 3

3x × × 12) 2

= 22 : 32 : 21 Total remaining profit after excluding the amount spend on maintaining business = 93750 ×

(100–225) 100

= 72656.25 Rs.

Required difference = 72656.25 ×

(22–21) 75

= 968.75 Rs. 19. (c); Let investment of A, B, C and D is a, b, c and d respectively. A B C 𝐷 Now in firt year → a × 12

:

b × 12

:

c × 12

4b 6c × 12 : × 12 3 5 6c In 3rd year × 12 : 𝑑 × 12 5 4 A : B : C : D ⇒ (a × 12 + 2a × 12) : (b × 12 + b × 12) : c × 12 + In 2nd year

→ 2a × 12 :

3

6

2 c × 12 : d × 12 5

3a :

7𝑏 3

:

17 5

𝑐 : d = 12 : 14 : 17 : 8

⇒a:b:c:d=4:6:5:8 Difference between B and C initial investment = 1150 Total Investment of A and D together = 20

1150 1

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20. (a); Ratio of investment of A, B and C 5 4 6 = ∶ ∶ = 25 ∶ 16 ∶ 24 4 5 5 Ratio of profit after one year A : B : C 25  3  35  3  42  6 : 16  4  20  4  25  4 : 24  3  30  4  36  5

= 432 : 244 : 372 = 108 : 61 : 93 30130 A’s share in profit = × 108 = 12,420 262

21. (a); Let investment of A, B and C = x : 1.25x : 1.5x = 4x : 5x : 6x ATQ— Profit share of A : B : C = 4x × 6 : (5x × 6 + 2.5x × 3) : (6x × 6 +7x × 6) = 24x : 37.5x : 78x Given→ C – (A + B) = 12375 78x – (24x + 37.5x) = 12375 x = 750 Profit share of C = 750 × 78 = 58500 Rs. 22. (d); Let Veer, Sameer and Satish invested Rs. x initially Ratio of profit of Veer, Sameer and Satish x 3x x = (x × 4 + × 8) : (x × 4 + × 8) : (x × 4 + × 8) 2 2 2 = 8x ∶ 16x ∶ 8x =1:2:1 1 Profit share of Veer = 8000 × = 2000 Rs. 4 Profit share of Sameer 2 = 8000 × = 4000 Rs. 4 Let both invested for n years 4000 (1 + 2000 (1 + 10

n

n

10

100 10 n

100

10

100

) – 2000 (1 +

n

) = 2420

) = 2420

(1 + ) = 1.21 100 n = 2 years 21

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23. (c); Lets P, Q and Q investment be 3x, 5x and 8x profit share of P:Q:R = (3x × 3 + 2x × 6 + 4x × 3) : (5x × 3 + 2.5x × 6 + 10x × 3) : (8x × 9 + 10x × 3) = 33x : 60x : 102x = 11 : 20 : 34 ATQ— 34 – 20 => 7000 14 => 7000 1 => 500 Profit share of P = 11 × 500 = Rs. 5500 24. (b); Initial investment of Veer, Sameer and Divyaraj =

1806 129

∶

1806 301

∶

1806 602

= 14 : 6 : 3 Let Veer, Sameer and Divyaraj initial investment be 14x Rs., 6x Rs. & 3x Rs. respectivelyProfit ratio of Veer, Sameer and Divyaraj 2

= (14x × 6 + 14𝑥 × × 6 ) : (6x × 6 + 9𝑥 × 6) ∶ [3𝑥 × 6 + (3𝑥 + 3

1800) × 6] = (84x + 56x) : (36x + 54x) : (18x + 18x + 10800) ATQ – 140𝑥 90𝑥+36𝑥+10800 𝑥

= (21𝑥+1800)

=

10500 20850−10500

1 23

2x = 1800 x = 900 Rs. Required difference = 14× 900 − 6 × 900 = 7200 Rs. 22

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25. (d); Let A invested 2x and C invested 3x. 80 Then B invested = × (3x + 2x) =

80 100

100

× 5x = 4x

Let 100Y be their total profit. According to question, 1% is given as donation and B gets 22% as salary of total profits. Now ratio of their shares is 2x × 6 : 4x × 1 : 3x × 4 : 12x 12x : 4x : : 3 1 3 3 A gets × [100Y– 23Y] 3

7

= × 77Y = 33Y 7

3

C gets = × 77Y = 33Y 7 1

B gets = × 77Y + 22Y = 33Y 7

Hence, ratio of their profit shares is 33Y : 33Y : 33Y =1:1:1

23

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1

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Chapter

Average and Ages

4

BEST APPROACH TO SOLVE THE QUESTIONS INTRODUCTION: We all know that definition of average is sum of magnitude (weight, age, marks or any measurable quantity) of all quantities divided by total number of all quantities. Always remember, if you are calculating average you are making all the quantities equal. For eg. three brothers have number of candies with them is 9, 11 and 16. But their mom took all the candies and distribute 12 to each. Average in some questions referred as ‘Arithmetic mean’. Questions from this topic are extremally easy and doesn’t require as much time or sharp concept, mostly questions are asked in the form of average of ages. Example: Marks obtained by seven students are 70, 85, 95, 86, 74, 83 and 67. What is average marks obtained by them. Solution: Sum of magnitude of all the observations is 70 + 85 + 95 + 86 + 74 + 83 + 67 = 560. Required average =

560 7

= 80.

NOTE: Please avoid short cut tricks mentioned in many books for such small calculations. Those tricks are more time consumable. There is other very simple and only trick that you must apply. Look at next question. Example: Average age of 4 members of a family is 28 years and average increases by 2 years when one of their relative joins them. Calculate the age of relative. 2

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Solution: Easy one! Let age of relative is x years. Total age of family before joining the relative = 4×28=112 After joining the relative

112+x 5

= 28+2, which gives us x = 38 years.

Right approach? Or wrong? Try this approach, Imagine, they all have 28 rupees/candies/marbles before their relative joins them. When he comes, they all have Rs 30 each. What does it mean? It means their relative gave Rs 2 to all 4 of them and even then, he was left with Rs 30, as average means amount is distributed equally. How much he had given to them? Rs. 8, Right? Therefore, he initially had total of Rs 38 with him. Why to waste paper and time if we can do it mentally. Try next question. Example: Average marks scored by a student in 4 subjects is 75. But when marks of English are added to it, overall average became 70. How much did he score in English? Solution: Let 4 men have 75 gold coins each and when 5th one joins them, each of those 4 men/exams gave 5 coins/marks to 5th one. And after taking 5 coins from 4 persons, 5th have total 70 coins. It means he initially had 50 coins and that the answer. Example: Average marks scored by a student in 5 subjects was 79. But when average is calculated for 4 subjects (without Hindi) it was 4 lesser than the average of 5 subjects. What was the marks scored by him in Hindi? Solution: Again let 5 men have 79 candies each and when one of them leave, he took all candies belonging to him. It means each of these 4 had borrowed 4 candies from him. He took 79 candies that was with him, plus these 4 persons returned 16 of his. Therefore, he had 95 candies or marks whatever you assume.

3

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Example: Average run scored by Sachin in 19 innings of year 2007 is 49 runs while total run scored run scored by Afridi in 19 matches were 760. In 20th match, average of Afridi decreased by 2 runs/ match and average run/match increased by 1 for Sachin. What is the difference b/w the scored runs in 20th match by them? Solution: Average runs by Afridi in 19 innings

760 19

= 40 runs.

His average decreased by 2, which means all the 19 innings/persons have given 2 runs/coins to 20th inning/person. New inning/person borrowed 19 × 2 = 38 runs/coins to become equal with others. Therefore, he himself have no money and we can conclude Afridi scored 0 in that match. While for Sachin, his 19 previous innings got 1 run each and even then, his 20th inning remains with 50 runs. Therefore, total runs by Sachin in 20th inning are 69 runs. Required difference is 69 runs. This may look longer when you read it, but once you get the concept you can solve these questions within 30 seconds. AGES The questions on ages are of two types either the average age of a group of people is asked or we are given the ratios of ages of friends and we are asked to calculate their individual age or average age. See it through a question. Example: Ratio of present age of Rahul and his father is 5:8 and 14 years earlier this ratio was 1:2. What is present age of Rahul Solution: 1 way to solve this is assume present age of Rahul and his father 5x and 8x. According to question 14 years earlier ratio of their age was 5x−14 8x−14

4

=

1 2

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(5x − 14) × 2 = 8x − 14 2x = 14 X=7 Present age of Rahul = 7× 5= 35 years. Or If you want to solve it mentally, or in a shorter time, observe carefully Present age ratio 5 : 8 14 years earlier 1 : 2 Note that as difference b/w ages of two persons is always same, so what we do? We will make difference b/w two ratio’s same. Multiply first ratio by difference of second and second ratio bye difference of first. First ratio when multiplied by difference of other i.e. (2-1=1) it becomes 5:8 When second ratio is multiplied by difference of first i.e. (8 - 5 = 3) we will get 1×3 : 2× 3 3 : 6    

Now think mentally, that 14 years ago age of Rahul was 3 units and now it is 5 unit. Therefore 2 units are equal to 14 years, which means 5 unit (age of Rahul) is 35 years. NOTE: Concept behind multiplying by difference is that, we must make both differences (b/w the ratios) equal. To make two numbers equal what we do? We take their LCM as common. If difference b/w two ratios are 4 and 6, we will multiply first ratio by 3 and second ratio by 2 to make their difference equal. Cross check it by your self for some quantities for better confidence. Example: Ratio of age of two friends is 7:9 while 10 years later it will be 19:23. Calculate their average age after 7 years from present. 5

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Solution: Difference b/w the ratios are 2 and 4. So to make them equal multiply first ratio by 2. After multiplying we will get 14 : 18. And second ratio we have is 19 : 23. Hence conclude mentally, that 14 units is present age of younger friend and 19 units is his 10 years later age. It means that 19 – 1 4 =5 units = 10 year 1 unit = 2 years. And their present average age is (28 + 36)/2 = 32 years. And 7 years later mean 7 is added to both of their age, which means average age is 32 + 7 = 39 years. Key Points to Remember for New Pattern Questions ➢ Calculating average means making each quantity equal. ➢ To make difference of two ratios equal, try to think through ‘LCM of difference’ approach. ➢ If the average age of a group of people is X years, Y years later it will be X + Y years.

6

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Practice Exercise Based on new Pattern

1.

In a group of 4 friends, ratio of present age of A and D is 4:5 and that B to C is 3:4. Calculate the present average age of A, B and C, if 4 years ago, A was 20 years younger than C and at present C is twice of age of A. (a) 20 years (b) 50 years (c) 40 years (d) 25 years (e) 30 years

2.

Ratio of age of Vikas to age of Rohit 2: 3 and average age of Rohit, Vikas and Rahul is 23. When Arjun joins the group average age of these 4 persons is 25 years. Calculate sum of age Arjun and Rahul, if Rahul is 13 years older than Vikas? (a) 40 years (b) 47 years (c) 60 years (d) can’t be determined (e) None of these

3.

S₁ is a series of five consecutive multiple of three, whose sum is 180 and S₂ is the series of four consecutive multiple of four whose second smallest number is 13 more than second highest number of S₁ series. Find the average of smallest number of S₁ series and highest number of S₂ series. (a) 51 (b) 49 (c) 47 (d) 45 (e) 43

4.

A student finds the average of five two digits numbers. If One number is reversed and the average is taken again then the average increase by 5.4. If all five digits are consecutive multiple of four, then find the number which is reversed?

7

(a) 58

(b) 36

(d) 48

(e) None of these Adda247 Publications

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5.

Average weight of a group is 20 kg. When 2 people include in the group average weight becomes 22 kg after that when 4 another people removed from the group, having weight half of the weight of 2 person included previously, average weight become 25. Find the strength of group initially (a) 18 (b) 20 (c) 22 (d) 16 (e) 14

6.

Four years ago, Ratio of Ram’s to Shyam’s age is 9 : 11 and Shyam’s age four years ago is same as Ram’s present age. If Rahul’s present age is average of present age of Ram and Shyam, then find Rahul’s age 2 year ago. (a) 20 (b) 18 (c) 24 (d) 22 (e) 26

7.

Ratio of Ayush Age to his first son’s age is 20 : 9 and Ayush wife’s age to Ayush 2nd son’s age is 3 : 1. If elder son of Ayush is 6 year elder than another son (ayush has only two son and no daughter) and average age of family is 26.5 year then find the age of Ayush’s wife. (a) 24 (b) 36 (c) 40 (d) 38 (e) 28

8.

Ritu take four, two-digit numbers and take average of these number which is 52.5. If she reversed all the digits of 4 numbers she found that their average is now 22.5 less than the previous one and they formed A.P having common difference 4. Find the largest number. (a) 82 (b) 80 (c) 36 (d) cannot be determined (e) 64

8

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9.

The average age of a group of six children is 15 years. From the group, two children, whose ages were 3 years more and 5 years more than the average age, left. 4 new children, whose average age is 4 years more than the given average age, join the group. Find the new average age. (a) 15 years (b) 16 years (c) 17 years (d) 18 years (e) 12 years

10. If age of P and R are added to twice the age of Q, the total becomes 59. If the ages of Q and R are added to thrice the age of P, the total become 68. And if the age of P is added to thrice the age of Q and thrice age of R, the total becomes 108, what is the total of ages of P, Q and R? (a) 45 years (b) 47 years (c) 49 years (d) 42 years (e) 44 years 11. Sweta and Neha profess to tell their present ages as 25 and 20 years respectively. (Not original age). Ratio of their original ages 5 year ago is 5 : 4. Sum of ages of both 5 years hence is

400 9

% more than the sum

of present ages of both professed by them. Find the ratio of their present original age. (a) 4:7 (b) 5:6 (c) 6:5 (d) 4:5 (e) 5:7 12. Average age of a man, woman and their son is 30 years. Man’s age is 1

two year more than his wife and age of son is th the sum of age of his 4

mother and father.When two other family members were added, new average becomes 27 years. If difference between age of two new member are one year then find difference between son and the new member who is elder. (a) 7 years (b) 8 years (c) 4 years (d) 2 years (e) 5 years 9

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13. Present average age of A, B, C and D is 25 years. Sum of age of A and B is 150% of sum of age of C and D. Ratio of age of B to age of C is 3 : 5. Calculate 10 years later age of A, if B and D are of same age. (a) 45 years (b) 40 years (c) 35 years (d) 55 years (e) 50 years 14. Ratio of present age of A and B is 4 : 5, while that of C and D is 6 : 7. If 10 years ago Ratio of age of A and C is 1 : 2 and that of B & D is 3 : 5, Find average of present age of A, B and C together. (a) 27.5 (b) 25 (c) 27 (d) 26.5 (e) 24.5 1 1 1

15. The ratio of ages of three friends Veer, Sandeep and Satish is : : . 3 2 4

The average age of Veer and his wife is 33 years whereas the average 3

age of Sandeep and his wife is 39 years. The age of Satish’s wife is th 8

of the sum of age of Veer’s wife and Sandeep’s wife and the ratio of Veer’s age to that of age of Satish’s wife is 4 : 3 then find the sum of ages of all the three friends. (a) 104 yr (b) 112 yr (c) 117 yr (d) 91 yr (e) 120 yr 16. There are some teachers in a college with average age of 45 years. Sum of maximum age and minimum age is 108 years and difference of maximum age and minimum age is 27 years. If fourteen teachers left the college with the average age of 32.5 years and six new teachers joined college. Out of six new teachers four ages are equal to maximum age teacher and two teachers age equal to minimum age teacher. If new average becomes 49 teachers initially? (a) 54 (d) 68 10

4 7

years, then find the number of

(b) 60 (e) 72 Adda247 Publications

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17. The ratio between present age of Veer & Saurbh is 3: 4 and between present age of Divyaraj & Gopal is 2: 3. If sum of present age of Veer, Saurbh, Divyaraj & Gopal is 128 years and ratio between sum of age of Veer & Saurbh to sum of age of Divyaraj & Gopal is 63: 65, then find the sum of age of Veer & Gopal after eight years. (a) 90 years (b) 78 years (c) 86 years (d) 84 years (e) 82 years 18. 2n years ago, the age of Raju was four times that of his son and n years ago, the age of Raju was thrice that of his son. If n years later, the sum of the ages of Raju and his son will be 80 years, then the difference in the ages of Raju and his son is (a) 20 years (b) 40 years (c) 24 years (d) 30 years (e) 34 years 19. ‘2n’ years ago ratio of Amit’s age to Inder’s age is 5 : 4. ‘n’ years ago ratio of Inder’s age to Satish’s age is 9 : 7. Difference between present age of Amit to Satish’s present age is 12 years. Find the sum of present ages of all three if ratio of Amit’s age to Satish’s age after ‘n’ year will be 13 : 9. (a) 81 (b) 84 (c) 87 (d) 90 (e) Cannot be determined 20.

P1 is the series of four consecutive numbers divisible by 4 with an average of 34, while P2 is the series of five consecutive numbers divisible by 9, and whose second lowest number is equal to second highest number of P1 series. If P3 is the series of seven consecutive 1

numbers, whose lowest number is 𝑟𝑑 𝑜𝑓 lowest number of P2 series, 3

then find the sum of average of P2 & P3 series? (a)53 (b)51 (d)55 11

(c)57

(e)49 Adda247 Publications

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21. Average of five two digit numbers, in which 2 numbers say A and B are reverse of each other, is 25.4. If all three numbers except A and B multiplied by (–2) then the average of five numbers decreased by 43.2. What could be the greatest value among A and B. (a) 32 (b) 23 (c) 50 (d) 41 (e) 14 22. Average of any 200 consecutive natural numbers is 499.5. If next 1000 numbers more add in it then find the new average. (a) 1035.5 (b) 1299.5 (c) 1199.5 (d) 1099.5 (e) 999.5 23. Average age of a group of people is four times of the number of people in the group. Sakshi leaves the group and the average age is still four times of the number of people in the group. After that Sheetal leaves the group and the average age is still four times of the number of people in the group. If ratio between Sakshi’s age to Sheetal’s age is 21 : 19, then find the average age of the group if Ritu leaves the group whose age is 20 years (a) 36 years (b) 37 years (c) 38 years (d) 39 years (e) 40 years Directions (24-25): Rakesh distributed Rs.50000 into (X, Y, Z) 3 groups having 50 people in total. Number of people in group ‘X’ is equal to the number of people in group ‘Z’ and total money got by group ‘Y’ is equal to the total money got by group ‘Z’. Average money got by group X and Y 3500 together is Rs. and average money got by group Y and Z is Rs.1000 3

24. Find the number of people in group Y? (a) 20 (b) 16 (d) 18 (e) 24

(c) 10

25. Find the amount got by each person in group Z. (a) 800 Rs. (b) 600 Rs. (c) 1000 Rs. (d) 750 Rs. (e) 500 Rs. 12

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Solutions

11. (e); Let the present age of B is 3K and that of C is 4K. Also assume present age of A is 4m and D is 5m. Note that age difference between A and C will always be 20 years. 4K – 4m = 20 K – m = 5 …(i) Also

4K 4m

=2

K = 2m …(ii) m = 5 years K = 10 years ∴ Average of A, B and C = 2.

20+30+40 3

= 30 years

(c); Let age of Arjun, Vikas, Rohit and Rahul is a, b, c and d respectively Now, b : c = 2 : 3 …(i) b + c + d = 23 × 3 = 69 …(ii) a + b + c + d = 25 × 4 = 100 ∴ a = 31 years. Also d – b = 13 d=b+13 Put this value in eq. (ii) 2b + c = 56 From (i) b = 16 years, d = 29 years c = 24 years ∴ a + d = 31 + 29 = 60 years. 13

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3.

(d); Let S₁ is a series consists 3x – 6, 3x – 3, 3x, 3x + 3, 3x + 6 ATQ, 3x – 6 + 3x – 3 + 3x + 3x + 3 + 3x + 6 = 180 ⇒ x = 12 S₁ series = 30, 33, 36, 39, 42 Second smallest no. of S₂ = 39 + 13 = 52 S₂ series = 48, 52, 56, 60 Required average =

30+60 2

=

90 2

= 45

4.

(b); Let the number which is reversed is “10a+b” If it is reversed then the number becomes “10b+a” ATQ, 10𝑏 + 𝑎 − 10𝑎 − 𝑏 = 5 × 5.4 ⇒ 9𝑏 − 9𝑎 = 27 ⇒𝑏−𝑎 =3 Numbers can be 14, 25, 36, 47, 58 and 69 but all the five digits are multiple of 4 ⇒ The number should be 36

5.

(a); Let initially strength = x ATQ, Total weight → 20x When two persons include Total weight = 22 (x + 2) Weight of two persons included = 2x + 44 4 persons excluded weight =

2x+44 2

= x + 22

ATQ, 22 (x + 2) – (x + 22) = (x + 2 – 4) × 25 x = 18 14

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6.

(d); Let age of Ram and Shyam 4 years age 9x and 11x respectively So ATQ, 11x = 9x + 4 x=2 their present age is 22 and 26 years Rahul’s age 2 year ago =

22+26 2

– 2 = 22

7.

(b); Let Ayush Age, his wife age, his first son’s age and his second son’s age is 20x, 3y, 9x and y respectively. ATQ, 9x – y = 6 …(i) 20x + 3y + 9x + y = 26.5 × 4 …(ii) Solving (i) & (ii) x=2 y = 12 his wife’s age = 12 × 3 = 36

8.

(a); Let four number is w, x, y and z after reversing they formed A.P. Sum of all number = (52.5 – 22.5) × 4 = 120 n=4 d=4 n Sn = [2a + (n– 1)d] 2

4

120 = [2a + (4 – 1)4] 2

a = 24 Reversed numbers 24, 28, 32, 36 Original number 42, 82, 23, 63 Largest number → 82

15

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9.

(b); Sum of the ages of 6 children = 15 × 6 = 90 When two children left, sum of the ages of 4 children = 90 − (18 + 20) = 52 Sum of the ages of New children = (15 + 4) × 4 = 76 ∴ Required average = =

128 8

76+52 8

= 16 years

10. (e); P + 2Q + R = 59 3P + Q + R = 68 P + 3Q + 3R = 108 Solving the equation, P = 12 years, Q = 15 years, R = 17 years. Sum of their ages = 44 years. 11. (c); Let age of Sweta 5 years ago = 5x Let age of Neha 5 years ago = 4x According to question (5x + 10) + (4x + 10) = (100% + 9x + 20 =

13 9

400 9

%) (25 + 20)

× 45

x=5 Sum of their present age = (5+ 4)×5+10 = 45 year + 10 years = 55 years Required ratio = 6 : 5 12. (e); Let age of man is x years Age of woman = (x – 2) years Age of son = ATQ—

16

(x+x–2) 4

=

(x–1) 2

years

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x+(x –2)+ 2 3 2x+2x–4+x –1 2

= 30 = 90

5x = 185 x = 37 years son age =

(37–1) 2

= 18 years

let age of two new members be y years and (y – 1) years ATQ— 90+(y+y–1) 5

= 27

2y – 1 = 135 – 90 y=

46 2

y = 23 Required difference = 23 – 18 = 5 years 13. (d); If average age of A, B, C and D is 25 years, then total age of A, B, C and D is 100 years. Also if C + D = 2x Then A + B = 3x [A + B is 150% of C + D] 5x = 100 x = 20 A + B = 60 …(i) C + D = 40 …(ii) Now, in question ratio of B : C is given as 3 : 5. This is also ratio of D : C, as B and D are of same age. From (ii) We can calculate C = 25 years D = 15 years & Hence B = 15 years ∴ A = 45 years 10 years later age of A is 45 + 10 = 55 years. 17

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14. (b); Let present age of A, B, C and D is 4x, 5x, 6y and 7y respectively. Then, ATQ, 4𝑥–10 6𝑦–10

1

= ⇒ 8x – 20 = 6y – 10 2

⇒ 8x – 6y = 10 …(i) Similarly 5x–10 7y–10

=

3 5

⇒ 25x – 50 = 21y – 30

⇒ 25x – 21y = 20 …(ii) Solving (i) and (ii) 25𝑥 – 21𝑦 = 20 16𝑥 – 12𝑦 = 20 – + – 9𝑥 – 9𝑦 = 0 𝑥=𝑦 Put this value in equation (i), we will get x=y=5. Therefore, average age of A, B and C together is 25 𝑦𝑒𝑎𝑟𝑠.

15 ×5 3

=

75 3

=

15. (a); Ratio of ages of Veer, Sandeep and Satish be 4 : 6 : 3 Let their age be 4x years, 6x years and 3x years respectively Age of veer’s wife = (66 – 4x) years Age of Sandeep’s wife = (78 – 6x) years Atq, Age of Satish’s wife 3

= × (66 − 4𝑥 + 78 − 6𝑥) 8 3

= × (144 − 10𝑥)𝑦𝑒𝑎𝑟s 8

Now, 4𝑥 3 (144−10𝑥) 8

=

4 3

⇒ 8x = 144 – 10x ⇒ 18x = 144 ⇒ x = 8 years Required sum = 8 × (4 +6 + 3) = 104 years 18

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16. (c); Let total teachers in a college = N Let maximum age = a year Minimum age = b years ATQ – a + b = 108 ------- (i) a – b = 27 --------(ii) From (i) & (ii) 2a = 135 year a = 67.5 years Ans, b = 40.5 years N× 45 − 14 × 32.5 + 4 × 67.5 + 2 × 40.5 + = (N – 14 + 6) 7(45N – 455 + 270 + 81) = 347N – 2776 315N − 728 = 347N – 2776 32N = 2048 N=

347 7

2048 32

N = 64 17. (e); Let present age of Veer & Saurbh be 3x and 4x respectively and present age of Divyaraj & Gopal be 2y and 3y respectively 3𝑥+4𝑥 2𝑦+3𝑦

=

63 65

x : y = 9: 13=9k:13k ATQ – 3x+4x+2y+3y=128 3× 9𝑘+4× 9𝑘+2×13k+3×13k=128k=128 K=1. Therefore, age of Veer is 27 years. Similarly, age of Gopal is 39 years. Sum of their present ages=66years After 8 years, sum of their ages = 66+16 = 82years. 19

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18. (d); Let, the present ages of Raju and his son be 𝑥 and 𝑦 respectively. 2n years ago, 𝑥 − 2𝑛 = 4(𝑦 − 2𝑛) 𝑥 = 4𝑦 − 6𝑛 ……….(i) 𝑛 years ago, 𝑥 − 𝑛 = 3(𝑦 − 𝑛) ⇒ 𝑥 = 3𝑦 − 2𝑛 …………..(ii) Solving (i) and (ii), 𝑦 = 4𝑛 And, 𝑥 = 4 × 4𝑛 − 6𝑛 = 10n N years later, 𝑥 + 𝑛 + 𝑦 + 𝑛 = 80 ⇒ 4𝑛 + 𝑛 + 10𝑛 + 𝑛 = 80 ⇒ 16𝑛 = 80 ⇒𝑛=5 Difference in their ages = 10𝑛 − 4𝑛 = 50 − 20 = 30 19. (d); Let, Present age of Amit, Inder and Satish be x, y and z respectively. ATQ, x –2n y –2n

5

=

4

⇒ 4x – 8n = 5y – 10n ⇒ 5y – 4x = 2n …(i) y –n z –n

=

9 7

⇒ 7y – 7n = 9z – 9n ⇒ 2n = 9z – 7y …(ii) Equating (i) & (ii) 5y – 4x = 9z – 7y 12y = 9z + 4x And, x – z = 12 x+n z+n

20

=

13 9

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⇒ 9x + 9n = 13z + 13n ⇒ 4n = 9x – 13z 9x – 13z = 18z – 14y 9x + 14y = 31z Now x = 12 + z 9 (12 + z) + 14y = 31z ⇒ 22z – 14y = 108 or 11z – 7y = 54 …(iii) and 10y – 8x = 9x – 13z 17x = 10y + 13z 17 (12 + z) = 10y + 13z 204 + 17z = 10y + 13z 4z + 204 = 10y …(iv) On solving (iii) and (iv) y = 30, z = 24, x =36 Required sum = 30 + 24 + 36 = 90 20. (c); Let four consecutive numbers of P₁ Series be a, (a + 4), (a + 8) and (a + 12) ATQ— 𝑎+(𝑎+4)+(𝑎+8)+(𝑎+12) 4

= 34

4a + 24 = 136 4a = 112 𝑎 = 28 P₁ series = 28, 32, 36, 40 Let P₂ series be b, (b + 9), (b + 18), (b + 27), (b + 36) Given, 𝑏 + 9 = 36 𝑏 = 27 P₂ series = 27, 36, 45, 54, 63 Let P₃ series be, 𝑐, (𝑐 + 1), (𝑐 + 2), (𝑐 + 3), (𝑐 + 4), (𝑐 + 5) and (𝑐 + 6) 21

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Given, 𝑐=

27 3

𝑐=9 P₃ series 9, 10, 11, 12, 13, 14, 15 Required average =

(27+36+45+54+63)

=

225 5

+

5 84 7

+

(9+10+11+12+13+14+15) 7

= 45 + 12 = 57

21. (d); Let A = 10a + b and B = 10b + a Let sum of other three number is = X X + 10a + b + 10b + a = 5 × (25.4) = 127 …(i) And –2X+10a+b+10b+a= 127 – 5 (43.2) …(ii) Solving (i) and (ii) We get a + b = 5 If 𝑎 would be → 0, 1, 2, 3, 4, 5 Then b would be → 5, 4, 3, 2, 1, 0 as 10a + b can’t be 50 because its reverse is 05 which is not a two digit number. Greatest 10a + b = 41 10a + b = 41 22. (e); Let 200 numbers are → a₁, a₂, a₅……, a₂₀₀ ATQ, 499.5 =

a1 +a2 +a3 +...+a200 200

⇒ 99900 = a1 + a2 + a3 +. . . +a200 Now, This form an A.P. a = a1 22

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d=1 n = 200 ⇒ 99900 =

200 2

[2a1 + (200– 1)1]

a1 = 400 Required sum for 1200 terms whose a1 = 400 =

1200 2

[2 × 400 + (1200– 1)1]

⇒ 1199400 Required Average =

1199400 1200

= 999.5

23. (c); Let, initial Number of people in the group be ‘n’. Let 21𝑥 𝑎𝑛𝑑 19𝑥 be ages of Sakshi and Sheetal respectively, A.T.Q., 4𝑛2 − 21𝑥 = 4(𝑛 − 1)2 ……..(i) And 4𝑛2 − 21𝑥 − 19𝑥 = 4(𝑛 − 2)2 4𝑛2 − 40𝑥 = 4(𝑛 − 2)2 ………….(ii) Solving (i) and (ii), 𝑥 = 4, 𝑛 = 11 Average age of group after Ritu leaves the group =

4×112 −21×4−19×4−20 11−3

=

304 8

= 38

Solution (24-25): Let number of people in X, Y and Z group is a 𝑎, 𝑏 and 𝑎 respectively. Let total money got by group X, Y and Z is h, k and k respectively. Average amount got by X and Y together ℎ+𝑘 𝑎+𝑏

=

3500 3

…(i)

Average amount got by Y and Z together 𝑘+𝑘 𝑎+𝑏

= 1000

k = 500 (𝑎 + 𝑏)…(ii) Put value of (𝑎 + 𝑏)from eq (ii) in eq (i) 23

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ℎ+𝑘 ℎ 𝑘 ℎ 𝑘

𝑘

=

7 3

+1= 4

7 3

= …(iii) 3

We know ℎ + 𝑘 + 𝑘 = 50,000 so, h = 20,000 and k = 15000 now ratio (i) becomes 𝑎 + 𝑏 = 30 And we know 𝑎 + 𝑏 + 𝑎 = 50 So, 𝑎 = 20 𝑏 = 10 24. (c); Number of people in Y group = 10 25. (d); Amount got by each person in group Z =

24

15000 20

= Rs. 750

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1

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Chapter

5

Percentage and Profit & Loss

BEST APPROACH TO SOLVE THE QUESTIONS A percentage is a number or ratio that represents a fraction of 100 i.e.

1 100

.

For faster and easy calculation one should use fractional equivalent of percentage. For example, it’s always easier to use

2 3

2

instead of 66 %. 3

It becomes easy to understand a question when one talks in terms of percentage, but while calculating, it is always the fractional equivalent that comes in hand. While solving the question one must try to stay away from unnecessary calculation. This happens in many of the questions related to profit and loss. Many times one starts calculating the absolute values of CP, SP etc. even when it’s not required. Let’s take a couple of sample questions. Example 1 : 'The CP of an item is Rs. 270. After giving a discount of Rs. 27 a shopkeeper earns a profit of 20%. What will be the profit percentage if he does not give the discount?’ Sol. Here, instead of finding out SP or MP, one should look at it the other way round. If discount of Rs. 27 is not given, then Rs. 27 will become further profit for the shopkeeper. 27 is 10% percent of 270. Profit of 20% is already there, so new profit% would become 20% + 10% = 30%. 2

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Example 2: ‘A man while buying cloth from a wholesaler uses a scale that measures 10% more than the actual length and while selling the same cloth he uses a scale that measures 10% less than the actual length. Find his overall profit or loss in this transaction.’ Sol. Here, one just needs to deal with the percentage the shopkeeper gains. While buying he gets 110 metre cloth on the cost of 100 metre coth and while selling he gains the cost of 100metre just by selling 90 metre cloth. 10 Multiplication factor while buying = 1+ =

11 10

Multiplication factor while selling = 1+ =

100−90 90

10 9

The final multiplying factor becomes = = 2

100

11 10 11 9

×

10 9 2

=1+

9

which clearly shows a profit of 22 %. 9

Practice Exercise Based on New Pattern

1.

Nishant bought an article at 20% discount on MRP, and claims to sell it at profit of 10% of MRP. When Nikhil offered him Rs. 500 banknote, he cheated again by giving him Rs. 125 instead of Rs. 225. Find overall profit% of Nishant. (a) 87.5% (b) 37.5% (c) 100% (d) 62.5% (e) None of these

2.

A shopkeeper marked up the price of a mobile phone by 40% of its cost price, if he increases the discount from 5% to 10%, the profit would decrease by 1400 Rs. How much profit shopkeeper would earn if he gives a discount of 20% on the marked price ? Assume that, he calculates discount only on MRP. (a) 1800 Rs. (b) 1200 Rs. (c) 2800 Rs. (d) 2200 Rs. (e) 2400 Rs. 3

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3.

A manufacturer of cricket balls wants to earn 25% profit on manufacturing cost after giving a discount of 23 ⅓% on MRP marked by him. But due to some reasons he lost 25% of balls & he decided to offer discount of 7 profit% or loss%. (a) 8% Profit (d) 8% Loss

9 13

% on MRP of remaining balls. Find his overall (b) 12% Profit (e) 5% Profit

(c) 12% Loss

4.

Rahul bought a cycle at a discount of 16⅔% on MRP. He earned half the amount of his CP by renting it for 200 days. After that he resells it at half of MRP. In this transaction he earned Rs. 200, find MRP of cycle (in Rs). (a) 1860 (b) 2490 (c) 2400 (d) 2280 (e) 2310

5.

A man purchased two items A & B and invested Rs 50 & Rs 75 on their repairing respectively. If he earns profit of 10% on A and 12% on B, overall profit earned by him is is Rs 84. But if he earns 20% on A & 10% on B, overall profit earned by him is 14% of total price of items. Find initial total purchasing price of both items. (a) Rs 675 (b) Rs 725 (c) Rs 750 (d) Rs 625 (e) Rs 775

6.

A shopkeeper sold two articles, if he marked up second article at 11 1 9

% above first article’s selling price and gives a discount of 20% on 1

that, then a loss of 20% occur on first one and 33 % profit on second 3

article. Find total selling price of both articles, if he made total loss of Rs.75 on both articles. (a) 4000 Rs (b) 4500 Rs (c) 5100Rs. (d) 4800 Rs (e) 5800Rs 4

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7.

Satish buy two articles i.e. type A at Rs 500 and type B at Rs 1500. He sold type A article at x% profit and mark up type B article 2x% above the cost price and gave x% discount at the time of the sale. By this Satish earn (x -6) % profit. Find the value of ‘x’. (a) 15% (b) 18% (c) 25% (d) 20% (e) 30%

Directions (8-10): A article is mark up above cost price such that markup percent is double of the profit percent. If discount is 12.5%, then profit 1

percent increased by 33 %. 3

8.

On selling 20 such article at 12.5% discount, profit is Rs.300. Find the M.P. of each article. (a) Rs.60 (b) Rs.160 (c) Rs.80 (d) Rs.240 (e) Rs.72

9.

If shopkeeper cheat his customer by giving 20% less quantity and reducing value of discount percentage by 20% then find the new profit percent. (a) 60%

(b) 75%

(d) 80%

(e) 70%

(c) 62.5%

10. Initially shopkeeper have 20 articles. Out of 20, 7 articles damaged and remains unsold. Marked Price should he labeled by how much percent more than cost price so that his overall profit does not change neither his discount percentage.

5

(a) 156%

(b) 146%

(d) 120%

(e) 125% Adda247 Publications

(c) 136%

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11. Thirty percent of total employees of Bankersadda are females, and sixty percent of female employees earn more than Rs. 17000 as monthly salary. If fifty five percent of total employees of Bankersadda earn more than Rs. 17000 as monthly salary. What fraction of total male employees of Bankersadda earn Rs. 17000 or less as monthly salary ? 13

11

(a)

(b)

20 33

(d)

100

(e)

(c)

20 69

51

100

100

12. A shopkeeper raised the marked price of an article by 60% and allow 1

2

2

7

three successive discounts of 12 %, 14 % and 20% on new MP and make a profit of 20%. If shopkeeper would allow only two discounts 1

of 12 % and 20% on new MP, he made a profit of Rs 800. Find the 2

cost price of article? (a) 1800 Rs. (d) 2400 Rs.

(b) 2000 Rs. (e) 2800 Rs.

(c) 1600 Rs.

13. Bankersadda started its publication with the 1500 DI books. Printing cost per book is 125 Rs, packing cost per book is 15 Rs and shipping cost per book is 40 Rs. if 50% of books sold on half of total cost price, then find on how much percent above should be remaining book sold to gets 20% profit on total outlay? (a) 50% (b)60% (c) 80% (d) 90% (e)70% 14. Adda247 publication sold a puzzle book in Rs. 475 and made profit of 25% on CP and sold a DI book in Rs. 575 and made a profit of 15% on CP. If publication sold puzzle book in Rs. 360, then find what price should DI book will be sold to made a profit of 30% on both the books ? (a) Rs.784 (b) Rs.874 (c) Rs.847 (d) Rs.748 (e) Rs.478 6

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Directions (15-16): Abhishek bought some chairs and tables from a shopkeeper. The marked price of a chair and a table were in the ratio 5 : 8. The shopkeeper gave discounts of 20% and 25% on the chair & the table respectively. The ratio of number of chairs and tables bought by Abhishek is 6 : 5. 15. If Abhishek sells each chair and table bought by him at discounts of 25% and 20% respectively after marking up the prices of both by 50% and gives one table free for every four chairs bought by a customer 2 and only rd of the total chairs are sold in bunch of four chairs, then 3 what is the net profit /loss % made by Abhishek after selling all of the items which he bought from the shopkeeper? 2 1 1 (a) 6 % (b) 3 % (c) 2 % 3 1

(d) 4 % 4

3

2

(e) 5 %

16. If the marked price of a table set by the shopkeeper was Rs.300 more than that of a chair and the total expenditure made by Abhishek in purchasing the chairs and table from the shopkeeper was Rs.108000, then how many chairs were purchased by Abhishek? (a) 150 (b) 60 (c) 120 (d) 90 (e) 80 Directions (17-18): A car ‘X’ is manufactured in a company ‘A’. Cost of raw material on a car is 25% of total cost and cost of labor is 20% of rest. 30% of total cost occur due to four engineers working there and remaining cost is due to machines and taxes which is in the ratio 7 : 3. Company sold the car ‘X’ at 28% profit. 17. If cost of raw material is increased by 30% and all their 4 engineers leave the company, so they hire 2 new engineers by giving 50% more salary as they give to previous engineer each. Find the % change in selling price so overall profit must remain same. (a) 0% (b) 100% (c) 2% (d) 50% (e) 3% 7

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18. Company A, started making car ‘Y’ in which same raw material used, labor cost is increased by 20%. If all other cost is same then find the ratio of selling price of car ‘X’ to car ‘Y’, given that car ‘Y’ sold at 20% profit. (a) 331 : 309 (b) 320 : 314 (c) 313 : 309 (d) 320 : 309 (e) 309 : 320 19. Marked price of two articles A and B are in the ratio of 7 : 9, at the time of selling shopkeeper gives discount of d% on article A and (d + 5)% on article B and made a profit of 25% on each article, if cost price of article A and B are in ratio 112 : 135. Then find percent of discount given by shopkeeper on both articles? (a) 12.5%, 17.5% (b) 25%, 30% (c) 10%, 15% (d) 15 %, 20% (e) 20%, 25% 20. Neeraj purchased two mobile from a shop. He sold first mobile at the price of Rs. 18750 and second mobile at the price of Rs. 14250. If the profit percent on first mobile is five times of the loss percent on second mobile, then find the overall profit made by Neeraj after selling both the mobile phones. (C.P of both mobile is same)? (a) 15% (b) 12% (c) 14% (d) 10% (e) 18% 21. There are three societies A, B and C. Ratio of total population of A, B and C is 2 : 3 : 5. In society A, 35% of total population is female. In 2

society B, number of females is 66 % less than number of males of 3

society B. In society C, ratio of male to female is 27 : 23. Find total number of females is what percent less than total number of males in all societies together ? 1

(a) 33 %

(b) 40%

(d) 66 %

(e) 75%

3 2 3

8

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(c) 60%

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22. The marked price of a bike in two stores A and B is same.In store A bike is available at 30% discount and in store B same bike is available at two successive discounts of d% and 8%. A man bought bike from store A at Rs. 49000. If man would have paid Rs. 7672 more, he could have bought the same bike from store B. Find the discount ‘d’ allowed by store B on bike ? (a) 8% (b) 10% (c) 12% (d) 16% (e) 15% 23. A shopkeeper gives a discount of 19% on marked price of article A 1

and selling price of article B is 11 % more than selling price at article 9

A. If shopkeeper made 20% profit on article B and selling price of article A was Rs. 840 more than cost price of article B, then find the cost price of article A. Shopkeeper sold article A at profit of 25%? (a) 9070 Rs. (b) 9060 Rs. (c) 9040 Rs. (d) 9020 Rs. (e) 9072 Rs. Directions (24-25): A sale conducted on Samsung mobiles store where Cost price of three types Samsung mobiles A, B and C in the ratio of 5 : 7 : 9. Store owner made mark price of mobile A and C, 30% above cost price and mobile B, 40% above cost price. Discounts allowed by store on mobiles A, B and C are 15

5 13

3

1

7

9

%, 21 % 𝑎𝑛𝑑 11 % respectively.

24. A man purchased fourteen type A mobile, twenty type B mobile and fifteen type C mobile in sale. If store owner made a total profit of Rs. 17500 on all type A mobile, purchased by man than find total profit made by store owner on type B and type C mobiles, which was purchased by man? (a) 82500 Rs. (b) 84500 Rs. (c) 78500 Rs. (d) 87500 Rs. (e) 88500 Rs.

9

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25. Store decided to give no discount on purchase of any type of mobile it will be given two mobiles free on purchase of six type A mobiles, one mobile free on purchase of five type B mobiles and three mobiles free on purchase of ten type C mobiles. A retailer come to purchase mobile and take away48 type A mobiles, 36 type B mobiles and 39 type C mobiles. Find overall loss percentage of store owner in this transaction?(2 marks) 28 28 76 (a) 10 % (b) 8 % (c)4 % (d) 14

281 28

281

%

(e) 9

281 28

281

281

%

26. Ratio between marked price of article A to article B is 4 : 5. Shopkeeper allowed d% discount on article ‘A’ and (d + 18)% discount on article ‘B’, so selling price of both articles become equal. If shopkeeper made a profit of 20% on article A and 25% on article B and profit made on article B is Rs. 384 more than that of article A, then find the cost price of article ‘A’ and article ‘B’ respectively? (a) 9000 Rs. 8400 Rs (b) 9600 Rs. 9216 Rs. (c) 9800 Rs. 9012 Rs. (d) 9600 Rs. 8488 Rs. (e) 9200 Rs. 9216 Rs. 27. Aman started a shoes manufacturing company, manufacturing cost of each shoes is Rs. 500 , Raw material cost is Rs 300 and transport cost of Rs 100 If company manufactured 150 Shoes in first order and sold 50% of shoes on half of its cost price. Then find at what price of should the company sold the remaining shoes to gets 50% profit on its total cost? (a) 100% (b) 150% (c) 75% (d) 125% (e) 120% 28. The ratio between marked price of a same watch in two shops P and Q is 7 : 5 . In shop P watch available at two successive discounts of 10% & 25% , while in shop Q watch available at two successive discounts of d% and 10%. If selling price of watch at shop Q is less than that of 44 selling price of watch at shop P by 𝑡ℎ of the selling price of watch at 105 shop P, then find the value of ‘d’ ? (a) 39% (b) 29% (c) 31% (d) 35% (e) 37% 10

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29. Cost price of a Samsung mobile is 20% more than cost price of a Mi mobile. Shopkeeper marked up Samsung mobile at 25% above its cost price and Mi mobile at 40% above cost price. If shopkeeper allowed discount of 20% on Mi mobile and 25% on Samsung mobile and difference between selling price of Samsung mobile and Mi mobile was 160 Rs. then find sum of marked price of one Samsung mobile and one Mi mobile? (a) 92840 Rs (d) 92800 Rs

(b) 92260 Rs (e) 98530 Rs

(c) 98460 Rs

30. In a city 60 percent are male and remaining are females. Out of total males, 20% of male are government employees. Out of remaining males, ratio between private employees to unemployed male is 5 :7. Out of total females, 40% of female are in government jobs and out of remaining females 40% of female work for private firms and remaining are unemployed. Find the total number of male and female, who are government employees, if difference between male and female who are unemployed is 3400? (a) 6000 (b) 5000 (c) 7000 (d) 4000 (e) 3000 31. A person ‘A’, invested 37½% of his savings in scheme XYZ, while B invested 62½% of his savings in the same scheme. If B invested 50% of his remaining savings in scheme PQR, and A thought to invest double of the amount invested by B in PQR scheme then find what percent of A’s remaining savings after investment in XYZ, A should invest in PQR. Given that amount invested by B in XYZ is twice the amount invested by A in same scheme. (a) 80% (b) 68% (c) 64% (d) 72% (e) 60% 11

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32. A man bought few kg’s of sugar of 3 variety (X, Y, Z) in the ratio of their quantity 2 : 3 : 5 respectively and started to sell them at profit of 20%, loss of 10% and profit of 32% respectively. He managed to sell only 80% of total sugar and rest was return at the same price as he bought. The price of these sugar is Rs. 30/kg, Rs. 40/kg and Rs. 50/kg respectively. Calculate his total profit amount, if total sugar bought was 50 kg and he sold all the sugar of X and Y types. (a) 320

(b) 300

(c) 0

(d) 180

(e) None of these

33. In an election survey, 83 ⅓% of total voters took part in survey, 50% then claims to votes for candidate A, 10% are uncertain & rest says to vote for B. If all of them voted according to their commitments on the day of election and those who were not part of survey, voted to A and B in ratio of 2 : 1. All the people who are uncertain in survey, voted to A : B in the ratio of 1 : 4 in election. If A won by 640 votes then find total number of votes in election. (a) 6000 (b) 5000 (c) 9000 (d) 7200 (e) 6750 2

34. In a village 60% are males in which 6 % males are above 60 years, in 2

3

remaining 14 % males are below 18 years, out of remaining 1

7

62 % are above 18 years but below 50 years and remaining males are 2

above 50 years but below 60 years. If difference between males above 60 years and males above 18 years but below 50 years are 7800, then find total female population of village? (a) 16000 (b) 18000 (c) 20000 (d) 12000 (e) 24000 12

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35. Divyaraj purchased jeans and shirt from a seller. Marked price of 2

jeans and shirt are in the ratio of 9 : 7 and seller offered 14 % 7

1

discount on shirt and 11 % on jeans and number of jeans and shirt 9

purchased by Divyaraj are in the ratio of 5 : 8. If Divyaraj marked up jeans and shirt 50% above their cost price and offered 25% discount 1

on jeans and 11 % on shirt, Find overall profit of Divyaraj in this 9

transaction. (a) 22

19

%

(b) 21

(d) 23 %

(e) 25

22 19

22

19 22 19 22

%

(c) 19

19 22

%

%

36. In 2016 a store sold some mobile phone at certain price. In 2017 store increase selling of mobile phone by reducing the price of mobile phone by 25%. The total revenue generated by selling mobile phone in 2017 is increased by 5% compared to previous year. The total revenue generated by store in 2017 is 94.5 lakh and number of mobile phones sold by store in the year 2017 is 90 more than that in 2016. Find initial cost of one mobile. (a) 32000 Rs. (b) 40000 Rs. (c) 36000 Rs. (d) 24000 Rs. (e) 20000 Rs. 37. Veer bought some Shirts & some Jeans from a store. The marked price of a shirt and a jeans is in the ratio 5 : 7 and store allows discounts of 20% and 25% on the shirt & the jeans respectively. The ratio between number of shirt and jeans bought by Veer is 9 : 8. Veer marking up the prices of both by 50% on the price at which he bought and sells each shirt and jeans bought by him at discounts of 25% and 20% respectively. If Veer gives one 1 shirt free for every 4 jeans bought by a customer, then what is the net profit made by Veer after selling all of the items which he bought from the store? (a) 2% (b) 7% (c) 6% (d) 5% (e) 8% 13

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38. A shopkeeper has two types Wheat i.e Type A & Type B and cost price per kg of Type A wheat is 10 times of cost price per kg of type B wheat. Shopkeeper cheated a costumer by saying that he will give him 200 kg of Type A wheat but shopkeeper gives 200 kg of mixture of Type A & Type B wheat on 10% above the cost price per kg of Type A 7 wheat. If shopkeeper made a total profit of 71 % in this transaction, 8 then find ratio between quantity of Type B wheat to Type A wheat in sold mixture? (a) 2 : 5 (b) 2 : 3 (c) 1 : 3 (d) 2 : 7 (e) 1: 5 1

39. In a store marked price of a shirt is 33 % less than marked price of a 3 jeans. There are two discounts schemes A & B. In Scheme A, if someone buy one jean and one shirt together, then store gives overall discount of 35%, while in scheme B if someone buy one shirt and one jeans separately, store give 20% discount on shirt and 25% on jeans. If difference between selling price of one jeans and one shirt in scheme A and B is Rs. 384 then find cost price of one jeans & one shirt, 1 given that store made of 33 % on shirt and 25% on jeans, when it 3 sold one jeans and one shirt in scheme B ? (a) 720 Rs, 1150 Rs (b) 768 Rs. 1152 Rs. (c) 786 Rs. 1168 Rs. (d) 796 Rs. 1144 Rs. (e) 790 Rs. 1108 Rs. 40. The Hero company manufactured cycles, 40% of the total manufactured cost is on raw material, 20% on labour charges, 20% on fixed charges and the rest on transportation. The cycle is sold at a profit of 25%. The price of the raw material increased by 15% and the labour charges increased by 20% and the cost on the transportation increased by 50% while the fixed costs remained unchanged. If the 7 manufacturer wants a 13 % profit, then by what percentage should 11 company reduced its expenditure on raw materials (at the increased price), the selling price remaining the same ? 17 17 17 (a) 19 % (b) 21 % (c) 19 % (d) 23 14

23 17 23

%

(e) 27

23 17 23

23

%

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

Solutions

1.

(a); Nikhil gave Rs. 500 note, & Nishant have to actually pay him Rs. 225. Hence Nishant sold it at Rs. 500 – 225 = 275 to Nikhil. Now, Let MRP is 100x. If he will get 10% profit, then SP will becomes 110x. 110x = 275 x = 5.5 ∴ MRP is Rs. 250 80

And CP for Nishant is

100

× 25 = Rs. 200.

He got Rs. (500 – 125) = Rs. 375 from Nikhil. Hence Profit % = (

375–200 200

) × 100

7

= × 100 = 87.5% 8

2.

(e); CP of Mobile M.P. mobile 100%

140%

S.P. of mobile 133%(1st discount) 126%(2nd discount)

(133 – 126)% = 1400

7% = 1400 C. P. =

1400 7

× 100

= 20000 Rs. S. P. after 20% discount = 20000 ×

112 100

= 22400 Profit = 2400 Rs. 15

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3.

(b); Let he manufacture 𝑛 balls and manufacturing cost of each ball is 𝑦 And 𝑥 is marked price. Now, ATQ 𝑥 [1– ] = 𝑦 ×

7

5

30

4

23

5

⇒𝑥×

30

=𝑦×

4

5

30

4

23

⇒𝑥 =𝑦× ×

=𝑦×

75 46

…(i)

3

1

12

4

13

13

After accident he sells 𝑛 balls on a price of 𝑥 (1– ) = 𝑥 × 3

3

12

4

4

13

Total SP of 𝑛 balls = 𝑛. 𝑥. 3

75

4

46

= ( × 𝑛) (𝑦 × = 𝑛𝑦 ×

12

)( ) 13

675 598 675

Hence profit % = =

4.

77 598

𝑛𝑦×598−𝑛𝑦 𝑛𝑦

× 100

× 100 ≅ 12% profit

(c); Let the MRP of cycle is 6𝑥. ∴ discount is

50 3

% of 6x = x

∴ Rahul bought this cycle for 5x He earns 2.5x by renting it and resells it at 3x. ATQ 3x + 2.5x – 5x = 200 0.5x = 200 ⇒ x = 400 ∴ MRP = 6x = 6 × 400 = Rs. 2400 16

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5.

(d); Let the purchasing price of items A is B be Rs x – 50, Rs y – 75 respectively. When he applied Rs 50 & Rs 75 on them, their actual CP becomes x & y ATQ 10 100

12

x+

y = 84

100

…(i)

Also, 20 100

10

x+

y=

100

14 100

(x + y)

⇒ 20x + 10y = 14x + 14y ⇒ 6x = 4y 2 x= y 3 Put this value in equation (i) 10 (2y) 100 3 20y+36y 300

⇒y=

+

12 100

y = 84

= 84 84×300 56

= Rs 450

2

x = 450 × = Rs 300 3

Initial purchasing price is = (300 + 450) – (75 + 50) = Rs 625 6.

(c); Let marked price of 10x Rs. Selling price of first article be 9x Rs.

second

article

be

Selling price of second article = 10x ×

8 10

= 8x Rs.

C.P. of fisrt article = 17

9x 4

× 5 = 11.25x Rs. Adda247 Publications

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

3

C.P. of second article = 8𝑥 ×

4

= 6x Rs.

Total C.P. of both articles = 11.25x + 6x = 17.25x Rs. Total S.P. of both articles = 9x + 8x = 17x Rs. ATQ— 17.25x – 17x = 75 x=

75 0.25

= 300

selling price of both articles = 17 × 300 = 5100 Rs. 7.

(d); ATQ 500 [ 1 +

𝑥 100

] + 1500 [1 +

500 + 5𝑥 + 1500 [1 +

2𝑥 100

500 + 5𝑥 + 1500 + 15𝑥 − 2000 + 20𝑥 − 3𝑥 2 10

3𝑥 2 10

2𝑥 100

−

] [1 −

𝑥 100

3𝑥 2 10

−

𝑥 100

] = 2000 [1 +

2𝑥 2 100×100

𝑥−6 100

] = 2000 [1 +

]

𝑥−6 100

]

= 2000 + 20(𝑥 − 6)

= 2000 + 20𝑥 − 120

= 120

x² = 400 ⇒ x =20%

18

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Solutions (8-10): Let cost price = 100 Profit percent = 𝑥% So, selling price ⇒ 100 + 𝑥 M.P. ⇒ 100 + 2𝑥 Now if discount is 12.5% 1

4

4

3

3

3

Then profit percentage increases by 33 % ⇒Profit percent = 𝑥 × = 𝑥 ATQ, 4

7

3

8

100 + 𝑥 = (100 + 2𝑥) 𝑥 = 30 So, if cost price = 100 Selling price = 130 M.P. = 160 Profit = 30 Discount % =

30 160

× 100 = 18.75%

8.

(c); Profit on one articles = 30 Profit on 20 articles = 30 × 20 ⇒ 600 600 → 300 1 → 0.5 M.P. → 160 → 160 × 0.5 ⇒ Rs.80

9.

(e); Let total quantity ⇒ 1000 He gives → 800 → for → 100 Cost price of → 800 → 80 Now initial discount = Reduced discount = S.P. =

160×85

19

× 100 = 18.75%

160 18.75×4 5

= 15%

= 136

100 (136–80)

Profit =

30

80

× 100 = 70% Adda247 Publications

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

10. (b); Let cost price of 20 articles → 20 × 100 = 2000 Actual profit on 20 articles → 20 × 30 = 600 S.P. → 2600 S.P. of each undamaged article ⇒

2600

= 200

20–7

Initial discount =

30 160

× 100 = 18.75%

So M.P. of each article should be =

200 81.25

× 100 ≈ 246

Approximately markup% = 146% 11. (d); Let total employee in Bankersadda = 100 Females employee = 30 Male employee = 70 Employee who earned more than Rs. 17000 as monthly salary = 55 Employee who earned less than Rs. 17000 as monthly salary = 45 Total

female

employee

earned

less

than

Rs. 17000 as monthly salary = 30 ×

40 100

= 12

Total male employee earned Rs. 17000 or less as monthly salary = 45-12 = 33 Required fraction = 20

33 100

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12. (b); Let original marked price be Rs 100x. Then,

New

marked

price

of

article

= Rs 160x. Selling price of article 7

6

4

8

7

5

= 160𝑥 × × × = 96x Rs.

C.P. of article = 96x ×

5 6

= 80x Rs. 2nd selling price 7

4

8

5

= 160𝑥 × ×

= 112x Rs.

Given 112x – 80x = 800 32x = 800 x = 25 cost price = 80 × 25 = 2000 Rs. 13. (d); Total CP of one DI book = 125 + 15 + 40 = 180 Rs. C.P. of 1500 books = 1500 × 180 SP of 750 book =

750×180 2

Let required profit percent = x%

21

∴

750×180

=

120

2 100

+(

100+x 100

) × (750 × 180)

× 180 × 1500 Adda247 Publications

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67500 + 135000 + 1350x = 324000 1350x = 324000 – 202500 x=

121500 1350

= 90%

14. (a); CP of puzzle book =

475 125

× 100

= 380 Rs. CP of DI book =

575 115

× 100

= 500 Rs. For 30% profit on both = (380 + 500) ×

130 100

= 1144

DI book should be sold = 1144 – 360 = 784 Rs. 15. (b); Let the MP of a chair and a table be Rs.5x and Rs.8x respectively. And, the number of chairs and tables bought be 6y and 5y respectively. CP of a chair for Abhishek = (100 – 20)% of 5x = Rs.4x CP of a table for Abhishek = (100 – 25)% of 8x = Rs.6x Total CP for Abhishek = 4x × 6y + 6x × 5y = 24xy + 30xy = 54xy SP of a chair for Abhishek = (100 – 25)% of (100 + 50)% of 4x = 4.5x SP of a table for Abhishek = (100 – 20)% of (100 + 50)% of 6x = 7.2x Number of chairs sold in bunch of four by Abhishek 22

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2

= rd of 6y = 4y 3

So,

number

of

table

sold

for

free

by

Abhishek

1

= th of 4y = y 4

Total SP for Abhishek = 4.5x × 6y + 7.2x × (5y – y) = 27xy + 28.8xy = 55.8xy Profit % = =

1.8xy 54xy

55.8xy – 54xy 54xy 1

× 100

× 100 = 3 % 3

16. (c); According to the question, MP of a table = 300 + MP of a chair ⟹ 8x = 300 + 5x ⟹ x = 100 Total CP for Abhishek = 108000 ⟹ 54xy = 108000 ⟹ 54 × 100 × y = 108000 ⟹ y = 20 Number of chairs purchased by Abhishek = 6y = 120 Solutions (17-18): Let total cost of car X = 100𝑥 Cost of raw material = 25𝑥 Cost of labor =

(100–25𝑥)×20 100

= 15𝑥

Cost of engineers = 30𝑥 Cost of machine = 21𝑥 Cost of taxes = 9𝑥 Profit = 28𝑥 So, proposed selling price = 128𝑥 23

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17. (a); New cost of raw material =

25𝑥×13 100

Now cost of each engineer =

30𝑥 4

= 32.5𝑥

= 7.5𝑥

Cost on new engineer = 2 × 7.5𝑥 ×

150 100

= 22.5 So new CP = 100𝑥 + 7.5𝑥 – 7.5𝑥 = 100𝑥 So, % change in S.P. = 0%

18. (d); Cost on labor for car Y =

15𝑥×120 100

= 18𝑥

Cost price of car Y = 103𝑥 Required ratio =

128𝑥 103𝑥×120

=

320 309

19. (e); Let cost price of article A Rs. 112x and cost price of article B Rs. 135x ATQ, Selling price of article A =

112𝑥 4

×5

= 140x Rs. Selling price of article B =

135𝑥 4

×5

= 168.75x Rs. Let mark price of article A is 7y And article B is 9y 140x = 7y ×(

100−𝑑 100

168.75x = 9y ×( 24

)

100–(𝑑+5) 100

________(I) ) ________(II)

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From (I) and (II) _________ 100–𝑑

140𝑥 168.75𝑥

=

140×9 168.75×7 16 15

7𝑦×( 100 ) 95–𝑑

9𝑦×( 100 )

=

100–𝑑 95–𝑑

100–𝑑

=

95–𝑑

16d – 1520 = 15d – 1500 First discount d = 20% Second discount = (20 + 5) = 25% 20. (d); Let cost price of each mobile = 100x Rs. Percent

profit

18750−100𝑥

= [(

100𝑥

Loss 100𝑥

mobile

second

mobile

𝑥

on

)]× 100 =

first

18750−100𝑥

percent

100𝑥−14250

= [(

)] × 100 =

on

100𝑋−14250 𝑥

ATQ – 18750−100𝑥 𝑥

=5×

100𝑋−14250 𝑥

18750 − 100𝑥 = 500𝑥 − 71250 600x = 90000 ⇒ x = 150 Rs. Total cost price of both mobile = 2 × 15000 = 30000 Rs. Total selling price of both mobile = 18750 + 14250 = 33000 Rs. Profit percentage =

25

33000−30000 30000

×100 = 10%

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21. (b); Let population of society A, B and C be 200x, 300x and 500x respectively 35

Female in society A =

100

× 200x = 70x

Male in society A = 200x – 70x = 130x In society B Let ‘a’ = Number of males ATQ, 1 3

4𝑎

a + a = 300x ⇒

3

= 300𝑥

⇒ a = 225x = Number of males Number of females = 300x – 225x = 75x In society C, Number of females = Number of males =

23 50

27 50

× 500x = 230x

× 500x = 270x

Total number of female = 70x + 75x + 230x = 375x Total number of male = 130x + 225x + 270x = 625x Required% = =

250 625

625x –375x 625x

× 100

× 100 = 40%

22. (c); Let marked price of bike = 100x Rs. Selling price for store A = 100x ×

(100 –30) 100

= 70x

Given, 70x = 49000 ⇒ x = 700 marked price of bike = 70000 Rs. 26

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ATQ— Selling price of bike in store B= (100−𝑑)

Marked price of bike×

100

×

(100−8) 10

= 𝑠𝑒𝑙𝑙𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 𝑜𝑓 𝑏𝑖𝑘𝑒 𝑖𝑛 𝑠𝑡𝑜𝑟𝑒 𝐴+7672 70000 ×

(100 –d) 100

×

(100 –8) 100

= (49000 + 7672) 644d = 64400 – 56672 d=

7728 644

⇒ d = 12 %

23. (e); Lets MP of article A = 100x Rs. Selling price of article A = 100x ×

(100−19) 100

= 81x Rs.

Selling price of article B = 81𝑥 ×

10 9

= 90x Rs. Cost price of article B =

90𝑥

Cost price of article A =

81𝑥

120 125

× 100 = 75x × 100

= 64.8x Rs. Given, 81x – 75x = 840 6x = 840 ⇒ x = 140 Rs. Cost price of article A = 64.8 × 140 = 9072 Rs. 27

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Solutions (24-25): Lets cost price of A, B and C type of mobile be Rs. 5x, Rs. 7x and Rs. 9x respectively M.P. of type A mobile = 5𝑥 ×

130

M.P. of type B mobile = 7𝑥 ×

140

M.P. of type C mobile = 9𝑥 ×

130

100 100 100

= 𝑅𝑠. 6.5𝑥 = 𝑅𝑠. 9.8𝑥 = 𝑅𝑠. 11.7𝑥

S.P. of type A mobile = 6.5𝑥 × (100–

200 13

)×

1 100

= 6.5𝑥 ×

11 13

= 𝑅𝑠. 5.5𝑥

S.P. of type B mobile = 11.7𝑥 × (100–

150 7

)×

1 100

= 𝑅𝑠. 7.7𝑥 8

S.P. of type C mobile = 11.7 × = 𝑅𝑠. 10.4𝑥 9

24. (d); Total

cost

price

of

Fourteen

type

A

mobile

= 14 × 5x = Rs. 70x Total selling price of fourteen type A mobile = 14 × 5.5x = 77x Given total profit = Rs. 17500 7x = 17500 ⇒ x = Rs. 2500 Total profit made by store owner on type B and type C mobile = (7.7 × 20 – 7× 20) × 2500 + (10.4 × 15 – 9 × 15) × 2500 = 14 × 2500 + 21 × 2500 = 35000 + 52500 = Rs. 87500

28

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25. (c); In the transaction — Given, In 48 type A mobile, 12 mobile are free. In 36 type B mobile, 6 mobile are free. In 39 type C mobile, 9 mobile are free. Total profit of store owner on selling of 48 type A mobile = 48 × 5x – 36 × 6.5x = 240x – 234x = 6x Total loss of store owner on selling of 36 type mobile = 36 × 7x – 30 × 9.8x = 252x – 294x = 42x Total loss of store owner on selling of 39 type C mobile = 9x × 39 – 11.7x × 30 = 351x – 351x = 0 Total loss of store owner in this transaction = 42x- 6x = 36x Total cost price = 48 × 5x + 36 × 7x + 39 × 9x = 240x + 252x + 351x = 843x Required% =

36𝑥 843𝑥

× 100 = 4

76 281

%

26. (b); Let marked price of article A and B be 400x and 500x respectively ATQ— 400x ×

(100–d) 100

= 500x ×

(100–d−18) 100

400 – 4d = 410 – 5d ⇒ d = 10% 90

Cost price of article A = 29

400x×100 120

× 100

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= 300x Rs. Cost price of article B =

(100–28) 100

500x×

125

× 100

= 288x Rs. ATQ— (500x ×

72 100

– 288x) – (400x ×

90 100

– 300x) = 384

72x – 60x = 384 ⇒ x = 32 Cost price of article A = 32 × 300 = Rs.9600 Cost price of article B = 32 × 288 = Rs.9216 27. (b); Total CP of one shoes = (500 + 300 + 100) = 900 CP of 150 shoes = 900 × 150 CP of 75 shoes =

900×75 2

Let’s required percentage =x ∴

900×75

=

150

2 100

+(

100+𝑥 100

) × (900 × 75)

× 900 × 150

= 33750 + 67500 + 675x = 202500 = 675x = 202500 – 101250 675x = 101250 𝑥=

30

101250 675

⇒ x = 150%

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28. (a); Let marked price of watch at shop P & Q be Rs. 140x and Rs. 100x respectively Selling price of watch at shop P = 140𝑥 ×

90 100

Selling

×

75

= 94.5𝑥 𝑅𝑠.

100

price

= 100x ×

(100−𝑑) 100

×

of

watch

at

shop

Q

90 100

ATQ – 94.5x - 100x ×

(100−𝑑) 100

×

90 100

= 94.5𝑥 ×

44 105

9450 – 9000 + 90d = 3960 90d = 3510 d=

3510 90

⇒ d = 39%

29. (d); Let cost price of one Mi mobile be Rs. 100x and cost price of one Samsung mobile be Rs.120x Marked price of Samsung mobile = 120x ×

125 100

= 150x Rs.

Marked price of Mi mobile = 100x ×

140 100

= 140x Rs.

Selling price of Samsung mobile = 150x ×

75 100

= 112.5 x Rs.

Selling price of Mi mobile = 140x ×

80 100

= 112x Rs. 31

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ATQ– 112.5 x – 112x = 160 ⇒ x = 320 Rs. Marked price of one Samsung & one Mi mobile = (150 × 320) + ( 140 × 320) = 48000 + 44800 = 92800 Rs. 30. (c); Male 60% Government Employees = 60 ×

20 100

= 12%

Private employees = 48 ×

5 12

= 20%

Female 40% Government employees 40

= 40 ×

= 16%

100

Private employees = 24 ×

40

= 9.6%

100

Unemployed = 48 – 20 =28% Unemployed = 24 – 9.6 = 14.4% Given — 28% – 14.4% = 3400 ⇒ 13.6% = 3400 Government (male + female) employees = 12% + 16% = 28% Government (male + female) employees =

3400 13.6

× 28 = 7000

31. (d); Let the savings of A & B be Rs 8 x and Rs 8y respectively. Then amount invested by A in XYZ scheme = Amount invested by B in XYZ =

62.5 100

37.5 100

× 8𝑥 = 3x

8y = 5y

ATQ, 32

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⇒ 6x = 5y 5

6

6

5

x = y or y = x Now

B

invested

50%

of

(8y

–

5y)

=

1 2

×

3y

= 1.5y in scheme PQR. A wants to invest 2× 1.5y = 3y 6

18 𝑥

5

5

=3× x=

His remaining savings = 8x – 3x = 5x % he could invest in PQR = =

1800 25

18x 5

5x

× 100

% = 72%

32. (e); He bought total of 50 kg sugar. And ratio of quantity of both is given as 2:3:5. Therefore, he have 10 kg of X, 15 kg of Y and 25 Kg of Z. And cost price of X, Y & Z is Rs. 30/kg, Rs. 40/kg and Rs. 50/kg Profit earn on 1kg of X is Loss earn on 1 kg of Y is

20 100 10

100

× 30 = Rs. 6

× 40 = Rs. 4

Profit earn on 1 kg of Z is 32/100] × 50 = Rs. 16 But he sold all of X & Y and 40 – (10 + 15) =15 kg of Z. ∴ Total amount of profit earned is 6 × 10 – 4 × 15 + 16 × 15 = 60 – 60 + 240 = Rs. 240 33

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33. (d); Let total votes be 6𝑥 People who took part in survey = 83⅓% × 6𝑥 = 5𝑥 2.5𝑥 claims to vote for A, 0.5𝑥 are uncertain. People who didn’t took part in survey = 6𝑥 – 5𝑥 = 𝑥 2

And vote for A=

3

2

×𝑥 = 𝑥 3

Vote for A, from people who are uncertain. 1

= × 0.5𝑥 = 0.1𝑥 5

2

Total votes for A = 2.5x + 0.1x + 𝑥 3

=

25

=

75𝑥+3𝑥+20𝑥

10

x+

1 10

30

x+

20 30

=

x

98𝑥 30

Votes for B = 6𝑥–

98𝑥 30

=

82𝑥 30

ATQ, 98𝑥 30

–

82𝑥 30

=

16𝑥 30

⇒

16𝑥 30

= 640

⇒ 𝑥 = 40 × 30 = 1200 Hence total votes are 6 × 1200 = 7200 34. (d); Let total population of village = 100x Male population = 60x Male above 60 year = 60x × 34

1 15

= 4x

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1

Male below 18 years = (60x – 4x) × = 8x 7

Male above 18 but below 50 years = (56x – 8x) ×

5 8

= 30x

Male above 50 year but below 60 years = (56x – 8x – 30x) = 18x ATQ— 30x – 4x = 7800 26x = 7800 ⇒ x = 300 Female population in village = 300 × 40 = 12000 35. (d); Let marked price of Jeans be Rs. 9x and shirt be Rs. 7x for Divyaraj 8

CP of jeans for Divyaraj = 9x × = 8x Rs. 9

6

CP of shirt for Divyaraj = 7x × = 6x Rs. 7

Let total number of jeans purchased by Divyaraj be 5y and shirt be 8y Total cost price for Divyaraj = 8x × 5y + 6x × 8y = 88xy SP of jeans on, which Divyaraj sold 3

3

2

4

= 8x × × = 9x SP of shirt on, which Divyaraj sold 35

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3

8

2

9

= 6x × × = 8x Total S.P. on which Divyaraj sold all items = 9x × 5y + 8x × 8y = 45xy + 64xy = 109xy Required% = = 23

19 22

109xy –88xy 88xy

× 100

%

36. (b); Let cost of each mobile be Rs. ‘a’ in 2016 and number of mobile sold in 2016 be ‘b’ Total revenue generated by store in 2016 = ab Rs. In 2017 — Cost of each mobile = 0.75a Rs. Total revenue generated by store in 2017 =

105 100

× ab = 94.5 lakh

ab = 90 lakh …(i) Total number of mobile sold by store in 2017 =

1.05ab 0.75a

= 1.4b

Given, Number of mobile sold in 2017 – number of mobile phone sold in 2016 = 90 1.4b – b = 90 b = 225 Cost price of each mobile in 2016 = 36

90,00,000 225

= 40000 Rs. Adda247 Publications

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37. (d); Let the marked price of a Shirt = 500 Rs And the marked price of a Jeans = 700 Rs Also, no. of shirt bought by Veer = 9x And no. of jeans bought by Veer = 8x 4

C.P. of shirt for Veer = × 500 = 400 Rs. 5

3

C.P. of jeans for Veer = × 700 = 525 𝑅𝑠. 4

Total C.P. of shirt and jeans for Veer = 9𝑥 × 400 + 8𝑥 × 525 = 7800𝑥 Rs. Total S.P. of shirt and jeans for Veer = 7𝑥 × 400 × 1.5 × 0.75 + 8𝑥 × 525 × 1.5 × 0.8 = 3150𝑥 + 5040𝑥 = 8190𝑥 Rs. Profit % = =

390 7800

8190𝑥−7800𝑥 7800𝑥

× 100

× 100 = 5%

38. (b); Let cost price per kg of Type B wheat = x Rs. And, Cost price per kg of type B wheat = 10x Rs. Selling price of mixture of Type A & Type B wheat = 11x Rs. Cost price per kg of mixture of Type A & Type B wheat = 11x ×

32 55

= 6.4𝑥 𝑅𝑠.

ATQ – Let N kg of type B wheat in mixture of 200 kg of Type A & Type B of wheat 37

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6.4𝑥 × 200 = N × 𝑥 + 10x (200 − N) 1280 = N + 2000 – 10N 9N = 920 N = 80 kg Required ratio = =

2

80 120

3

39. (b); Let marked price of jeans is 300𝑥 Rs. while marked price of shirt is 200𝑥. Selling price of a jeans & a shirt in scheme A = (300𝑥 + 200𝑥) ×

65 100

= 325𝑥 Selling price of one jeans & one shirt in scheme B = 300𝑥 ×

75 100

+ 200𝑥 ×

80 100

= 225𝑥 + 160𝑥 = 385𝑥 Rs. Given, 385𝑥 – 325𝑥 = 384 Rs. 60𝑥 = 384 Rs. 𝑥 = 6.4 Rs. Cost price of a shirt = 160𝑥 ×

3 4

3

= (160 × 6.4 × ) 4

= 768 rs. Cost price of jeans = 225𝑥 × =225 × 6.4 ×

4 5

4 5

= 1152 Rs. 38

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40. (b); Let, total manufactured cost of a cycle = 100x Rs. Then S.P a cycle = 125x Rs. Expenditure

on

raw

materials

= 100x × 0.40 = 40𝑥 𝑅𝑠. Expenditure on labour = 100𝑥 × .20 = 20𝑥 𝑅𝑠. Fixed charge = 100𝑥 × .20 = 20𝑥 𝑅𝑠. Expenditure on transportation = 100x − (40𝑥 − 20𝑥 − 20𝑥 ) = 20𝑥 𝑅𝑠. New expenditure on raw materials = 1.15 × 40x = 46x Rs. New expenditure on labour = 1.2 × 20x = 24x Rs. New expenditure on transportation = 1.5 × 20x = 30x Rs. To get a profit of 13

7 11

% at a selling price of Rs. 125x, the C.P. of

an item must be equal to 125𝑥 ×

1100 1250

= 110𝑥 Rs. We need to reduce C.P. (or expenditure) by 10, and this 10 has to be reduced from expenditure on raw materials. Required % = = 21

39

17 23

10𝑥 46𝑥

× 100

%

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1

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

Chapter

6

Simple Interest and Compound Interest

BEST APPROACH TO SOLVE THE QUESTIONS The interest can be defined as amount paid by the borrower to the lender in addition to the amount which he had borrowed. In competitive exams, interest is classified into Simple interest (SI) and Compound Interest (CI). Easiest example, when we borrow money from bank, it doesn’t say pay Rs. 100 as interest every year no matter whether you took a loan of Rs 1000 or Rs 100000. They offer us a rate generally in % at which we have to pay interest. This rate is known as Rate of interest (ROI). For eg. If I borrowed Rs. 500 for 1 year at a rate of 8%/annum. Find the interest that I have to pay. Always remember, this 8% we have to calculate on the amount we borrowed. 8% of Rs. 500 means Rs. 40, that I have to pay as an interest. Note: Different forms of ROI may be given or we have to change, as 3% for 3 months or 24% for 18 months. For easier calculation we can normalize these in 12 month/year rate. In above rates, former means 12%/annum and later means 16%/annum. Now, if I asked what would be amount of interest that I have to pay, If I borrow Rs. 500 for 3 years at same rate? Here arise difference b/w SI and CI.

2

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SIMPLE INTEREST: Simple interest simply means that we have to pay same amount of interest every time. If I have to pay interest on Rs. 500 at ROI of 8%/annum for 3 years, it means I will pay 40 × 3 = RS 120. Formula to calculate SI on an amount is PRT SI = 100 Here, P is the amount borrowed and generally it is referred as Principal. R is ROI in %. (100 in equation is used to because of %) And T is time in years. NOTE: If it is asked in a question, that what is interest obtained at the ROI of 10%(or any value) for 5 years on amount of Rs. Z. In order to save time first calculate mentally 10 × 5 = 50%. Total interest will be 50 % of Principal amount. COMPOUND INTEREST: The one liner difference b/w SI and CI is that interest paid for every year is not same. Consider the above example. Interest accrued in 1st year = Rs. 40 In CI principal amount changes, after 1year principal amount will be Rs. 540 not Rs. 500. 8 After second year interest obtained on Rs. 540 is × 540 = Rs. 43.2 100 And interest after 3 years will be calculated on Rs 540+Rs 43.2= Rs 583.2 i.e 8 × 583.2= Rs. 46.65 100 And total interest will be 40 + 43.2 + 46.65= Rs 129.85 The formula for amount obtained after the given time period in CI 𝐑

𝐧

𝐀 = 𝐏 (𝟏 + ) 𝟏𝟎𝟎 Here A is amount obtained after interest, P is principal, n is number of years R is ROI. CI= A – P. 3

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In competitive exams, this approach is rarely used because it will be too calculative if time is 3 years and ROI is 17%. A = P (1 +

17 100

3

117 3 ) 100

) = P(

Imagine if you started calculating cube of 117 .

Best approach to use in CI is resultant ROI, it makes CI as easy as SI. Eg. If ROI on Rs.10000 in a scheme is Y%, calculate total interest after two years. Then resultant interest is Y + Y + (Y × Y)/100 = 2Y +

Y2 100

17×17

Resultant interest % on 17%= 2×17 + = 34+ 2.89= 36.89%. 100 It is more time saving than traditional method. Little use of mind will help you to solve all kind of problems. Note that if we can calculate resultant interest % for two years, we can also calculate it for three years. Actually general formula for calculating resultant % for two years, if ROI on first year is X% and second year is Y%, then resultant interest % will be X×Y X+Y+ 100

For calculating 3year resultant interest, first calculate resultant interest of first two years and then again use resultant formulae for adding resultant of first two year and ROI of 3rd year. REMARK: Check 3 years total interest % for ROI 0f 10% whether it comes 33.1%. DIRECT FORMULAE TO CALCULATE 3 YEARS RESULTANT INTEREST % 3r. 3 r² r³ __ Here, r is ROI Note that 4 underscores (_ _ _ _) represent 4 digits after decimal. Further, make it clear that _ _ (last two digits are for last two digits of value of r³] And starting two (_ _) are for value of 3r². Consider eg. 3r. 3 r² r³_ When r = 2 r³ = 08, r² = 4, 3r² = 12 4

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Resultant interest is for 3 years 6.1208 Now consider r = 8 r³ = 512, r² = 64, 3r² = 192 use last two digits of r³ for last two digits. i.e. 3r. _ _ 12, and carry on 5 to 3r². 3r² = 192 + 5 = 197. Here also, use only its last 2 digit and carry on 1. 3r. 9 7 1 2 3r = 24 + 1 → carry on from 197. Resultant interest % = 25.9712% Check for r = 10%. While solving a question, student should always keep in mind resultant interest %. Example: Amount of Rs 4000 becomes Rs 6600 in SI in 5 years. Calculate ROI. PRT

Solution: Though general formula of SI i.e. SI = is an easy approach, 100 But try mentally to solve RS 2600 is what % Rs 4000. You can mentally calculate that 2600 13 = which means 65%. 4000 20 And 65% in 5 years in SI means 13%/year. Example: Difference b/w interest obtained in SI and CI on an amount at the ROI of 10% is Rs. 310. Find the principal amount. Solution: Calculate resultant % for SI and CI for 3 years at ROI of 10%. Its 30% for SI and 33.1% for CI. But we know that we calculate interest on principal. It means that 3.1% of principal is equal to Rs 310. Hence principal can be calculated. POINTS TO REMEMBER ALWAYS: Difference b/w interest obtained on CI and SI for 2 years is PR2/100. Resultant interest % in CI for 3 years at 10% ROI is 33.1% and 15.7625% at ROI of 5%. 5

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Practice Exercise Based on new Pattern

1.

Ankur invested X Rs. at the rate of 15% per annum on compound interest for two years and gets total interest of 5805 Rs. if Ankur invest (X + 7000) Rs. for another two year at additional rate of 5%, then what will be compound interest on that investment ? (a) 10000 Rs. (b) 11000 Rs. (c) 12000 Rs. (d) 15000 Rs. (e) 18000 Rs.

2.

Shikha invested 32000 Rs. at simple interest for 2 years at the rate of R% and gets an interest of Rs.8000. If he invested total amount (Principle + Interest) in a scheme, which offered compound interest at the rate of (R%+2.5%) then find total compound interest obtained by Shikha after 2 years ? (a) 12600Rs. (b) 12800Rs. (c) 14400 Rs. (d) 12000 Rs. (e) 12900Rs.

3.

Ankit borrowed Rs. 12000 from Veer on C.I. at 10% per annum for three years and added some extra amount and lent to Arun on C.I. at 20% per annum for two year. If Ankit got Rs. 3948 more interest than interest got by Veer.Find how much extra amount was added by Ankit ? (a) 6000 Rs. (b) 4000 Rs. (c) 5000 Rs. (d) 8000 Rs. (e) 9000 Rs. Abhi invested some amount on scheme ‘P’ which offer CI at the rate of 15% p.a.. After 2 years he got Rs. 1032 as interest. Abhi invest the amount he got from scheme ‘P’ in scheme ‘Q’ which offer 15% p.a. at SI for 4 years. Find the total interest he earned from scheme P and scheme Q together? (a) 3571.2 (b) 3715.2 (c) 4232 (d) 3148 (e) 3379.2

4.

6

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5.

A man invests Rs. 1,200 at 10% p.a. At the end of the year he withdraws 30% of total amount and pays Rs. 24 as transaction fee. At the end of 2nd year he withdraws 30% of the amount and pays Rs. 93 as transaction fee. What is the balance at the end of the third year? (a) Rs. 660 (b) Rs. 825 (c) Rs. 500 (d) Rs. 770 (e) None of these

6.

A sum of Rs. 1000 after 3 years at compound interest of 10% p.a. becomes a certain amount that is equal to the amount that is the result of 3 year depreciation from Rs. 1728. Find the approximate difference between the rate of C.I. and rate of depreciation. (a) 3.33% (b) 0.66% (c) 4% (d) 2% (e) 6%

7.

A person invested a certain amount at simple interest at the rate of 6 per cent per annum earning Rs. 900 as an interest at the end of three years. Had the interest been compounded every year, how much more interest would he have earned on the same amount with the same interest rate after three years? (a) Rs. 38.13 (b) Rs. 25.33 (c)Rs. 55.08 (d) Rs. 35.30 (e) None of these

8.

Mayank borrowed Rs. 48000 from a bank at 12

1 2

% per annum

compound interest at the end of 1st, 2nd and 3rd year, he paid 14000, 13000 and 16000 respectively. If he wanted to clear his loan at the end of 4th year, what would he pay at the end of the fourth year to clear his loan ?

7

(a) 21000

(b) 22500

(d) 26000

(e) None of these Adda247 Publications

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9.

Rakesh lent Rs. P to Rahul on an agreement that for first two year, interest will be calculated on SI at 15% per annum and for next two years interest will be calculated on C.I at 5% additional rate. If Rahul paid total amount of Rs. 17971.2 at the end of four year, Find the amount borrowed by Rahul? (a) 9000 Rs. (b) 9800 Rs. (c) 9900 Rs. (d) 9600 Rs. (e) 9200 Rs.

10. Veer invested Rs 22500 for 2 year at the Rate of x% in scheme A at compound interest annually and gets a total amount of Rs 32400 If he added Rs 2600 in amount and invested total amount in scheme B at S.I. for 3 year at same rate . Then find the total simple interest veer gets from scheme B? (a) Rs 22500 (b) Rs 22000 (c) Rs 17500 (d) Rs 20000 (e) Rs 21000 11. Veer Invested X Rs. in SBI at the rate of 18% for 2 year and obtained a total simple interest of 6750 Rs. If he invested 2250 Rs. more for same 2

2

7

3

period time at the rate of 14 % for first year and at 16 % for 2nd year on compound annually. Then find the total compound interest obtained by Veer after 2 year ? (a) 6000 Rs. (b) 5000 Rs. (c) 3000 Rs. (d) 7000 Rs. (e) 5500 Rs. 12. Satish invested 16000 Rs. in simple interest for 2 years on certain rate and gets an interest of 4800 Rs, if he invested total amount (Principle + Interest) in a scheme, which offered compound interest on 5% more interest rate as earlier rate. Then find total interest gets by Satish after 2 years ? (a) 9252 Rs. (b) 9225 Rs. (c) 9512 Rs. (d) 925 Rs. (e) 9152 Rs. 8

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13. Bhavya and Veer invested their principle in two different schemes, Bhavya invested X Rs. on compound interest for two year at rate of 20% annually and Veer invested 4000 Rs. more than Bhavya on simple interest for three year at 15% annually, if both gets total interest of Rs. 9632, then Find the amount invested by Veer ? (a) 12900 (b) 12400 (c) 8800 (d) 12800 (e) 12600 14. A man invested Rs. x, y and z in three difference schemes which offers 10% p.a., 20% p.a. and 30% p.a. rate respectively and after one year he earns Rs. 2000 interest. If x, y and z are in arithmetic progression and sum of x, y and z is Rs. 9,000. Find the difference between ‘x’ and ‘z’. (a) 1500 (b) 2000 (c) 2500 (d) 2250 (e) 1750 15. P invested Rs. X in a scheme for 2 year which offered simple at the rate of 15% per annum and Q invested Rs. (X + 2500) in another scheme for same period of time, which offered compound interest at the rate of 20% per annum. If from both scheme P and Q got total interest of Rs. 32550, then find the value of X ? (a) 41500 (b) 42500 (c) 40500 (d) 40000 (e) 38250 16. Rajat have some money in his hand. He invested 20% of the money in scheme ‘A’ for 4 year at 6% p.a., 30% of the amount in scheme ‘B’ for 6 year at 12% p.a., remaining in scheme ‘C’ 2 year at 15% p.a. If total amount Rajat received from scheme ‘A’, ‘B’ and ‘C’ is 11,355, then find the difference of sum invested in scheme ‘B’ and scheme ‘A’. (a) 1800 (d) 750 9

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17. Satish borrowed Rs. 24000 from a SBI at the rate of 12

1 2

%

compounded annually. If at the end of 1st, 2nd and 3rd year, he paid Rs. 7000, Rs. 6500 and Rs. 8000 respectively. If Satish wanted to clear his loan at the end of 4th year, what would he pay at the end of the fourth year to clear his loan ? (a) 11500 Rs. (b) 11250 Rs. (c) 10500 Rs. (d) 9500 Rs. (e) 14500 Rs. 18. Divyaraj invested an amount into two parts in the ratio of 4 : 3 on compound interest for two years at the rates of 20% & 15% respectively. If he exchange rate of interests, then he will get Rs. 705 less interest than earlier interest, then find how much Simple interest he will get, if he invest total amount at the rate of 17.5% for two years? (a) 14500 Rs. (b) 14700 Rs. (c) 14900 Rs. (d) 14800 Rs. (e) 14900 Rs. 19. Rahul took a loan of Rs.80000/- at rate of of 5% p.a. at SI & lent it at 5% pa at CI. After 3 years, he cleared all his debts and invested the profit earned in a scheme which offers C.I. After two years he got Rs.128.1/- as interest from the scheme. Find the rate of interest offered by scheme. (a) 8% (b) 7% (c) 6% (d) 10% (e) 12% 20. A man invested his total savings in two parts. He invested Rs. 30000/in a scheme which offers CI at ROI of 10% per annum and remaining in SI at ROI of 5%per half year. After two years the amount obtained by him in former scheme is half of later, then calculate amount invested in later scheme. (a) Rs. 242000 (b) Rs. 60500 (c) Rs. 12100 (d) Rs. 80525 (e) None of these 10

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21. Satish saved 55% of his monthly income, which he further invested in two different schemes A & B in the ratio of 7 : 5 on CI compounded annually at the rate of 10% p.a. and 20% p.a. respectively. If after two years, Satish gets total interest of Rs. 8074 from both the schemes, then find the total annual saving of Satish? (a)316,600 Rs. (b)316,400 Rs. (c)316,000 Rs. (d)316,200 Rs. (e)316,800 Rs. 22. Aman invest Rs. (x – 2000) in scheme ‘P’ which offers 20% CI pa while Rs. (x + 2000) in scheme ‘Q’ which offers 8% SI pa. After 2 year difference between interest earn from scheme P and Q is 1600, then amount invested by Aman in scheme ‘Q’ is what percent more/less than amount invested by him in scheme ‘P’ ? 1

(a) 75%

(b) 133 %

(d) 50%

(e) 33 %

1

3

(c) 150%

3

23. Ritu invested her total saving in three different FD schemes A, B and C in the ratio of 5 : 4 : 6 on CI for two years at the rate of 10%, 15% and 20% respectively.If interest is calculated annually and interest from scheme B is Rs. 744 more than interest from scheme A then, find difference between interest received from scheme C and scheme B by Ritu? (a) Rs. 4185 (b) Rs. 4175 (c) Rs. 3840 (d) Rs. 4580 (e) Rs. 3250 24. Raghv invested Rs. 8400 and Rs. 9600 at the rate of R% and (R + 5)% respectively on simple interest for two years and gets total interest of Rs. 6360. If Raghv invested Rs.(8400 + P) and Rs.(9600 + P) at the rate of R% and (R + 5)% respectively on compound interest for two year, then he would get total interest of Rs. 8153. Find value of P? (a) 1800 Rs. (b) 1600 Rs. (c) 2200 Rs. (d) 2400 Rs. (e) 2600 Rs. 11

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Directions (25-26): A person invested Rs. 20000 in a bank which is offering 10% per annum simple interest. After two years he withdrew the money from the bank and deposited the total amount in another bank which gives an interest rate of r% p.a. compounded annually. After 2 years he received an amount of Rs. 2460 more than what he had invested in that bank. 25.

What is the value of r? (a) 10% (d) 12%

(b) 15% (e) None of these

(c) 5%

26. If the person had invested Rs. 50,000 instead of 20000 in the bank that offered simple interest, what would have been his net profit after following the same procedure as given above? (a) Rs. 16,800 (b) Rs. 16,150 (c) Rs. 16,350 (d) Rs. 16,000 (e) None of these 27. A bank offers 10% p.a. at CI in scheme ‘A’, 20% p.a. at CI in scheme ‘B’ and 40% p.a. at CI in scheme ‘C’. Veer have same amount in his hand. He invested 32% amount in scheme ‘A’, 20% in scheme ‘B’ and remaining in scheme ‘C’. After two year he received 6600 as interest. Find the difference between amount invested by Veer on scheme ‘C’ to that of in scheme ‘B’. (a) 5,000 (b) 4,000 (c) 2,000 (d) 3,000 (e) 2,500 28. Bhavya borrowed Rs. 37500 on compound interest with agreement that, he will not pay amount after 2 year with interest then the rate of interest increase by 5% as earlier interest rate. If Bhavya paid Rs. 12093.75 as total compound interest after two year. Then find how much amount Bhavya have to pay after two year for completing his entire borrowing? (a) Rs. 58000 (b) Rs. 52000 (c) Rs. 42000 (d) Rs. 45000 (e) Rs. 54000 12

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29. A man invested ‘X’ amount in scheme A and ‘Y’ amount in scheme B. Scheme A offers 8% p.a. at S.I. scheme B offer 20%. p.a. at C.I. Man invested both schemes for 2 years. The sum of interest he earns after 2 years will be equal to the simple interest he can earn if he invested ‘Y’ amount in scheme A for 7 years. Find the total amount he has initially if difference between interest earned by him after 2 years is 2304. (a) 12,300 (b) 12,900 (c) 12,600 (d) 13,500 (e) 13,800 30. Shikha and Sameer have equal amount. Shikha invested on C.I. for two 2 years at the rate of 10% p.a. and Sameer invested 66 % of amount at 3 the rate of R% p.a. on simple interest and remaining amount at the rate of 6.5% p.a. on simple interest. If interest received by both at the end of two years are equal, then find the value of ‘R’? (a) 10% (b) 12.5% (c) 15% (d) 12% (e) 8% 31.

Veer invested Rs. x in SBI for three years on S.I. at the rate of 15% per annum and Sameer invested Rs. (x + 4000) in BOB for two years on C.I. at the rate of 8% per annum. If Divyaraj invested equal to sum of what Veer and Sameer invested in UBI for two years on C.I. at the rate of 20% per annum. Find the sum invested by Divyaraj, if they all got total interest of Rs. 20382.4? (a) Rs. 20000 (b) Rs. 18000 (c) Rs. 28000 (d) Rs. 30000 (e) Rs. 26000

32. Three employee Ankit, Veer and Sameer got Rs. P, Rs. (P+2400) and Rs. (P+4400) Diwali bonus. Ankit and Veer invested their half of bonus on CI at the rate of 10% and 20% respectively for two years. Sameer invested 60% of his bonus on simple interest at the rate of 15% p.a. for three years. If Sameer got Rs. 132 more as interest got by Ankit and Veer together, then find the bonus got by Sameer? (a) 12000 Rs. (b) 10000 Rs. (c) 14400 Rs. (d) 14000 Rs. (e) 18000 Rs. 13

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33. Manoj lend Rs. P for three years on S.I. at the rate of 15% per annum and Rajesh lend Rs. (P + 8000) for two years on C.I. at the rate of 8% per annum. Suresh borrowed sum equal to of what Manoj and Rajesh lend, for two years on C.I. at the rate of 20% per annum. If Suresh paid interest Rs. 5352 more than, what Manoj and Rajesh got total interest on their sums together. Find total sum borrowed by Suresh? (a) Rs. 34000

(b) Rs. 44000

(d) Rs. 46000

(e) Rs. 30000

(c) Rs. 32000

34. A person invested his savings of Rs. 32000 in two parts at equal rate of interest in CI and SI. After 3 years, he withdrew amount invested in CI, and put the cash in his almirah. After another 2 years, the amount invested in SI got doubled and he also withdrew them. If the final ratio of both the amount is 36:65, then find amount invested by him in SI (in Rs.) (a) 20000/-

(b) 18050/-

(c) 19500/-

(d) 21500/-

(e) can’t be determined

35. A man invested an amount into three schemes P, Q and R in the ratio of 5 : 6 : 9 at the rates of 20%, 15% and 10% respectively. Schemes P and R offered compound interest annually, while scheme Q offered simple interest annually. Total interest received by man from scheme P and Q together is Rs. 1899 more than total interest received from scheme R after two years. If man would invest total amount on C.I at the rate of 15% p.a., then find the total interest received by the man? (a) 5800 Rs. (b) 5805 Rs. (c) 5580 Rs. (d) 5590 Rs. (e) 5900 Rs 14

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Solutions

(b); ATQ, 5805 = X [(1 +

1.

15 100

2

) – 1]

129X = 5805 × 400 X = 18000 Rs. Now Total Interest=(18000+7000) ×[(1 +

(15+5) 2 100

) – 1]

11

= 25000 × ( ) 25 = 11000 Rs. 2.

(e); Let shikha invested at the rate of R% ATQ— 8000×100 R= R = 12.5% 32000×2 New Rate = 12.5 + 2.5 = 15% Total amount = 32000 + 8000 = 40000 Rs. Equivalent CI of two years at the rate of 15% 15×15 = 15 + 15 + 100 = 32.25% 32.25 Required compound interest = 40000 × = 12900 Rs. 100

10

(a); Interest got by Veer=12000[(1 + )3 − 1] 100 = 3972 20×20 Two year CI on 20% = 20 + 20 + 100 = 44% Atq, 44 (12000 + 𝑥 ) − 3972 = 3948 100 5280+.44x – 3972 = 3948 .44x = (3948 + 3972) – 5280 .44x = 2640 x = 6000 Rs

3.

15

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4.

(a); Let, sum Abhi have initially = 100x ATQ, 100x ×

115 100

×

115 100

– 100x = 1032

132.25x – 100x = 1032 ⇒x=

1032 32.25

= 32

Amount initially Abhi have = 3200 Interest earned from scheme Q =

4232×15×4 100

= 2539.2 Required amount = 2539.2 + 1032 = 3571.2 5.

(a); The man invests Rs. 1,200 at 10% p.a. At the end of 1st year the amount = Rs. 1,320 Withdrawal

30 100

× 1320 + 24 = Rs. 420

Amount at the end of second year = 900 × 1.1 = Rs. 990 Withdrawal =

30 100

× 990 + 93 = Rs. 390

∴ Amount at the end of 3 years = 600 × 1.1 = Rs. 660 6.

(d); The amount @ 10% C.I. could become Rs. 1331. Also, Rs. 1728 depreciated at R% has to become Rs. 1331. Thus, 1728 × [

100−R 3 100

] =1331(approximately).

The closest value of R = 8% Thus, the difference is 2%.

16

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900 ×100

7.

(c); P = 6×3 P = 5000

Required value = 955.08 – 900 = 55.08 8.

(b); 1st year CI 1 48000 × = 6000 8 Amount = 48000 + 6000 = 54000 2nd year principle = 54000 – 14000 = 40000 2nd year CI = 1 40000 × = 5000 8 Amount = 40000 + 5000 = 45000 3rd year principle = 45000 – 13000 = 32000 3rd year CI 1 = 32000 × = 4000 8 Amount = 32000 + 4000 = 36000 4th year principle = 36000 – 16000 = 20000 At the end of 4th year amount 1 = 20000 × + 20,000 8 = 2500 + 20000 = 22500

9.

(d); For first two years total S.I = 15 × 2 = 30% For next two years total C.I on (15+5)% 20×20 = 20+20+ 100 = 44% Amount after two year 17

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=P+ = Rs.

𝑃×30 100 13𝑃 10

Amount after 4 years 13𝑃 10

P=

×

144

= 17971.2

100 17971.2 1.872

P = Rs. 9600 10. (e); ATQ,

𝑅 2 ) 32400 = 22500 (1 + 100 324 100 + 𝑅 2 ) =( 225 100 18 100 + 𝑅 = 15 100 15R = 300 R = 20% 20×3 Simple interest = (32400 + 2600) × 100

= 21000 Rs 11. (d); Principle =

X×18×2

100 6750×100

= 6750

⇒X= 18×2 X = 18750 Rs. According to question — 1 1st year CI = (18750 + 2250) × 7 = 3000 1 2 year CI = (21000 + 3000) × 1

6

= 24000 × 6 = 4000 Total CI after 2 year = 3000 + 4000 = 7000 Rs. 18

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12. (e); Rate =

4800×100 16000×2

R = 15% New rate = 15 + 5 = 20% 2 year CI on 20% = 20 + 20 +

20×20 100

= 44% Interest gets Satish = (16000 + 4800) ×

44 100

= 9152 Rs. 13. (d); ATQ, (X+4000)45 100

+

44X 100

= 9632

0.45X + 1800 + 0.44x = 9632 0.89X = 9632 – 1800 X=

7832 0.89

X = 8800 Veer principle = 8800 + 4000 = 12800 Rs. 14. (b); x + y + z = 9,000 & 2y = x + z ⇒ y = 3,000 10 20 30 ×x+ × 3,000 + × z = 2000 100 100 100 0.1x + 0.3z = 2000 – 600 = 1400 and x + z = 6,000 0.1x + 0.3z = 1400 …(i) x + z = 6,000 …(ii) on solving (i) & (ii), we get x = 2,000 z = 4,000 Difference between x & z = 2,000 19

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15. (b); ATQ 15×2x 100

+ (x + 2500) [(1 +

20 100

2

) – 1] = 32550

0.3x + 0.44x + 1100 = 32550 0.74x = 31450 x=

31450 0.74

x = 42500 16. (d); Let the amount Rajat have = 100 ATQ, 20×4×6 30×6×12 50×2×15 100 + + + = 11,355 100 100 100 151.4 = 11355 ⇒ 100 = 7500 Required difference = = 750

7500[30 –20] 100

17. (b); 1st year CI 1 24000 × = 3000 Rs. 8 Amount = 24000 + 3000 = 27000 Rs. 2nd year principle = 27000 – 7000 = 20000 Rs. 2nd year CI = 1 20000 × = 2500 Rs. 8 Amount = 20000 + 2500 = 22500 Rs. 3rd year principle = 22500 – 6500 = 16000 Rs. 3rd year CI 1 = 16000 × = 2000 8 Amount = 16000 + 2000 = 18000 Rs 4th year principle = 18000 – 8000 = 10000 Rs. At the end of 4th year amount, which Satish have to pay 1 = 10000 × + 10,000 8 = 1250 + 10000 = 11250 Rs. 20

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18. (b); Let Divyaraj invested total amount = 7x Rs. Equivalent CI of two year at the rate of 20% 20×20 = 20 + 20 + = 44% 100 Equivalent CI of two year at the rate of 15% 15×15 = 15 + 15 + = 32.25% 100 ATQ – 44 32.25 32.25 44 4x × + 3𝑥 × − 4𝑥 × − 3𝑥 × = 705 100 100 100 100 (1.76x + .9675x) – (1.29𝑥 + 1.32𝑥) = 705 2.7275x – 2.61𝑥 = 705 0.1175x = 705 x =6000 Rs. Total amount invested by Divyaraj = 6000 × 7 = 42000 𝑅𝑠. 17.5×2 Required simple interest = 42000 × =14700 Rs. 100

19. (d); Profit earned in 3 years = Difference in interest obtained D=P r2 (300+r)

(100)3 80000×25×305

= 1000000 = Rs610 Let scheme offers R% rate of interest 𝑅

2

610 (1 + ) − 610 = 128.1 100 R = 10% 20. (b); Let he invested Rs. Y in second scheme 5% per half year means 10% per annum in SI Therefore, amount obtained by him after two years Y(2)(10) =Y+ 6𝑌

100

= 5 If he invested Rs. 30000/- in a scheme, which offers ROI of 10% per annum Amount obtained by him after two years is = 30000 (1 + 21

10 100

)

2

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= Rs. 30000 ×

11 10

×

11 10

= Rs. 36300 Now According to Q ⇒ 36300 = ⇒Y=

1 (6𝑌)

2 5 36300×5×2 6

= Rs. 60500/21. (e); Let total monthly income of Satish = 100x Rs. Amount invested in scheme A = 100𝑥 × =

385𝑥 12

55 100

×

7 12

Rs.

Amount invested in scheme B = 100𝑥 × =

275𝑥 12

55 100

×

5 12

Rs.

Equivalent CI of two years on 10% = 10 + 10 +

10×10 100

= 21% Equivalent CI of two years on 20% = 20 + 20 +

20×20 100

= 44% ATQ— 385𝑥

×

21

+

12 100 1617𝑥+2420𝑥

275 12

×

44 100

= 8074

= 8074 4037x = 8074 × 240 𝑥 = 480 Rs. Annually saving of Satish 55 = 48000 × × 12 100 = 316,800 Rs. 240

22

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22. (d); Compound interest rate for 2 years 20+20 = 20 + 20 + = 44% 100 ATQ, 44 (x+2000)×8×2 (x– 2000) × – = 1600 ⇒

44𝑥 100 28𝑥

–

16𝑥 100

100

100

– 880 – 320 = 1600

⇒ = 1600 + 880 + 320 = 2800 100 ⇒ x = 10,000 12000−8000 4 Required % = × 100 = × 100 = 50% 8000

8

23. (a); Let total saving of Ritu be x Rs. 5𝑥 Invested in Scheme A = 15

𝑥

= Rs. 3

Amount Invested in scheme B = Invested in scheme C =

2𝑥 5

4𝑥 15

Rs.

𝑅𝑠. 10×10

Two year C.I on 10% =10+10+ 100 = 21% 15×15 Two year CI on 15% =15+15 + 100 = 32.25 % 20×20 Two year CI on 20% =20+20+ 100 = 44% ATQ, 4𝑥 32.25 𝑥 21 × − × = 744 15 100 129𝑥 21𝑥

3

100

− = 744 300 24x = 744 × 1500 744×1500 x= 24 x = 46500 Rs. Required difference 46500×6 44 46500×4 32.25 = × − × 15 100 15 100 = 8184 – 3999 = 4185 Rs. 1500

23

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24. (b); ATQ— 8400×𝑅×2 1000

+

9600×(𝑅+5)×2 100

= 6360

168R + 192R + 960 = 6360 360R + 960 = 6360 360R = 5400 R = 15% Rs. (8400 + P) invested at 15% And Rs. (9600 + P) invested at 20% Equivalent CI of two years at the rate of 15% 15×15 = 15 + 15 + 100 = 32.25% Equivalent CI of two year at the rate of 20% 20×20 = 20 + 20 + = 44% 100 ATQ— 32.25 44 (8400 + 𝑃 ) × + (9600 + 𝑃 ) × = 8153 100 100 270900 + 32.25P + 422400 + 44P= 815300 76.25P = 122000 P = 1600 Rs. 25. (c); Amount withdrawn from bank which offered simple interest = 20000×10×2 20000 + = 24000 100 Compound interest accrued from another bank = 2460 2460 = 24000 [(1 +

𝑅 100

2

) − 1]

⇒r=5% 26. (b); Amount withdrawn after 2 years from bank offering S.I. = 5000×10×2 50000 + 100 = 60000 Amount withdrawn after another 2 years from another bank offering C.I. = 60000 (1 +

5 100

2

) = 66150

New profit = 66150 − 50000 = 16,150 24

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27. (d); Let, Total amount Veer have = ‘x’ ATQ, 11 11 12 12 14 14 0.32x [ × – 1] + 0.20x [ × – 1] + .48x [ × – 1] 10 10 10 10 10 10 = 6600 21 44 96 0.32x [ ] + 0.20x [ ] + 0.48x [ ] = 6600 6.72𝑥 100

+

⇒x=

100 8.8𝑥

+

46.08𝑥

100 100 6600×100

100

100

= 6600

61.6

Required difference =

6600×100 61.6

28. (e); ATQ, 12093.75 = 37500 [(1 + 12093.75

√

37500

49593.75

√

37500

+1 = 1+ =1+ r

r 100

× [. 48 – .20] = 3,000

2

) – 1]

r 100

r 100

1.15 – 1 = 100 r = 15% New rate = 15 + 5 = 20% Amount paid by Bhavya after next two more years 144 = 37500 × = 54000 Rs. 100

29. (c); ATQ 𝑥×8×2 100

+ 𝑦 [(1 +

20 100

2

) − 1] =

16x + 44y = 56y 16x = 12y 𝑦 4 3𝑦 = ⇒𝑥= 𝑥 3 4 44y – 0.16x = 2304 3 44y – 0.16 × 𝑦= 2304 4 0.44y – 0.12y = 2304 y = 7200 x = 5400 Required amount = 12,600 25

𝑦×8×7 100

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30. (b); Let Shikha and Sameer have Rs. 100𝑥 Equivalent CI for two years at the rate of 10% 10×10 = 10 + 10 + = 21% 100 ATQ— 21 2 R×2 1 6.5×2 100x × = 100x × × + 100x × × 100 4x×R

13x

3

100

3

100

21x = + 3 3 63x = 4x × R + 13x 4x × R = 50x 50x R= 4x R = 12.5% 31. (c); 2 year C.I. on 8% per annum 8×8 =8+8+ = 16.64% 100 2 year C.I. on 20% per annum 20×20 = 20 + 20 + = 44% 100 ATQ— (x+4000)16.64 (2x+4000)44 x×3×15 + + = 20382.4 100 100 100 0.45x + 0.1664x + 665.6 + 0.88x + 1760 = 20382.4 1.4964x = 20382.4 – 2425.6 17956.8 x= = 12000 Rs. 1.4964 Sum of Divyaraj invested = (2 × 12000 + 4000) = 28000 Rs. 32. (d); Bonus of Ankit = P Rs. Bonus of Veer = (P + 2400) Rs. Bonus of Sameer = (P + 4400) Rs. Equivalent CI at 10% for two years 10×10 = 10 + 10 + 100

= 21% Equivalent CI at 20% for two years = 20 + 20 +

20×20 100

= 44% 26

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ATQ— (P + 4400) × 27P+118800

60

×

15×3

100 100 65P+105600

P

21

2

100

– ×

−

(P+2400) 2

×

44 100

= 132

– = 132 200 54P + 237600 – 65P - 105600 = 26400 11P = 105600 P = 9600 Sameer bonus = (9600 + 4400) = 14000Rs. 100

33. (c); Three years SI on 15% = 15×3=45% 8×8 Equivalent two years CI on 8 %= 8 +8+ = 16.64% 100 20×20

Equivalent two years CI on 20% = 20+20+ ATQ – 44(2𝑃+8000) 45𝑃 16.64(𝑃+8000) ) = 5352 −( + 100

100

100

= 44%

100

.88P + 3520 - .45P – .1664P – 1331.2 = 5352 .2636P = 3163.2 3163,2 P= = 12000 𝑅𝑠. .2636 Suresh borrowed = 12000×2+8000 = 32000 Rs 34. (c); ATQ, The amount invested by him in SI, got doubled after 5 years. Therefore ROI was 20% per annum. Let amount invested by him in SI was x, then amount invested by him in CI is (32000 – x) Now, (32000–x)(

⇒

120 3 ) 100

2x (32000–x)216 2x×125 (32000–x)3

= = 1

36 65 36 65

⇒ = x×25 13 ⇒ 32000 × 39 = 64x x = 500 × 39 = 19500/27

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35. (b); Let man invested in scheme P, Q and R be Rs. 5x, Rs. 6x and Rs. 9x respectively Equivalent CI of two years on 20% = 20 + 20 + Equivalent CI of two years on 10% = 10 + 10 +

20×20 100 10×10 100

= 44% = 21%

ATQ – 5𝑥 ×

44 100

+ 6x ×

15 ×2 100

– 9x×

21 100

= 1899

2.2x + 1.8x – 1.89x = 1899 2.11x = 1899 x=

1899 2.11

x = 900 Rs. Total amount = 900 × (5 + 6 + 9) = 18000 Rs. If man invested total amount on C.I at the rate of 15% p.a. Equivalent CI of two years on 15% = 15 + 15 + 32.25

Required interest = 18000×

28

100

15×15 100

= 32.25

= 5805 Rs.

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1

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Chapter

7

Time & Work and Pipe Cistern

BEST APPROACH TO SOLVE THE QUESTIONS • Questions asked from this topic generally to test mental ability of a student calculate the time in which a man/team complete the given work while efficiency is given. • While doing the questions from this topic, a most common approach that student must use is that ‘if a man completes 1 unit of work in N days, then amount/part of work done by him in 1 day is 1/𝑁,’ Assume that work of man is to eat 100 chocolates and he eats all of them in 10 days, then amount of work done (chocolates eaten) in 1 100 day is = 10 𝑐ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒𝑠 𝑒𝑎𝑡𝑒𝑛. 10 Or If Rashmi can make 10 cakes in 5 hours, then amount of work done 10 (Cake made) by her in 1 hour is = 2 cakes. 5 Another easy way to solve questions from this topic is % approach i.e ‘if a man can complete a work in 20 days, then amount of work done 100 by him in 1 day is % = 5%. 20 Consider an example: Rahul can complete a work in 10 days and Arun can complete the same work in 20 days. Find the time in which the work will be completed if both of them work together. There are three approaches to solve this: (i) TRADITIONAL APPROACH If Rahul can complete the work in 10 days, then amount of work done by him in 1 day 1 = 𝑢𝑛𝑖𝑡𝑠. 10 Similarly amount of work done by Arun in 1 day 1 = 𝑢𝑛𝑖𝑡𝑠. 20 Hence when both of them work together, total work done is 2

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1 10

+

1 20

=

3 20 3

𝑢𝑛𝑖𝑡𝑠.

Hence if 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑤𝑜𝑟𝑘 𝑖𝑠 𝑑𝑜𝑛𝑒 𝑖𝑛 1 𝑑𝑎𝑦 20 Then 1 unit of work will be completed in 20/3 days. (ii) LCM APPROACH In this method, assume the LCM of days as a number of chocolates. LCM of 10 and 20 is 20. Now assume that there were 20 chocolates, and If Rahul take 10 days to eat them all, it can be concluded that he ate 2 chocolates per day. Similarly, Arun can eat 1 chocolate in a day. Therefore, They both will eat 3 chocolates in 1 day. 20 Time taken to complete whole work(to eat all of them)= 𝑑𝑎𝑦𝑠. 3 NOTE: Our answer will be same, we consider any number of chocolates, but for easy calculation purpose we use a number that is easily divisible by number of days given. (iii) % APPROACH This method is kind of same as (i). Consider the work unit as 100% work. Now note that if Rahul take 10 days to complete 100% of work, then work done by him in 1 day is 10%. Similarly work done by Arun in 1 day is 5%. Therefore, both of them working together will complete 15% of work in 1 day. 100 20 100% of work will be completed in = 𝑑𝑎𝑦𝑠. 15

3

In all these approaches, 1 common thing that occurs is to calculate amount/part/percentage of work done in 1 day. LCM approach is best used in the questions in which efficiency of pipe or volume of tank is given. Consider eg. Example 1: Pipe A and Pipe B when both opened together can fill a tank in 4 minutes. Pipe A alone can fill it in 16 minutes. Find the time in which pipe B alone will fill the tank. 3

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In traditional way: 1 Sol. Tank filled by Pipe A alone in 1 minutes = 𝑢𝑛𝑖𝑡𝑠. 16 Let pipe B alone can fill it in x minutes. 1 Then tank filled by pipe B alone 1 minute = 𝑢𝑛𝑖𝑡. 1

1

𝑥

16 16

Therefore, + 1 𝑥

=

3 16

, 𝑥=

=

1

𝑥

4

3

LCM APPROACH Let total units of tank are 16 liters. It means A can fill 1 liter in a minute. Assume that B fills x unit in a minute. Tank was filled in 4 minutes’ means 4(𝑥 + 1) = 16 𝑥+1 = 4 x = 3 units per minute 16 time taken is minutes. 3 All this could be done was mentally through LCM method. Give it a try! MAN-DAYS-HOUR-WORK THEOREM According to this theorem if more than 1 man of equal efficiency are working together on a same work, then amount/part of work done by them is directly proportional to the time given to work by number of man. In more simple words work done depends on efficiency and total time given. 𝐦 𝐝 𝐡 𝐦 𝐝 𝐡 ⇒ 𝟏 𝟏 𝟏= 𝟐 𝟐 𝟐 𝐰𝟏

𝐰𝟐

Here m₁, m₂ are number of men d₁, d₂ are number of days h₁, h₂ are working hours in given days w₁, w₂ are part/amount of work done. For eg. 10 men in 5 days can do same amount of work as 5 men in 10 days or if we explore it more we can see 10 men in 6 days working 4 hours a day can do same amount of work as 30 men in 4 days working 2 hours a day. Please verify above two examples, by assuming a man can eat 1 chocolate in 1 hour and try to solve given problem. Example 2: 12 Men working 4 hours a day can complete 48% of work in 8 days. Calculate how many men are required to complete 2 units of a work, if they have to complete it in 10 days, 8 hours a day. 4

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Sol. In this type of problems, by default we have to assume that all men have equal efficiency and they do the work with uniform speed. Applying Man-day-hour formulae m1 d1 h1 w1

=

m2 d2 h2 w2

Here, m₁ = 12 d₁ = 8 h₁ = 4 w₁ = 48

m₂ = ? d₂ = 10 h₂ = 8 w₂ = 200

Note that, here w₂ = 200 is used because, 2 units of work means 200% of work. m2 =

12×8×4×200 48×8×10

m2 = 20 Another type of concept that is introduced here is when people with unequal efficiencies are working, for eg. 4 women can do the same amount of work in a given time as 3 men. For these type of questions, we use ratio approach, we simply write 4W= 3M =>

𝑊 𝑀

3

= . 4

And then we consider efficiency of a men as 4 units and that of women as 3 units throughout the problem. Few points to remember: If the ratio of efficiencies of two persons is x:y then ratio of respective time taken by them is y:x. If X is N times efficient than Y, time taken by X to complete a work is 1 𝑁

𝑡𝑖𝑚𝑒𝑠 𝑜𝑓 time taken by Y.

Total work done by a man of efficiency X in N days is 𝑁𝑋 units. 5

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Practice Exercise Based on New Pattern

1.

Ratio between efficiency of Arun, Yash and Rana is 6 : 4 : 5. All three starts to work together with same efficiency. But Rana, destroys his 60% of work in every evening, due to which they have to work 20 more days then estimated time. Find the estimated days by them to complete the work. (a) 80 (b) 76 (c) 84 (d) 72 (e) 90

2.

Pipe P can fill a tank in 24 minutes, pipe Q can fill the same tank in 36 minutes and an outlet pipe can empty the fully filled tank in 48 minutes. Find in how many minutes tank will be filled, if for first minute only pipe P & Q opened together and in second minute all three pipes opened in tank alternatively? 4 4 4 (a) 15 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 (b)13 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 (c)18 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 5 4

(d) 16 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 5

5 4

5

(e)12 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 5

Direction (3-4): One day earning of P, Q and R together is Rs.1026 to do a work. ‘P’ earns more than R which is same as ‘R’ earns more than ‘Q’. Efficiency of R and Q is 19 : 18. 3.

Find the amount earned by R and P together in 5 days to do the same work? (a) Rs. 3620 (b) Rs. 3430 (c) Rs. 3510 (d) Rs. 3310 (e) Rs. 3710

4.

S, who’s efficiency is average of efficiency of P and Q, can complete a work in 36 days. If P, Q and R work together, and complete that work then find the total wage of Q (a) 3888 Rs. (b) 4104 Rs. (c) 4320 Rs. (d) 3666 Rs. (e) 4520 Rs. 6

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5.

(X+4) men can complete a work in 2X days while (X+12) women can complete same work in (X+8) days. If ratio of efficiency of men to women is 5 : 4 then find in how many days 12 men and 15 women together can complete the same work? (a) 32 days (b) 16 days (c) 48 days (d) 64 days (e) 80 days

6.

A cistern can be filled completely by pipe A and B together in 12 hours. If pipe A works with twice speed while pipe B work with 50% more speed than cistern can be filled completely in 7 hours. Find the capacity of cistern if flow of water through pipe A is 2.5ℓ/ minute. (a) 2800 liter (b) 3150 liter (c) 3300 liter (d) 3650 liter (e) 4200 liter

7.

Ratio of efficiency of A and B in completing a work is 3 : 4. Both started to work together but A left after 2 days. Another person C joins B and they together complete the remaining work in 6 days. If A and B together can complete the work in 8 days then C alone can complete the work. 27 56 41 (a) days (b) days (c) days (d)

4 28 3

days

(e)

3 49 3

3

days

8.

Pipe A can fill a tank in 45 hr, pipe B is 50% more efficient than A and pipe C can fill the same tank in 7.5 hr less than B. A and B opened together for X hr and closed after that and pipe C fill remaining tank in (X + 9) hr, if the ratio between tank filled by (A + B) together to tank filled by pipe C is 1 : 2. Find the value of X ? (a) 3 hr (b) 4 hr (c) 6 hr (d) 8 hr (e) 7 hr

9.

A and B can do a piece of work in 72 days and 64 days respectively. C 2 can do the same work in 2 more days as A & B take together to 17 complete. If first day A & B work together and second day B & C work together alternatively, then in how many days work will be completed? 13 13 13 (a) 22 𝑑𝑎𝑦𝑠 (b) 27 𝑑𝑎𝑦𝑠 (c) 32 𝑑𝑎𝑦𝑠 (d) 25 7

25 13 25

𝑑𝑎𝑦𝑠

(e) 29

25 13 25

𝑑𝑎𝑦𝑠

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10. Four persons started to do a work together. ‘A’ works only in starting two days after that B, C and D works alternately starting from B. Ratio of time taken by A, B, C and D if they work alone is 4 : 3 : 2 : 5. If the work is completed in 12 days then in how many days A and C can complete the work if they work together ? (a) 6 days (b) 12 days (c) 10 days (d) 8 days (e) 4 days 11. A person C can complete 21% of work in 10 days while working with 1 233⅓% of his efficiency. B is 11 % more efficient than C. A, while 9 working with his half efficiency can complete the work in half time as compared to time taken by B. Find the time taken by A & B together to complete the 50% of whole work. (a) 15 days (b) 10 days (c) 20 days (d) 25 days (e) 22 days 12. Two pipes A and B can fill a cistern in 15 hours and 10 hours respectively. A tap can empty the full cistern in 30 hours. All the three taps were open for 2 hours, when it was remembered that the emptying tap had been left open. It was then closed. How many hours more would it take for the cistern to be filled ? (a) 4hr 30 min. (b) 4hr 12 min. (c) 4 hr24 min. (d) 4hr 35 min. (e) 4hr 54 min. Directions (13-15): ‘B’ and ‘C’ together started to do a work ‘X’. After 6 days ‘B’ is replaced by ‘A’ who can do work ‘X’ alone in same time in which ‘B’ and ‘C’ together can do. After 4 days more ‘A’ left the work and 1 remaining work is completed by ‘C’ alone in 2 days. ‘B’ did 33 % of work ‘X’ 3 in total. ‘C’ and ‘D’ can complete the same work ‘X’ in 20 days while ‘D’ is 60% less efficient than ‘B’. Efficiency of A, B, C and D remain same for work ‘X’, ‘Y’ and ‘Z’ 13. Find the ratio between work done by ‘A’ in 4 days to work done by ‘B’ in 9 days? (a) 3 ∶ 4 (b) 1 ∶ 2 (c) 1 ∶ 3 (d) 2 ∶ 3 (e) 3 ∶ 1 8

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14. Find in how many days ‘A’ can complete another work ‘Y’ if ‘C’ and ‘D’ together can complete work ‘Y’ in 26 days? (2 Marks) (a) 14.4 days (b) 15.6 days (c) 18.2 days (d) 16.6 days (e) 19.5 days 15. ‘E’ who is 60% more efficient than ‘B’ can complete another work ‘Z’ 1 in 12 days. Find in how many days ‘A’ and ‘C’ together can complete 2 work ‘Z’? (2 Marks) (a) 20 days (b) 15 days (c) 12 days (d) 10 days (e) 8 days 1

16. Veer and Sameer alone can do 25% & 33 % of a task in 16 days 3 individually. If Sameer & Satish together can do same task in 16 days, then find in how many days Veer, Sameer and Satish will complete the whole task, if they do it on alternate days, starting with Satish and followed by Sameer & Veer respectively? (a)38 days (b)42 days (c)36 days (d)32 days (e)30 days 17. Working alone, A can complete a task in ‘a’ days and B in ‘b’ days. They take turns in doing the task with each working 2 days at a time. If A starts they finish the task in exactly 10 days. If B starts, they take half a day more. How long does it take to complete the task if they both work together? 1 1 5 (a) 5 days (b) 5 days (c) 5 days (d) 5

3 5

11

7

days

9

(e) None of these

18. Monika can do 40% of a work in 24 days and Anshika can do 12.5% of same work in 4.5 days. Shikha can complete the same work in equal time as Monika and Anshika can do together. If all three starts work alternatively, starting with Shikha and followed by Monika and Anshika respectively, then find how many days will be required to complete the work? 1 1 1 (a) 30 𝑑𝑎𝑦𝑠 (b) 24 𝑑𝑎𝑦𝑠 (c) 33 𝑑𝑎𝑦𝑠 3 1

(d) 28 𝑑𝑎𝑦𝑠 3

9

3 1

2

(e) 20 𝑑𝑎𝑦𝑠 3

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19. One day efficiency of A is 20% more than B and that of C is 20% less than B. D can do a piece of work in twice of time as A, B & C complete same work in together. If first eight days D work alone after that he left the work and A, B & C complete the remaining work in 12 days. Find in what time work will be completed, if B and D work in rotation on each day, starting with D? 1 1 1 (a) 41 𝑑𝑎𝑦𝑠 (b) 38 𝑑𝑎𝑦𝑠 (c) 36 𝑑𝑎𝑦𝑠 3 1

(d) 44 𝑑𝑎𝑦𝑠 3

3 1

3

(e) 46 𝑑𝑎𝑦𝑠 3

20. 5 inlet pipes (same capacity) can fill a tank in same time in which 3 outlet pipes (same capacity) can empty it. If 2 inlet and one outlet pipe is opened for first minutes and 5 inlet and 2 outlet pipe open for 2 nd minute and process continues till tank is completely filled in 30 minutes. Find the time in which 2 outlet pipe can empty the completely filled tank. (a) 9 minute (b) 12 minutes (c) 6 minutes (d) 10 minutes (e) 15 minutes Direction (21-22): P, Q, R and S four pipes can fill a cistern ‘N’ in 7 minutes if all are opened together. P is 25% less efficient then Q which is 100% more efferent than R. ‘S’ filled 25% of the cistern. P and Q together can fill the cistern in ‘x’ minutes while P and S can fill half cistern in ‘y’ minutes. 21. Pipe A and B together can fill another cistern ‘M’ in (x+12) minutes while pipe B and C together can fill cistern ‘M’ in (y+5) minutes. If pipe 6𝑥 A, B and C together can fill cistern ‘M’ in ( ) minutes, then find 𝑦

efficiency of ‘B’ is what % more/less than efficiency of ‘A’ ? (a) 50% (b) 75% (c) 100% (d) 150% (e) 200% 22. ‘D’ men can complete a work in (x+9) day while (y+7) men can complete same work in ‘E’ days. If y men can complete same work in (E+12) days then find in how many day (E–D) men can complete the same work? (a) 21 days (b) 42 days (c) 35 days (d) 14 days (e) 84 days 10

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Directions (23-24): These questions are based on the information given below. Each of A, B, C and D need a unique time to do a certain work. A can do the work in 𝑥 days and B can do the work in 2𝑥 days. A started the work and do 2

it for 22 days then he is replaced by B and B completed remaining work in 9

same time as C and D together can complete the whole work. The ratio of the efficiency of C and D is 4 : 5. If C and D work for alternative 1

days starting from C then they can do the total work in 44 days. 2

23. What is the value of 𝑥? 2

(a) 66

3 2

(d) 14

7

1

(b) 33

3

2

(c) 16

3

(e) none of these

24. A and B together can compete 225% of the work in how many days? 2

(a) 66 days

(b) 60 days

(d) 25 days

(e) none of these

3

25.

(c) 50 days

1

P can do 50% of a work in 32 days, Q can do 37 % of same work in 2

1

27 days, while R can do 62 %of same work in 30 days. Q and R 2

together started and worked for x days. After x days Q left the work and P joined R and both completed remaining work in (x +4) days. If ratio between work done by (Q + R) together to work done by (P + R) is 5 : 7 then in 2x days R will complete what fraction of work?

11

(a)

1

(d)

1

3 2

(b)

1

(e)

1

4

(c)

2 3

5

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26. Veer and Sameer can do a work in 64 days and 72 days respectively. Satish and Ankit together take 48 days to complete the same work. Ankit is 40% more efficient than Satish. If Veer and Sameer work for 18 days together, after that both left the work, then find in how many days remaining work will be completed by Satish and Ankit if they work on alternated days and Ankit starts the work? 6 6 6 (a) 40 𝑑𝑎𝑦𝑠 (b) 42 days (c) 48 𝑑𝑎𝑦𝑠 7 6

(d) 44 𝑑𝑎𝑦𝑠 7

27.

7 4

7

(e) 49 𝑑𝑎𝑦𝑠 5

Veer takes 50% more time to complete a task than that of Satish takes 2 to complete the same task, while Mahendra takes 2 days less than to 5 complete same task as Veer and Satish takes together. If Satish 1 complete 𝑟𝑑 of task in 8 day, then find in how many days work will 3 be completed, if Veer and Mahendra work in rotation, starting with Mahendra? (a) 16 days (b) 18 Days (c) 24 days (d) 12 days (e) 22 days

28. There are 3 pipes A, B and C. A can fill a bucket in 8 minutes. C can fill 18 3 buckets in 18 minutes & pipe B can fill buckets in a minute. These 5 20 pipes are opened in a tank alternatively, 1 minute each starting from A, then C, then B. If tank is filled after 2 hours, then find the volume of tank if the capacity of bucket is 5 liters. (a) 88 liters (b) 95 liters (c) 90 liters (d) 105 liters (e) 92 liters 29. Efficiency of Ram is 25% more than Shyam who completes a task in 2

60 days. Ghanshyam takes 6 days less than the days taken by Ram 3

and Shyam together to complete the work. If Ram and Shyam work for 16 days, after that both left the task then find in how many days Ghanshyam will complete remaining work? (a) 6 days (b) 8 days (c) 4 days (d) 5 days (e) 10 days 12

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1

30. A child can do th of work in same time as his father can do the whole 5

work and efficiency of child’s mother is equal to half the sum of father and child efficiency. If they complete 70% of work in 24 days, when they work alternatively, starting with father and child together on first day followed by father and mother together on second day. Find in how many days mother can complete 85% of work alone? (a) 48 days (b) 68 days (c) 60 days (d) 56 days (e) 54 days 31. Ankit and Apoorv together can finish a work in 8

4

𝑑𝑎𝑦𝑠 while Ankit

7

can do it individually in 15 days. Both work on another task for (x + 4) days and (x + 20) days respectively. If remaining

90 7

% of the second

task is completed by Bhavya in 18 more days with the efficiency of 2 units work/day. Then, find what portion of the second task will be completed in x days, if all three work together? (a) (d)

29 35 23 35

(b)

26

(e)

27

35

(c)

31 35

35

32. One day efficiency of Raj is equal to one day efficiency of Divyaraj and Veer together. Divyaraj is 25% more efficient than Veer and starts with 20% more of his efficiency with Veer and works for x days, after that Divyaraj left the work and remaining work will be completed by Veer and Sameer together in (x + 8) days. If ratio of work done by Divyaraj and Veer together to work done by Veer and Sameer is 5 : 7 and efficiency of Sameer is 25% less than Veer. Then find Raj and Sameer complete work together in how many days? (a) 14 days (b) 12 days (c) 16 days (d) 18 days

13

(e) 20 days

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33. Two pipes P and Q can fill tank A in 28 minutes and 56 minutes respectively and empty pipe M can empty same tank in 42 minutes. If all three pipes opened in tank B for (x – 24) minutes together they filled 90 liter of the tank which is 25% of the quantity tank B. Find in ‘x’ minutes what portion of tank B filled, if all pipe P and Q and M opened alternatively starting with P, followed by Q and M respectively? 7 7 5 (a) (b) (c) 12

(d)

36

3

(e)

35

36

7 41

34. There are two pipes A & B, pipe A is for filling the swimming pool and pipe B is to empty the swimming pool. Capacity of swimming pool is 5040 m3 and volume of pipe B is 8 m3/minute more than that of pipe 1 A. If pipe A takes 11 more minutes to fill same swimming pool, than 4

time taken by B to empty the same swimming pool. If pipe B can empty second swimming pool in 112.5 minutes, then find the capacity of second swimming pool? (a) 7200 m3 (b) 6400 m3 (c) 5600 m3 (d) 7800 m3

(e) 8400 m3

35. Two pipes P and Q can fill tank A in 28 minutes and 56 minutes respectively and empty pipe M can empty the tank in 42 minutes. Tank A have the capacity of 168 liters. If all three pipes opened in tank B for (x – 24) minutes together they filled 90 liter of the tank which is 25% of the quantity of tank B. Find in x minutes what portion of tank B filled, if all pipe P and Q and M opened alternatively in each minute starting with P, followed by Q and M respectively? 5 7 9 (a) (b) (c) 36

(d)

14

7 38

36

(e)

38

7 39

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Directions (36-38): Working efficiency of A is 20% more than that of B. B can complete a work ‘X’ in 36 days. B and C together started to complete the work ‘X’ and after 10 days they both left the work and then remaining work is done by A alone in 15 days. A and C together started to complete another work ‘Y’ and after working for 12 days they both left the work. Remaining work is done by B alone in 16 days. D first completed work ‘X’ and then completed work ‘Y’ in total 38 days. It is given that efficiency of all, in completing work ‘X’ and work ‘Y’ is same. 1

36. A, B and C working together completed rd of work ‘X’, and then A and 3 C are replaced by D. Now remaining of work ‘X’ is completed by B and D together. For how many days B worked? (a) 12 days (b) 10 days (c) 15 days (d) 4 days (e) 8 days 37. A, C and D working simultaneously completed work ‘X’ in ‘n’ days and A, B, C and D working simultaneously completed work ‘Y’ in ‘m’ days. Find the value of (𝑚 + 𝑛). (a) 15 days (b) 10 days (c) 12 days (d) 8 days (e) 17 days 38. A person E starts the work ‘X’ and leave after 12 days, then B and C complete the remaining work in 8 days. What is the ratio of number of days taken by A and E together to complete the work ‘X’ to the number of days taken by D, B and C together to complete the both work ‘X’ and ‘Y’ . (a) 3 : 5 (b) 5 : 3 (c) 8 : 7 (d) 1 : 2 (e) 2 : 3 Directions (39-40): Ankur & Shubham can do a piece of work 64 days together and Shubham &Kartik can do the same work in 72 days together. First 22 days Ankur & Shubham work together, after that Ankur left the work and Kartik joined Shumbham. Kartik and Shubham work for next six days, after that Shubham left the work and remaining work complete by Kartik & Anup in 30 more days. Efficiency of Anup is 75% more than efficiency of Kartik. 15

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39. Kartik and Anup work for x days, Ankur and Shubham work for next 2 1 (x+2 ) days, If remaining 12 % 𝑜𝑓 𝑡𝑜𝑡𝑎𝑙 work complete by Prabhat 3 2 in 6 more day. Prabhat will do, what portion of total work in (x+12) days? 2 1 3 (a) (b) (c) 3 2

4 2

(d)

4

(e)

3

5

40. Ankur and Shubham start work alternatively and work for y days. 𝑦 After that Kartik and Anup replace them and work for next days 2

75

alternatively. If % of total work still remain, which complete by 8 Abhimanyu in 9 more days. Find if Abhimanyu work for (y+16) days, then in how many days remaining work will be completed by Anup and kartik? (y is divisible by 4) 48 34 (a) days (b) days (c) 20 days 11

(d) 28 days

16

11

(e) 32 days

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Solutions

1.

(a); Ratio of efficiency of Arun, Yash and Rana is 6 : 4 : 5 Total work done by them in 1 day = (6 + 4 + 5) units = (15) units. Let they estimate 𝑥 days to complete the work. Then total work = 15𝑥 40 But Rana’s 1-day work is only 40%, i.e. × 5 = 2 unit 100 The work done by them in actual = (6 + 4 + 2) (x + 20) = 12(x + 20) ∴ 15𝑥 = 12𝑥 + 240 3𝑥 = 240 𝑥 = 80 Hence, estimated days are 80.

2.

(d);

First minutes (P + Q) = (6 + 4) = 10 unit Second minutes (P + Q – R) = (6 + 4 – 3) = 7 unit 17 unit in 2 minutes So, 17 × 8 = 136 unit in 16 minutes Remaining by (P + Q) = 4

(144 –136) 10

= minutes 5

Total time = 16 + 4

4 5

= 16 minutes 5

17

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Solution (3-4): Let R earns = y And P earns = y + x So, Q earns = y – x ATQ, y – x + y + y + x = 1026 y = 342 Each person earns according to their efficiencies so 342 Q earns → × 18 = 324 19

342

An P earns = × 20 = 360 19 Ratio of efficiency of P, R and Q = 20 : 19 : 18 Money per unit work = Rs.18 3.

(c); Amount earned by R and P together in 5 days Work done = (20 + 19) × 5 = 195 Money earned = 195 × 18 = Rs.3510 20+18 (a); Efficiency of S = = 19 2 Total work → 19 × 36 Days taken by P, Q and R to do that work 19×36 ⇒ = 12 day

4.

(20+19+18)

Q’s earning ⇒ 12 × 324 = Rs.3888 5.

(b); (X+4) men can complete work in 2X days ⇒ 1 man can complete same work in 2X(X+4) days (X+12) women can complete work in (X+8) days ⇒ 1 woman can complete same work in (X+8)(X+12) days Now ratio of efficiency of men to women is 5 : 4 ⇒ Ratio of days taken by men to women 4:5 4

2X(X+4)

5

(X+8)(X+12) 2

So, =

is

⇒ 2(𝑋 + 20𝑋 + 96) = 5𝑋(𝑋 + 4) ⇒ 2𝑋 2 + 40𝑋 + 192 = 5𝑋 2 + 20𝑋 ⇒ 3𝑋 2 − 20𝑋 − 192 = 0 ⇒ 3𝑋 2 − 36𝑋 + 16𝑋 − 192 = 0 18

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⇒ 3𝑋(𝑋 − 12) + 16(𝑋 − 12) = 0 ⇒ (3𝑋 + 16)(𝑋 − 12) = 0 16 ⇒ 𝑋 = 12, − 3

16×24

12 men can complete work in 12 = 32 𝑑𝑎𝑦𝑠 24×20 15 women can complete work in 15 = 32 𝑑𝑎𝑦𝑠 32×32 Required time = = 16 𝑑𝑎𝑦𝑠 32+32

6.

(e); Let, pipe A and pipe B alone can fill the tank in x and y hours respectively. ATQ, 1 x

1

1

y 2

12 1.5

x

y

+ =

and, +

… (i) 1

= … (ii) 7

On solving (i) and (ii) x = 28 hour ⇒ y = 21 hour Capacity of cistern = 28 × 60 × 2.5 = 4200 liter 7.

(b); Let A and B can do 3𝑥 and 4𝑥 unit of work in one day. So, Total work = (3𝑥 + 4𝑥 ) × 8 = 56𝑥 (A + B) two day work = 7𝑥 × 2 = 14𝑥 Remaining work = 42𝑥 In 6 days B will complete = 6 × 4𝑥 = 24𝑥 units So, remaining 18𝑥 units are completed by C in 6 day So, 56𝑥 unit will be completed in =

19

56𝑥 18𝑥 6

=

56 3

days

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8.

(c); A = 45 hr A : B = 100 : 150 = 2 : 3 Total capacity of tank = 45 × 2 = 90 liter C=

90 3

– 7.5 = 22.5 hr

C efficiency =

90

=4

22.5

ℓ hr

According to question ⇒

5X 4(X+9)

=

1 2

⇒ 10X – 4X = 36 X = 6 hr 9.

(b);

576

C takes = (8+9) + Efficiency of C =

36

=

17 576 367

612 7

= 36 days

= 16 units/days

When First day (A and B) and second day (B and C) work alternatively Two day work = (A and B) one day work and (B and C) one day work = (8 + 9) + (9 + 16) = 17 + 25 = 42 units In 26 day =

26 2

× 42 = 546 units

(A and B) on 27th day = 17 units After 27 days remaining work = (576 – 546 – 17) = 13 units 13 units work done by (B and C) on 28th day = Total time = (27 +

20

13 25

) = 27

13 25

13 25

days

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10. (d); A, B, C and D worked for 2 days together after that A leave and B, C and D worked alternatively for 10 days starting from B ∴ B worked for 4 days, C for 3 days, and D for 3 days. Total days A worked = 2 Total days B worked = 4 + 2 = 6 Total days C worked = 3 + 2 = 5 Total days D worked = 3 + 2 = 5 Let, their alone time to complete the work is 4x, 3x, 2x and 5x respectively. 2 6 5 5 ∴ + + + =1 4x 3x 2x 5x 30+120+150+60

⇒

=1

60𝑥 360

⇒𝑥= =6 60 ‘A’ can complete the work in 4 × 6 = 24 days ‘C’ can complete the work in 2 × 6 = 12 days 12×24 12×24 Required time = = = 8 days 12+24

36

11. (b); Let the efficiency of C [is C units/days] total work is 100a Units ATQ, 7 c × 10 = 21a 3

1

7

3 9𝑎

3

[233 % = ] C=

10

𝑈𝑛𝑖𝑡𝑠 /𝑑𝑎𝑦 1

Also B is 11 % more efficient means B=

10 9 10

9

of C 9𝑎

B = × Units/day 9 10 = a Units / day A which while working with half efficiency total half time. ∴ A : B = 4 : 1 [efficiency ratio] Efficiency of A = 4a Work done by then together is 4a + a = 5a/day Time take to complete 50 a Units 50𝑎 = = 10 days 5𝑎

21

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12. (c);

In 2 hours, part of the tank filled = 2(4) = 8 units Now required time = =

22 5

(30−8) 5

= 4hour 24mins.

Solution (13-15): Let, efficiency of A, B, C and D be ‘a’, ‘b’, ‘c’ and ‘d’ respectively Total work = 6b + 6c + 4(a +c) + 2c And also, a = b + c ⇒ Total work = 10b + 16c ATQ, 1

B did of work in 6 days 3

⇒ B can complete whole work in 18 days And, (10b + 16c) = 18b ⇒ 16c = 8b 𝑏

2

𝑐

1

⇒ =

⇒ ‘C’ can complete whole work ‘X’ in 36 days A can complete whole work ‘X’ in

18×36

D can complete whole work ‘X’ in

18

18+36 2

= 12 day.

× 5 = 45 days.

Ratio of efficiency of A, B, C and D =

1 12

∶

1 18

∶

1 36

∶

1 45

= 15 : 10 : 5 : 4 22

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13. (d); Required ratio =

4×15 9×10

=

60 90

=

2 3

14. (b); Let efficiency of A, B, C and D be 15x, 10x, 5x and 4x respectively Total work ‘Y’ = (5x + 4x) × 26 = 9x × 26 =234x 234𝑥 ‘A’ can complete work ‘Y’ in = 15.6 day 15

15. (d); Efficiency of ‘E’ = 16x 25 Total work ‘Z’ = 16𝑥 × = 200x 2 ‘A’ and ‘C’ together can complete work ‘Z’ in 200𝑥 200𝑥 = (15+5)𝑥 = = 10 days 20𝑥

16. (a); Veer can complete the whole task alone = 16 × 4 = 64 𝑑𝑎𝑦𝑠 Sameer can complete the same task alone = 16 × 3 = 48 𝑑𝑎𝑦𝑠 Total work = 192 𝑢𝑛𝑖𝑡𝑠(LCM of 64 and 48) Efficiency of Satish 192 192 = − = 8 𝑢𝑛𝑖𝑡𝑠/𝑑𝑎𝑦 16 48 If all three work alternatively First day by Satish = 8 units Second day by Sameer = 4 units Third day by Veer = 3 units Total work in three days = 8 + 4 + 3 = 15 units 36 In total 36 days = × 15 = 180 𝑢𝑛𝑖𝑡𝑠 3 Satish on 37 days = 8 units Remaining work after 37 days = 192 – 180 – 8 = 4 units On 38 days remaining work 4 = = 1𝑑𝑎𝑦𝑠 4 Total time = 38 days 23

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17. (b); If each works 2 days at a time alternately starting with A, the work is completed in exactly 10 days. ∴ A works for 6 days and B worked for 4 days. 6 4 + =1 ………….(i) a b If B starts, the work is completed in 10.5 days. ∴ B works for 6 days and A worked for 4.5 days. 6 4.5 + =1 ………….(ii) b a By solving (i) and (ii) a = 9 days And, b = 12 days Time taken by A and B working together to complete the work 1 1 =1 1 =1 1 + 36

a b

=

7

+ 1

9 12

= 5 days 7

18. (c); Monika can do whole work alone 24 = × 100 = 60 days 40 Anshika can do whole work alone 4.5 = × 100 = 36 days 12.5 Let Total work = 180 units (LCM of days taken by Anshika and Monika) 180 Efficiency of Monika = = 3 units/day 60 180

Efficiency of Anshika = = 5 units/day 36 Shikha take to complete whole work alone 180 = (3+5) = 22.5 days Efficiency of Shikha = 8 units/day when Shikha, Monika & Anshika work alternatively— Work done on First day by Shikha = 8 units Work done on First day by Monika = 3 units Work done on First day by Anshika = 5 units In three days = (8 + 3 + 5) = 16 units 24

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In total 33 days =

33 3

× 16 = 176 units

Remaining work complete by Shikha =

(180 –176) 8

1

= days 2

Total time = 33 + 1

1 2

= 33 days 2

19. (b); Let efficiency of B = 100 So, efficiency of A = 120 And efficiency of C = 100 × Efficiency of D =

100+120+80 2

80 100

= 80

= 150

Ratio of efficiency of A, B, C and D = 6 : 5 : 4 : 7.5 Let one day work of A , B , C & D be 6x units, 5x units, 4x units & 7.5x units respectively ATQ – Total work = 7.5x × 8 + (5𝑥 + 6𝑥 + 4𝑥) × 12 = 240x units When B & D work alternatively First day by D = 7.5x units Second day be B = 5x units Two day work of = 7.5x + 5x = 12.5x units In 38 days total work =

38 2

× 12.5𝑥 = 237.5𝑥 units

Remaining work = 240x – 237.5x = 2.5x Remaining work by D on 39 days = 1

2.5𝑥 7.5𝑥

= 𝑑𝑎𝑦𝑠 3

1

Total time = 38 𝑑𝑎𝑦𝑠 3

25

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20. (a); Let efficiency of one inlet pipe = 3x Efficiency of 5 inlet pipes = 5 × 3x = 15x Efficiency of 1 outlet pipe =

15x 3

= –5x

ATQ, 15 (2 × 3x + 1 × (–5x)) + 15 (5 × 3x + 2 × (–5x)) = Tank capacity Tank capacity = 90x Required time =

90x 10x

= 9 minutes

Solution (21-22): Let efficiency of R is 2 units/ minute ⇒ Efficiency of Q is 4 units/ minute So Efficiency of P is 3 units/ minute P + Q + R together completed 75% of work ⇒ 75% of work = 9 units ⇒ 100% of work = 12 units Efficiency of S is 3 units/minute Total work = 7 × (12) = 84 unit Number of days taken by P, Q, R and S individually: -

P and Q together can fill the cistern = =

28×21 49

28×21 28+21

⇒ x = 12 min. 28×28

P & S together can fill half cistern = 28+28 y = 7 min. 26

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21. (c); A and B together can fill cistern ‘M’ in (x+12) minutes = 12+12=24 minutes B and C together can fill cistern ‘M’ in (y+5) minutes = 7 + 5 = 12minutes 6𝑥

A, B and C together can fill cistern ‘M’ in ( ) minutes 𝑦

=

6×12 7

=

72 7

minutes

A’s efficiency = 49 – 42 = 7 units/minute C’s efficiency = 49 – 21 = 28 units/ minute B’s efficiency = 49 – 7 – 28 = 14 units/minute Required % = 7

14−7 7

× 100

= × 100 = 100% 7

22. (b); ATQ, D × (12+9) = (7+7) × E 3

⇒D× =𝐸 2

And (7+7) × E = 7 × (E+12) ⇒ 2E = E + 12 ⇒ E = 12 ⇒D=8 Let (E – D) men can complete same work in ‘a’ days ATQ, (12−8) × a = 8 × (12+9) a = 2 × 21 = 42 days 27

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23. (b); Let the C do in one day = 4𝑦 work Let the D do in one day = 5𝑦 work 2 day work of C + D = 9𝑦 In 44 day they will complete 9𝑦 × 22 = 198𝑦 1 In another days 2𝑦 work will be done 2

200y

C will take =

4y

days = 50 day

D will take = 40 days. C and D will complete work together in 50×40 200 = = 90 9 According to condition 200 200 + =1 9𝑥 9 × 2𝑥 400+200 =1 18𝑥

⇒ 𝑥 = 33

1 3

1

24. (c); A – 33 𝑑𝑎𝑦𝑠 3 2

B − 66 days 3

(A+𝐵)−

200 9

𝑑𝑎𝑦𝑠

Required Time −

200 9

9

× = 50 𝑑𝑎𝑦𝑠 4

25. (d); P can do whole work alone = 64 days Q can do whole work alone =

27 3

× 8 = 72 days

R can do whole work alone =

28

30 5

× 8 = 48 days

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ATQ— (8+12)𝑥

= (9+12)(𝑥+4)

5 7

140𝑥 = 105𝑥 + 420 𝑥=

420 35

𝑥 = 12 days In 2𝑥 days work done by R = 2 × 12 × 12 = 288 units Required portion =

288 576

=

1 2

26. (d);

Let efficiency of Satish is 5x unit/day and that of Ankit is 7x unit/day ATQ— 5x + 7x = 12 unit x = 1 unit/day Efficiency of Satish = 5 unit/day Efficiency of Ankit = 7 unit/day Work done by Veer and Sameer in 18 days = (9 + 8) × 18 = 306 unit Remaining work = 576 – 306Ru = 270 unit When Ankit and Satish work alternatively Then two days’ work = (7 + 5) = 12 units In 44 days total work done = = 264 unit

44 2

× 12

On 45th days remaining work done by Ankit = 6

(270 –264) 7

6

= days 7

Total time = 44 days 7

29

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27. (b); Let Veer take 3𝑥 day and Satish take 2𝑥 days to complete the task So, efficiency of Veer and Satish be 2𝑥 units/day & 3𝑥 units/day respectively. Whole work completed by Satish in 24 days So, while task done by Veer alone =

24 2𝑥

× 3𝑥 = 36 days

Total task = 72𝑥 units Veer and Satish do it together = 2

72𝑥 5𝑥

= 14 days 5

Mahendra do same task alone =

72 5

–

12 5

= 12 days

Efficiency of Mahendra =

72𝑥 12

= 6𝑥 unit/day

When Veer and Mahendra do it alternatively — First day task = 6𝑥 unit Second day task = 2𝑥 units In two day = 8𝑥 unit 72𝑥

Task completed =

8𝑥

× 2 = 18 days.

28. (b); When these pipes are opened for two hours, means each of them open for 40 min. individually. Pipe A can fill 1 bucket in 8 minutes & 5 buckets in 40 minutes …(i) Similarly pipe C can fill

18×40 5×18

= 8 bucket in 40 min. …(ii) And Pipe B can fill

3 20

× 40 = 6 buckets in 40 minutes …(iii)

Hence volume of tank = 5 + 6 + 8 = 19 buckets = 19 × 5 = 95 liters 30

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29. (b); Efficiency of Ram :Shyam = 125 : 100 =5:4 Ram takes =

60 5

× 4= 48 days

Let total work = 60×4= 240 unit Ram and Shyam takes together =

240 (5+4)

=

80 3

days

Ghanshyam takes =

80 3

20

−

3

= 20 days

Efficiency of Ghanshyam =

240 20

= 12 unit/day

Rams and Shyam 16 days work 16 × (5 + 4)= 144 unit Remaining work by Ghanshyam = =

96 12

240−144 12

= 8 days

30. (b); Let efficiency of child and father be x unit/day and 5x unit/day Efficiency of mother =

x+5x 2

= 3x unit⁄day ATQ, They start work alternatively— First day efficiency =(5x + x) = 6x Second day efficiency= (5x + 3x) = 8x Total work = =

168x 70

6x×12+8x×12 70

× 100

× 100

= 240x Mother complete 85% of work in 85

= =

240x×100 3x 204x 3x

= 68 days 31

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36×100×7

31. (e); Total work = = 280 units 90 ATQ 4 (x + 4) + 3 (x + 20) + 18 × 2 = 280 4x + 16 + 3x + 60 = 280 – 36 7x = 168 x = 24 days They all together in 24 days = 24 (4 + 3 + 2) = 216 units Required portion = =

27

216 280

35

32. (c); Ratio of efficiency of Divyaraj : Veer = 125 : 100 =5:4 Efficiency of Sameer = 4 × = 3 units/day ATQ, 120 +4)x 100

(5×

(4+3)(x+8) 10x 5 7x+56

=

=

3 4

5 7

7

70x – 35x = 280 x = 8 days Total work = 10 × 8 + 7 × 16 = 80 + 112 = 192 Raj efficiency = 9 units/day Sameer and Raj efficiency = (3 + 9) = 12 units/day (Sameer + Raj) = = 16 days 32

192 12

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33. (b); ATQ For tank A —

ATQ For tank B — 6 (x – 24) + 3 (x – 24) – 4 (x – 24) = 90 6x – 144 + 3x – 72 – 4x + 96 = 90 5x = (90 + 144 + 72 – 96) 210 x= = 42 minutes 5 Total quantity of tank B = 90 × 4 = 360 liter Alternatively (P + Q – M) for 42 minutes, means each work pipe for 14 minutes— All three in 14 minutes (P + Q – M) = 14 × 6 + 14 × 3 – 14 × 4 = 70 liter 70 7 Filled portion = = 360

36

34. (a); Let capacity of pipe A = y m3 So, capacity of pipe B = y + 8 m3 Required time to filled the swimming pool 5040 = minutes 𝑦

5040

Required time to empty the swimming pool = minutes (𝑦+8) ATQ – 5040 5040 45 – (𝑦+8) = 𝑦 112

112

1

4

− (𝑦+8) = 4 2 3584 = y + 8y y2 + 8y −3584 = 0 y2 + 64y – 56𝑦 − 3584 = 0 y (y + 64) – 56(𝑦 − 64) = 0 (y + 64) (𝑦 − 56) = 0 Y = 56 m3 Capacity of second swimming pool = (56 + 8) × 112.5 = 7200 m3 𝑦

33

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35. (b); ATQ For tank A —

ATQ For tank B — 6 (x – 24) + 3 (x – 24) – 4 (x – 24) = 90 6x – 144 + 3x – 72 – 4x + 96 = 90 5x = (90 + 144 + 72 – 96) 210 x= = 42 minutes 5 Total quantity of tank B = 90 × 4 = 360 liter Alternatively (P + Q – M) for 42 minutes, means each pipe for 14 minutes— All three in 14 minutes (P + Q – M) = 14 × 6 + 14 × 3 – 14 × 4 = 70 liter 70 7 Filled portion = = 360

36

Solutions (36-38):

Lets efficiency of Kartik is x unit/day and Anup is 1.75x unit/day ATQ— 22 days (Ankur + Shubham) + 6 days (Shubham + Kartik) + 30 day (Kartik + Anup) = 576 22 × 9 + 6 × 8 + 30 (x + 1.75x) = 576 30 (2.75x) = 576 – 246 330 2.75x = 11

30

x= ⇒ x = 4 unit/day 2.75 efficiency of Anup = 4 × 1.75 = 7 unit/day 34

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36. (a); Working efficiency of A = 120% of working efficiency of B B can complete ‘X’ work = 36 days A can complete ‘X’ work = 30 days Let ‘X’ work = 180 unit A working efficiency = 6 unit/day B working efficiency = 5 unit/day 15 × 6 + 10 × 5 + 𝐶 × 10 = 180 so, C working efficiency = 4 unit/day For work ‘Y Total of work ‘Y’=(6 + 4) × 12 + 16 × 5 = 200 𝑢𝑛𝑖𝑡𝑠 Total units of both work ‘X’ and work ‘Y’ = 180+200= 380 units 380 so,D working efficiency = 38 = 10 unit/day Now: A, B and C work together in ‘x’ work 1 = × 180 unit = 60 unit 3 A + B + C = (6 + 4 + 5) unit per day 60 = = 4 days 15 Remaining work = 120 unit (B + D) ⇒ (5 + 10) unit per day 120 = = 8 days. 15 B work for = 4 + 8 = 12 days. 37. (e); ‘X’ work = 180 unit A + C + D ⇒ 6 + 4 + 10 = 20 unit/day Days =

180 20

= 9 days

Work = 200 unit (A + B + C + D) ⇒ (6 + 5 + 4 + 10) = 25 unit/day =

200 25

= 8 days

Total time = (9 + 8) = 17 days 35

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38. (a); Let efficiency of E is Z unit/day he work for 12 days work complete = 12Z unit B and C work for 8 days = (5 + 4) × 8 unit = 72 unit Remaining work = 180 - 72 = 108 unit 108 Efficiency of E = = 9 unit⁄day 12 Now, A and E completed work ‘X’ 180 = = 12 days 15 D, B and C completed both work ‘X’ and ‘Y’ 200+180 = = 20 days 19 Ratio = 12 : 20 ⇒ 3 : 5 2

39. (c); (Kartik + Anup) × x days + (Ankur + Shubham) × (𝑥 + 2 ) day = 576 ×

3

87.5 100

3x+8

7

) = 576 × (4 + 7) × x + 9 × ( 3 8 11x + 9x + 24 = 504 20x = 480 x = 24 days efficiency of Prabhat =

1 8

576× 6

= 12 unit/day

Prabhat will do in (x + 12) days = (24 + 12) × 12 = 432 unit Required portion =

432 576

=

3 4

40. (a); If Ankur and Shubham work alternatively, then both work for

𝑦 2

days individually 𝑦 Same, if Kartik and Anup work for day alternatively both work 𝑦

2

for day individually 4

36

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ATQ— y y y y Ankur × + Shubham × + Kartik × + Anup × = 2

75

2

1

4

4

576 × (100 – ) × 8 100 or y y (Ankur + Shubham) × + (Kartik + Anup) × = 522 𝑦

2

9 × 𝑦 + (4 + 7) × = 522 ∗ 2 2 y = 72 days Efficiency of Abhimanyu 75

4

1

576× 8 ×100

= = 6 unit/day 9 Abhimanyu work did in (y + 16) days = (72 + 16) × 6 = 528 unit 576 –528 48 Remaining work done by Anup and Kartik in = = days 11

37

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Chapter

8

Speed Time and Distance

BEST APPROACH TO SOLVE THE QUESTIONS Speed is defined as the distance covered in per unit time. Mathematically, 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 Speed= . 𝑇𝑖𝑚𝑒

The whole concept of TSD (Time, speed and distance) lies within this formula. Three most logical points in TSD are: 1. When speed is constant, Distance is directly proportional to time i.e. D ∝ 𝑇 2. When distance is constant, 1 Speed is inversely proportional to time i.e. S ∝ 𝑇 3. When time is constant, Speed is directly proportional to distance i.e. S ∝ 𝑇 Another important term that we must know while solving questions in this chapter is RELATIVE SPEED. Relative speed is defined as the net speed of two entities when they are in motion. If the two entities are moving in opposite direction to each other, net/relative/resultant speed is sum of the speeds. And when they move in the same direction, their net/relative speed is the difference of their speeds. Example: ‘A and B stay next to each other and study in the same school. A and B start walking towards their school at speeds of 2 km/hr and 3 km/hr respectively. The faster of the two reaches the school first, turns around and starts walking back. If A and B meet 200 m away from the home. What is the distance between their residences and their school?’ 2

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Solution: Let the distance where they met be ‘x’ metres from the school. Distance travelled by A = 200m Distance travelled by B= 200+x+x=200+2x As the ratio of their speed is 2:3, and time is same, hence the ratio of distance travelled by the them is also 2:3. 200 200+2𝑥

=

2 3

Solving, x= 50 Distance between their residences to school= 200+ x= 250m Example: A goods train leaves a station at a certain time. After 6 hours, an express train leaves the same station and travels in the same direction at a speed of 90 km/hr. If the express train catches up the goods train in 4 hours, what is the speed of the goods train? Solution: Distance covered by express train after leaving the station = 90× 4 = 360𝑘𝑚 Goods train covers this distance in 6+4=10 hours. Hence speed of goods train=

360 10

= 36 𝑘𝑚/ℎ𝑟

Key Points to Remember for New Pattern Questions ➢ Average speed is calculated by dividing the total distance travelled by total time taken. ➢ Always be careful that units of speed, time and distance for two or more entities should be same while solving a problem. ➢ To change speed of km/hr into m/second multiply the former by

3

5 18

and by

18 5

𝑖𝑛 𝑣𝑖𝑐𝑒 𝑣𝑒𝑟𝑠𝑎 𝑐𝑎𝑠𝑒.

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Practice Exercise Based on new Pattern

1.

In how much time a train can cover a distance of 600 km if speed of train is 25% more than the speed of car and speed of car is 20 % more than the speed of a truck which cover 640 km in 20 hours in which he takes 40 minute break after every 100 km. (a) 9 hours (b) 8 hours 20 mins (c) 10 hours (d) 8 hours (e) None of these

2.

A train travelling at 144 km/hr crosses another train, having 30 meter less length and travelling in opposite direction at 126 km/hr in 6 seconds. If longer train cross a railway platform in 20 second then find smaller train will cross same platform in how many seconds? (a) 22 seconds (b) 24 seconds (c) 28 seconds (d) 32 seconds (e) 30 seconds

3.

Train A running at the speed of 108 km/hr crosses a man, who running in the opposite direction at the speed of 12 km/hr in 7.2 sec. If speed of train A increased by 25% and it takes 48 seconds to cross another train B, which running at the speed of 90 km/hr in same direction. Find the length of train B? (a) 280 meters (b) 360 meters (c) 180 meters (d) 160 meters (e) 220 meters

4.

Train P leaves Delhi at 7 PM and is travelling at a speed of 80 kmph, 3 hours later another train Q leaves Delhi and is travelling in same direction as train P in how many time (in minutes) train Q will be 18 km ahead train P, if speed of train Q is 120 kmph? (a) 383 minutes (b) 387 minutes (c) 380 minutes (d) 377 minutes (e) 375 minutes 4

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5.

A bus and a car start from a place A toward B simultaneously at 10:00 am. Bus reach at B in 4 hours and return for C, which is exactly in the middle of A and B, and meet the car after 1 hour after leaving B. Find the time at which car reach C after returning from B. (a) 6:00 pm (b) 8:00 pm (c) 10:00 pm (d) 8:00 am (e) 12:00 pm

6.

Two trains A and B running at the speed of 108km/hr and 162km/hr crossed another train ‘P’, which is standing stationary of length 240 meters in 14 sec and 16 sec respectively. How much time both train A and B will take to cross each other, if both running in opposite direction? (a) 8.4 𝑠𝑒𝑐 (b) 8.6 𝑠𝑒𝑐 (c) 8.8 𝑠𝑒𝑐 (d) 9 𝑠𝑒𝑐 (e) 8.2 𝑠𝑒𝑐

7.

There was a race of 3000 meters between A & B on a circular track of 750 meters. First time they meet during the race is after 5 minutes of starting of race. Find the time taken by B to complete the race, if he runs at half the speed of A. (a) 20 minutes (b) 18 minutes (c) 15 minutes (d) 10 minutes (e) None of these

8.

Two cities Rampur and Dhamnagar are 300 km apart, Bhavya starts from Rampur at 8 : 24 am and an hour later Abhi starts from same city. After travelling for one hour Abhi reaches at city which Bhavya had passed 40 min earlier. The city is in the way of Rampur and Dhamnagar. If they reaches city Dhamnagar at same time. Find their speed. (a) 50 km/hr, 75 km/hr (b) 75 km/hr, 75 km/hr (c) 100 km/hr, 100 km/hr

(d) 75 km/hr, 100 km/hr

(e) None of these 5

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9.

A 240 meters long train crosses a 210 meters long train running in opposite direction in 6 sec. Ratio between speed of longer train and smaller train is 7 : 8. If faster train crosses a platform in 9 sec, then find time taken by slower train to cross a bridge, which is 60 meters more long than platform? (a) (d)

142 7 148 7

𝑠𝑒𝑐

(b)

𝑠𝑒𝑐

(e)

136 7 142 7

𝑠𝑒𝑐

(c)

90 7

𝑠𝑒𝑐

𝑠𝑒𝑐

10. Train P leave place A by is travelling at a speed of 120 kmph. 4 hours later another train Q leaves place by travelling in same direction as train P. Train Q will be 36 km ahead of train P in T hours and speed of train Q is 160 kmph, then find a bus travel what distance in (T + 3.1) hr, if speed of bus is 75% of speed of train P? (a) 1360 km (b) 1440 km (c) 1540 km (d) 1280 km (e) 1720 km 11. Satish started from point A in a boat to reach point B. After 6.5 hours he covered only 20% of the distance and reach at point M. Now, Satish started from point M reached at mid point of A and B and came back to M in 29.25 hours. In what time Satish can cover the distance between B and A if he started from B? (a) 58.5 hours (b) 32.5 hours (c) 65 hours (d) Can’t be determined (e) 50 hours 12. Train X having length 130 m and train Y having length 145 m moving in opposite direction. They enter into a tunnel which have length equal to the sum of length of both trains. Trains meet after 10 second of entering in the tunnel. What percent of train X part is leave out the tunnel when it meet train Y if they have there speed in the ratio of 5 :6. (a) 2

11 13

(d) 5% 6

%

(b) 3

11 13

%

(e) 8%

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(c) 4 % 3

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13. A car and a bus starts from point ‘A’. After ‘T’ hours bus is 48 km ahead of car, while after ‘8’ hours distance between bus and car is same as distance covered by bus in one hour. If relative speed of car and bus if they move towards each other is 240, then how much 1 distance can bus cover in (𝑇 – ) hour. 2 (a) 320 km (b) 280 km (c) 384 km (d) 336 km (e) 256 km 14. Point A to Point B is a downstream journey of 300 km on a stream which flows at a speed of 5 km/hr. Two boats P and Q starts from point A and Point B respectively with speed of 25 km/hr and 15 km/hr in still water. After reaching the opposite point they return to their starting points, find after how much time will they meet second time? (a) 7.5hour (b) 15hour (c) 20hour (d) 10hour (e) None of these 15. Distance between Delhi and Jaipur is 300 km. Aman starts from Delhi and Rajiv from Jaipur at same time. After two hours, Aman realized he was travelling slow and therefore increased his speed by 25% and meet Rajiv at a point 108 km from Delhi. Find the increased speed of Aman, if Rajiv derived at a constant speed of 75 km/hr. (a) 40 km/hr (b) 50 km/hr (c) 60 km/hr (d) 55 km/hr (e) 65 km/hr Directions (16-17): Time taken by train A to cover a distance is 1.5 times the time taken by train B to cover the same distance. Train A and train C starts simultaneously from a station in same direction. Train B, start after 30 min and overtake train C in 1.5 hour later than it cross train A. 16. If speed of train A is 80 km/hr, then what is the speed of train C. (2 Marks) (a) 60 km/hr (b) 100 km/hr (c) 125 km/hr (d) 75 km/hr (e) 80 km/hr 7

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17. If train A double its speed, then speed of train A is what % more than the speed of train C. (1 Mark) (a) 60% (b) 100% (c) 80% (d) 50% (e) None of these Directions (18-19): A consignment should deliver on time to customer for which Mahendra start his journey with uniform speed, after 2 hours customer call Mahendra to deliver his order 1 hour before the decided time. Mahendra increased his speed by 50% to deliver it 1 hour before the decided time. 18. Find the total time taken by Mahendra to deliver the consignment? (a) 4 hours (b) 5 hours (c) 6 hours (d) 7 hours (e) 3 hours 19. If initially, customer alto start moving toward Mahendra at speed of 60 km/hr and take his consignment after 3 hours then find the total distance between Mahendra and Customer initially. (a) 256 km (b) 150 km (c) 500 km (d) 450 km (e) 300 km Directions (20-21): A man decided to run 15 rounds of a circular track of 400 m in certain time with certain speed. He starts running but after completing some round around the track he reduced his speed by 40% due to which he takes 4 min extra as scheduled. But if he reduced his speed by completing 3 more rounds he would have reached 2 min earlier than the time he actually reached. 20. Find the original speed of man? (a) 240 m/min (b) 160 m/min (d) 400 m/min (e) 350 m/min

(c) 200 m/min

21. Find the number of rounds at which he decided to reduce his speed? (a) 7 (b) 9 (c) 10 (d) 8 (e) 11 8

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22. A cyclist left point A for point B and travelled at the constant speed of 25 km/h. When he covered the distance of 25/3 km, he was overtaken by a car that left point A twelve minutes after the cyclist and travelled at a constant speed too. When the cyclist travelled another 30 km, he encountered the car returning from B. Assume that the car did not stop at point B. Find the distance between A and B. (a) 39.5833 km (b) 41.0833 km (c) 60.833 km (d) 43.33 km (e) 50 km 23. Two person left simultaneously two places A and B. One of them left A for B while the other left B for A. Both travel at uniform speed . The first person on reaching B returns to A and then again travels back to B and so on. Similarly the second person on reaching A returns to B and then travels back to A and so on. What will be the distance covered by the first person when they meet for the third time given the ratio of the speed of the first person to that of the second person is 3 : 2 and the distance between A and B is 500 m? (a) 1000 m (b) 1500 m (c) 2500 m (d) 1200 m (e) 1800 m 24.

A train M running at the speed of 108 km/hr crosses a man running at 12 km/hr in opposite direction of train in 12 second and crosses a platform in 32 sec. If a train N is Standing on same platform and length of platform is 140 meter more than length of train N. Find in what time train N will cross train M running in same direction, if train N passes a pole in 12 second? (a) 168 sec (b) 164 sec (c) 154 sec (d) 186 sec (e) 172 sec

Directions (25-26): Satish and Bhavya starts from point P and Q respectively at same time to reach at point B. Satish overtakes Bhavya after 10 hours at point A. If Bhavya starts 4 hours earlier than Satish then they will meet at point B after 16 hours when Satish starts. Ratio between distance between Q and A to Q and B is 1 : 2. Point P, Q, A and B lies on a straight road. 9

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25. If distance between point A and point B is 180km then find time taken by Bhavya to reach point ‘C’ which is 432km ahead of point ‘B’? (a) 36 hours (b) 44 hours (c) 24 hours (d) 32 hours (e) 40 hours 26. If Satish and Bhavya started to move towards each other from P and Q at 8:00 a.m, then at what time they will meet? (a) 9 : 00 a.m (b) 9 : 30 a.m (c) 10 : 00 a.m (d) 10 : 30 a.m (e) 11 : 00 a.m 27. A bus (X) start its journey from P to Q simultaneously another bus (Y) start from Q to P they reach their respective destination after 3 hour. Bus X travel with a speed of 40 km/hr in first hour 50 km/hr in second hour and 60 km/hr in 3rd hour. Bus Y travel with a speed of 60 km/hr in first hour, 40 km/hr in second and 50 km/hr in last hour. Find out the distance between Q and the point where the buses cross each other. (a)87 (b)88 (c)89 (d)82

2

(e) 87

9

28. Speed of a car is

175 9

7 9 2

𝑚/𝑠 and speed of bus is 14 % less to speed of car, 7

bus and car covered different distance in different interval of time and the sum of the time taken by them to cover their respective distances is10 hours but distance covered by bus is 80 km more than distance covered by car. If a train covered total distance which covered by car 2

and bus together in 6 ℎ𝑜𝑢𝑟𝑠, then find speed of train? 3

10

(a) 96km/hr

(b) 84km/hr

(d) 72km/hr

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(c) 78km/hr

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20

29. Two trains running in opposite directions cross each other in sec. 3 Longer train cross shorter train in 60 sec when both train running in same direction. If length of shorter train increased by 50% than longer train cross shorter train in 72 sec, when both running in same direction. If difference between length of both trains is 60 meters, then find the time in which longer train will cross a platform whose 7 length is 77 % more than length of longer train? 9 (a) 20 sec (b) 10 sec (c) 15 sec (d) 14 sec (e) 18 sec 30. Two trains P and Q started from two points A and B towards each other to cover 300 km distance. Speed of train Q is twice of speed of train P. If train P runs at its normal speed and train Q reduces its speed to 50% then it will take 2.5 hrs. more to meet each other. Find the time (in sec) in which train P can cover a platform of 150m length if the length of train P is given as 50m. (a) 18 seconds (b) 36 seconds (c) 72 seconds (d) 54 seconds (e) 24 seconds 27

31. A train P, 180 meter long train passed a pole in sec and also passed 4 two trains Q and R in 9 sec and 39 sec respectively, where train Q running in opposite direction of train P and train R is running in same direction of train P. If length of train Q and R is 240 meter and 210 meter respectively, then in what time train Q will pass train R, if both runs in opposite direction ? 7 3 (a) 35 sec (b) 9 sec (c) 12 sec 11 11 (d) 15 sec (e) 55 sec 32. Total distance between Delhi to Lucknow is 480 km. A train starts running with an average speed of 60 km/hr from Delhi to Lucknow while another train starts Journey after 120 minutes of first train and reaches Lucknow 30 minutes before first train. If first train stops for 5 minutes on each station and second train did not stop at any station ,then find the ratio between speed of first train to speed of second train given that total number of station between Delhi and Lucknow are nine.? (a) 25 : 33 (b) 25 : 32 (c) 35 : 25 (d) 25 : 31 (e) 21 : 31 11

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33. The ratio between length of two trains is 3 : 2 and speed of both trains is 72 km/hr & 90 km/hr respectively. Faster train crosses slower trains in 60 sec running in same direction. If slower train crosses a goods train, which is running at the speed of 108 km/hr in 7.2 sec running in opposite direction. Then find the time taken by faster train to cross goods train running in same direction? (a) 40 sec (b) 64 sec (c) 60 sec (d) 56 sec (e) 50 sec 34. Two athletes ‘A’ and ‘B’ practice running on 800 meters’ track. If they start running at their usual speeds, in same direction, at same time, from same point, first time they meet is in 100 seconds. During warm2 up, ‘A’ runs at 18 % of his usual speed and completes 400 meters in 11 50 second. Find the time taken by athlete ‘B’ to complete 9 rounds of 5 400-meter track, if he runs at of his usual speed. Assume athlete ‘A’ 4 is faster than ‘B’. (a) 100 second (b) 450 second (c) 90 second (d) 80 second (e) 85 second 35. The speed of car is 25% more than the speed of bus. The time difference between them to cover a certain distance D, is 1 hour. On particular day, driver noticed that if they are (D- 40) km apart from each other then they can meet in 2 hours driving in opposite direction at their normal speed. Find 150% of the speed of bus ? (a) 150 km/hr (b) 90 km/hr (c) 120 km/hr (d) 105 km/hr (e) None of these 36. Time taken by car A to cover distance from point X to point Y is 6½ hrs including ½ hour of stoppage. B also started from point X and stopped 90 km earlier from point Y after travelling for 6½ hrs. Had they started from same point and at same time, the difference between their distance covered in 7.5 minutes will be 2.5 km. Find the distance between two cars, if they start a race for the distance which is twice the distance between X and Y , assuming that both cars stops when one of the car finish the race (a) 360 km (b) 240 km (c) 180 km (d) 300 km (e) 280 km 12

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37. Two trains A and B with their length difference 17 m cross each other in 2.6 seconds while running in opposite direction. Train A crosses train B in 13 seconds while they are running in same direction. If train A takes 2 seconds to cross a pole, then find the sum of speed of both the trains, assuming that train B is longer than train A. (a) 34 ms¯¹ (b) 51 ms¯¹ (c) 68 ms¯¹ (d) 85 ms¯¹ (e) 102 ms¯¹ 38. ‘Rajdhani express’, 240 m long train crosses a pole in 6 sec and it also 120 crosses two trains ‘Shatabdi express’ and ‘Duranto express’ in 𝑠𝑒𝑐 17 & 42 sec respectively, where ‘Shatabdi express’ is running in opposite direction of ‘Rajdhani express’ and ‘Duranto express’ running in same direction as that of ‘Rajdhani express’. If length of ‘Shatabdi express’ & Duranto express’ is 360 m and 180 m respectively, then find in what time ‘Shatabdi express’ will cross ‘Duranto express’ while running in same direction? (a) 24 sec (b) 28 sec (c) 32 sec (d) 36 sec (e) 48 sec Directions (39–40): There are two trains A and B, both train starts travelling to each other from stations P and Q respectively. Train A reaches the station Q in total 4 hours while train B reaches the station P in total 4 hours 48 minutes. Speed of train A is 120 km/hours. 39. Train A and B running on two different railroads parallelly in opposite direction to two different trains C and D respectively. Train A passed train C, which running at the speed of 60 km/hr. in 6 sec. and train B passed train D, which running at the speed of 110 km/hr. in 9 sec. if ratio between length of train A to train C is 3: 2, and that of between train B to train D is 4 : 3. Find in what time train D passed train C, if both running in same direction ?(2 marks) (a) 22.64 sec (b) 24.84 sec (c) 24.24 sec (d) 28.84 sec (e) 20.84 sec 13

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40.

Ratio between speed of train A to train E is 6: 5.Train E starts a journey from Kolkata to Delhi, which is 600 km at 9.45pm and train A starts same journey after 30 minutes of train E.Train A stop for x minutes on each three stoppages and train E stop for (x+10) minutes each on three stoppages between Kolkata to Delhi.Train A reaches Delhione hour earlier than train E, which reaches at 4.30 am.Findinwhat time train B a total distance of 720 km cover between Lucknow to Jaipur. If train B stop for (x+10) minutes on four stoppages between Lucknow to Jaipur?(2 marks) (a) 4 hours (b) 8 hours (c) 8.2 hours (d) 10 hours (e) 12 hours Solutions

1.

(c); Stop of truck = 6 × 40 = 240 mint or 4 h. 640 640 Speed of truck = = = 40 𝑘𝑚/ℎ Speed of car =

20−4 ℎ. 40×120

= 48 𝑘𝑚/ℎ

Speed of train =

100 48×125

Required time = 2.

16

100 600 60

= 60 𝑘𝑚/ℎ

= 10 ℎ𝑜𝑢𝑟𝑠

(a); Let length of both trains be L meters and (L – 30) meters respectively ATQ, 5 𝐿+ (L – 30) (144 + 126)× = 18 6 450 = 2L – 30 L = 240 Smaller train length = 210 meters Let length of platform be P meters ATQ, 5 240+𝑃 144 × = 18 20 P = 800 – 240 = 560 meters Let required time = T ATQ, 5 210+560 126 × = T = 14

18 770 35

𝑇

⇒ T = 22 sec Adda247 Publications

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3.

(b); Let length of train A be L meters (108 + 12) ×

5 18

=

𝐿 7.2

L = 240 meters New speed of train A = 108 ×

125 100

= 135 𝑘𝑚/ℎ𝑟 Let length of train B be S meters (135 – 90)×

5 18

=

240+𝑆 48

S = 360 meters 4.

(b); Relative speed of both train = (120 – 80) = 40 kmph Distance covered by train P from Delhi when train Q was not moving = 80 × 3 = 240 km Train Q should be covered (240 + 18) km from Delhi for ahead 18 km to train P from Delhi Required time =

258 40

km

= 6.45 = 6.45 × 60 = 387 minutes 5.

(b);

Car travel → 3/4 part of one side distance in 5 hour. 1 part =

20 3

hour

Car can travel 1.5 part in 20 3

× 1.5 = 10 hour

Time to reach C after returning form point B → 10:00 + 10 = 8:00 pm

15

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6.

(c); Let speed of train A and B be ‘x’ meters and ‘y’ meters respectively ATQ, 5 x+240 108 × = m⁄s 18 14 30 × 14 = x + 240 x = 180 meters And, 5 𝑦+240 162 × = 18 16 45 × 16 = y + 240 y = 480meters Let required time be T sec 5 480+180 (108 + 162) × = 18 T 75 𝑇 = 660 T = 8.8 sec

7.

(a); Let speed of A is 2x & speed of B is x According to question, 750 = 5 min a−b ⇒ a – b = 150 meters/min ⇒ 2x – x = 150 meters/min ⇒ x = 150 meters/min It is speed of B hence 3000 3000 Time taken = = = 20 minutes x

8.

150

(d); Let the city in way of Rampur and Dhamnagar is C.

Time Speed Bhavya 80 3x 60 4x Abhi 16

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ATQ, 300 300 – =1 3x 25

4x

= 1 ⇒ x = 25 ∴ Bhavya Speed = 3 × 25 = 75 km/hr Abhi speed = 4 × 25 = 100 km/hr x

9.

(c); Let speed of longer train and smaller train be 7𝑥 and 8𝑥 respectively. ATQ— (240+210) (7𝑥 + 8𝑥 ) = 6

90𝑥 = 450 𝑥=5 Speed of longer train = 5 × 7 = 35 m/s Speed of smaller train = 5 × 8 = 40 m/s Let length of platform be l meter 𝑙+210 40 = 9 40 × 9 = 𝑙 + 210 l=150 Then, length of bridge = 150 + 60 = 210 meters Let time taken by slower train to cross bridge be T sec. T=

210+240 35

35T = 660 T=

90 7

sec.

10. (b); Relative speed of both trains = (160 – 120) = 40 kmph Distance covered by train P from place A when train Q was not moving = 120 × 4 = 480 km Train Q should cover (480 + 36) km from place A in T hrs Required time, T = 3

516 40

= 12.9 hr

Bus speed = 120 × = 90𝑘𝑚𝑝ℎ 4

Bus covered in ( T+3.1) hours = ( 12.9 + 3.1) × 90 = 1440 km 17

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11. (c); Let total distance from A to B = ‘D’ ATQ, Satish cover 20% distance in 6.5 hours So, he can cover 30% distance (M to mid-point of A and B) in 6.5 × 3 = 9.75 hr. 2 Time taken by Satish to come back from mid-point to M = 29.25 – 9.75 = 19.5 hr 30% distance covered by Satish in 19.5 hr. 19.5 100% distance covered by Satish in × 10 = 65 hr 3

12. (b); Speed of train X and Y 5x and 6x respectively. Length of tunnel → 130 + 145 = 275 m. Speed of trains per second 275 = = 27.5 𝑚/s 10 5x + 6x = 27.5 x = 2.5 speed of train X → 12.5 m/s Speed of train Y → 15 m/s Distance cover by train X in tunnel = 12.5 × 10 = 125 m. Length of train X leaves out = 130 – 125 = 5m 5 Required% = × 100 =3

11 13

%

130

13. (a); Let, speed of car = x km/hr Speed of bus = y km/hr ATQ, Distance covered by bus in one hour = Distance between bus & car in 8 hour ⇒ y = 8 (y – x) [ ∵ y > x] y = 8y – 8x 8x = 7y …(i) and x + y = 240 …(ii) On solving (i) & (ii) x = 112, y = 128 After T hour bus is 48 km ahead of car ⇒ (128 – 112) T = 48 48 ⇒T= =3 16 Required distance = 2.5 × 128 = 320 km 18

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14. (c); Time when P and Q meet first time 300 300 = = = 7.5 hr 25+15 40 Distance travelled by P = 7.5 × (25 + 5) = 225 km Distance travelled by Q = 7.5 × (15 – 5) = 75 km 75 P reach at point B in i.e. 2.5 hour 30 In 2.5hr Q travelled 2.5 × 10 = 25 km Now Boat P returns and relative distance b/w P and Q is 100 km and relative speed is 10 km i.e. difference of speed of P (25 -5) = 20 km/hr (upstream) and speed of Q (15 – 5) = 10 km/hr (upstream) 100 Time taken by P and Q to meet = = 10 hr 10 Total time when P and Q meet second time = 7.5 +2.5 + 10 = 20 hr 15. (b); Let initial speed of Aman is x km/hr. Distance travelled by him in 2 hours is 2x km. While distance travelled by Rajiv in these 2 hours is 75 × 2 = 150 km When both of them meet, Aman had travelled a distance of 108 km. Distance travelled by Aman with 25% increase in his speed = (108 – 2x) km 125 5x And his increased speed = x( ) = km/hr 100 4 If Aman had travelled 108 km, Rajiv had travelled 192 km. 192 – 150 = 42 km after 2 hours Time taken by Rajiv to travel 42 km 42 14 = = hours 75 25 This is equal to time taken by Aman to travel (108 – 2x) km (108–2x) 14 = 5 4

25 14×5x

x

108– 2x = 108 = 27

7 10

25×4

x + 2x

x = 108 ⇒ x = 40 km/hr. Therefore, increased speed of Aman was 5 = × 40 = 50 km/hr 10

4

19

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Solutions (16-17); Time ratio between train A and train B = 1.5 : 1 So speed ratio between train A and train B is =1 : 1.5 or 2 : 3 Let speed of train A = 2x km/hr Speed of train B = 3x km/hr Relative speed = 3x – 3x = x km/hr Distance travel by train A in 30 m 2𝑥 ⇒ = x km 2

𝑥

Train B cross train A ⇒ = 1 hour 𝑥 So, Train B cross train C ⇒ 1 + 1.5 = 2.5 hour Let speed of train C = y km/h ATQ 2.5 × 3x = 3y x:y⇒2:5 Ratio of speed of A, B and C 4:6:5 16. (b); Speed of train A = 80 km/hr 80 Speed of train C = × 5 = 100 km/hr 4

17. (a); Let speed of train A = 4x km/hr So speed of train C = 5x km/hr (2×4𝑥–5𝑥) Required % = × 100 = 60% 5𝑥

18. (b); Let speed of Mahendra = 2𝑥 km/hr So, Distance covered by him in 2 hours = 2𝑥 × 2 = 4𝑥 Let he decide to deliver his consignment in 𝑡 hour So total distance ⇒ 2𝑥 × 𝑡 = 2𝑥𝑡 ATQ ⇒ 2𝑥𝑡 = 2𝑥 × 2 + (𝑡– 3)3𝑥 3𝑥 → increase speed 𝑡– 3 = time for which he travelled after speed is increased On Solving equation → 5 ℎ𝑜𝑢𝑟𝑠 = 𝑡 20

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19. (d); Distance covered by Mahendra in 3 hours → 2𝑥 × 3 = 6𝑥 Total distance ⇒ 5 × 2𝑥 = 10𝑥 Remaining distance ⇒ 10𝑥 – 6𝑥 = 4𝑥 = Distance covered by customer and 4𝑥 = 60 × 3 𝑥 = 45 km Total distance = 45 × 10 = 450 km Solutions (20-21): Total distance = 15 × 400 = 6000 m Let his original speed = x meter/min Decided time = t min Let after ‘n' round he reduced his speed 1st decided condition ⇒t= 3x 5

6000 x

…(i)

is the reduced speed after n rounds

Now 2nd condition 400×n x

+

6000–400×n 3x 5

= t + 4 …(ii)

3rd → supposed condition 400(n+3) x

+

6000–400×(n+3) 3x 5

= t + 2 …(iii)

Putting value of ‘t’ from (i) into (ii) and (iii) ⇒ 3x + 200n = 3000 …(iv) ⇒ 3x + 400n = 4800 …(v) Solving (iv) and (v) x = 400 m/min n=9 20. (d); Original speed of man = 400 m/min 21. (b); Number of rounds after which he reduced his speed = 9

21

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25

1

22. (c); Time taken by Cyclist to reach 25/3 km = = hr = 20 min 25 × 3 3 Car has taken to reach 25/3 km = 20 – 12 = 8 min Speed of Car =

25 3  60 = 62.5 km/h 8

Now time taken by cyclist to go further 30 km 30 6 = = hr = 72 min 25

5

Car will go in 72 min =

72  62.5 = 75 km 60

Now, according to question, distance between first meeting and second meeting is 30 So, distance between first meeting and point B will be 75+30 = = 52.5 km 2 Required answer = 52.5 + 8.33 = 60.833 km 23. (b); When they meet for the third time they together cover 5 × 𝐴𝐵 = 5 × 500 = 2500𝑚 The first person will cover 3 × 2500 = 1500 𝑚 2+3

24. (b); Lets length of train M = L meter ATQ, (108+12)×5 18

=

𝐿 12

L = 400 m Let length of platform = Lp 𝐿𝑝 +400 108×5 = 18 32 𝐿𝑝 = 960– 400 𝐿𝑝 = 560 meter Length of train N = 560 – 140 = 420 meter Speed of train N = 𝑆𝑛 𝑆𝑛 = 22

420 12

= 35 m/s Adda247 Publications

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Relative speed of train M and N, when both running in same direction 5 = 35 −108 × = 5 𝑚/𝑠𝑒𝑐 18 Let required time is t ATQ, 400+420 5= 820

𝑡

t= 5 t = 164 sec Solution (25-26)

Let speed of Satish = ‘a’ And, speed of Bhavya = ‘b’ Let distance between P and Q = x And, QA and QB be y and 2y respectively. After 10 hrs Satish overtakes Bhavya at point A. 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 ⇒ 𝑆𝑝𝑒𝑒𝑑 = 𝑇𝑖𝑚𝑒 y= 10 × b and, (a - b) × 10 = x …(i) If Bhavya starts 4 hour earlier than Satish, then Satish overtakes Bhavya at point B after 16 hours Distance travelled by Bhavya in 4 hours = 4b So, (x + 4b) = (a – b) × 16 …(ii) On solving (i) & (ii) 20𝑏 2𝑦 𝑥= ⇒𝑥= …(iii) 3 3 Satish can travel and reach point B in 16 hours ⇒ 16a = 2y + x …(iv) On solving (iii) & (iv) y = 6a 3𝑥 𝑥 3𝑥 So, 𝑦 = , 𝑎 = , 𝑏 = 2

23

4

20

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25. (b); ATQ, 3𝑥 = 180 2 ⇒ x = 120 km/hr Required time = =

3×120+432

=

792 18

3𝑥 3𝑥 + +432 2 2 3𝑥 20

18

= 44 hours

26. (d); Required time = 𝑥 =

𝑥

3𝑥 + 4 20

𝑥 5𝑥+3𝑥 20

20

= = 2 ℎ𝑜𝑢𝑟 30 𝑚𝑖𝑛𝑢𝑡𝑒 8 Satish & Bhavya meet at 10 : 30 am 27. (d); Distance between P and Q is 150 km. Now X bus cover 40 km in 1 hour Y bus cover 60 km in 1st hour Remaining distance = 50 km 𝑅𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑡ime to cross each other 50 50 5 = = = ℎ𝑟 40+50

90

9

5

Distance which is covered by ′Y ′ in hr 5

200

9

= 40 × = km 9 9 Distance between Q and the point where buses crosses each other 200 2 = 60 + = 82 km 9 9 24

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28. (a); Speed of bus = 50

175 9 18

50

6

25×6

7

9

× =

= m⁄s = × = 60 km⁄hr 3 3 5 Let distance travel by car and bus be D and (D + 80) respectively. D

+

(D+80)

= 10

70 60 6D+7D+560

= 10 13D = 4200 – 560 13D = 3640 D = 280 km 420

Speed of train = = 96 km⁄hr

[280+(280+80)]×3 20

29. (a); Let length of shorter train be L meters and length of longer train be (L + 60) meters And also speed of shorter and longer train be V m/s and U m/s respectively. When both trains travelling in opposite direction (L+L+60)3 20 3L+180

= U+V

= U + V …(i) When both train running in same directions— L+L+60 = U– V 20

60 (2L+60)

= U– V …(ii) When length of shorter train increased and both train running in same direction — 60

(1.5L+L+60) 72 2.5L+60

= U –V

= U– V …(iii) From (i) and (ii) 3L+180 2L+60 2U = + 72

U= U= 25

20 11L+600 120 7L+480 120

60

…(iv)

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After solving (iii), (iv) and (v) L = 120 m, U = 25 m/s and V = 20 m/s Length of longer train = 120 + 60 = 180 meter 7 Length of platform = 180 + 180 × = 320 meters 9 Time taken by longer train to cross platform 320+180 500 = = = 20 sec 25

25

30. (b); Let, speed of train P = x Speed of train Q = 2x Let, Normal time taken by train to meet each other = ‘y’ ATQ, (x + 2x) y = (x + x) (y + 2.5) 3y = 2(y + 2.5) y = 5 hr. 300 Now, x + 2x = = 60 5 3x = 60 x = 20 km/hr 50 = m/sec 9

(150+50)

Required time = 50 200 = × 9 = 36 sec 50

×9

31. (c); Lets speed of train P, Q and R be S₁, S₂ and S₃ respectively 180 80 m Speed of train P (S1 ) = 27 m⁄s = Speed of train Q (S2 ) 80 240+180 + 𝑆2 = 3

420

80

4

3 s

9

S2 = – 9 3 S2 = 20 m⁄s Speed of train R (𝑆3 ) 80 210+180 – S3 = 3

S3 = 26

80 3

39

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50

S3 = m⁄s 3 Lets required time be T sec 50 240+210 Required time = 20 + = 110 3

=

450

T = 12

T 3 11

⇒T =

450×3

3

T

110

sec

32. (b); Speed of first train = 60 km/h Total distance = 480 km Time taken by first train to cover that distance without stoppage 480 = = 8ℎ 60 9 station × 5 minute = Total stoppage time = 45 minute Train took total time to reach Lucknow 35 = 8 hr 45 minute = ℎ𝑜𝑢𝑟 4 2nd Train reach 30 min. before it start 2 hour late from Delhi So, Time taken by 2nd train = 8 h 45 min – 2h – 30 min. = 6 hour 15 minute 1 25 = 6 = hour 4

4

480×4 384

Speed of 2nd Train=

25

=

5

km/h 60

Required Speed ratio of train = 384 =25:32 5

33. (c); Let length of two trains be 3L meter and 2L meter ATQ— 5 3L+2L (90– 72) × = 18 60 300 = 5L L = 60 meter Length of faster train = 120 meter Length of slower train = 180 meter Let length of goods train be G meter 27

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5

180+𝐺

(108 + 72) × = 18 7.2 50 × 7.2 = 180 + G G = 360 – 180 G = 180 meter Let faster train cross good train in T sec — 5 120+180 (108 – 90) × = 18 T 5T = 300 T = 60 sec 34. (d); Let the speed of athletes ‘A’ and ‘B’ is 𝑎 and 𝑏 respectively From 1st condition 800 = 100 ⇒ 𝑎– 𝑏 = 8 ms –1 𝑎–𝑏 From 2nd condition 2 When athlete ‘A’ will run at 18 % of his usual speed = =

200 1100 400 2 𝑎 11

×𝑎=

2 11

11

𝑎

= 50 ⇒ 𝑎 = 44 ms –1

∴ 𝑏 = 36 ms –1 5 If 𝑏 runs at of his speed, which means with 45 ms –1 4

Time taken by him =

4×400 45

= 80 seconds.

35. (c); Let speed of car is 5𝑥 𝑘𝑚/ℎ𝑟 and speed of bus is 4𝑥 𝑘𝑚/ℎ𝑟. According to first condition, 𝐷 𝐷 5𝐷–4𝐷 – =1⇒ =1 4𝑥 5𝑥 20𝑥 ⇒ 𝐷 = 20𝑥 …(i) Also if they are travelling in opposite direction, Then 𝐷–40 = 2 ⇒ D – 40 = 18x …(ii) 4𝑥+5𝑥 From (i) & (ii) 18𝑥 + 40 = 20𝑥 ⇒ 𝑥 = 20 Hence speed of bus = 4 × 20 = 80 km/hr 150 150% of speed of bus = × 80 = 120 km/hr 100

28

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36. (b); Let the Distance b/w X to Y be D km and speed of car A and B be A km /hr & B km/hr respectively 13 6A – B = 90 – (i) 2 & according to second condition 7.5 7.5 A– 𝐵 = 2.5 60 60 A – B = 20 – (ii) Solving (i) & (ii) A = 80 km, B = 60 km Distance b/w X & Y is = 6 × 80 = 480 km 960 Time take by A to cover 960 km is = 12 hours 80 Distance covered by B in 12 hours = 12 × 60 = 720 km Distance b/w them = (960–720) km = 240 km 37. (d); Let the length of train A be L m and speed is a m/s. & the length of train B will be L + 17 m and speed is b m/s. ATQ, 2𝐿+17 = 2.6 …(i) 𝑎+𝑏 2𝐿+17

= 13 ...(ii) For train A 𝐿 = 𝑎 ⇒ 𝐿 = 2𝑎 2 Put L=2a in (i) and (ii) 4𝑎+17 = 2.6 …(iii) 𝑎–𝑏

𝑎+𝑏 4𝑎+17

= 13 .....(iv) On solving (iii) and (iv), a= 51m/s and, b=34m/s Hence sum of speed =51+34= 85 𝑚𝑠 –1 𝑎–𝑏

38. (d); Let speed of ‘Rajdhani express’ be S m/s ATQ – 240 S= 6 S = 40 m/s Speed of ‘Shatabdi express’ be V m/s 29

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17(240+360)

(40 +V) = 120 4800 + 120V = 10200 120V = 5400 V = 45 m/s Speed of Duranto express’ be u m/s 240+180 (40 − U) = 42 40 − U = 10 U = 30 m/s Let ‘Shatabdi express’ will cross ‘Duranto express’ running in same direction in T sec 360+180 (45 − 30) = 𝑇 15T = 540 T = 36 sec Solutions (39-40): Speed of train A = 120km/h Distance between P and Q = 120 × 4 = 480 480 ×5 Speed of train B = = 100 km/hr 24

39. (b); Relative speed of train A and C, when both running in opposite direction 5 = (120 + 60) × 18 = 50 m/sec Lets length of train A and C is 3L meter and 2L meter respectively. ATQ— 3L+2L 50 = 6 5L = 300 L = 60 meter Length of train C = 60 × 2 = 120 meter Relative speed of train B and D, when both running in opposite 5 175 direction = (100 + 110) × = m/sec 18 3 Lets length of train B and D is 4L meter and 3L meter respectively ATQ— 175 4L+3L = 3

30

9

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7L = 525 L = 75 Length of train D = 75 × 3 = 225 meter Relative speed of train C and D when both train running in same direction 5 = (110 – 60) × =

125 9

18

meter/sec

Required time =

(120+225)×9 125

= 24.84 sec

40. (c); Speed ratio of train A and Train E = 6 : 5 So, ratio of time taken by train A and train E (without stoppage) =5:6 Now → Train E start = 9 : 45 Halt → 3 × (x+ 10) ⇒ 3x + 30 min Train A start →9 : 45 + 30 → 10 : 15 Halt time ⇒ 3 × x = 3x min If halt time add, before starting journey Train E ⇒9 : 45 + 30 + 3x ⇒10 : 15 + 3x Train A ⇒10 : 15 + 3x So, time difference on reaching on Delhi is have same proportion 6 – 5 = 1 hour Train A take → 5 hour Train E take→ 6 hour Train A speed = 120 km/h Train E = 100 km/h For x – (3.30 am – 10 .15 pm) – 3x = 5hours 3x = 5. 15 – 5 X = 5minutes Speed of train B = 100 km/hr Train cover total distance of 720 between Lucknow to Jaipur 720 = + 4(𝑥 + 10) 100 = 7.2 hours + 4(5+10) = 7.2 hour + 60 minutes = 8.2 hours 31

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1

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

Chapter

9

Boat and Stream

BEST APPROACH TO SOLVE THE QUESTIONS The questions based on boats and streams are simply based on the concept of relative speed. Generally, two cases arise when a boat moves. Either the boat moves along the flow of the stream or the boat moves in the direction opposite to that of stream. 1. 2.

If boat moves in the direction of stream. The net speed is the addition of the speed of stream and boat. If boat moves in the direction opposite to that of stream, the net speed is the difference of their speeds (speed of boat being more than that of stream).

In questions related to boat and streams, usually it happens that one may need to solve complex calculations but using a bit of intelligent guess work one may save a good amount of time. A sample problem shows one way of approach: Example: ‘A man can row 24 km in upstream and 36 km in downstream in 9 hours. The same man can row 8 km in upstream and 48 km in downstream in 6 hours. Calculate the speed of the stream.’ Sol. The basic approach to this question involves formation of two equations, which are If we assume speed of stream as r and speed of boat as x. 24 𝑥−𝑟 8 𝑥−𝑟

+ +

2

36 𝑥+𝑟 48 𝑥+𝑟

=9 =6

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It’s cumbersome to solve these set of equations. So we need to look at distance travelled in downstream or in upstream. In majority of cases it’s better to take HCF of distance travelled in downstream in both cases. HCF of 36 and 48 is 12. Now we have to break 12 in two parts such that, adding or subtracting them, completely divides all the four distance. Try (10,2) (9,3) (8,4) and check whether these fulfill the condition. Or check from option. In short we can only check this by hit and trial method. Here, speed of stream=4km/hr and that of boat in still water is 8 km/hr. Practice Exercise Based on New Pattern

1.

A boat goes 28 km downstream and while returning covered only 75% of distance that covered in downstream. If boat takes 3 hr more to cover upstream than downstream then find the speed of boat in 5 still water (km/hr) if speed of stream is m/sec ? 9 (a) 8 km/hr (b) 2 km/hr (c) 5 km/hr (d) 4 km/hr (e) 3 km/hr

2.

The ratio of time taken by boat A and boat B to swim a certain distance downstream in a river is 3 : 4 respectively. The time taken by boat B to cover some other distance in upstream is 50% more than the time taken by it to cover the same distance in downstream. What is the ratio of speed of boat A to that of boat B in still water? (a) 7 : 5 (b) 7 : 4 (c) 7 : 3 (d) 7 : 9 (e) 7 : 2

3.

There are three points P, Q and R in a straight line in a river such that point Q is equidistant from point P and Point R. Boat travelled from point P to R downstream in 8 hours and from point Q to P upstream in 12 hours. Find the ratio between speed of boat in still water to speed of current? (a) 3:1 (b) 2:1 (c) 1:2 (d) 5:2 (e) 2:3 3

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4.

A boat cover 60 km upstream and 60 km downstream in 22.5 hr with its usual speed. If boat double its speed then new upstream speed is 150% more than the usual upstream speed. Find the time taken by boat to cover 80 km in downstream with usual speed. (a) 12 hr (b) 20 hr (c) 5 hr (d) 16 hr (e) 10 hr

5.

A boat covers total distance of 360km downstream in three equal parts with the speed of boat be 4x km/hr, 5𝑥 km/hr and 7𝑥 km/hr respectively. If speed of stream is x km/hr and boat takes total 29.5 hours, then find boat will cover total given distance upstream in three equal parts with three givens speeds? (a) 36 hours (b) 40 hours (c) 45 hours (d) 48 hours (e) 54 hours

6.

The speed of two boats A and B in the still water is in the ratio of 3 : 4 and the speed of current is 4km/hr. A start from point P ,45 minutes earlier than B in downstream direction. If B catch boat A in three hours then, find boat B takes how much time to cover 120 km distance each in downstream and in upstream. (a) 12 hours (b) 16 hours (c) 18 hours (d) 14 hours (e) 20 hours

7.

A boat has to travel from point P to point Q in upstream and reach Q by 5.00 PM. If boat starts form point P at 2.00 PM with the speed of 20 km/hr and after travelling one hour the speed of current is increased by 25% of its initial speed. If distance from point P to Q is 36 km and speed of current is 8 km/hr, then find by what percent boat has to increase its speed to reach point Q in time? (a) 5% (b) 15% (c) 12% (d) 10% (e) 17.5%

Directions (8-9): A boat ‘A’ start from point X, toward point Y in upstream simultaneously second boat ‘B’ start from point Y to X having speed more than first boat ‘A’. Difference between speed of both boats in still water is equal to the speed of stream and they cross each other first time after 2 hours. Given that both boats start their respective return journey by reaching their destination points. 4

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8.

If distance between X and Y is 56 km and speed of stream is 4 km/hr then find the distance from ‘Y’ when boats cross each other 2nd time. (a) 2.8 km (b) 2.4 km (c) 5.6 km (d) 3.2 km (e) 4.8 km

9.

Find the time taken by boats to cross each other on 2 nd time from the first time if speed of second boat ‘B’ is 1.5 times of the speed of first boat ‘A’ in still water? (a) 5 hours (b) 3 hours (c) 2 hours (d) 6 hours (e) 4 hours

10. Point A to Point B is a downstream journey of 300 km on a stream which flows at a speed of 5 km/hr. Two boats P and Q starts from point A and Point B respectively with speed of 25 km/hr and 15 km/hr in still water. After reaching the opposite point they return to their starting points, find after how much time will they meet second time? (a) 7.5hour (b) 15hour (c) 20hour (d) 10hour (e) None of these 11. Speed of current is 10 km/hr and speed of a motor boat is 80% more than speed of current. Motor boat travels 280 km downstream with its usual speed, after that it’s increased speed by ‘s’ kmph and travelled for another 280 km then it returns and covers 560 km in upstream. If boat complete whole journey downstream to upstream in 45 hr, then find the value of ‘s’? (a) 10 km/hr (b) 8 km/hr (c) 6 km/hr (d) 12 km/hr (e) 4 km/hr 12. Speed of current is 5 km/hr and speed of boat in still water is 80% more than speed of current.A boat travels 50% of total distance of 280 km, in downstream. After that the speed increase by x kmph because of a man who added an engine. If boat complete whole journey downstream and upstream in 45 hr, then find the value of x? (a) 6 km/hr (b) 8 km/hr (c) 9 km/hr (d) 4 km/hr (e) 5 km/hr 5

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13. A lake in which speed of water current is zero has a stretch of 144 km. Two boats, whose sum of speed in still water is 28 km/hr, starts from two ends of lake. A fish starts swimming simultaneously with one boat toward another boat and reach the second boat in 4 hours and then returns toward first boat and touch it in 48 minutes. Find the speed of fish. (a) 12 km/hr (b) 16 km/hr (c) 24 km/hr (d) 40 km/hr (e) 8 km/hr 14. A boat goes certain distance downstream and then return distance upstream.It takes

3 2

3th 4

of the

of the time in upstream than in 1

downstream. If boat increases its speed by 33 % and cover a distance 3 of 60 km in downstream and then return upstream in 16 hours, find increased speed of boat? (a) 6 km/hr (b) 14 km/hr (c) 16 km/hr (d) 12 km/hr (e) 8 km/hr 15. Vikas can swim in still water with twice the speed as that of speed of water. Time difference to cover a certain distance in upstream and in downstream is 4 hours at his usual speed. But if he had doubled his usual speed, then time difference would reduce by 80% of its usual time difference to cover same distance. Calculate total time taken by Vikas to cover 6 km upstream and 2 km downstream, at his usual speed. 3 5 (a) 1 hour (b) hour (c) hour 4

4

(d) hour 3

4

(e) Can’t be determined

16. Speeds of three motor boats A, B and C are equal and all cover 8 km of distance upstream in 48 minutes. Ratio between speed of motor boats and speed of stream is 6 : 1. On first day A starts in downstream from point P to Q, which shifts 9 km away from point P each day. On second day B starts from point P in downstream and reach at point Q in 4.5 hours, then find time take by C on third day to reach at point Q, (consider speed of stream same on all three days)? 1 1 1 (a) 4 ℎ𝑜𝑢𝑟𝑠 (b) 5 ℎ𝑜𝑢𝑟𝑠 (c) 3 ℎ𝑜𝑢𝑟𝑠 7 1

(d) 2 ℎ𝑜𝑢𝑟𝑠 7

6

7 1

7

(e) 9 ℎ𝑜𝑢𝑟𝑠 7

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17. Usual speed of a boat is S km/hr. Boat rowing in a lake (neglect the current speed) and covered distance of 360 km in three equal parts with three different speeds in total 7.5 hours. If boat covered first part with usual speed and second part with 20% increased speed and third part with further increase of 25% of latest speed. Find the speed of boat at which it covered third part of journey? (a) 56km/hr (b) 36km/hr (c) 40km/hr (d) 60km/hr (e) 48 km/hr 18.

Ratio between speed of two boats A & B in still water is 4 : 3 and speed of current is 4 km/hr .Boat B starts from point P in downstream, two hours earlier than boat A and it is found that boat A is16 km ahead of boat B after 12 hours of its start. If boat A start from point Q in upstream and boat B start from point P in downstream at same time and distance between P and Q is 108 km, then find in what time boat A & B meet each other? 6 6 6 (a) 4 ℎ𝑜𝑢𝑟𝑠 (b) 5 ℎ𝑜𝑢𝑟𝑠 (c) 9 ℎ𝑜𝑢𝑟𝑠 7 6

(d) 3 ℎ𝑜𝑢𝑟𝑠 7

7

6

7

(e) 11 ℎ𝑜𝑢𝑟𝑠 7

19. When two ships A & B are travelling in opposite direction crosses each other in 2 second. Speed of ship B is 50% more than the speed of ship A in still water. On a particular day, ship A was ahead of ship B by 9 meters and both of them were travelling in downstream. If Ship B overtook Ship A in 11 seconds and length of ship B is 54 meters then find the speed of ship A. (a) 56m/s (b) 36m/s (c) 40m/s (d) 60m/s (e) 18m/s 20. A boat goes a certain distance downstream and then returns and covers 40% of distance covered in downstream. Ratio of time taken in covering downstream and upstream distances is 3 : 2. If speed of boat in still water is reduced by 50% then it covers 60 km downstream in 10 hours. Find the speed of boat in still water. (a) 9 km/hr (b) 8 km/hr (c) 6 km/hr (d) 10 km/hr (e) 12 km/hr 7

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Solutions 5

1.

18

(c); Speed of current = × 9 5 = 2 km⁄hr Let’s still water speed = x km/hr ATQ, 28×

3 4

28

– =3 (x–2) (x+2) 21x + 42 – 28x + 56 = 3x² – 12 –7x + 98 = 3x² – 12 3x² + 7x – 110 = 0 x = 5 km/hr 2.

(a); Let total distance be D km. And, speed of boat A in x km/hr Speed of boat B in still water be y km/hr Speed of stream be r km/hr ATQ 𝐷 𝐷 3 ∶ = 𝑥+𝑟

still

water

be

𝑦+𝑟 4 𝑦+𝑟 3

Or, = ……..(i) 𝑥+𝑟 4 Also, for a distance D1, 𝐷1 𝑦−𝑟

3

= (

𝐷1

2 𝑦+𝑟

)

or, 2𝑦 + 2𝑟 = 3𝑦 − 3𝑟 or, 𝑦 = 5𝑟 Putting this in (i), 6𝑟 𝑥+𝑟

=

3 4

or, 3𝑥 + 3𝑟 = 24𝑟 or, 𝑥 = 7𝑟 Ratio of speed of boat A and B in still water= 7r : 5r =7:5 8

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3.

(b); Let still water speed x km/hr and Speed of current y km/hr — Downstream speed = (x + y) km/hr Upstream speed = (x – y) km/hr Let distance between P to R is 2ℓ and Q to P is ℓ km ATQ— 2ℓ =8 … (i) (x+y)

and ℓ

(x–y)

… (ii)

= 12

From (i) and (ii) x 2 = y

4.

1

(e); Let usual speed of boat in still water = x River speed = y ATQ, (𝑥−𝑦)250 = (2𝑥 − 𝑦) 100 5𝑥 − 5𝑦 = 4𝑥 − 2𝑦 x = 3y Now, 60 60 + = 22.5 x –y x+y 60 60 2y

+

4y

= 22.5

y = 2 km/hr x = 6 km/hr 80 Required time = 6+2 = 10 hr 5.

(c); ATQ— 120 120 120 + + = 29.5 (4𝑥+𝑥) (5𝑥+𝑥) (7𝑥+𝑥) 120

+

120

+

120

5𝑥 6𝑥 8𝑥 5760+4800+3600 240×29.5

= 29.5 =𝑥

𝑥=2 120 120 120 Required time = + + 1

1

3𝑥 1

6

8

12

= 120 ( + + 9

4𝑥

6𝑥 3

) = 120 × = 45 ℎ𝑜𝑢𝑟𝑠 8 Adda247 Publications

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6.

(b); Let speed of boat A and B in still water be 3𝑥 km/hr and 4𝑥 km/hr respectively Downstream speed for boat A = (3𝑥 + 4) km/hr Downstream speed for boat B = (4𝑥 + 4) km/hr Relative speed of boat A and B = (4𝑥 + 4) – (3𝑥 + 4) = 𝑥 km/hr ATQ— 45 60

×

(3𝑥+4) 𝑥

=3

9𝑥 + 12 = 12𝑥 𝑥 = 4 km/hr Speed of boat B = 4 × 4 = 16 km/hr 120

120

Required times = (16+4) + (16–4) = 6 + 10 = 16 hours 7.

(d); Boat travels total distance in the hours Upstream speed of boat = 20 – 8 = 12 km/hr In one hour, boat will cover a distance of = 12 × 1 = 12 𝑘𝑚 Remaining distance = 36 – 12 = 24 km To reach Q at 5.00 PM boat has to cover remaining distance at speed =

24 2

= 12 𝑘𝑚/ℎ𝑟

New speed of stream = 8 ×

125 100

= 10 𝑘𝑚/ℎ𝑟

New upstream speed of boat =20 – 10 = 10 km/hr For 12 km/hr speed boat has to increase its speed by 2 km/hr Required percentage =

10

2 20

× 100 = 10%

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Solutions (8-9): Let speed of first boat ‘A’ = a km/hr Speed of stream = b km/hr So, Speed of 2nd boat ‘B’ = 𝑎 + 𝑏 km/hr Distance XY ⇒ 2 × (a – b + a + b + b) = 4a + 2b 8.

(d); Given XY = 56 = 4𝑎 + 2𝑏 2𝑎 + 𝑏 = 28 Now speed of stream = 𝑏 = 4 km/hr 2𝑎 + 4 = 28 𝑎 = 12 Speed of first boat ‘A’ in still water = 12 Speed of 2nd boat ‘B’ in still water = 16 First time crossing Time taken by 2nd boat ‘B’ to reach at X after first crossing 16 4 = = hour 20

5

4

Distance travelled by first boat ‘A’ in hour 4

5

32

= × 8 = km 5 5 Now, Distance between first boat ‘A’ and 2nd boat ‘B’ 32 16 + ⇒ 22.4 km 5 On return journey of second boat ‘B’, speed of boat = 12 km/hr Time taken by first boat to reach Y 56–22.4 = = 4.2 hours 8 Distance remaining between both boat after reaching of first boat ‘A’ at Y ⇒ 56 – 4.2 × (12 – 8) = 5.6 km Time taken by both boat to cover this distance 5.6 = = 0.2 hours 12+16 Distance from Y = 0.2 × 16 = 3.2 km 11

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9.

(b); If speed of first boat ‘A’ = 𝑎 km/hr 3 Speed of second boat ‘B’ ⇒ 𝑎 km/hr 2

3

𝑎

So speed of stream ⇒ 𝑎– 𝑎 = km/hr 2 2 Distance travelled by first boat till first crossing 𝑎 ⇒ (𝑎– ) × 2 2 ⇒ 𝑎 km 𝑎 Speed of first boat in upstream = km/hr 2 nd Speed of 2 boat in downstream = 2𝑎 km/hr 2nd boat reached at X ⇒ 1

𝑎 2𝑎

1

= ℎ 2 1

𝑎

𝑎

2

2

4 𝑎

In hour first boat cover ⇒ × = km 2

Distance between both boats ⇒ 𝑎 + 5𝑎

4

⇒ km 4 Relative speed of boat ‘A’ and ‘B’ when they cross each other 3𝑎 𝑎 𝑎 𝑎 = ( – – )= 2

2

2

2

Time taken to cross 5𝑎×2

5𝑎 4

⇒ = 2.5 hour 4×𝑎 So total time for 2nd crossing ⇒ 2.5 + 0.5 = 3 hours 10. (c); Time when P and Q meet first time 300 300 = = = 7.5 hr 25+15 40 Distance travelled by P = 7.5 × (25 + 5) = 225 km Distance travelled by Q = 7.5 × (15 – 5) = 75 km 75 P reach at point B in i.e. 2.5 hour 30 In 2.5hr Q travelled 2.5 × 10 = 25 km 12

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Now Boat P returns and relative distance b/w P and Q is 100 km and relative speed is 10 km i.e. difference of speed of P (25 -5) = 20 km/hr (upstream) and speed of Q (15 – 5) = 10 km/hr (upstream) Time taken by P and Q to meet 100 = = 10 hr 10 Total time when P and Q meet second time = 7.5 +2.5 + 10 = 20 hr 11. (d); speed of boat in still water = (10 + 10 × = 18 km⁄hr ATQ— 280

80 100

) km⁄hr

280

560

+ (18+10)+s + (18–10)+s = 45 (18+10) 280 28+s 8 28+s

+ +

560 8+s 16 8+s

= 35 =1

64 + 8s + 448 + 16s = 224 + 28s + 8s + s2 s² + 12x – 288 = 0 s = 12 km/hr 12. (a); Still water speed = (5 + 5 × = 9 km⁄hr ATQ— 140

140

80 100

) km⁄hr

280

+ (9+5)+x + (9–5)+x = 45 (9+5) 140 14+x 4 14+x

+ +

280 4+x 8 4+x

= 35 =1

12x + 128 = 56 + 14x + 4x + x² x² + 6x – 72 = 0 x = 6 km/hr 13

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12. (a); Still water speed = (5 + 5 × = 9 km⁄hr ATQ— 140

140

80 100

) km⁄hr

280

+ (9+5)+x + (9–5)+x = 45 (9+5) 140 14+x 4 14+x

+ +

280 4+x 8 4+x

= 35 =1

12x + 128 = 56 + 14x + 4x + x² x² + 6x – 72 = 0 x = 6 km/hr 13. (c);

Let speed of boat 2 and speed of fish is x and y respectively. Boat 2 and fish meet in 4 hour So, (x + y) × 4 = 144 …(i) Now ATQ, Speed of boat 1 = 28 – x In 4 hour boat 1 cover = 4 × (28 – x) 48 × (𝑦 + (28 – 𝑥 )) = 144 – [4𝑥 + 4(28 − 𝑥 )] 60 4 5

(𝑦 − 𝑥 ) + 22.4 = 144 − 112

𝑦 − 𝑥 = 12 … … . (𝑖𝑖) Sovling (i) & (ii) x = 12 km/hr y = 24 km/hr

14

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14. (e); Let distance travelled by boat in downstream be D and speed of boat in still water be x km/hr and speed of current be y km/hr ATQ, 3 ×𝐷 4

(𝑥−𝑦) 1

3

= × 2

=

2(𝑥−𝑦)

1

𝐷 (𝑥+𝑦)

𝑥+𝑦

x = 3y 1 New sped of boat = 3y + 3y × 3 = 4 y km/hr 60 60 + = 16 (4𝑦+𝑦) 12 20 𝑦

+

𝑦 32

(4𝑦−𝑦)

= 16

y = = 2 𝑘𝑚/ℎ𝑟 16 Increased speed of boat = 4×2 = 8 km/hr 15. (e); Let the speed of water is x km/hr, then speed of vikas in still water is 2x km/hr From 1st condition, 𝐷 𝐷 – =4 2𝑥–𝑥 𝐷

2𝑥+𝑥 𝐷

⇒ – =4 𝑥 3𝑥 ⇒ D = 6𝑥 …(i) From 2nd condition, when speed of Vikas is 4𝑥 km/hr 𝐷 𝐷 20 – = ×4 4𝑥–𝑥 4𝑥+𝑥 100 𝐷 𝐷 4

⇒ – = 3𝑥 5𝑥 5 ⇒ D = 6𝑥 …(ii) Hence both equations are same so answer can’t be determined. 16. (b); Speed of motor boats in upstream 60 = 8 × = 10 km/hr 48 ATQ, Let speed of motor boats be 6x km/hr and speed of stream be x km/hr 6x – x = 10 15

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x = 2 km/hr Downstream speed of all boats = (6 × 2 + 2) = 14 km/hr Let distance between point P to Q on first day = y km Second day distance = (y + 9) 𝑦+9 14 = 4.5 y= 63 – 9 y = 63 – 9 y= 54 km Distance travelled on third day = 54 + 9 × 2 = 72 km Total time taken by boat C on third day to reach point Q 72 1 = = 5 hours 14

7

17. (d); Three equal parts of journey 360 = = 120 km 3 ATQ— 120 120 120 + S + 6S 6S 1 = 7.5 S

120

+

S+5 600

+

+ × 5 5 4 120×10

= 7.5

S 6S 12S+3S 3600+3000+2400 15 9000 30S

30S 15

=

2

=

2 9000×2

S= 30×15 S = 40 km/hr Speed of which boat which it covered third part of Journey 20 20 1 = (40 + 40 × ) + (40 + 40 × ) × = 60 km/hr

100

100

4

18. (d); Let speed of boat A & B in still water be 4x km/hr and 3x km/hr respectively Downstream speed of boat A = (4x + 4) km/hr 16

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Downstream speed of boat B = (3x + 4) km/hr AT Q – 12 (4x + 4) – 16 = 2 (3x + 4) + 12 (3x + 4) 48x + 48 − 16 = 6x + 8 + 36x + 48 6x = 24 x = 4 km/hr Speed of boat A in still water = 16 km/hr Speed of boat B in still water= 12 km/hr Relative speed = (16 – 4) + (12 + 4) = 28 km/hr 108 Required time = 6

28

= 3 ℎ𝑜𝑢𝑟𝑠 7

19. (e); Let speed of ship A in still water is 2x m/s , and its length is ℓ m And speed of ship B is 3x m/s and its length is 54 and speed of water is Y m/s. ATQ, when both of them are travelling in opposite direction. Downstream speed of ship A = (2x + Y) m/s [assume ship A is travelling in downstream and ship B in upstream] Upstream speed of ship B = (3x – Y) m/s Their relative speed = 2x + Y + 3x – Y = 5x m/s ATQ, 54+ℓ = 2 …(i) 5x Similarly when both are travelling in downstream their relative speed is = 3x + Y – 2x –Y= x m/s ATQ, 54+9+ℓ = 11 𝑥 63 + ℓ = 11x …(ii) From (i) ℓ = 10x – 54 put this value in eqn. (ii) = 63 + 10x – 54 = 11x x = 9 m/s Hence speed of ship A in still water = 2 × 9 = 18 m/s 17

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20. (b); Let speed of boat in still water and speed of current be respectively x km/hr and y km/hr And Let time taken to cover given distance in downstream and upstream be = 3m hours and 2m hours 40% of distance in upstream is covered in 2 m hour 100% of distance upstream is covered in 5m hours According to question x−y 3 = x+y

5

5x - 5y = 3x + 3y x 4 ⇒ = y

1

And 60

x (2+y)

x

= 10

+y=6

2 4𝑦

+𝑦=6 6y = 12 y = 2 km/hr so, x = 8 km/hr 2

18

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1

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Chapter

10

Mensuration

BEST APPROACH TO SOLVE THE QUESTIONS Questions related to mensuration usually don’t involve any short-cut methods for solving a problem. The reason being how can one use short tricks when all one has to do is to deal with formulae. To solve questions faster and in a lucid manner, one must avoid doing unnecessary and multiple calculations. Rather one must calculate only where it is needed. In case, ratio or percentage is asked in the question, then it becomes more easy to do away with hefty calculations and arrive at the answer by using basic elimination methods. ‘The length, breadth and height of a room are in the ratio 3:2:1. If the breadth and height are halved while the length is doubled, then the percentage change in total area of the four walls of the room.’ Area of four walls of the room= 2(lh+bh), where l,b and h are length, breadth and height of the room respectively. As we have been asked the percentage change, let’s take length, breadth and height of the room to be 6,4 and 2 respectively. Then changed values of length, breadth and height would be 12, 2 and 1. Initial area of four walls= 2 (6 × 2 + 4 × 2) = 40 Final area of four walls= 2 (12× 1 + 2 × 1)= 28 28 is 70% of 40, which means there has been a reduction of 30%. ‘A solid cylinder has height 10 meters and radius R meters. Some part of it is melted and casted into two cones. One of the cones has radius 3cm and height 4cm. The other cone has height 8 cm and radius 6cm. If the ratio of initial curved surface area of cylinder to the sum of curved surface areas of two cones is 4:1, then find the radius of the cylinder.’ 2

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Slant height of cone having radius 3cm is 5cm (PLEASE SOLVE IT PROPERLY) Slant height of cone having radius 6 cm is 10 cm Curved surface area of a cone is πrl and that of cylinder is 2πrh 2𝜋×𝑅×10 4 ATQ, = 𝜋(3×5+6×10)

➢ R=15cm

1

Practice Exercise Based on new Pattern

1.

The ratio of the radius and height of a cone 2 5 : 12. Its volume is 314 cm. Then its slant height is ? 7 (a) 9 cm (b) 10 cm (c) 11 cm (d) 12 cm (e) 13 cm

2.

If the perimeter of a rectangle and a square are equal and the ratio of 2 adjacent sides of a rectangle is 2:1. Then find the ratio of area of rectangle and area of square. (a) 1:1 (b) 1:2 (c) 2:3 (d) 8:9 (e) 9:4

3.

The parameter of a square is equal to the parameter of a rectangle having dimensions 24 cm × 16 cm. Find the circumference of a circle whose diameter is equal to the side of square ? 360 410 470 (a) cm (b) cm (c) cm (d)

4.

7 440 7

cm

(e)

7 460 7

is

7

cm

Total surface area of a cylinder mounted with a hemispherical bowl on one end is 2552 cm². If height of cylinder is 8 cm then find the volume of the solid body? 1 1 1 (a) 10443 (b) 10677 (c) 10547 (d) 10977 3

3 1 3

(e) 10787

3 1

3

3

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5.

Sum of Area of circle and a rectangle is equal to 622 sq cm. The diameter of the circle is 14 cm, then what is the sum of circumference of the circle and the perimeter of the rectangle if the length of rectangle is 26 cm? (a) 142cm (b)132cm (c)152cm (d) 140cm (e)134cm

6.

A copper sphere of diameter 42 cm is drawn into a wire of diameter 70 mm. Then, the length of the wire is? (a) 243 cm (b) 343 cm (c) 443 cm (d) 972 cm (e) None of these

7.

A cylindrical roller rolls over a square field. It takes 500 rounds to cover the complete field. If the radius of the cylindrical roller is 0.63 meter and length is 7.7 m then find out the side of square (approximately). (a) 135 m (b) 125 m (c) 100 m (d) 150 m (e) 110 m

8.

A cuboid is cut along its base such that, now it become a cube. Cube again cut along its base, now ratio of height to length becomes 1 : 𝑎. Find the value of ‘𝑎’. (a) 𝑎 > 1 (b) 𝑎 = 1 (c) 𝑎 ≥ 1 (d) 𝑎 < 1 (e) 𝑎 ≤ 1

9.

Volume of a cone and cylinder, having same base, are in ratio of 4 : 3. If height and radius of cone is 4 : 3 then find the ratio of total surface area of cone and that of cylinder. (a) 1 : 1 (b) 3 : 1 (c) 5 : 3 (d) 2 : 1 (e) 4 : 1

10. If the volume and curved surface area of a cylinder 616 m3 and 352m2 respectively, what is the total surface area of the cylinder (in m2 )? (a) 429 (b) 419 (c) 435 (d) 421 (e) 417 4

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11. Area of a rectangle is 252 cm². If length of the rectangle is equal to side of a square, whose perimeter 72 cm then find the volume of cone, whose radius is equals to breadth of rectangle and height equals to length of rectangle? (a) 4752 cm2 (b) 3586 cm2 (c) 4026 cm2 (d) 4246 cm2 (e) 3696 cm2 12. Sum of length, breadth and height of cuboid is 12 cm and length of its diagonal is 5√2. Then find the total surface area of cuboid. (a) 94 cm² (b) 84 cm² (c) 72 cm² (d) 64 cm² (e) 90 cm² 13. Circumference of a circular garden is 66 cm and area of circular garden is 25% more than perimeter of a square hall. Find the area of square hall ? (a) 4802.49 sq cm (b) 4704.48 sq. cm (c) 4820.49 sq.cm (d) 4822.49 sq.cm (e) none of these 14. Internal radius of pipe is 3 cm and the external radius of pipe is 5 cm and the length of pipe is 15 cm then find total surface area of pipe ? (a) 278 π cm² (b) 275 π cm² (c) 268π cm² (d) 272 π cm² (e) none of these 15. A solid cube is melted to form 3 spheres, with radius, 1 cm, 2 cm and 3 cm and additional irregular shape was formed. Calculate the volume of irregular shape (upto two decimal place) if diagonal of cube is 6√3 cm. (a) 65.14 𝑐𝑚3 (b) 70.35 𝑐𝑚3 (c) 54.75 𝑐𝑚3 (d) 50.00 𝑐𝑚3 (e) 78.90 𝑐𝑚3 16. 20% increase in both radius and height of the cylinder increases the total surface area of cylinder by 677.6 cm², If the ratio of radius to height is 1 : 4, then find the radius of cylinder (a) 21 cm (b) 10.5 cm (c) 3.5 cm (d) 14 cm (e) 7 cm 5

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17. A Cone cylinder and hemisphere have equal radius and height. Find the ratio of total surface area of cylinder, cone and hemisphere. (a) 4 ∶ √2: √6 (b) 4 ∶ √2 ∶ 3 (c) √2 + 1 ∶ 4 ∶ 3 (d) 3 ∶ 4 ∶ √2 + 1 (e) 4 ∶ √2 + 1 ∶ 3 18. Sum of circumference of a circle and perimeter of a rectangle is 220 1 cm while area of circle is 1386 sq. cm. If length of rectangle is 33 % 3 more than radius of the given circle then find the area of rectangle? (a) 408 𝑐𝑚2 (b) 418 𝑐𝑚2 (c) 428 𝑐𝑚2 (d) 448 𝑐𝑚2 (e) 438 𝑐𝑚2 19. Four identical coins are placed in a square. For each coin, area of each coin is equal to circumference of each coin. Then, find the area of the square that is not covered by the coins

(a) 16(π − 1) π (d) 16 (4 − ) 2

(b) 16(8 − π) (e) None of these

(c) 16(4 − π)

20. The surface area of a sphere is 423.5 cm2 less than total surface area of a hemisphere. If ratio between radius of hemisphere and sphere is 3 : 2, then find the radius of hemisphere? (a) 5.5 cm (b) 5 cm (c) 4 cm (d) 7 cm (e) 10.5 cm 21. A cylindrical jar, whose base has a radius of 15 cm is filled with water up to a height of 20 cm. A solid iron spherical ball of radius 10 cm is dropped in the jar to submerge completely in water. Find the increase in the level of water (in cm) is 17 5 8 (a) 5 (b) 5 (c) 5 (d) 5 6

27 25 27

7

(e)

9

7 5 27

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22. The dimensions of a field are 20 m and 9 m. A pit 10 m long, 4.5 m wide and 3 m deep is dug in one corner of the field and the earth (mud) removed has been evenly spread over the remaining area of the field. What will be the rise the height of field as a result of this operation ? (a) 1.5 m (b) 2 m (c) 3 m (d) 4m (e) 1 m 23. Difference between perimeter of two rectangles is 8 cm. If breadth of both rectangle is 16 cm and length are in the ratio of 7 : 6 then find the total surface area of cylinder whose radius is half of the length of larger rectangle and height equal to length of smaller rectangle? (a) 3344 cm2 (b) 3668 cm2 (c) 3444 cm2 (d) 3468 cm2 (e) 3462 cm2 Directions (24-25); There are three toys which are in cylindrical, conical and spherical shape. Height of all the toys is same. Ratio between the height of cylinder to radius of cylinder is 4 : 3. 24. Find the ratio between total surface area of cylinder to that of sphere. (a) 17 : 15 (b) 21 : 16 (c) 28 : 23 (d) 21 : 8 (e) 15 : 4 25.

Volume of cone is what percent of the volume of sphere if radius of cone is equal to that of cylinder (a) 112.5% (b) 100% (c) 125% (d) 75%

8

(e) 88 % 9

26. A well whose shape is of equilateral triangle having side 8 cm has to be dug 44√3 cm deep. The earth taken out of the well is used to fill a cylindrical tank having diameter 8 cm. Find the height up to which earth is filled in the tank. (a) 10.5 cm (b) 21 cm (c) 42 cm (d) 84 cm (e) 32 cm 7

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27. A circle is inscribed in a square. If the difference between area of the square and circle is 262.5 cm², then find the area of the rectangle whose perimeter is same as that of circle while length of rectangle is 20% more than the breadth of rectangle (in cm²) (a) 1500 (b) 1400 (c) 700 (d) 750 (e) 3000 28. Inside a square plot a circular garden is developed which exactly fits in the square plot and the diameter of the garden is equal to the side of the square plot which is 28 metre. What is the area of the space left out in the square plot after developing the garden? (a) 98 m2 (b) 146 m2 (c) 84 m2 (d) 168 m2 (e) 68 m2 29. The area of a rectangle gets reduced by 9m2 if its length is reduced by 5 m and breadth is increased by 3m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 67 m2. The length of the rectangle is : (a) 9 m (b) 15.6 m (c) 17 m (d) 18.5 m (e) 19 m 30.

A toymaker makes a toy in which a cone is mounted on the base of a hemisphere. If the total surface area of the toy is 858 cm² then find the volume of the toy, given that the diameter of the toy is 14 cm? 2

(a) 1950 cm3 3

3

(d) 1500 cm

2

(b) 1250 cm3 3

(c) 1400 cm3

(e) 1200 cm3

31. If sum of circumference of a circle and perimeter of rectangle is 204 cm and perimeter of rectangle is 50% of magnitude of area of square and side of square is 12 cm .If length of rectangle is 8 cm more than that of breadth, then find sum of area of circle and rectangle ? (a) 1694sq cm (b) 1450sq cm (c) 1550sq cm (d) 1750sq cm (e) 1460sq cm 8

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32. A sphere is melted into ‘n’ number of small spheres. Total surface area of these small spheres is what percent more or less than total surface area of large sphere, if the ratio of radius of small sphere to radius of large sphere is 1 : 3. (a) 50% (b) 100% (c) 150% (d) 200% (e) 75% 33. Ratio of height of cylinder to that of diameter is 2: 3. This cylinder is formed by melting a sphere having same radius as the radius of circle. Ratio of magnitude of area of circle to magnitude of its circumference is 21: 2. Find height of cylinder. (a) 14 units (b) 21 units (c) 42 units (d) 28 units (e) 35 units 34. If the curved surface area of first cone is thrice that of second cone and slant height of the second cone is thrice that of the first, find the ratio of the area of their base. (a) 81 : 1 (b)9 : 1 (c)3 : 1 (d)27 : 1 (e) None of these 35. There are two cylindrical vessels with same capacity and same dimensions. If radius of one vessel increased by 7 cm and height of second vessel increases by 15 cm, then capacity of both vessel increased by equality ‘Q’ cm3. If actual height of both vessels is 12 cm, then find value of Q? (a) 9840 cm3 (b) 9240 cm3 (c) 9460 cm3 (d) 9160 cm3 (e) 9780 cm3 36. There are four light poles in four corners of a park, four at midpoint of each side and one pole in the middle of that square park. Area of park is 19600 cm2. If side of park is 50 times of radius of each light pole, then find the total base area covered by these poles if pole’s base is circular in shape? (a) 197.12 cm2 (b) 172.48 cm2 (c) 246.4 cm2 (d) 220.76 cm2 (e) 221.76 cm2 9

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37. Area of a square is 306.25 cm2 and breadth of rectangle is 3.5 cm less than side of square. Area of rectangle is 252 cm2 and height and radius of a cylindrical vessel equal to length & breadth of rectangle respectively. If vessel filled, contains mixture of milk and water in which milk is 75%, then find the volume of water contained by cylindrical vessel? (a) 2572 cm3 (b) 2472 cm3 (c) 2772 cm3 (d) 2872 cm2 (e) 2878 cm2 Directions (38-39); A hemispherical bowl is filled with hot water to the brim. The contents of the bowl are transferred into a cylindrical vessel whose radius is 50% more than its height. 38.

If diameter of the bowl is the same as that of the cylindrical vessel, then the volume of the hot water in the cylindrical vessel is (a) 60% of the cylindrical vessel (b) 80% of the cylindrical vessel (c) 100% of the cylindrical vessel (d) 50% of the cylindrical vessel (e) none of these

39. This hemispherical bowl is joined at one end of the cylindrical vessel and the solid obtained by combining these figure is filled completely with water. If a drain pipe is connected to it then pipe will empty 539 cm³ of water is one minute. If radius of the solid is 21 cm, then in what time whole of the water will be emptied from the solid. Given that diameter of cylinder and hemisphere are same. (a) 72 min (b) 60 min (c) 90 min (d) 75 min (e) None of these 40. A cylinder whose height is equal to its radius is full of milk. Its milk is poured into a hemi spherical bowl of same radius as of cylinder and remaining milk is poured into another hemi-spherical bowl of same volume as of previous one. What percentage of the volume of second hemispherical bowl remained empty? (a) 40% (b) 66⅔% (c) 33⅓% (d) 50% (e) 100% 10

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41.

In a house, there is a semicircular corridor, a circular hall and a circular garden. Sum of perimeter of corridor and hall is 102 m.If radius of hall is 50% more than radius of corridor then find area of circular garden, given radius of garden is 300% more than sum of radius of corridor and radius of hall together. (a) 6300𝜋 sq.m (b) 5400𝜋 sq.m (c) 5700𝜋 sq.m (d) 4200𝜋 sq.m (e) 4900𝜋 sq.m

42. Height of a cylindrical vessel is equal to side of a square, whose area is 256 cm². Radius of cylindrical vessel is equal to average of radius of a sphere and hemisphere. If ratio between the radius of hemisphere and that of the sphere is 2 : 1 and difference between their total surface area is 4928 cm², then find volume of cylindrical vessel ? (a) 21,288 cm³ (b) 20,176 cm³ (c) 22,176 cm³ (d) 26,176 cm³ (e) 24,288 cm³ 43. Radius of a cylinder is half of the side of square, whose perimeter is 6 cm more than perimeter of rectangle. If ratio between length to breadth of rectangle is 7 : 2 and between length of rectangle to side of square is 3 : 2. Find volume of cylinder, given height of cylinder is equal to breadth of rectangle? (a) 24,938 cm3 (b) 24,948 cm3 (c) 24,848 cm3 (d) 24,996 cm3 (e) 24,918 cm3 44. A solid sphere of some radius is melted and re-casted into a number of small cylinders and cones. The height of the cone is 33⅓% less than the height of the cylinder which is 6 cm and the ratio of radius of cone to the radius of cylinder is √3 : 1 and the ratio of number of cylinder to the number of cone is 2 : 3. Find the total surface area of the sphere if the number of cylinders are 12 and the radius of cylinder is 4 cm. 1

(a) 1810 cm² (d) 905 11

7 11 14

cm²

2

(b) 1810 cm² (e) 905

9

7

14

4

(c) 1810 cm² 7

cm²

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45. Radius of a cylindrical vessel is 1.5 cm less than breadth of rectangle, while height of cylindrical vessel is equal to length of rectangle, whose area is 168 cm2. If ratio between length & breadth of rectangle is 7 : 6, then find the think ness of 1 disc, if these discs are kept one above another in cylindrical vessel and 10 discs get completely fit in cylindrical vessel.(Let radius of discs equal to radius of cylindrical vessel)? (a) 1.2 cm (b) 1.6 cm (c) 1.8 cm (d) 1.4 cm (e) 1.0 cm 46. The ratio between radius of two hemispheres solid tin pieces is 2 : 3 and difference between volume of both is 836/21 cm3. These two 𝜋 hemispheres are melted into a cylindrical vessel and used 74 cm3 3 extra tin material for polishing the vessel. If ratio between height & radius of cylindrical vessel is 3 : 4, then find the total surface area of cylindrical vessel? (a) 154 cm2 (b) 132 cm2 (c) 176 cm2 (d) 208 cm2 (e) 198 cm2 Solutions

1. (e);

Let, radius = 5X and height = 12X cm. 2200 Volume = 1

22

7

= × × 12X × 5X × 5X = 3 7 X=1 r = 5 and h = 12 ℓ = √52 + 122 = √25 + 144 = √169 = 13 cm 2. (d);

2200 7

let side of a rectangle=a, 2a, side of a square=b 2 2(a+2a)=4b , a= b 3

8

Area of rectangle=a×2a=2a2 or ⌈ ⌉b2 9 2 Area of square=b Ratio = 8:9 12

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3.

(d); Parameter of square = parameter of rectangle = 2 (24 + 16) 4a = 80 cm Side of square = 20 cm 20 Radius of circle = = 10 cm 2 Circumference of circle 22 440 2πr = × 20 = cm 7

4.

7

(b); T.S.A = 2πr² + πr² + 2πrh = 2552 3πr² + 2πr × 8 = 2552 2552 3r² + 16r = ×7 22 3r² + 16r = 812 3r² + 16r – 812 = 0 3r² + 58r – 42r – 812 =0 r(3r² + 58) – 14(3r + 58) = 0 58 r = 14, − 3

2

Required volume = πr 3 + πr 2 h 2

22

3

7

= ×

× (14 1

)3

+

3 22 7

× (14)2 × 8

= 5749 + 4928 3

= 10677

1 3 14

5.

(b); Radius of circle = = 7 2 Area of circle 22 = × 7 × 7 = 154 sq. cm 7 Rectangle area = 622 – 154 = 468 sq. cm Breadth of rectangle 468 = = 18 cm 26 Required sum = 2πr + 2(ℓ + b) 22 = 2 × × 7 + 2(26 + 18) 7

= 44 + (44) × 2 = 132 cm. 13

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6.

(e); Let length of the wire = h 70 Radius = = 35 mm = 3.5 cm 2

4

Volume of the wire = πr 2 h = π𝑟 3 4

3

and, volume of sphere = π × (21)3 2

4

3

π𝑟 h = π × 21 × 21 × 21 3 h = 1008 cm 7.

(b); Radius of cylinder = 0.63 m Length = 7.7 m Let side of square is ‘a’ meter ATQ, Curved surface area of cylinder × 500 = Area of field 22 2 × × 0.63 × 7.7 × 500 = a2 7 a = 123.47 m approximately (a) ≈ 125 𝑚

8.

(a); Ratio of length, breadth and height of cuboid → 𝑎∶𝑎∶1 As, a cuboid is cut into a cube this means after cutting ratio become 1 : 1 : 1 or a : a : a In starting height is more than length and breadth. We conclude this because of the condition that base is same, and cuboid is cut along its base, so height is reduced. After cutting this cube again along its base a : a : a becomes a : a : 1⇒𝑎 >1

9.

(a); Same base → Same radius Let radius = r ATQ— 1 2 πr H 3 πr2 h

4

= 3 H and h are the height of cone and cylinder respectively. H 4 = h 1 Ratio of height and radius of cone H:r=4:3 14

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h=x h = 4x r = 3x slant height of cone = 5x Ratio of T. S. A = 𝜋𝑟 2 ℎ

πrℓ+πr2

2πrh+2πr2

1

= ⇒1∶1 1

616

10.

(a); = 2𝜋𝑟ℎ 352 𝑟 = 3.5 m 𝜋𝑟 2 ℎ = 616 616 ℎ= = 16 m 11×3.5 Total S.A. = 2𝜋𝑟ℎ + 2𝜋𝑟 2 = 2𝜋𝑟(ℎ + 𝑟) 22 = 2 × × 3.5(3.5 + 16) 7 = 429 m2

11.

(e); Let, length and breadth of rectangle be ‘l’ and ‘b’ respectively Given (ℓ × b) = 252 cm² and, 4𝑙 = 72 cm 𝑙 = 18 cm length of rectangle = 18 cm 252 breadth of rectangle, 𝑏 = = 14 𝑐𝑚 18 Radius of cone = 14 cm Height of cone = 18 cm 1 1 22 Volume of cone = 𝜋𝑟 2 ℎ = × × 14 × 14 × 18 = 3696 cm² 3

12.

15

3

7

(a); Length + breadth + height = 12 cm and √ℓ2 + 𝑏 2 + ℎ2 = 5√2 ℓ² + b² + h² = 50 cm 𝑙 + 𝑏 + ℎ = 12 Square both sides (𝑙 + 𝑏 + ℎ)2 = 122 𝑙 2 + 𝑏 2 + ℎ2 + 2 (ℓb + bh + hℓ) = 122 12² = 50 + 2 (ℓb + bh + hℓ) 2 (ℓb + bh + hℓ) = 94 cm² Adda247 Publications

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13. (a); Radius of circular garden 66×7 21 = = cm 2×22 2 Area of circular garden (πr²) 22 21 21 = × × = 346.5 sq cm 7

2

2

4

Perimeter of square = 346.5 × 5 =277.2 cm 4a = 277.2 cm a = 69.3 cm Area of square hall = 69.3 × 69.3 = 4802.49 sq cm 14. (d); r = 3 cm, R = 5 cm, h = 15 cm Total surface area of pipe = 2πrh + 2πRh + 2(πR² – πr²) = (2π × 3 × 15) + (2π × 5 × 15) + 2(5²π – 3²π) = 90π + 150π + 32π = 272π cm² 15. (a); If the diagonal of cube is 6√3 cm Then side of cube is 6cm. Now let volume of irregular shape is x cubic centimeter. 4 216 = π((1)3 + (2)3 + (3)3 ) + x 4

3

216 = × 3

22 7

𝑥 = 216 − 𝑥=

456 7

× [36] + x 1056 7

= 65.14 cubic centimeter

16. (e); Let the radius and height of cylinder be are r & h respectively Now 20% income in both mean 20 𝑛𝑒𝑤 𝑟𝑎𝑑𝑖𝑢𝑠 = (1 + ) 𝑟 = 1.2 𝑟 100 Also 1.2h. 2π × 1.2 r(1.2r + 1.2h) 44 × 2πr (h + r) = 67760 16

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r(h+r)= 245 Let radius and height be x and 4x 𝑥 × 5𝑥 = 245 x = 7 cm 17. (e); Let ratio of cone, cylinder and hemisphere = r Height = r (Because height of hemisphere is equal to its radius) Required ratio → 2πr(r + h) : πr (ℓ +r) : 3πr² ⇒ 4r : √2r+r : 3r 4 : √2 +1 : 3 18. (d); 2πr + 2 (ℓ + b) = 220 cm πr² = 1386 sq. cm r2 =

1386×7 22

r = 21 cm Length of rectangle 4

= 21 × = 28 cm 2×

22 7

3

× 21 + 2(28 + b) = 220

132 + 56 + 2b = 220 b=

32 2

= 16 cm

Area of rectangle = (28 × 16) = 448 𝑐𝑚2 19. (c);

Let r be the radius of each circle. Then by given condition, πR2 = 2πR ⇒ R = 2 ∴ The length of the side of the square = 8 Now the area covered by 4 coins = 4 × π (2)2 = 16 π And area of the square = 64 ∴ The area which is not covered by the coins = 64 – 16π = 16 (4–π)

17

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20. (e); Total surface area of sphere = 4πr² Total surface area of hemisphere = 3πr² Let radius of hemisphere and sphere be 3𝑥 cm And 2𝑥 cm respectively. ATQ— 3πr³ – 4πr² = 423.5 cm² 22 22 3 × × (3𝑥 )2 – 4 × × (2𝑥 )2 = 423.5 7 7 𝑥 = 3.5 cm 21 Radius of hemisphere = cm =10.5 2

21. (d); Let level of water will be increased by h. 4 π × (15)2 × h = π(10)3 4

3 10×10×10

3

15×15

⇒h= ×

=5

25 27

cm

22. (e); Volume of earth removed = 10 × 4.5 × 3 = 135m3 Remaining area = [20 × 9 − 10 × 4.5] = 180 − 45 = 135 Let, rise in height be h m Therefore, ℎ × 135 = 135 ℎ = 1m 23. (a); Lets length of two rectangles be 7x cm 6x cm respectively ATQ 2 (7x + 16) – 2 (6x + 16) = 8 cm 2x = 8 cm x = 4 cm Radius of cylinder =

4×7 2

= 14 cm

Height of cylinder = 4 × 6 = 24 cm Total surface area of cylinder = 2πr (r + h) 22 = 2 × × 14(14 + 24) 7

= 88 × 38 = 3344 cm² 18

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Solutions : (24-25); Let height of cylinder = 4𝑥 So radius of cylinder = 3𝑥 Height of cone = 4𝑥 4𝑥 Radius of sphere = = 2𝑥 2

24. (d); T.S.A of cylinder = 2𝜋(𝑟𝑎𝑑𝑖𝑢𝑠)(𝑟𝑎𝑑𝑖𝑢𝑠 + ℎ𝑒𝑖𝑔ℎ𝑡) T.S.A of sphere = 4𝜋(𝑟𝑎𝑑𝑖𝑢𝑠)2 2𝜋×3𝑥×(4𝑥+3𝑥) 21 Required ratio = = = 21 : 8 2 4𝜋(2𝑥)

8

1

25. (a); Volume of cone = × 𝜋(3𝑥 )2 × 4𝑥 3

4

Volume of sphere = 𝜋(2𝑥)3 3

Required % =

1 ×𝜋×36×𝑥 3 3 4 ×𝜋×8×𝑥 3 3

× 100 = 112.5%

26. (c); Volume of Equilateral triangle well = volume of cylindrical tank filled. Let h = height of cylinder ATQ, √3 4

× 8 × 8 × 44√3 =

⇒ h = 42 cm

22 7

×4×4×h

27. (d); Let, the side of square = 2a ⇒ Diameter of circle = 2a ATQ, 4a2 – ⇒

22

× a × a = 262.5

7 28a2 –22a2 7

= 262.5

⇒ 6a² = 262.5 × 7 ⇒ a² = 306.25 ⇒ a = 17.5 cm Perimeter of circle = 2 × =2× 19

22 7

22 7

×a

× 17.5 = 110 Adda247 Publications

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Let, Breadth of rectangle = x ⇒ length of rectangle = 1.2x ATQ, 2(1.2x + x) = 110 ⇒ 𝑥=

55 2.2

= 25

Required area = 1.2 × 25 × 25 = 750 cm² 28. (d); Area of the space left out = Area of square – Area of circle

= (28)2 −

22

× 14 × 14

7 11×14

= 28 (28 −

7

)

= 28(28 − 22) = 28 × 6 = 168 m2 29. (c); (ℓ − 5)(𝑏 + 3) = ℓ𝑏 − 9 ……..(i) (ℓ + 3)(𝑏 + 2) = ℓ𝑏 + 67 ……..(ii) Solving (i) and (ii), ℓ𝑏 + 3ℓ − 5𝑏 − 15 = ℓ𝑏 − 9 or, 3ℓ − 5𝑏 = 6 ………….(iii) ℓ𝑏 + 2ℓ + 3𝑏 + 6 = ℓ𝑏 + 67 or, 2ℓ + 3𝑏 = 61 ………(iv) Solving (iii) and (iv), ℓ = 17m 20

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30. (a); Total surface area of the toy = C.S.A of cone + C.S.A of Hemisphere Let, slant height of cone be l cm. πrℓ + 2πr² = 858 cm2 πr(ℓ + 2r) = 858 cm2 ℓ =25 cm height of cone = 24cm volume of the toy 1 2 = πr 2 h + πr 3 3 1

3

2(

= πr h + 2r) 3

2

= 1950 cm3 3

31. (a); 2πr + 2(ℓ + b) = 204 cm 2 (ℓ + b) = 144 × ½ 2 (ℓ + b) = 72 ℓ + b = 36 (b + 8) + b = 36 b = 14 ℓ = 22 Area of rectangle = 14 × 22 = 308 sq. cm Circle radius 2πr = 204 – 72 πr = 66 r = 21 cm 22 Area of circle = × 21 × 21 7 = 1386 sq. cm Required sum = (308 + 1386) sq. cm = 1694 sq. cm 32. (d); Let the radius of large sphere is 3𝑥 and small sphere is 𝑥. Then if large sphere is melted into n small spheres. 4 4 𝜋(3𝑥 )3 = 𝑛 𝜋(𝑥 )3 3

27𝑥 3

3

𝑛 = 3 = 27 𝑥 Surface area of large sphere = 4𝜋(3𝑥 )2 = 36𝜋𝑥 2 21

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Surface area of 27 small spheres = 27 × 4 𝜋 (𝑥 )2 = 108𝜋𝑥 2 Required %=

108𝜋𝑥 2 –36𝜋𝑥 2 36𝜋𝑥 2

× 100 = 200%

33. (d); Let the radius of circle is R Atq, πR2 2πR

=

21 2

⇒ R = 21

Volume of sphere of same radius 4

= πR3 3

Let, radius of cylinder is r and height is h 4 3

πR3 = πr 2 h 4

r 2 h = × 21 × 21 × 21 3

= 4 × 7 × 21 × 21 r 2 h = 28 × 21 × 21 …(i) In question h : D = 2 : 3 [D is diameter of cylinder] h : r = 2 : 3/2 = 4: 3 4x: 3x Put this in equation (i) 36x3 = 28 × 21 × 21 x³ = 7 × 7 × 7 x=7 height = 28 units 34. (a); Let radius of 1st cone be r1 and that of second cone be r2. 𝜋𝑟1 ℓ = 3(𝜋𝑟2 . 3ℓ) or, 𝜋𝑟1 ℓ = 9𝜋𝑟2 ℓ ⇒

𝑟1 𝑟2

=

9

1

Ratio of areas = 81 : 1 22

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35. (b); Let initial radius of both vessels r cm and volume is V cm3 V = 𝜋𝑟 2 12 For first vessel, 𝑉1 = 𝜋(𝑟 + 7)2 12 Second Vessel, 𝑉2 = 𝜋(𝑟 + 7)2 12 Second Vessel, 𝑉2 = 𝜋𝑟 2 (12 + 15) Given, 𝑉1 = (𝑉 + 𝑄 ) 𝑐𝑚3 _______(I) 𝑉2 = (𝑉 + 𝑄 ) 𝑐𝑚3 ________(II) From (i) and (ii) We get 𝑉1 = 𝑉2 𝜋(𝑟 + 7)2 12 = 𝜋𝑟 2 (27) r = 14 cm So increased capacity (Q) 22 22 = × (14 + 7)2 × 12 − × 14 × 14 × 12 7 7 = 16632 – 7392 = 9240 cm3 36. (e); Given, area of square park = 19600 cm² a² = 19600 𝑎 = 140 cm 140 Radius of each light pole = = 2.8 cm 50 Area covered by One pole in park 22 = × 2.8 × 2.8 = 24.64 cm² 7 Total area covered by these nine poles = 9 × (24.64) = 221.76 cm² 37. (c); Given, Area of square = 306.25 cm2 Side of square = 17.5 cm Breadth of square = 17.5 − 3.5 = 14 cm Length of rectangle =

252 14

= 18 𝑐𝑚

Radius & height of cylindrical vessel be 14 cm & 18 cm respectively Volume of cylindrical vessel = 23

22 7

× 14 × 14 × 18 ×

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38. (c); let height of vessel = h 3

Let radius of bowl = h = radius of vessel 2

2

Volume of hot water = 𝜋ℎ3 3

Required percentage 2 3 3 𝜋( ℎ) 3 2 3 2 𝜋(2ℎ) ×ℎ

=

× 100

= 100% 39. (a); Let radius be 3r then height will be 2r Value of figure formed 2

= π × (3r)3 + π(3r)2h 3 2

= π × 27r 3 + π9r 2 × 2r 3

= 36 πr³ = 3r = 21 r=7 = 36 ×

22 7

×7×7×7

= 36 × 22 × 49 cm³ Required time =

36×22×49 539

= 72 minute 40. (d); Let radius of cylinder and hemisphere is ‘r’ Also height of cylinder = ‘r’ Total milk = πr² × r = πr³ Milk poured in 1 hemi-sphere 2

= πr 3 3

1

Remaining milk = πr 3 3

When this milk poured in 2nd sphere then it remained empty 2 3

1

1

3

3

πr 3 – πr 3 = πr 3

Required% = 50% 24

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41. (e); Perimeter of corridor and hall = 102 m Let radius of hall is R m and corridor is r m ATQ— 2πR + πr + 2r = 102 m 2×

22

204r

= 102

7

14

×

3r 2

+

22r 7

+ 2r = 102 m

r=7m 21 m 2 Radius of circular garden So, R = 21

=( =

2 35 2

+ 7) ×

400 100

× 4 = 70 m

Area of Garden = πr² = 70 × 70 𝜋 = 4900 𝜋 sq. m 42. (c); Given, height of cylindrical vessel = side of square a2 = 256 cm² a = 16 cm Let radius of hemisphere is 2r cm and sphere is r cm. ATQ— 3π(2r)² – 4πr² = 4928 cm² 12r² – 4r² = 1568 r² = 196 r = 14 Radius of cylindrical vessel =

28+14 2

= 21 cm

Volume of cylindrical vessel =

22 7

× 21 × 21 × 16

= 22176 cm³ 25

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43. (b); Ratio between length, breadth and side of square = 21 : 6 : 14 Let length and breadth of rectangle be 21x cm and 6x cm respectively. And, side of square be 14x cm ATQ – 4(14x) – 2 (21x + 6x) = 6 56x – 54x = 6 x = 3 cm 14×3 Radius of cylinder = = 21 cm 2 Height of cylinder = 6 × 3 = 18 𝑐𝑚 Volume of cylinder = πr2h 22 = × 21 × 21 × 18 7 = 24948 cm3 44. (b); Let the radius of sphere be x cm. 2 4 1 ⇒ 𝜋(𝑥 )3 = 18 ( 𝜋 × (4√3) × 4) + 12(𝜋 × (4)2 × 6) 3 4

)3

3

⇒ (𝑥 = 1152 + 1152 cm³ 3 ⇒ x = 12 cm Then, total surface area of sphere = 4πx² 22 = 4 × × (12)2 7

2

= 1810 cm² 7

45. (d); Let length & breadth of rectangle be 7x & 6x respectively Given area of rectangle = 168 cm2 7x × 6𝑥= 168 cm2 42x2 = 168 x2 = 4 x = 2 cm Radius of cylindrical vessel = 6 × 2 − 1.5 = 10.5 𝑐𝑚 Height of cylindrical vessel = 7 × 2 = 14 𝑐𝑚 Volume of cylindrical vessel = 𝜋 × 10.5 × 10.5 × 14 = 1543.5π cm3 Let think ness of each discs be x cm So, 10 × 𝜋 × x × 10.5 × 10.5 = 1543.5π cm3 1543.5 x= 10.5×10.5×10 x = 1.4cm 26

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46. (c); Let the radius of two hemisphere be 2x & 3x ATQ, 2 3

2

836

3

21

π(3x)3 – π(2x)3 = 2

836

3

21

⇒ π[19x3 ] =

cm3 [∵ volume of hemisphere = ⅔ π r³)

⇒x=1 ∴ radius are 2 cm & 3 cm. Now, 2

2

74

3

3

3

[ π(2)3 + π(3)3 ] + [ πcm3 ] = πR2 H 2

2

3

3

⇒ π[8 + 27] + π 37 = πR2 H 2

⇒ 𝜋[35 + 37] = 𝜋𝑅2 𝐻 3

⇒ R²H = 48 ⇒ R : H = 4 : 3 (given) Let R = 4a, H = 3a ⇒ 16a². 3a = 48 𝑎=1 ∴ R = 4, H = 3 T.S.A. of cylinder = 2πR(R + H) =2×

22 7

× 4(7)

= 176 cm²

27

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1

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Chapter

11

Permutation, Combination and Probability

BEST APPROACH TO SOLVE THE QUESTIONS The best way to define probability is number of favorable outcomes divided by total number of possible outcomes. Eg. When we conduct an experiment of throwing a dice and we have to calculate probability of getting 2. Now number of favorable outcome is 1 i.e. ‘2’ And total number of possible outcomes is 6 i.e it may appear ‘1, 2, 3, 4, 5 or 6’ on throwing a dice. Hence probability of getting a ‘2’ is =

1 6

If we were asked what is the probability of getting an even number. Number of possible outcome = 3 (2, 4 or 6) 3

1

6

2

Probability of getting an even number on throwing a dice = =

But the probability in bank exam is not this much easy, we encounter with way more complicated questions. Example 1: A C++ program shuffles all the letters of a word. When the word INDIA was entered, a new word was given as an output. What is the probability that new word formed is having all the vowels together? Solution: We need to calculate two things (i) Total number of possible outcomes (ii) Total number of favorable outcomes Imagine how many different words can be formed by rearranging the words of INDIA. And then how many different possible words can be formed when ‘IIA’ of ‘INDIA’ are together. To solve these kind of questions we have PERMUTATION AND COMBINATION. 2

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Permutation and Combination: Basic definition: Permutation means different possible arrangement of things taken all or some of them together. While Combination means different way of selection of things taken all or some of them together. Eg. Of arrangement/permutation: There were two chairs numbered 1 and 2. In how many ways A, B and C could sit on them? SOLUTION: AB or BA, AC or CA and BC or CB. (First person denotes to sit on chair named A). Hence there are 6 possible ways. Eg. Of selection/ Combination There were two chairs and three persons. In how many different ways they can sit on chairs? Solution: Either AB, BC or AC will sit on two chairs. Hence there are 3 possible ways. The point is arrangement when sequence matters is known as permutation or selecting some or all the persons is called combination. Before further discussion, let us revise few terms of permutation and combination. ! is called factorial. N! where N is a natural number is N × (N-1) × (N-2)….. ×1 0! = 1 N! N Pr = (N–r)!, it is used in permutation cases N

Cr =

N!

, it is used in combination cases.

C!(N–r)!

E.g. In previous cases. 3! 3×2×1 3 P2 = (3–2)! = = 6 arrangements. 3

C2 =

3!

2!(3–2)!

1

= 3 cases of selection.

NOTE: Though permutation and combination in itself are too big, here we will discuss few of the types of questions that are asked in exam. Look at the previous example of INDIA. 3

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Example 2: What is the probability that after rearranging the word ‘INDIA’ a new word is formed in which all vowels are together? SOLUTION: When n things in which p are of one type, q are of one type , then total no. of arrangements of those n things is =

n! p!q!

Taking two Is and one A as a single entity, we have in all 3 entities (N,D and the group of is and A). Total no. of ways of arranging these three entities is 3!. Two Is and one A can be rearranged among themselves in

3! 2!

ways. Hence, total no. of ways in which all vowels are together is equal to 3!×

3!

=18.

2!

Total no. of ways in which all letters of word INDIA can be 5!

arranged = =60. 2!

Hence required probability=

18

=

3

60 10

Example:

What is the probability that a word formed after rearranging “ARRANGEMENT’’ such that all vowels are together? SOLUTION: Following the same method as that of previous question, Total no. of ways of rearranging the letters of word ARRANGEMENT=

11! 2!2!2!2!

=

11! 16

No. of ways of arranging letters such that all vowels are together=

8! 2!2!

×

4! 2!2!

=

8!4! 16

Required probability=

8!4! 16 11! 16

=

4 165

KEY POINTS: Number of permutation of n things taken all at a time is n! N Pr = NCr × r!

4

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Practice Exercise Based on New Pattern

1. There are five mangos and six oranges in a bucket. What will be probability of Picking up four fruits which contains at least two orange? 53 43 59 (a) (b) (c) (d)

66 49

66

(e)

66 3

66

5 1 2

3

2. Three shooter having probability of shot a target , and respectively. 3 5 7 What is the probability that exactly one of them did not hit the target? 12 29 19 (a) (b) (c) (d)

35 41

105

(e)

105 1

35

5

3. Rahul, Sachin and Sanjeev speaks 2 times, 5 times and 3 times truth out of 5 times, 6 times and 8 times respectively. Find the probability that they contradict each other when asked to speak on a fact? 3 5 7 (a) (b) (c) (d)

16 13 16

(e)

16 9

16

16

4. Probability of choosing one boy from a group of 26 girls and ‘x’ boys is 3 . If one more boy joined the group then find the probability of choosing 5 one boy and one girl from that group? 8 6 4 (a) (b) (c) (d)

33 14 33

(e)

11 16

11

33

5. In how many ways word “EATEN” can be arranged with or without meaning in which at least two vowels will come together? (a) 48 (b) 54 (c) 72 (d) 108 (e) 90 6. Sum of digits of a 5-digit number is 43. Find the probability that such a number is divisible by 11? (a) 1/3 (b) 1/5 (c) 2/5 (d) 2/15 (e) 2/3 5

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7.

8.

9.

Three-unit squares are chosen at random on a chessboard. What is the probability that two of them are of same color while remaining one is of another color? (a) 16/21 (b) 8/21 (c) 5/21 (d) 4/7 (e) 32/63 Two dice are thrown, find the probability of getting an even number after multiplying numbers obtained from the dice. (a) ¼ (b) ¾ (c) ⅔ (d) ½ (e) ⅙ In how many ways 9 hindi books are placed between 10 English books such that no two Hindi book comes together? (a) 60 (b) 55 (c) 45 (d) 96 (e) 75

10. Fourteen applications apply for a job out of which there are 9 men and 5 women. If only three applications selected for the job then find the probability that at least one of the selected application is of a woman? (a) (d)

60 91 40 91

(b) (e)

10 13 50

(c)

80 91

91

11. In a bucket there are three type of fruits i.e, mango, banana and papaya. 1 Probability of selecting one mango out of total fruits in bucket is and 3

3

the probability of selecting one papaya out of total fruits in bucket is . 8 If number of banana in the bucket is seven, then find the number of papaya in the bucket. (a) 4 (b) 5 (c) 11 (d) 9 (e) 7 12. A bucket contains seventeen rectangular boxes, from number 1 – 17. If three boxes drawn at random from bucket without repeated, then what will be probability of that all three rectangular boxes represent even number boxes? 9 6 7 (a) (b) (c) (d) 6

85 11 85

(e)

85 21

85

85

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13. Two boxes contain 4 and 16 balls respectively. Two balls in the first box and four in the second box are black. If a box is chosen randomly and two balls are drawn at random from it, what is the probability that at least one ball is black ? 11 43 77 (a) (b) (c) (d)

20 9

20

(e)

120 7

120

23

14. Probability of getting one Red ball from a bag full of balls is 2/13 and Black balls in the bag is 5. If white ball is 30% less than Brown ball and 40% more then the black balls then find the number of Red balls. (a) 5 (b) 4 (c) 6 (d) 10 (e) 13 15. Probability of choosing one boy from a group of 26 girls and ‘x’ boys is 3 . If one more boy joined the group then find the probability of choosing 5 one boy and one girl from that group? 8 6 4 (a) (b) (c) (d)

33 14 33

(e)

11 16

11

33

16. In a meeting of HCL company, in the month of June, there are 6 CEO’s , 5 Y MD’s and 7 HR’s. Probability of calling one MD on stage is . When 18 the meeting is conducted in next month (Y + 1) CEO’s, (Y + 3) MD’s and (Y + 2) HR’s attended the meeting. If two representatives are called together on stage, then what will be probability of both being CEO’s? 1 1 1 (a) (b) (c) 16 1

(d)

18

(e)

14 1

12

20

17. There are three groups of children in a school i.e. A, B and C. In group A, there are (x + 2) girls and x boy. In group B, there are 2 boys and 2 girls, whereas in group C, there are 1 girl and 2 boys respectively. One child is selected at random from each group. Find the value of x, if the 3 probability of selecting 1 girl and 2 boys is . 8 (a) 2 (b) 3 (c) 1 (d) 4 (e) 5 7

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18. In a room, 10 people sit in which 4 left hander and 6 right hander. Find the probability that left hander shake hand with only left handers. (a) (d)

1 15 5 9

(b) (e)

7 45 2

(c)

17 45

15

19. In a cold drink cartoon 3 pepsi, 5 limca and 7 sprites bottels. If one bottle is drawn at random from the cartoon what is the probability that it is either sprite or pepsi ? (a) (d)

5 7 8 9

(b)

3 4

(c)

2 3

(e) None of these

20. Find the difference between number of words formed from ALAIS when all vowels are together, and words formed from ENGLISH when no vowels are together? (a) 3528 (b) 3618 (c) 3572 (d) 3592 (e) 3582 21. In a box some Munch chocolate, some Dairy milk chocolate and Kit Kat chocolate. The probability of selecting one Munch chocolate out of box is 2 5

3 10

and the probability of selecting one Kit Kat chocolate out of box is

. The number of Dairy milk chocolate in box is 6. Find probability of

selecting three chocolate from the box such that at most two chocolate are same? (a) (d)

64 85 87 95

(b) (e)

64 95 67

(c)

42 85

85

22. Find the probability of forming the word from ANANAB such that N never comes together. (a) (d) 8

2 3 2 5

(b) (e)

4 5 4

(c)

3 5

5

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23. A bag contains 5 red, 4 green and 3 black balls. If three balls are drawn out of it at random, find the probability of drawing exactly 2 red balls? (a) (d)

5

(b)

22 2

(e)

11

7 24 7

(c)

6 25

22

Directions (24-25): A man can send a message by using six flags. He had 4 black flags, 3 blue flags and 1 green flag. 24. In how many ways a message can be transmitted? (a) 108 (b) 185 (c) 20 (d) 136 (e) 110 25. In how many ways a message can be transmitted by using all three colored flags? (a) 150 (b) 90 (c) 100 (d) 70 (e) 125 Directions (26-27): A box contains two types of ball i.e., red and Blue. Probability of choosing one red and one blue ball is choosing two red balls is

3

1 2

while probability of

.

20

26. Find the probability of selecting two balls without replacement from the box such that at least one of them is red? (a) (d)

9 20 17 20

(b) (e)

11 20 3

(c)

13 20

4

27. If 5 Red balls and 6 blue balls are added in the box, then find the probability of choosing two blue balls from the box at present? (a) (d) 9

1 2 1 5

(b) (e)

1 3 1

(c)

1 4

6

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Directions (28-30): There are three sections A, B and C in a class. Every section has some boy and some girl students in it. Probability of a girl being 2 selected when one student is selected randomly from section A is , that 4

5

9

9

5

from section B is and that from section C is . 28. If the ratio of total number of students in sections A, B and C is 10 : 12 : 9, then what is the probability of a girl being selected when one student is selected randomly from the students from all the three sections together? (2 Marks) 14 11 13 (a) (b) (c) (d)

31 43 93

23

31

(e) Cannot be determined

29. If the number of girls in sections A is same as the number of boys in section C, then what is the ratio of number of boys in section A to the number of boys in section C? (2 Marks) (a) 2 : 3 (b) 3 : 4 (c) 3 : 2 (d) 4 : 3 (e) 3 : 5 30. If 20 girls leaves section B, then the probability of a boy being selected 5 when one student is selected randomly from this section will be . What 8 is the number of boys in section B? (2 Marks) (a) 120 (b) 100 (c) 75 (d) 60 (e) Cannot be determined Directions (31-32): There are some horses, hens and elephants in a park. Ratio between number of heads in park to number of legs in park are 7 : 20. 31. Find the probability of choosing a hen over all the animals. 3 2 3 (a) (b) (c) (d)

7 4 7

(e)

7 2

8

9 9

32. If probability of choosing 2 horses over all the animals are , then find 119 the number of elephant present in park. (Number of horses is double than that of the number of elephants in park). (a) 5 (b) 10 (c) 20 (d) 15 (e) 25 10

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Directions (33-34): There are two bags, one of which contains 3 black and 4 white balls, while the other contains 4 black and 3 white balls. A dice is cast, if the face 1 or 3 turns up, a ball is taken from the first bag and if any other face turns up, a ball is chosen from the second bag. 33. The probability of choosing a black ball is: 11 11 (a) (b) 24 11

(d)

(e)

29

21 3

(d)

23

(e)

(2marks)

27

8

34. The probability of choosing a white ball is: 11 11 (a) (b) 21 14

(c)

11

29 11

(2marks) 10 (c) 21

23

Directions (35-37) : India wins every 6 matches out of 10 matches against England. Virat Kohli (Indian Player) make a century in every 3 matches out of 5 matches and in which he makes century, India wins 80% of the matches. Alastair cook (England’s player) make 1 century in every 4 matches and England wins 60% of that matches. 35. Find the probability of India’s winning against England when Kohli and Cook both make centuries. 20 17 36 (a) (b) (c) (d)

623 24

625

(e)

625 18

625

625

36. Find the probability that India wins against England when Kohli failed to make a century. 8 3 6 (a) (b) (c) (d)

25 4

25

(e)

5 12

25

25

37. Find the probability of winning England against India when Kohli make a century but cook does not. 9 16 2 (a) (b) (c) (d) 11

250 14

250

(e)

250 21

5

250

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Direction (38-39): In a stationary shop there are four types of colored sheets of red, blue, green and white colors. The probability of selecting one 1 red sheet out of the total sheets is ,the probability of selecting one blue sheet out of the total sheets is

2 7 1

3

and the probability of selecting one white

sheet out of the total sheets is . The number of green sheets in the bag is 4 22. 38. If all the sheets are numbered as 1,2,3,….and so on and one sheet is picked up at random, then find the probability of picking up a sheet which is numbered as a multiple of 13 or 17. (2 marks) 1 1 5 (a) (b) (c) (d)

8 9

10

7

8

(e) none of these

39. What is the total number of sheets in the bag? (a) 117 (b) 168 (d) 120 (e) none of these

(1 marks) (c) 154

Direction (40-41) : A coach invited 9 players of the team on dinner. All 10 persons seated in a circular table. 40. If in how many ways captain and vice-captain seat opposite to each other? (a) 80640 (b) 5040 (c) 3624 (d) 40320 (e) 10080 41. In how many ways team can seat, if caption and vice-captain seat either side of coach. (a) 80640 (b) 5040 (c) 3624 (d) 40320 (e) 10080 Directions (42-44): A joker have 10 balls, 10 ribbon and 10 plates of red, yellow and blue color in ratio 2 : 3 : 5, 5 : 2 : 3 and 3 : 5 : 2 respectively. He does his performance by using one item of each category. 42. What is the probability that he performs with only red color? (a) 0.3 (b) 0.02 (c) 0.25 (d) 0.03 (e) 0.15 12

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43. What is the probability that he chooses yellow color for ribbons? (a) 0.0333 (b) 0.025 (c) 0.2 (d) 0.25 (e) 0.03 44. If he chooses 3 balls, 5 ribbons and 1 plate then find the probability that all items are of same color. (a) 1 (b) 0 (c) 0.05 (d) 0.5 (e) 0.002 Direction (45-47): - Kishan have two trays of eggs i.e. tray A and tray B. Tray A have same number of fresh eggs as tray B have rotten eggs. Rotten eggs in tray A is 40% less than that of in tray B. Fresh eggs in tray B is 14 more than that in tray A. Total number of eggs Kishan have is 140. 45. Fresh eggs in tray B is what percent more than that in tray A? (a) 140% (b) 80% (c) 60% (d) 40% (e) 20% 46. Find the probability of choosing two eggs from tray ‘A’ such that at least one egg should be fresh? 19 17 27 (a) (b) (c) (d)

22 25

(e)

44

22 21 22

47. Find the probability of choosing one rotten egg? 23 7 (a) (b) (d)

48 25

(e)

48

44

16 9

(c)

19 48

16

Direction (48- 50)- There are three bucket P, Q & R contains three fruits Mangos, Bananas & Oranges. In bucket P, contains x Mangos and y Bananas. Number of bananas are 3 more than the number of Mangos. If Ram selected one fruits for eating, then 3 1 the probability of getting orange is . The value of x is 37 % less than y. 16 2 In bucket Q, number of Mangos is 40% more than that of bucket P. If Satish selecting two fruits at random for eating then the probability of getting both 18 Bananas is . Total number of fruits in bucket Q is 29. 203

13

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In bucket R, the ratio of number of mangos to number of orange is 3 : 4. Total number of bananas & orange is 33 . If Veer selecting one fruit for 3 eating the probability of getting mangos is . 14

48. What is the total number of bananas present in all the three buckets? (a) 30 (b) 38 (c) 36 (d) 32 (e) 28 49. If all mangoes of bucket P is placed in the bucket Q then find the probability of choosing a mango from bucket Q? 5 9 3 (a) (b) (c) (d)

17 7

(e)

17

17 6

17

17

50. If 25% oranges of bucket R are spoiled and thrown away then find the probability of choosing a good orange from bucket R? 3 5 7 (a) (b) (c) (d)

13 6

(e)

13

13 2

13

13

Solutions

1. (a); Favorable case = (20, 2M) or (30, 1M) or 40 ∴ Probability = =

6C × 5C 2 2 11C 4

15×10 330

+

+

6C × 5C 3 1 11C 4

20×5 330

+

6C

+ 11 4

15

330

=

C4 265 330

=

53 66

2. (b); Required probability 1 2 4 1 3 3 2 3 2 = × × + × × + × × = 3

5

7

3

7

5

5

7

3

29 105

3. (d); Probability that Rahul speaks truth =

2 5 5

Probability that Sachin speaks truth =

6

Probability that Sanjeev speaks truth = 14

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Rahul, Sachin and Sanjeev do not contradict with each other if they all are correct or they all are incorrect means if they all speak truth or they all speak lie. 2

5

3

1

5 1

6 5

8 1

8

Probability that they all speak truth = × × = 3

Probability that they all speak lie= × × = 1

1

8

16

Required probability = 1 – –

=

5 6 8 16 16 –2 –1 13 16

=

16

4. (e); ATQ, x 26+x

3

=

5

⇒ 5x = 26 × 3 + 3x ⇒ 2x = 26 × 3 ⇒ x = 39 Total no. of students now in group = 26 + 39 + 1 = 66 Required probability = 2 × 5. (b); Total number of ways =

5! 2!

26 66

×

40 65

=

16 33

= 60 ways

Selection of vowels can occur in three pair i.e. ( AE, EE, EA) Number of ways in which two vowels will come together =

3×4!

= 3 × 4 × 3 = 36

2!

Above, we had not multiplied 2 in numerator because further arrangement of vowels selected i.e. (AE, EE, EA) will give the same arrangement Number of ways in which three vowels will come together =

3!×3! 2!

= 3 × 2 × 3 = 18

Required no. of ways = 36 + 18 = 54 Alternate, No. of ways in which no vowel will came together =2×

3! 2!

=6

Required no. of ways = 60 – 6 = 54 15

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6. (b); Let, numbers in 5-digit number is a, b, c, d, e. In order to get sum of 43, the combinations exist are 9, 9, 9, 9, 7 = 5 numbers 9, 9, 9, 8, 8 = 10 numbers Total 15 numbers can be formed having sum 43. For number which are divisible by 11 main condition is (a + c + e) – (b + d) = 11 …(i) and (a + b + c + d + e) = 43 …(ii) On solving (i) and (ii) a + c + e = 27, b + d = 16 a, c and e can be 9, 9, 9 while b, d can 7, 9 or 8, 8 or 9, 7 respectively So, there are three numbers which are divisibly by 11 having sum 43 i.e., 97999, 99979, 98989. 3 1 Required probability = = 15

5

7. (a); Total number of ways = ⁶⁴C₃ = 41664 Favorable ways = 2 × ³²C₂ × ³²C₁ = 2 × 496 × 32 = 31744 31744 16 Required probability = = 41664 21 (³²C₂ × ³²C₁ is taken because if two are black, then one will be of white or if two are white, then one will be of black) 8. (b); Favourable case → (Dice 1 → even and dice 2 → odd) (Dice 1 → even and dice 2 → even) (Dice 1 → odd and dice 2 → even) 3 3 3 3 3 3 3 = × + × + × = 6

6

6

6

6

6

4

9. (b); Let us denote English book by E and Hindi books by H The possible arrangements is HEHEHEHEHEHEHEHEHEHEH There are → English book, H can occupy 10 + 1 = places So no. of ways → ¹¹C₉ 11×10 = = 55 ways 2

16

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10. (b); Total number of ways to select three persons = 14C3 Required case = 2 men 1 women or 1 man 2 women or 3 women = ( 9C2 × 5C1 ) + ( 9C1 × 5C2 ) + ( 5C3 ) = (36 × 5) + (9 × 10) + (10) = 180 + 90 + 10 = 280 14C3 = 14C3 = 364 280 10 Required probability = = 364

13

11. (d); Let number of papaya and mango in bucket be ‘P’ and ‘M’ respectively Number of banana in bucket = 7 1 Probability of selecting one mango = 3

3

and selecting one papaya = 8 So, M 1 = (7+M+P) 3

3M = 7 + M + P 2M – P = 7 …(i) Similarly— P 3 = (7+M+P) 8

8P = 21 + 3M + 3P 5P – 3M = 21 …(ii) Solving (i) and (ii) we get Number of Mango = 8, Banana = 7, Papaya = 9 Total number of papaya in bucket = 9 12. (c); Total even number boxes between 1 to 17 = 8 (2, 4, 6, 8, 10 …..16) Total boxes = 17 Probability of first box represent even number box =

8 17

Probability of second box represent even number box = Probability of third box represent even number box = Required probability = 17

8 17

×

7 16

2

7

5

85

× =

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13. (c); At least one black can be chosen in three ways: 1. first one is black, second is non-black 2. first one is non-black, second is black 3. both are black. 1 Probability of selecting a box is 2 Now, probability of choosing at least one black ball from first box 1

2

2

2

2

2

1

5

2

4

3

4

3

4

3

12

= ×( × + × + × )=

Probability of choosing at least one black ball from 2nd box 1

4

2

16

= ×(

×

12 15

+

12 16

×

4 15

Required probability =

+ 5 12

4 16

+

× 27

3 15

120

)=

=

27

120 50+27 120

=

77 120

14. (b); Let number of red balls is 2x Total balls = 13x Now ATQ Black balls = 5 5×140 White balls = =7 100 7

Brown balls = × 100 = 10 70 Now→ 2x + 5 + 7 + 10 = 13x x=2 Red balls = 4 15. (e); ATQ, x 3 = 26+x 5 ⇒ 5x = 26 × 3 + 3x ⇒ 2x = 26 × 3 ⇒ x = 39 Total no. of students now in group = 26 + 39 + 1 = 66 26 40 16 Required probability = 2 × × = 66 65 33 16. (b); In June — Total CEO’s = 6 MD’s = y ⇒ HR’s = 7 Now, 18

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ATQ, y 13+y

=

5

⇒ y=5

18

In July — CEO’s ⇒ 5 + 1 = 6 MD’s ⇒ 5 + 3 = 8 HR’s ⇒ 5 + 2 = 7 6C 2 21C 2

Required probability =

=

6×5 21×20

=

1 14

17. (c); Possible cases Girl Boy Boy or Boy Girls Boy or Boy Boy Girl 3 8

2

2

2x+2 4 (2x+2)×4×3

3

=(

⇒ ⇒

x+2

× × )+(

2×8 (x+1)×9 2

x

2

2

× × )+(

2x+2 4 3 2(x+2) 2x x

=(

3

+

3

x

2x+2

2

1

4

3

× × )

+ ) 3

= 5x + 4

⇒ 9x + 9 = 10x + 8 ⇒ x=1 18. (e); Total no. of ways in which hand shakes =

10(10 –1) 2

= 45

No. of ways in which left hander shakes hand with left handers =

4(4 –1) 2

=6

Required probability =

6 45

=

2 15

19. (c); Total number of bottle in cartoon = 3 + 5 + 7 = 15 ∴ Probability of either sprite bottle or pipe bottle = 19

7C + 3C 1 1 15C 1

=

7+3 15

=

2 3

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20. (e); AAILS when all vowels are together = 3! ×

3! 2!

= 18 words

Total words from ENGLISH = 7! = 5040 Words from ENGLISH when no vowels are together = 5040 – [when all vowels are together] = 5040 – [6! × 2] = 3600 ∴ Required difference = 3600 – 18 = 3582 21. (d); Let total chocolate in box be 10c. Munch chocolate be a and kit-kat chocolate be b . Probability of selecting Munch chocolate =

a 10c

=

3 10

a = 3c ... (i) Probability of selecting Kit Kat chocolate =

b 10c

=

2 5

b = 4c … (ii) ATQ, a + b + 6 = 10c ⇒ 3c + 4c + 6 = 10c ⇒ c = 2 So, total chocolate = 20 Required probability = 1 − =1−

(20+56+20) 1140

=

1044 1140

=

6C + 8C + 6C 3 3 3 20 C 3

87 95

22. (a); Total number of words possible from ANANAB =

6! 3!2!

= 60

No. of word in which N is always together =

5! 3!

= 4 × 5 = 20

No. of in which N is never together = 60 – 20 = 40 Required probability = 20

40 60

=

2 3

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23. (e); Exactly 2 red balls can be selected in two ways. (i) 2 red ball & 1 green ball (ii) 2 red ball & 1 black ball Total no. of possible outcome = ¹²C₃ =

12×10×11 3×2

= 220 ways

Total favourable outcome = ⁵C₂ × ⁴C₁ + ⁵C₂ × ³C₁ =

5×4 2

×4+

5×4 2

Probability =

× 3 = 40 + 30 = 70 ways.

70

220

=

7 22

24. (b); Case I: When all 4 black flag selected Two ways could be 1 green flag + 1 blue flag or 2 blue flags No. of ways to arrange =

6! 4!

+

6! 4!2!

Case II: When all 3 blue flag selected 3 ways could be 2 black flags + 1 green flag or 3 black flags No. of ways =

6! 3!2!

+

6! 3!3!

Case III: When one green selected ⇒ 3 black + 2 blue [other cases already included] Adding all ⇒ 30 + 15 + 60 + 20 + 60 = 185

6! 3!2!

25. (a); All colored used ⇒ 1 green flag + 2 black flags + 3 blue flags 1 green flags + 3 black flags + 2 blue flags 1 green flag + 4 black flags + 1 blue flag ⇒

6! 2!3!

+

6! 3!2!

+

6! 4!

⇒ 60+ 60 + 30 = 150 21

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Solution (26-27): Let, Red balls = x Blue balls =y ATQ, 2×

x

×

(x+y)

y (x+y−1)

=

1

... (i)

2

And, x×(x−1) (x+y)(x+y−1)

=

3

... (ii)

20

Dividing (ii) by (i) ⇒

x−1 y

=

3 5

⇒ 5x – 5 = 3y And, from (i) 4xy = (x + y) (x + y – 1) From (iii) and (iv) 5x−5

4x ( ⇒

20x 3

3

)=[

…(iii) ... (iv)

5x−5+3x

] [x +

3 8x−5

(x − 1) = [

3

][

5x−5

8x−8 3

− 1]

3

]

⇒ 60x² – 60x = 64x² - 64x – 40x + 40 ⇒ -4x² + 44x – 40 = 0 ⇒ x² – 11x + 10 = 0 ⇒ x²– 10x – x + 10 = 0 ⇒ x(x – 10) – 1 (x-10) = 0 ⇒ x = 1, 10 x = 1 x = 10 ⇒ y = 0 ⇒ y = 15But y can’t be zero. ⇒ Number of red balls = 10 Number of blue balls = 15 26. (c); Required probability y=

10 C + 10 C 15 C 2 1 1 25 C 2

=

45+150 300

27. (b); Required probability = 22

=

195 300

21 C 2 25+11 C 2

=

=

13 20

21×20 36×35

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28. (d); Probability of a girl being selected from a section =

Total girls in the section Total students in the section

Let the number of girls, number of boys and total number of students respectively: For section A: 2x, 3x and 5x. For section B: 4y, 5y and 9y. For section C: 5z, 4z and 9z. According to the question, Ratio of total number of students in the three sections: ⟹ 5x : 9y : 9z = 10 : 12 : 9 ⟹x:y:z=6:4:3 Let the values of x, y and z be 6k, 4k and 3k respectively. Total number of girls in all the three sections = 2x + 4y + 5z = 12k + 16k + 15k = 43k Total number of students in all the three sections = 5x + 9y + 9z = 30k + 36k + 27k = 93k Probability of a girl being selected from the students from all the three sections together =

Total girls in all sections Total students in all sections

=

43k 93k

=

43 93

29. (c); According to the question, Number of girls in sections A = Number of boys in section C ⟹ 2x = 4z ⟹ x = 2z Number of boys in section A : Number of boys in section C = 3x : 4z = 6z : 4z = 3 : 2 30. (b); Probability of a boy being selected from this section B after 20 girls left the section = ⟹ ⟹

5

8 Number of boys in section B

Total number of students in section B−20 5y 5 9y − 20

=

=

5 8

8

⟹ 40y = 45y – 100 ⟹ y = 20 Number of boys in section B = 5y = 100 23

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Solution (31-32): Let number of hens, horses and elephant in park is — a, b and c respectively. Let number of heads and number of legs be 7x and 20x respectively So, a + b + c = 7x b + c = 7x– a …(i) 2a + 4(b + c) = 20x …(ii) Solving (i) and (ii) 2a + 4 × (7x– a) = 20x 2a + 28x – 4a = 20x a = 4x 4x

31. (d); Probability of choosing hen over all the animals =

7x

=

4 7

32. (a); No. of hens = 4x No. of horses and elephants together = 7x– 4x = 3x If number of horses is double than that of the number of elephants then Number of horses = 2x Number of elephant = x ATQ 2xC 2 7xC 2

=

9

119 2x(2x–1) 9 7x(7x–1)

=

119

x=5 2

3

1

6 4

7 4

7 8

6

7

21

33. (b); Probability of choosing a black ball from bag 1 = × = Probability of choosing a black ball from bag 2 = × = 1

8

7

21

Total probability = +

=

11 21

2

4

6 4

7 3

6

7

34. (c); Probability of choosing a white ball from bag 1 = ×

=

Probability of choosing a white ball from bag 2 = × = Total probability = 24

10

8

42 12 42

21

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6

35. (e); Probability that India wins against England = 10 Condition applied 3 Probability that Kohli makes century = 5

Probability that India win when Kohli makes century = 1

4 5

Probability of Cook’s century = 4 Probability of other country wining when Cook make century 3 2 = 1– = 5

5

Combining all =

6 10

3

4

1

2

18

5

5

4

5

625

× × × × =

2

36. (c); Probability when Kohli failed to make a century = 5 So, Probability that India wins against England when Kohli failed to 6 2 6 make a century = × = 10

5

25

37. (a); Probability that Kohli will make a century =

3 5

Probability that India lose when Kohli make century = Probability that Cook doesn’t make century =

3 4

Probability that of Winning of England against India = Overall probability =

4 10

3

3

1

4

5

5

× × ×

=

9

1 5 4 10

250

38. (a); Let, total no. of sheets be 84x, 1 Then no. of red sheets = × 84x = 28x 3

2

no. of blue sheets = × 84x = 24x 7

1

no. of white sheets = × 84x = 21x 4 Now, 28x + 24x + 21x + 22 = 84x or, 11x = 22 or, x = 2 Total no. of sheets = 84x = 168 There are 12 multiples of 13 and 9 multiples of 17 from 1 to 168. 25

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Also, there is no multiple of 13 and 17 from 1 to 168. (because, 13× 17 = 221) 21 1 Hence, required probability= = 168

8

39. (b); Total no. of sheets = 168 40. (e);

When two places fixed so we can arrange other 8 in → (8!) ways and captain and vice-captain Total ways ⇒ 8! ⇒ 40,320 41. (e);

3 places fixed ⇒ So total place ⇒ 10 – 3 = 7 Number of way to arrange ⇒ 7! Captain and vice-captain change their place in two ways So, total ways ⇒ 7 ! × 2 = 10,080 Solutions (42-44) Balls = 10 Red = 2 Yellow = 3 Blue = 5 26

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

42. (d); Performance with only red color One ball of red color, one ribbon of red color and one plate of red color Required Probability =

2 10

×

5 10

×

3 10

=

3 100

= 0.03

43. (c); Condition is yellow color for ribbon while plate and ball can be of any color so, probability of choosing ball and plate is 1 while probability of choosing one yellow color ribbon is =

2 10

= 0.2

44. (b); Now Joker choose → 3 balls, 5 Ribbon and 1 plate Condition → all are of some color ⇒ 5 ribbons of same color is only of red color But, 3 red ball cannot be possible so Required Probability = 0 Solution (45-47) : - Let fresh eggs in tray A = 100x = Rotten eggs in tray B Rotten eggs in tray A = 60x Fresh eggs in tray B = 100x + 14 Total eggs = 100x + 100x + 60x + 100x + 14 = 140 360x = 126 ⇒ x = 0.35 Tray A Fresh Rotten 35 21 45. (d); Required % =

49−35 35

Tray B Fresh Rotten 49 35

× 100 =

46. (a); Required probability =

35 56

×

35

34 55

1

21

2

56

47. (c); Required probability = × 27

14

× 100 = 40%

+2×

35×21 56×55

1

35

2

84

+ ×

=

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=

3 16

+

17

+

44 5 24

21 44

=

=

19 22

19 48

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Direction (48– 50): In bucket P – Given, y = x + 3 … (i) 3y x= … (ii) 8 By solving (i) & (ii) we get Mangos (x) = 5 Bananas (y) = 8 Let the number of Orange be z. z 3 Then, = 13+z 16 16z = 39 + 3z z=3 In bucket Q – 140 Mangos = 5 × =7 100 Let the number of Bananas be ‘a’ and number of Orange be ‘b’ 7 + a + b = 29 ⇒ a + b = 22 aC 18 a×(a−1) 18 2 = ⇒ = 29C 2×406 203 2 203

a × (a − 1) = 72 Bananas (a) = 9 ⇒ Orange (b) = 12 In bucket R – Let Mangos & Orange be 3x & 4x respectively 3xC 1 33+ 3xC1

=

3 14

⇒

3x 33+3x

=

3 14

14x = 33 + 3x ⇒ x = 3 Mangos = 9 ⇒ Orange = 12 Bananas = 33 – (9 + 12 ) = 21 48. (b); Total bananas in all three buckets = 8 + 9 + 21 = 38 49. (e); Required probability =

5+7 29+5

=

6 17 3

50. (a); In bucket R Good orange = 12 × = 9 4 Total fruits = 9 + 21 + 9= 39 9 3 Probability = = 39

28

13

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1

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

Chapter

12

Inequality

BEST APPROACH TO SOLVE THE QUESTIONS In examination, we come across questions in which we need to find two quantities and compare them to reach at the correct answer. Usually such questions are asked in two ways. In one type of questions, a common data is provided based on which we are supposed to find both quantities. In other type of questions each quantity comes up with some data pertaining to it only. In these questions, it’s important not to mingle or use data of one quantity in finding other quantity in cases where separate data is given for both the quantities. Where common data is provided for both quantities, we have to use it for calculating both the quantities. After calculating, we compare both quantities and choose the correct answer from the options available. Example: A and B starting a business together. B invested 60% more than that of A. Ratio of time period of investment of A and B is 8 : 9. Quantity I: A’s share of profit out of total profit of Rs. 25,200. Quantity II; B’s share of profit out of total profit of Rs. 14,000. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation Solution: (e); Ratio of profit share of A and B x×8 x×1.6×9 5 : 9 5 Quantity I: A’s profit = 25,200 × = 9000 14 9

Quantity II: B’s profit = 14,000 × = 9000 14 Quantity I = Quantity II 2

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Over the years, questions that test a candidate’s basic algebraic knowledge have been and are being asked. In these questions, two algebraic equations are provided. The equation may be linear, quadratic or cubic. At times, we may come across equations containing higher powers of variable, but they aren’t too difficult to solve. These equations generally contain one variable. From the given couple of equations we need to find the values of two variables. Usually one of the equations gives us the value of one of the variables and the other gives the value of the other variable. After finding the values of two variables we compare them and select the correct option accordingly. Example:

I. 21x² – 22x + 5 = 0 II. 63y² – 54y + 11 = 0

Solution (a); I. 21x² – 22x + 5 = 0 ⇒ 21x² – 15x – 7x + 5 = 0 ⇒ 3x (7x – 5) – 1 (7x – 5) = 0 ⇒ (3x – 1) (7x – 5) = 0 1

5

3

7

⇒ 𝑥 = or

II. 63y² – 54y + 11 = 0 ⇒ 63y² – 21y – 33y + 11 = 0 ⇒ 21y (3y – 1) – 11 (3y – 1) = 0 ⇒ (21y – 11) (3y – 1) = 0 ⇒𝑦 =

11 21

or

1 3

∴ no relation can be established.

3

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Practice Exercise Based on new Pattern PART A 1.

B is thrice as efficient as C. B and C can compete a work together in 45/2 days. A takes 50% more days than the days taken by A and B to complete the same work together. Quantity 1: No. of days taken by fastest among them to complete the work alone. Quantity 2: Time taken by A and C to complete the work together. (a) Quantity I > Quan tity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation

2.

Quantity I→ The profit earned by selling an item (in Rs) if the 2

difference b/w the SP and the CP is 117 % of 600. 3

Quantity II→ The cost price of an article (in Rs) if the selling price of the article is 1000 Rs and he got 25% profit after selling the item. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I  Quantity II (d) Quantity I = Quantity II (e) No relation 3.

The largest possible right circular cylinder is cut out from a wooden cube of edge 7 cm. Quantity I: volume of the cube left over after cutting out the cylinder Quantity II: Surface area of cube remained after cutting out the cylinder. Note: compare the magnitudes of both quantities. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 4

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4. Quantity I- ‘Area of rectangle’ Sum of circumference of circle and perimeter of rectangle is 220 cm. and area of circle is 1386 sq. cm. If 1 length of rectangle is 33 % more than radius of circle. 3

Quantity II- ‘Perimeter of square’ Circumference of a circle is 132 cm and area of circle is 710 sq. cm more than area of square (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 5. Quantity I – Veer takes two times more than the time taken by Sameer to complete a work. Sameer and veer started to work alternatively starting with Sameer on first day and Veer on second day. If both completed work in 36 days, then find in how many days Sameer and Veer will complete the work together, if both work with their double efficiency? Quantity II – Tap P can fill a swimming pool in 8 days and tap Q can fill the same swimming pool in 24 days, while tap R can empty the swimming pool in 16 days. If all three taps are opened together in swimming pool, then in how much time swimming pool will filled completely? (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 6. Quantity I: P alone can do a piece of work in 72 days and Q alone can do same work in 48 days, R is 50% more efficient than Q then find in how many days total work will be completed if all three work on alternate days starting with P and ending with R ? Quantity II: A alone can do a task in 48 days B is 25% less efficient than A and C can do the same task in 40 days lesser than B does. Then find in how many days all three will complete the task together? (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 5

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7.

Quantity I: A bucket carried one dozen of mangoes. One third of mangoes become bad, if three mangoes taken out from the bucket at random. Then what is probability that one mango out of three mangoes picked up is good? Quantity II: Find the probability of selecting a face card from a well shuffled pack of card given that 2 additional Joker card also counts as a face card? (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation

8.

Quantity I – A train running at the speed of 72 km/hr passed a man in 14 sec and a platform in 32 sec. Find the length of platform ? Quantity II – A train moves at the speed of 108 km/hr, passes a platform and a bridge in 15 sec and 18 sec respectively. If the length of platform is 50% of length of bridge, then find the length of train. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation

9.

Quantity I — Difference between the speeds of P and Q if 2 places A and B are 60 km apart. P and Q start from A at same time & meet 1st time at a place 12 km from B & they reach A after immediate return from B. The speed of slower person is 48 km/hr. Quantity II — Average speed of train if a distance of 600 km is to be covered in 2 parts. In 1st phase 120 km is travelled by train and rest by car and it took total of 8 hrs, but if 200 km is covered by train and rest by car it takes 20 min more. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 6

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10. Quantity I — Value of 𝑥, if ABCD is a rectangle and AB= 10 unit, AD= 6 unit

Quantity II — value of 𝑦, if volume of the cone is 16𝜋 unit3 Radius = 4 unit

(a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 11. Quantity I – What is value of (A + B) if 25% of 48% of 55% of A is 1188 and 21% of 35% of B is equal to 5 more than square of 32. Quantity II – Ratio of male population to female population of a town in the year 2016 was 8 : 9. If in the year 2017 male population increased by 25 % and female population increased by 40% as compared to previous year so total population of town in the year 2017 was 45200. What was the total population of town in 2016. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 7

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12. Quantity I – A man have two bucket which contains some fruits, first bucket contains 5 mango and 7 banana and second bucket contains 4 mango and 5 banana. Also man have a pack of six card with numbering 1 - 6, if a card is taken out and it is even number, then he choose a fruit from first bucket but if card is odd number, then man choose a fruit from second bucket. Find the probability of choosing a mango. Quantity II – In egg tray carried two dozen of eggs. One third of eggs are rotten, if three eggs taken out from the tray at random. What is probability that two eggs out of three eggs picked up is good? (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 13. ‘A’ can complete a work alone in 12 days. A and B together can complete 2 the work in 6 days. C takes 5 more days than B to complete the work 3

alone. C is 50% more efficient than D. Quantity I= ‘x’: A, B and C worked for alternate days starting from A and finish the work in ‘x’ days Quantity II= ‘y’: B and C worked together for 6 days and left and then D completes the remaining work. ‘y’ is total time taken to complete the work. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 14. Quantity I= ‘x’: A shopkeeper mark up an article 40% above its cost price. At the time of sale, he gave 20% discount. By this, shopkeeper earn Rs 168. ‘x’ is the difference between mark up price and cost price. Quantity II= ‘y’ : Satish invested Rs.12000 in scheme ‘A’ at 12% S.I P.A. for 4 years and Rs.25,000 in scheme B at 11% S.I P.A for 2 year. ‘y’ is the difference between the interest earn from both schemes. 8

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(a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 15. ABCD is a rectangle and O is the midpoint of the line EF and diagonal AC. AO=20cm and AB=24cm

Quantity I: Area of shaded region Quantity II: 98 cm² (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 16. Quantity I= ‘x’: Veer distributed Rs.80,000 among his wife, two daughter and three sons. His wife got 300% more than the amount every son got. If 30% of the total amount given to the sons then ‘x’ is the amount every daughter got Quantity II= ‘y’: Yogesh start a business with Rs 10800. After 4 months veer join him with Rs 37800. After 2 months Amit join them with Rs 28800. After 1 year they got total Rs 54880 as profit. ‘y’ is the share of Yogesh in profit. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 9

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17. Quantity I: Profit share of ‘A’ out of total annual profit of Rs. 56,500. A, B and C enter into a partnership. ‘A’ invests Rs. 4000 for the whole year, ‘B’ puts in Rs. 6000 at the first and increasing to Rs. 8000 at the end of 4 months, whilst C puts in at first Rs. 8000 but withdraw Rs. 2000 at the end of 9 months. Quantity II: Amount which when lend on C.I. at 20% interest being compounded annually for 3 years, gives total interest equal to Rs.9100 (a) Quantity I > Quantity II (b) Quantity I ≥ Quantity II (c) Quantity I < Quantity II (d) Quantity I = Quantity II or No relation (e) Quantity I ≤ Quantity II 18. ‘A’, ‘B’ and ‘C’ together can complete a work in 48 days if they work 1 alternatively. ‘A’ is 25% more efficient than ‘B’ who is 33 % less 3 efficient than ‘C’. Quantity I: Difference between days taken by ‘A’ alone and ‘C’ alone to complete the work. Quantity II: Days in which ‘A’, ‘B’ and ‘C’ together can complete half of the work. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 19. X started from a point A towards point B. After 2 hours. Y started from B towards A. By the time X travelled one-fifth of the total distance, Y had also travelled the same. Y’s speed is thrice of that of X’s speed. Quantity I: Difference in time (in hours) taken by X and Y to reach their respective destinations. Quantity II: 12 hours (a) Quantity I = Quantity II or No relation (b) Quantity I ≤ Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I < Quantity II (e) Quantity I > Quantity II 10

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𝑋

20. 5X men can complete a work in days while 2Y men can complete same work in

4𝑌 5

2

days.

Quantity I: Value of ‘Y+20’. Quantity II: Value of ‘1.25X' (a) Quantity I ≥ Quantity II (b) Quantity I = Quantity II or No relation (c) Quantity I > Quantity II (d) Quantity I < Quantity II (e) Quantity I ≤ Quantity II 21. Total surface area of a cylinder is 200% more than that of its curved surface area. Volume of cylinder is 2156 𝑐𝑚3 Quantity I: Volume of cone whose base radius and height is same as that of radius and height of cylinder respectively. Quantity II: Volume of hemisphere whose radius is same as that of radius of cylinder. (a) Quantity I > Quantity II (b) Quantity I = Quantity II or No relation (c) Quantity I ≥ Quantity II (d) Quantity I < Quantity II (e) Quantity I ≤ Quantity II 22. When two ships A & B are travelling in opposite direction crosses each other in 2 second. Ship B is 50% faster than ship A in still water. On a particular day, ship A was ahead of ship B by 9 meters and both of them were travelling in downstream. if Ship B overtook Ship A in 11 seconds. ‘X’ is the speed of ship A and if length of ship B is 54 meters. Quantity I: 16 m/s Quantity II: Value of ‘X’ in m/s (a) Quantity I = Quantity II (b) Quantity I ≥ Quantity II (c) Quantity I ≤ Quantity II (d) Quantity I > Quantity II (e) Quantity I < Quantity II 11

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23. Quantity I= 𝒙

(o is the center of the circle) Quantity II = 𝒚

(a) Quantity I ≥ Quantity II (c) Quantity I ≤ Quantity II (e) Quantity I < Quantity II

(b) Quantity I > Quantity II (d) Quantity I = Quantity II

24. ‘B’ kg of Rs. 36 per/kg rice that mixed with 8 kg of Rs. 42 per/kg rice, in order to earn profit of 10% while selling the mixture at Rs. 44 per/kg. Quantity I: Value of ‘B+20’. Quantity II: Value of ‘6B' 25. Quantity I –Vessel A contains (Q + 36) liter mixture of milk & water in the ratio of 7 : 2, while vessel B contains (2Q + 42) liter mixture of milk & water in the ratio of 2 : 3. If 40% & 46% of mixture from vessel A and B taken out respectively, then remaining mixture in vessel B is 150% of remaining mixture in vessel A. Find the total initial quantity of milk in mixture of vessel A & vessel B together? 12

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Quantity II – Two vessels contains mixture of mango juice and orange juice in the ratio of 5 : 3 and 5 : 4 respectively. If 40 liter mixture from first vessel taken out and mixed in second vessel, so new ratio of mango juice and orange juice in second vessel becomes 25 : 19. Find initial quantity of mixture in second vessel? (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 26. Quantity I – Center of semicircle B is O1 and radius is 28 cm. find area of shaded figure. BO1 is perpendicular to AC

Quantity II – Length of a rectangle is two times of side of square, while ratio between side of square to breadth of rectangle is 8 : 7. If perimeter of rectangle is 28 cm more than perimeter of square, then area of rectangle. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 27. 5y² + 21y + 18 = 0 and 16(𝑥+2) ÷ 4(𝑥+3) = 64(𝑥+3) × 4(𝑥+4) Quantity I: Value of y. Quantity II: Value of x. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 13

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28. Quantity I: Find the cost of B per kg if a product C is made of two ingredients A and B in the proportion of 2 : 5. The price of A is three times that of B. The overall cost of C is Rs. 5.20 per kg including labour charges of 80 paisa per kg. Quantity II: Find the cost of manufacturing each article if a manufacturer estimates that on inspection 12% of the articles he produces will be rejected. He accepts an order to supply 22,000 articles at Rs. 7.50 each. He estimates the profit on his outlay including the manufacturing of rejected articles, to be 20%. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 29. Quantity I: Value of x

Quantity II: Value of y (20)3 +32 ×103 +25×50 54

=𝑦+1

(a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 30. Quantity I: What is probability of getting two red balls when they are drawn from a bag which contains 3 Red balls, 5 blue balls and 4 white balls. Quantity 2: Value of x. 1 2 1 1 3 11 5 +5 −5 = +5 − 2

7

14

𝑥

14

2

(a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 14

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31. Total surface area of cube and sphere are equal. Quantity I: Volume of cube ÷ √𝜋 Quantity II: volume of sphere (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 32. Figure I and II shows two semicircles of radius 14 cm with O and P as their centres respectively.

Quantity I: Area of shaded region in figure I Quantity II: Area of triangle in figure II (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 33. Quantity I – A can complete a task in 24 days and B can do same task in 1

18 days. another two persons C & D complete 58 % of the same task in 3

7 days and efficiency of D is 40% more than that of C. Find in how many days A, B and D will be complete the task together? Quantity II – Ankit is 60% less efficient than Satish and complete a piece of work in 22.5 days. Ankit and Saatish start work together and after 4.5 days both left the work, if veer complete the remaining of work in 4.5 days then find in how many days the whole work will be completed, if all three work together ? (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 15

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34. Quantity I – A shopkeeper has two articles A & B. Marked price of article B is 20% more than marked price of article A, shopkeeper sold article A at 25% discount and article B at 20% discount. He made 20% 2 loss on article A and 6 % 𝑝𝑟𝑜𝑓𝑖𝑡 on article B. If total loss of shopkeeper 3 was Rs. 765, then find marked price of article B? Quantity II – A shopkeeper gives a discount of 24% on marked price of Shirt and cost price of Jeans is 25% more than selling price of Shirt. If shopkeeper sold Jeans at 10 % profit and selling price of Jeans was Rs. 1140 more than selling price of Shirt , then find the cost price of article Jeans ? (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 35. Quantity I – A bag contains four green balls, three red balls and five blue balls. If three balls taken out at random what is probability of at least one ball is green and at least one ball is blue color. Quantity II – There are five red toys and six green toys in a cartoon. What will be the probability of selection of four toys which contains at least two green toys? (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 36. Quantity I – Six years ago ratio between age of Rohit and Prakash was 1 1 7 : 8, while six years hence ratio between 𝑡ℎ of Rohit age and 𝑟𝑑 of 6 3 Prakash age will be 9 : 20. The age of Rohit two years hence will be. Quantity II – Ratio between age of A, B and C is 16 : 9 : 7. Three years hence average of all three age will be 35 years. The age of A two years hence will be? (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 16

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37. A diagram is given below. AC is the diameter of circle. AD || BC and ∠𝐴𝐶𝐷 = 90° ∠𝐴𝐶𝐵 ≤ 45°

Quantity I: -. ∠𝐴𝐷𝐶 Quantity II: - 45° (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 38. Quantity I: ‘x’: There is an interview conducted by HCL. Out of total applicants six are female. Probability of selecting two candidates such 7 that at least one is male is . ‘x’ is total number of males. 8

Quantity II: ‘y’ : A, B and C together start a business. Ratio between investment of A and B is 15 : y while ratio between investment of B & C is 9 : 11. A left business after 8 months of starting of business and B left before 7 months of completion of year. Out of total profit of Rs. 2080, A got Rs. 900. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 39. Quantity I: ‘x’: Two mutually perpendicular chords PQ and EF meet at a point S inside a circle. PS = 12 cm, SQ = 8 cm and FS = 6 cm. ‘x’ is the area of circle. Quantity II: ‘y’: In a circle, two parallel chords are drawn on opposite sides of diameter. Distance between these chords is 14cm. Length of one chord is 16cm while length of another chord is 12cm. ‘y’ is the area of circle 17

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(a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 40. Quantity I: ‘x’ :A dishonest shopkeeper sales his good by weighting 15% less and cheat his wholesaler by giving 15% less amount. ‘x’ is his approximate total profit % if he sells it at cost price. Quantity II: ‘y’ : A dishonest milkman, add 20 litre water in pure milk, after selling half of mixture. He again added 20 liter of water in remaining mixture now milk and water ratio in mixture becomes 4 : 3. ‘y’ is his overall profit % on selling total mixture. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I = Quantity II or No relation (d) Quantity I ≥ Quantity II (e) Quantity I ≤ Quantity II 41. Quantity I: ‘x’ : A and B can do a work in 24 days together, in which 40% of work is done by B. C who is 25% more efficient than A, do the same work alone in ‘x’ days. Quantity II: ‘y’ : P, Q and S work for x, x+1 and x+2 days and their efficiency are in the ratio of x : x+1 : x+2 respectively. If P got 16 Rs for his work, then ‘y’ is the total money earned by all three in Rs.. (Compare Magnitude of x and y) (a) Quantity I > Quantity II (b) Quantity I ≥ Quantity II (c) Quantity I = Quantity II or No relation (d) Quantity I < Quantity II (e) Quantity I ≤ Quantity II 42. Quantity I:’x’ -. Two circles are concentric with center ‘O’. Their radii are 8cm and 10cm respectively. ‘B’ and ‘C’ are the points of contact of two tangents drawn from bigger circle to smaller circle from point ‘A’ lying on bigger circle. ‘x’ is area of quadrilateral ABOC formed in 𝑐𝑚2 18

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Quantity II: - ‘y’ - The lengths of two parallel sides of a trapezium are 6 cm and 8 cm. If the height of the trapezium be 6 cm, then its area is ‘y’ 𝑐𝑚2 (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 43. Quantity I: ‘x’ -. Kishan invested Rs. 20,000 in a bank offering 22% p.a. at Simple interest. After 2 years he withdraws his money and invested in another bank which is offering ‘x%’ p.a. at compound interest. After 3 years, interest earned by him is Rs.1350 less than amount invested by him in this bank. 5

Quantity II: - ‘y’ – Bhavya sell a diary at Marked price and earns 85 % 7

profit while if he gives 'y%’ discount on Marked price then he will earn ‘y%’ profit. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 44. Quantity I —‘x’ : B alone can do the work in ‘x’ days. A can complete a work in 5 more days than B while A does the same work in 9 more days than C. If A and B can complete the whole work in same time as time taken by C alone to do the whole work. Quantity II —‘y’ : ‘y’ is the days taken by 8 men and 14 women to reap 7 12

𝑝𝑎𝑟𝑡 of 360-hectare land by working 7 hrs per day if 6 men and 10

women can reap

5 12

part of the land in 15 days by working 6 hrs per day.

It is also given that work of 2 men is equal to that of 3 women. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 19

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45. Quantity I — ‘x’ : ‘x’ is the difference between the speeds of P and Q. Distance between A and B are 60 km. P and Q start from A at same time & meet 1st time at a place 12 km from B. They return to A immediately after reaching B. The speed of slower person is 48 km/hr. Quantity II —‘y’ : ‘y’ is the average speed of train if a distance of 600 km is to be covered in 2 parts. In 1st phase 120 km is travelled by train and rest by car and it took total of 8 hrs, but if 200 km is covered by train and rest by car it takes 20 min more. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 46. Quantity I: ‘x’ : ‘𝒙′ is the number of days taken by A alone to complete a work. A and B together can complete the same work in 16 days while A 1

and C can complete the same work in 15 days. Ratio of efficiency of B 3

and C is 3 : 4. Quantity II: ′𝒚′: P, Q and R takes ‘y’ days to complete a work. Ratio of efficiency of P, Q and R are 2 : 3 : 4 respectively and P and Q together takes 18 days to complete that work. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 47. Quantity I: ‘x’ : A article is sold at 100 Rs. profit whose cost price is ‘Rs.𝒙’. If article is sold at 20% more than its S.P. then S.P becomes 2𝑥. Quantity II: ′𝒚′ : Article is sold at y% profit whose M.P. is 400% above its C.P. and discount given on M.P. is 60%. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 20

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48. Quantity I: ‘x’: Sum of ‘𝒙′ consecutive number is 4040. First number is 11. Quantity II: ′𝒚′: Sum of ′𝒚′ consecutive even numbers is 7120. First number is 10. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 49. Quantity I: ‘x’: A man invested 𝑹𝒔. ′𝒙′ in two scheme which offers 40% S.I. for 2 years and 44% S.I. for 5 years respectively. Difference between interest earned from both scheme is 18,900. Quantity II: ′𝒚′: Interest earned in a scheme is ′𝑦′ which offers 12% rate of interest for 3 years on a sum. 30% of the sum is 12000. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 50. The largest sphere that can fit in a cube of edge 7 cm. Quantity I: ‘x’: ‘x’ is the volume of vacant space left over after putting the sphere.(𝑖𝑛 𝑐𝑚3 ) Quantity II: ′𝒚′: ’y’ is the surface area of the sphere. (in 𝑐𝑚2 ) (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation PART B Directions (Q1-5): In each of these questions, two equations I. and II. are given. You have to solve both the equations and give answer 1. I. 3𝑥+5 . 92𝑥−4 = 95𝑥−14 II. 2𝑦 2 − 15𝑦 − 28 = 3𝑦 2 − 23𝑦 − 13 (a) x < y (b) x ≥ y (c) No relation can be established between x and y. (d) x ≤ y (e) x > y 21

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25

2. I. II.

x² 40 y²

−

15 x

+ 1=

+2=0 13 y

(a) x < y (b) x ≥ y (c) No relation can be established between x and y. (d) x ≤ y (e) x > y 3. I. 2(y+4) . 12(2y+4) = 9(y+2) . 16(y+4) II. x 2 – 10x + 24 = 0 (a) x ≤ y (b) x > y (c) No relation can be established between x and y. (d) x < y (e) x ≥ y 4. I.

48 x2

II.

−

45 y²

14 x 1

+1=0

+ =2 y

(a) No relation can be established between x and y. (b) x ≤ y (c) x < y (d) x > y (e) x ≥ y 5. I. (𝑥 − 4)2 = 9 II. (2𝑦 + 3)2 = 25 (a) No relation can be established between x and y. (b) x < y (c) x ≤ y (d) x ≥ y (e) x > y 22

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Directions (6-10): In each of these questions, two equations I. and II. are given. You have to solve both the equations and give answer (a) if x>y (b) if x≥y (c) if x
II.

2y² – 13y + 20 = 0

7. I. x² – 7x + 12 = 0

II.

2y² – 11y + 15 = 0

8. I. 2x² + 11x + 15 = 0

II.

2y² + 9y + 10 =0

9. I. 3x² + 7x – 40 = 0

II.

5y² – 29y + 42 = 0

10. I. 3x² – 23x + 42 = 0

II.

2y² – 19y + 45 = 0

Directions (11-15): In each of these questions, two equations I. and II. are given. You have to solve both the equations and give answer (a) if x>y (b) if x≥y (c) if x
II.

𝑦 2 + 5𝑦 + 6 = 0

12. I. 2𝑥 2 − 𝑥 − 15 = 0

II.

3𝑦 2 − 23𝑦 + 42 = 0

13. I. 𝑥 2 − 15𝑥 + 54 = 0

II.

𝑦 2 + 15𝑦 − 54 = 0

14. I. 𝑥 2 + 14𝑥 + 40 = 0

II.

𝑦 2 − 5𝑦 − 24 = 0

15. I. 𝑥 2 − 225 = 0

II. 𝑥 2 + 𝑦 2 = 306

23

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Directions (16-20): In each of these questions, two equations I. and II. are given. You have to solve both the equations and give answer (a) if x>y (b) if x≥y (c) if x
II.

3y² – 19y + 28 = 0

17. I. x² – 13x + 36 =0

II.

3y² – 19y + 30 = 0

18. I. x³– 120 = 96

II.

y–

19. I. x² – 2x – 48 = 0

II.

y² – 15y + 54 = 0

20. I. 2x² + 25x + 72 = 0

II.

3y² + 22y + 40 = 0

24

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

Solutions 1. (a); Suppose B takes ‘x’ days, Then C will take ‘3x’ days 1 1 1 + = 3𝑥

𝑥

3𝑦

30

22.5

Or, x = 30 Now, Let A and B together take ‘2y’ days, Then A will take ‘3y’ days, 1 1 1 + = 2𝑦

Or, y = 5 Hence, A will take 3y = 3 × 5 = 15 days. Quantity I = 15 90 Quantity II = = 90/7 6+1

Quantity I > Quantity II 2. (b); Quantity I→

117×3+2

Quantity II→ 𝑥 + 𝑥=

100 125

× 600 = 706 Rs

300 25𝑥 100

= 1000

× 1000 = 800 Rs

∴ Quantity I < Quantity II 3. (b); Quantity I: Volume of cube left 7 2

3

=7 −𝜋( ) ×7 = 343 −

2 22 7

×

49×7 4

= 343 − 269.5 = 73.5 cm3 Quantity II: Surface area of cube left 2

7 2

7

2

2

= 6 × 7 − 2. 𝜋 ( ) + 2𝜋 ( ) 7 = 294 − 77 + 154 = 371 cm2 Quantity II > Quantity I 25

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4. (a); Quantity I – 2πr + 2 (ℓ + b) = 220 cm πr² = 1386 sq. cm r2 =

1386×7 22

r = 21 cm Length of rectangle 4

= 21 × = 28 cm 2×

22 7

3

× 21 + 2(28 + b) = 220

132 + 56 + 2b = 220 b=

32 2

= 16 cm

Area of rectangle = (28 × 16) = 448 cm Quantity II–2πr = 132 r=

132×7 2×22

r = 21 cm πr² – a² = 710 22 7

× 21 × 21 – a2 = 710

a² = 1386 – 710 a² = 676 cm a = 26 cm Perimeter of square = 4 × 26 = 104 cm Quantity I > Quantity II 5. (b); Quantity I— Let Veer take 3𝑥 days and Sameer take 𝑥 days. Efficiency of Veer and Sameer be 𝑥 unit/day and 3𝑥 units/day respectively Total work = 3𝑥 × 18 + 𝑥 × 18 = 72x units If both do with double efficiency Then, 72𝑥

= (3𝑥×2+2𝑥) = 9 days 26

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Quantity II—

When all three opened together Total work in one day = (6 + 2 – 3) = 5 units 48 3 Required time = = 9 days 5

5

Quantity I < Quantity II 6. (a); Quantity I -

R efficiency = 3 ×

150 100

= 4.5 w⁄d

3 days work = (P + Q + R) = 2 + 3 + 4.5 = 9.5 work Required days = 9.5 × 15 = 142.5 work Remaining work = 144 – 142.5 = 1.5 1.5 3 Next day by P = = 2

4 3

3

4

4

Total days = 15 × 3 + = 45 days Quantity II. - A : B = 100 : 75 = 4 : 3 Total work = 4 × 48 = 192 192 B alone = = 64 days 3

C alone = 64 – 40 = 24 days 192 C efficiency = = 8 w⁄d 24

(A + B + C)together =

192

= (4+3+8)

192 15

4

= 12 days 5

So, Quantity I > Quantity II 27

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7. (b); Quantity I - No. of selection of 3 mango out of one dozen mangoes = ¹²C₃ = 220 No. of selection of 2 bad mango out of 4 bad mangoes = ⁴C2 = 6 No. of selection of a good mango from 8 good mangoes = 8 8×6 12 Required probability = = 220

Quantity II - Face card = 12 Joker = 2 14 7 Probability = = 54

55

27

So, Quantity I < Quantity II 8. (e); Quantity I - Let length of train be L meter ATQ— 5 Speed = 72 × = 20 m/sec 20 =

18

L 14

L = 280 meter Let length of platform be P meter P+280 20 = 32

P = 640 – 280 P = 360 meter 𝐐𝐮𝐚𝐧𝐭𝐢𝐭𝐲 𝐈𝐈 – Let’s length of train is X meter and length of bridge and platform are 2L and L meter respectively 5 Speed of train = 108 × = 30 m⁄s 18

When train passed the bridge Distance covered = X + 2L X + 2L = 540 …(i) When train passes the platform Distance covered = X + L X + L = 450 …(ii) From (i) and (ii) L = 90 meter X = 360 meter So, Quantity I = Quantity II 28

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9. (b);

Let P is faster than Q Then P covers 72 km distance in the same time as Q covers 48 km distance Ratio of the speed = 72 : 48 = 3 : 2 48 ∴ Speed of faster train i.e., P = × 3 = 72 km/hr 2

Quantity 1→ Difference between P and Q = 72 – 48 = 24 km/hr. Let speed of train = T km/hr Let speed of car = C km/hr 120 480 ∴ + = 8 ………..(i) 𝑇 200 𝑇

+

𝐶 400 𝐶

1

= 8 ………(ii) 3

On solving (i) and (ii) T = 60 km/hr ∴ Quantity I < Quantity II 6

10. (b); Quantity I → = 3 2

1

And, 𝜋𝑟 2 ℎ = 16𝜋 3

ℎ=3 ∴ 𝑦 = √42 + 32 = √16 + 9 𝑦=5 ∴ Quantity I < Quantity II 11. (a); Quantity I— 𝐴 × A=

25

100 1188×100×100×100

×

48 100

×

55 100

= 1188

25×48×55

A = 18000 21 35 𝐵× × = (32)2 + 5 𝐵× 29

100 21 100

×

100 25 100

= 1024 + 5 Adda247 Publications

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𝐵=

1029×100×100 21×25

B = 19600 (A + B) = 18000 + 19600 = 37600 Quantity II— Let male and female population of town in 2016 be 8x and 9x respectively 125

8𝑥 ×

+ 9𝑥 ×

100 63𝑥

10𝑥 +

5

140 100

= 45200

= 45200

113𝑥 = 226000 𝑥 = 2000 Total population of town in 2016 = 16000 + 18000 = 34000 Quantity I > Quantity II 12. (a); Quantity-I : Probability of choosing Mango 1

5

2

12

= ×

1

4

31

2

9

72

+ × =

Quantity-II : Total no. of cases to choose 3 egg = ²⁴C₃ Favourable cases = ¹⁶C₂ × ⁸C₁ Probability =

16C × 8C 2 1 24C 3

=

15 253

Quantity I > Quantity II 13. (e); A can complete work in= 12 days ‘B’ 1 day work is =

3 20

−

1 12

=

9−5

B can complete the work in =

60 60 4

=

4 60

= 15 days

C can complete the work in ‘20’ days D can complete the work in

20×150 100

= 30 days

Quantity I: ‘A + B + C three days’ work =

1 12

+

1 15

+

1 20

=

5+4+3 60

=

12 60

=

1 5

‘A + B + C’ can complete the work in total ’15 days 30

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Quantity II: Remaining work for D =1− 3 10

6 15

−

6 20

=

60−24−18 60

=

3 10

of work is completed by D in

30×3 10

= 9days Total time = 6 + 9 = 15 days Quantity I = Quantity II 14. (a); Quantity I: let C.P = 100 Mark Up price = 140 Selling price = 140 ×

80 100

= 112

ATQ, 12 → 168 x = 40 → 560 x = 560 Quantity II: Interest from scheme A =

12000×12×4 100

= 5760

Interest from scheme B =

25000×11×2 100

= 5500

y= 5760 – 5500 = 260 x>y Quantity I > Quantity II 15. (b); Quantity I: OF=√202 − 12²=√400 − 144 = √256 = 16 Area of shaded region Area of ∆AFE – Area of ∆ AFO 1

1

2

2

= × 12 × 32 − × 12 × 16 = 192 – 96 = 96 cm² Quantity II > Quantity I 31

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16. (a); Quantity I: Total amount sons got =

30 100

× 80000 = 24000

Each son got =

24000 3

Wife got = 8000 ×

= 8000

400 100

= 32000

Total amount daughters got 80,000 – 24000 – 32000 = 24000 Each daughter got =

24000 2

= 12000

x = 12,000 Quantity II: Ratio of investment or profit = 10800 × 12 : 37800 × 8 : 28800 × 6 =3:7:4 Share of Yogesh in profit =

54880 14

×3

y = 11760 x>y Quantity I > Quantity II 17. (c); ATQ, Quantity I: Ratio of profit share A : B : C (4000×12) (6000×4)+(8000×8) (8000×9)+(6000×3) 24 : 44 : 45 Share of A =

24 113

× 56,500 = Rs. 12,000

Quantity II: 9100 + P = P (1 +

20 3 100

)

⇒ 9100 + P = 1.728P or, P = Rs. 12,500 𝐐𝐮𝐚𝐧𝐭𝐢𝐭𝐲 𝐈𝐈 > 𝐐𝐮𝐚𝐧𝐭𝐢𝐭𝐲 𝐈 32

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18. (e); ‘A’, ’B’ and ‘C’ alternatively can complete a work in 48 days ⇒ ‘A’, ‘B’ and ‘C’ together can complete same work in 16 days Ratio between efficiency of A, B and C is 5 : 4 : 6 Ratio between days taken by A, B and C alone to complete the same work is 12 : 15 : 10 Let A, B and C alone can complete work in 12x, 15x and 10x days respectively. ⇒

1 12𝑥

+

1 15𝑥

+

1 10𝑥

=

1 16

⇒𝑥=4

Quantity I: Required difference = (12 − 10)4 = 8 𝑑𝑎𝑦𝑠 Quantity II: A, B and C together can complete same work in 16 days, so they can complete half work in 8 days. Quantity I = Quantity II 19. (d); Let the speed of X be x kmph. Distance travelled by X in 2 hours = 2x km. Suppose X takes ‘t’ hours to travel Y would take (t–2) hours to travel

1th

of the distance AB.

th

of the distance AB.

5 1 5

As Y’s speed is thrice that of X’s speed. 𝑡−2 𝑡 1𝑡ℎ 5

=

1 3

⇒ t=3

of the distance AB = 3x km.

AB =15x km Time taken by x to cover 15x km = Time taken by Y to cover 15x km =

15𝑥 𝑥 15𝑥

= 15 hours

3𝑥

= 5 ℎ𝑜𝑢𝑟𝑠. ∴ Difference in the time = 10 hours. Quantity I : Difference in the time = 10 hours. Quantity II : 12 hours Quantity I < Quantity II 33

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𝑋

4𝑌

2

5

20. (c); Total work = 5𝑋 × = 2𝑌 × ⇒

𝑋2 𝑌2

=

16 25

⇒

𝑋 𝑌

=

4 5

Let X = 4a and Y = 5a Quantity I: Y+20 = 5a+20 Quantity II: 1.25X = 5a Quantity I > Quantity II 21. (b); T.S.A of cylinder = 2𝜋𝑟(𝑟 + ℎ) C.S.A of cylinder = 2𝜋𝑟 2 ATQ, 3 1

=

2𝜋𝑟(𝑟+ℎ) 2𝜋𝑟 2

⇒ ℎ = 2𝑟 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝜋𝑟 2 ℎ = 2156 ⇒ 𝑟 = 7𝑐𝑚, ℎ = 14𝑐𝑚 1

2156

3

3

Quantity I: Volume of cone = 𝜋𝑟 2 ℎ = Quantity II: Volume of hemisphere 2

2156

3

3

= 𝜋𝑟 3 =

Quantity I = Quantity II 22. (e); Let speed of ship A in still water is 2x m/s , and its length is ℓ m And speed of ship B is 3x m/s and its length is 54 and speed of water is Y m/s. ATQ, when both of them are travelling in opposite direction. Downstream speed of ship A = (2x + Y) m/s [assume ship A is travelling in downstream and ship B in upstream] Upstream speed of ship B = (3x – Y) m/s Their relative speed = 2x + Y + 3x – Y = 5x m/s ATQ, 54+ℓ 5x

=2

…(i)

Similarly when both are travelling in downstream their relative speed is = 3x + Y – 2x –Y= X m/s 34

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ATQ, 54+9+ℓ 𝑥

= 11

63 + ℓ = 11x …(ii) From (i) ℓ = 10x – 54 put this value in eqn. (ii) = 63 + 10x – 54 = 11x x = 9 m/s Hence speed of ship A in still water = 2 × 9 = 18 m/s Quantity I = 16 m/s Quantity II = 18 m/s Quantity II > Quantity I 23. (b); 𝑥 = 180 − (90 + 30) = 90 − 30 = 60° 𝑦 = 180 − (60 + 40) (angles subtended by same arc in the same segment are equal) = 80° ∴ Quantity I < Quantity II 24. (e); Selling mixture at Rs. 44/kg with 10% profit means, the actual price of mixture is Rs. 40/kg Let B kg of Rs. 36/kg are mixed Then 36𝐵+8×42 𝐵+8

= 40

36B + 336 = 40B + 320 4B = 16 B=4 Quantity I = B + 20 = 24 Quantity I = 𝟔 × 𝑩 = 𝟐𝟒 Quantity I = Quantity II 35

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25. (b); Quantity I – ATQ – 60 150 (Q + 36) × × = [(2Q + 42) × 100

100

54 100

]

3(60Q + 2160) = 2 (108Q + 2268) 180Q + 6480 = 216Q +4536 36Q = 1944 Q = 54 l Total initial quantity of milk in mixture of vessel A & vessel B 7 2 = (54 + 36) × + (2 × 54 + 42) × = 70 + 60 = 130 l 9

5

Quantity II – ATQ – Let mixture of mango juice and orange juice in second vessel be 5y liter and 4y respectively 5

5𝑦+40×8 3 4𝑦×40× 8

=

25 19

⇒

5𝑦+25 4𝑦+15

=

25 19

19y + 95 = 20y + 75 y = 20 l Initial quantity of mixture in second vessel = 20 × 9 = 180 𝑙 Quantity I < Quantity II 26. (e); Quantity I–Area of semicircle – area of triangle 22 28×28 1 = × − × 56 × 28 = 1232 – 784 = 448 cm² 7

2

2

Quantity II – Ratio of Length, breadth of a rectangle & side of square = 16 : 7 : 8 Let Length, breadth of a rectangle & side of square be 16x cm, 7x cm & 8x cm respectively ATQ – 2(16x + 7x) – 4 × 8𝑥 = 28 𝑐𝑚 46x – 32𝑥 = 28 x = 2 cm Area of rectangle = (16× 2 ) × (2 × 7) = 448 cm2 Quantity I = Quantity II 36

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

27. (c); Quantity I: 5y² + 21y + 18 = 0 5y² + 5y + 15y + 18 = 0 Y (5y + 6) +3 (5y + 6) = 0 (y + 3) (5y + 6) = 0 –6 y = –3 or 3

Quantity II: 16(x+2)÷4(x+3) = 64(x+3)×4(x+1) 24x+8–2x–6 = 26x+18+2x+2 8x + 20 = 2x + 2 ∴ x = –3. ∴ Quantity I ≥ Quantity II 28. (b); Quantity I: Let the price of B per kg be Rs. X. Then, the price of A per kg = Rs. 3X 1kg of C contains 2/7 kg of A and 5/7 kg of B Price of 1 kg of C = (2/7) × 3X + (5/7)X = (11/7)X By the given condition, 11X/7 = 5.20 – 0.80 = Rs. 4.40 ⇒ X = 4.40 × (7/11) = Rs. 2.80 Hence the price of B per kg = Rs. 2.80. Quantity II: If 100 articles are manufactured then 12 will be rejected Total selling price of 88 articles = 88 × 75 660×100 Total cost price = = 550 120

cost of manufacturing per article = 5.5 Rs Quantity II > Quantity I 29. (b); Quantity I: ∠CAD = 90° – ∠ACD = 62 (x) = ∠CBD = 90° – ∠CAD x = 28° Quantity II:

8000+9000+1250 625

=𝑦+1

y = 28.2 Quantity II > Quantity I 37

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams 3C

30. (b); Quantity I: 12 2 = C2

1 22

1

1

2

1

2

2

7

14

Quantity II: 5 ( + + − 1

1

𝑥

5

5= ⇒𝑥 =

−

3 14

)=

1 𝑥

Quantity II > Quantity I 31. (b); Let side of cube and radius of sphere be a and r respectively. ⇒ 6a² = 4πr² 2

a=√ π r 3

Quantity I: Volume of cube =a×a×a 2

2

= π × √ π r 3 ÷ √𝜋 3

3

3 2

=(2/3 ) π r 3 Quantity II : volume of sphere

4 3

πr 3

∴ Quantity II > Quantity I 32. (a); Quantity I: Area of shaded region Area of circle – Area of triangle Base of triangle = 2 × 14 = 28 =

22 7

×

14×14 2

1

– × 28 × 14 2

2

⇒ 112 cm

Quantity II: Radius = 14 1

Area of triangle = × 14 × 14 = 98 cm² 2

Quantity I > Quantity II 38

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33. (a); Quantity I – Total work = 72 units (Lcm of days taken by A & B) 72 Efficiency of A = = 3 𝑢𝑛𝑖𝑡𝑠/𝑑𝑎𝑦 24 72

Efficiency of B = = 4 𝑢𝑛𝑖𝑡𝑠/𝑑𝑎𝑦 18 Let efficiency of C = x unit/day 175 1 7x + 7(1.4x) = 72 × = 42 𝑢𝑛𝑖𝑡𝑠 3 100 x = 2.5 units Efficiency of D = 1.4 × 2.5 = 3.5 𝑢𝑛𝑖𝑡𝑠 72 6 (A + B + C) together = = 6 𝑑𝑎𝑦𝑠 (3+4+3.5)

7

Quantity II – Ratio of efficiency of Satish : Ankit = 100x : 40x = 5x : 2x Total work = 22.5 × 2x = 45x units ATQ, (Ankit + Satish) × 4.5 = 4.5 (5x + 2x) = 31.5x work Remaining work = 45x– 31.5 x = 13.5x 13.5 Veer efficiency = = 3 units⁄day 4.5

(Ankit + Satish + Veer) =

45x (5x+2x+3x)

= 4.5 days So, Quantity I > Quantity II 34. (a); Quantity I – Let marked price of article A = 100x Marked price of article B = 120x 3 S. P. of article A = 100x × = 75x 4

4

S. P. of article B = 120 × = 96x 5

5

C. P. of article A = 75 × = 93.75x 4

15

C. P. of article B = 96x × = 90x 16 ATQ— (93.75x + 90x) – (75x + 96x) = 765 183.75x – 171x = 765 12.75x = 765 x = 60 M.P. of article B = 60 × 120 = 7200 Rs. 39

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Quantity II – MP of shirt = 100x Rs 76 SP of shirt = 100 × = 76𝑥 Rs 100

5

Cost price of jeans = 76𝑥 × = 95𝑥 Rs 4

Selling price of jeans = 95𝑥 ×

110 100

= 104.5x Rs ATQ – 104.5x - 76x = 1140 Rs 28.5x = 1140 Rs. X = 40 Rs. Cost price of jeans = 95×40 = 3800 Rs. So, Quantity I > Quantity II 35. (b); Quantity I – Required cases = (1 green, 1 red, I blue) or ( 2 green , 1 blue) or (1 green , 2 blue) =

4C × 3C × 5C 1 1 1 12C 3

+

4C × 5C 2 1 12C 3

+

4C × 5C 1 2 12C 3

=

3 11

+

3 22

+

2 11

=

13 22

Quantity II – Favorable case = (2G, 2R) or (3G, 1R) or 4G ∴ Probability = =

6C × 5C 2 2 11C 4

15×10 330

+

+

6 C × 5C 3 1 11C 4

20×5 330

+

+

15

330

6C 4 11C 4

=

265 330

=

So, Quantity I < Quantity II

53 66

36. (e); Quantity I – Let age of Rohit and Prakash six years ago be 7x years and 8x years respectively ATQ – 𝟕𝒙+𝟏𝟐 𝟔 𝟖𝒙+𝟏𝟐 𝟑

=

𝟗 𝟐𝟎

140x + 240 = 144x + 216 x = 6 years Age of Rohit two years hence will be = (7× 6 + 8) = 50 𝑦𝑒𝑎𝑟𝑠 40

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Quantity II – Let age of A, B and C be 16y, 9y and 7y respectively. ATQ – (16y + 9y + 7y) = 35 × 3 − 3 × 3 32y = 96 y = 3 years Age of A two years hence = 3 × 16 + 2 = 50 years So, Quantity I = Quantity II 37. (c); ∠𝐴𝐵𝐶 = 90° (Semicircle property) ∠𝐶𝐴𝐵 + ∠𝐴𝐶𝐵 + ∠𝐴𝐵𝐶 = 180° ∠𝐶𝐴𝐵 + ∠𝐴𝐶𝐵 = 90° As ∠𝐴𝐶𝐵 ≤ 45° So, ∠𝐶𝐴𝐵 ≥ 45° ∠𝐴𝐶𝐷 + ∠𝐶𝐴𝐷 + ∠𝐴𝐷𝐶 = 180° ∠𝐶𝐴𝐷 + ∠𝐴𝐷𝐶 = 90° But ∠𝐴𝐶𝐵 = ∠𝐶𝐴𝐷 (AD || BC) ∠𝐴𝐶𝐵 + ∠𝐴𝐷𝐶 = 90° As ∠𝐴𝐶𝐵 ≤ 45° So, ∠𝐴𝐷𝐶 ≥ 45° Quantity I ≥ Quantity II 38. (a); Quantity I: Let, no. of males = x ATQ, 2×

x x+6

×

12x+x2 –x

6

x(x–1)

x+5

= (x+6)(x+5)

+ (x+6)(x+5) =

7 8

7 8

88x + 8x² = 7 (x² + 5x + 6x + 30) x² + 11x – 210 = 0 x² + 21x – 10x – 210 = 0 x (x + 21) – 10 (x + 21) = 0 x = 10, –21 41

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Quantity II: Ratio between investment of A, B and C is 135 : 9y : 11y Profit sharing of A, B and C is 135 × 8 : 9𝑦 × 5 : 11𝑦 × 12 44𝑦 = 360 : 15𝑦 : ATQ, 900 360 = 2080

360+59𝑦

⇒ (360 + 59y) = 104 × 8 ⇒ 59y = 472 ⇒y=8 Quantity I > Quantity II 39. (a); Quantity I:

As, PQ and EF are two line that intersect at S. So, PS × SQ = ES × SF 12 × 8 = ES × 6 ES = 16 cm From center O draw OB ⊥ PQ and OA ⊥ EF So, PB = PQ =

(12+8) 2

= 10 cm

BS = OA = 2 cm EF = 16 + 6 = 22 cm 22 EA = = 11 cm 2

In triangle ∆𝑂𝐴𝐸 OA² + EA² = OE² 2² + 11² = OE² OE² = 125 42

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OE = √125 = 5√5 cm 2

Area of circle = πr²= 𝜋(5√5) = 125 ×

22 7

= 392.86 cm ²

Quantity II:

Let AB and CD are the chord drawn on opposite sides of diameter. Length of CD is 16 cm and length of AB is 12 cm. ATQ FC = AE =

𝐶𝐷 2 𝐴𝐵 2

= =

16 2 12 2

= 8𝑐𝑚 = 6𝑐𝑚

EF = 14 cm But AO = OC = radius of circle 𝑂𝐴2 = 𝐴𝐸 2 + 𝑂𝐸 2 = 𝑂𝐶 2 = 𝐶𝐹 2 + 𝑂𝐹 2 Let OE and OF be ‘a’ and ‘b’ respectively 𝑎2 − 𝑏2 = 82 − 62 = 28 And 𝑎 + 𝑏 = 14 So, 𝑎 = 8, 𝑏 = 6 Radius of circle = √62 + 82 = √36 + 64 = √100 = 10𝑐𝑚 Area of circle = 𝜋(10)2 = 314.28 𝑐𝑚2 Quantity I > Quantity II 43

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

40. (b); Quantity I: Let cost price is 1000 of 1000 gm. he pays to whole seller ⇒ 850 Rs. for 1000 gm. Effective cost price = 0.85 Rs. per gm. Now he sells 850 gm. instead of 1000 gm. Effective cost price of 850 gm. = 0.85 × 850 = 722.5 1000 –722.5 Profit(x) = × 100 = 38.40 ≈ 38% 722.5

Quantity II: Let milkman have → q litre of milk Initially he add → 20 litre of water 𝑞 Sale half Remaining mixture = + 10 2

Again add 20 litre water So,

𝑞 2

10+20

=

4 3

Total profit(y) =

40 80

× 100 = 50%

Quantity II > Quantity I 41. (d); Quantity I: If B do 40% of work thus A do 60% of work Ratio of efficiency of A and B = 3 : 2 Total work = 24 × (3 + 2) = 120 units 3×125 15 Efficiency of C = = units/day 100 4 120 × 4

Time taken by C =

15

= 32 days

Quantity II: P work for → x day Efficiency → x So, x × x ⇒ 16 Rs. x ⇒ 4 Rs. Money earned by P, Q and R ⇒ x² + (x + 1)² + (x + 2) ² = 4² + 5² + 6² ⇒ 77 Rs. Quantity II > Quantity I 44

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42. (a); Quantity I

OB = 8cm and OA = 10cm As, AB and AC are tangents on smaller circle ⇒ ∠𝐴𝐵𝑂 = ∠𝐴𝐶𝑂 = 90° ⇒ 𝐴𝐵 = √102 − 82 = 6 Area of quadrilateral ABOC = 2 × 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑎𝑖𝑛𝑔𝑙𝑒 𝐴𝐵𝑂 Area of triangle ABO 1

1

2

2

= × 𝑂𝐵 × 𝐴𝐵 = × 8 × 6 = 24𝑐𝑚2 Area of quadrilateral ABOC = 2 × 24 = 48𝑐𝑚2 Quantity II: Area of trapezium 1

= (sum of parallel sides) × perpendicular distance 2 1

= (6 + 8) × 6 = 42𝑐𝑚2 2

Quantity I > Quantity II 43. (b); Quantity I Interest earned by Kishan after 2 years =

20,000×22×2 100

= 8800

Amount invested by KIshan in another bank = 20,000 + 8800 = 28,800 ATQ, 45

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⇒ 28,800 [1 + −1350 ⇒ 28,800 [1 + ⇒ [1 + ⇒ [1 + ⇒ [1 +

𝑥 100 𝑥 100 𝑥 100

3

] =

3

𝑥 100

] = 28,800 + 28,800 3

𝑥

] = 56,250

100 56,250 28,800

3

] =[

125 64

]

5

] =[ ] 4

⇒ 𝑥 = 25% 5

Quantity II: Marked price is 185 % 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑠𝑡 𝑝𝑟𝑖𝑐𝑒 𝑜𝑓 𝑎𝑟𝑡𝑖𝑐𝑙𝑒 7

Let Cost price = 100x and profit % is y% after y% discount ⇒ 100𝑥 + 𝑥𝑦 = ⇒ 100𝑥 + 𝑥𝑦 =

1300𝑥 700 1300𝑥 700

(100 − 𝑦) (100 − 𝑦)

⇒ 700 + 7𝑦 = 1300 − 13𝑦 ⇒ 20𝑦 = 600 ⇒ 𝑦 = 30% Quantity II > Quantity I 44. (b);

A B C Time 𝑥 + 5 𝑥 𝑥 − 4 1 1 1 ∴ + = 𝑥+5

𝑥

𝑥−4

𝑥 = 10 ∴ Quantity I → 10 days Given 2M = 3W ∴

(8M+14W)×x×7

𝑥=

7 ×360 12

171 13

=

⇒ = 13

(6M+10W)×15×6 5 ×360 12

2 13

Quantity II → 13

2 13

∴ Quantity I < Quantity II 46

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45. (b);

Let P is faster than Q Then P covers 72 km distance in the same time as Q covers 48 km distance Ratio of the speed = 72 : 48 = 3 : 2 ∴ Speed of P =

48 2

× 3 = 72 km/hr

Quantity I→ Difference between P and Q = 72 – 48 = 24 km/hr. Let speed of train = T km/hr Let speed of car = C km/hr ∴

120

𝑇 200 𝑇

+

+

480

𝐶 400 𝐶

=8

=8

………..(i)

1

………(ii)

3

On solving (i) and (ii) T = 60 km/hr ∴ Quantity I < Quantity II 46. (a); 𝑥: Let efficiency of A = 𝑎 and efficiency of B and C be 3𝑦 and 4𝑦 Now 1

(𝑎 + 3𝑦) × 16 = 15 (𝑎 + 4𝑦) 3

𝑎 = 20𝑦 Time taken by A =

(20𝑦+3𝑦)×16 20𝑦

= 18.4 days

𝑦: Let efficiency of P, Q and R be 2𝑎, 3𝑎, 4𝑎 respectively Total work = (2𝑎 + 3𝑎) × 18 Required time =

5𝑎×18 9𝑎

= 10 days

Quantity I > Quantity II 47

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47. (a); 𝑥: Cost price = Rs. 𝑥 S.P. = 𝑥 + 100 S.P. after increasing it by 20% 120 (𝑥 + 100) 2𝑥 = 100

𝑥 = 𝑅𝑠. 150 𝑦: Let C.P. = Rs.100 So M.P. = Rs.500 500×40 S.P. = = 200 100

Profit % =

200–100 100

× 100 = 100%

Quantity I > Quantity II 48. (e); 𝑥 : 𝑎 = 11 𝑑=1 𝑛 𝑆𝑛 = [2𝑎 + (𝑛– 1)𝑑] 2

𝑥

4040 = [2 × 11 + (𝑥– 1)1] 2

𝑥 = 80 𝑦: 𝑎 = 10 𝑑=2 So, 𝑦 7120 = [2 × 10 + (𝑦– 1)2] 2

𝑦 = 80 Quantity I = Quantity II 49. (b); 𝑥 : Amount Invested = 𝑥 Time in first scheme = 2 years Rate = 40% 𝑥×2×40 4𝑥 S.I. = = 100

Time in 48

2nd

5

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Rate = 44% 𝑥×5×44 11𝑥 S.I. = = 100

5 11𝑥

4𝑥

Difference = – = 18,900 5 5 𝑥 = Rs. 13,500 𝑦: 𝑦 = Interest earned 12000 P= × 100 = 40,000 30

40000×3×12

Interest = = 14,400 100 Quantity II > Quantity I 50. (a); 𝑥 : Edge of cube = 7 cm 7 So, largest sphere radius = cm 2 Volume of vacant space 4 22 7 7 7 490 =7×7×7− × × × × = cm³ 3 7 2 2 2 3 𝑦 : Surface area of sphere = 4πr² 22 7 7 = 4 × × × = 154 𝑐𝑚2 7 2 2 Quantity I > Quantity II PART B 1. (b); I. 3𝑥+5 . 92𝑥−4 = 95𝑥−14 ⇒ 3𝑥+5 . 34𝑥−8 = 310𝑥−28 ⇒ 3𝑥+5+4𝑥−8 = 310𝑥−28 ⇒ 35𝑥−3 = 310𝑥−28 ⇒ 5𝑥 − 3 = 10𝑥 − 28 ⇒ 5𝑥 = 25 ⇒ 𝑥=5 II. 2𝑦 2 − 15𝑦 − 28 = 3𝑦 2 − 23𝑦 − 13 ⇒ 𝑦 2 − 8𝑦 + 15 = 0 ⇒ 𝑦 2 − 3𝑦 − 5𝑦 + 15 = 0 ⇒ 𝑦(𝑦 − 3) − 5(𝑦 − 3) = 0 ⇒ (𝑦 − 5)(𝑦 − 3) = 0 ⇒ 𝑦 = 5 ,3 ⇒ x ≥ y 49

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2. (d); I.

II.

25

15

− +2=0 x² x ⇒2x 2 − 15x + 25 = 0 ⇒2x 2 − 10x − 5x + 25 = 0 2x (x − 5) − 5(x − 5) = 0 (2x − 5)(x − 5) = 0 5 x = ,5 40 y²

2

+1= 2

13 y

⇒ y − 13y + 40 = 0 ⇒ y 2 − 8y − 5y + 40 = 0 ⇒ y(y − 8) − 5(y − 8) = 0 (y − 5)(y − 8) = 0 y = 5, 8 y ≥ x 3. (e); 𝟐𝐲+𝟒 . 32y+4 . 42y+4 = 32y+4 . 22y+16 𝒚 + 4 + 4𝑦 + 8 = 4𝑦 + 16 𝐲=4 𝒙𝟐 – 10 + 24 = 0 𝒙𝟐 – 6𝑥– 4𝑥 + 24 = 0 𝒙(𝑥– 6)– 4(𝑥– 6) = 0 (𝑥– 4)(𝑥– 6) = 0 𝒙 = 4, 6 𝒙≥𝑦 4. (d); I.

II.

48

14

y²

y 2

− +1=0 x² x 2 ⇒ x − 14x + 48 = 0 ⇒ x 2 − 8x − 6x + 48 = 0 ⇒ x(x − 8) − 6(x − 8) = 0 ⇒ (x − 8)(x − 6) = 0 x = 8, 6 45 1 + =2 ⇒ 2y − y − 45 = 0 ⇒ 2y 2 − 10y + 9y − 45 = 0

50

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⇒ 2y(y − 5) + 9(y − 5) = 0 ⇒ (2y + 9)(y − 5) = 0 9 𝐲 = 5, − 2 𝐱 > y 5. (d); I. (𝑥 − 4)2 = 9 ⇒ 𝑥 − 4 = ±3 ⇒ 𝑥 = 7, 1 II. (2𝑦 + 3)2 = 25 ⇒ 2𝑦 + 3 = ±5 ⇒ 𝑦 = 1, −4 ⇒x≥y 6. (c); I. 5x² + 3x – 36 = 0 5x² + 15x – 12x – 36 = 0 5x (x + 3) – 12 (x + 3) = 0 (5x – 12) (x +3) = 0 x = 12/5, –3 II. 2y² – 13y + 20 = 0 2 y ² – 8 y – 5 y + 20 = 0 2 y (y – 4) – 5(y – 4) = 0 (2 y – 5) (y – 4) = 0 y = 5/2, 4 y>x 7. (b); I. x² – 7x + 12 = 0 x² – 4x – 3x + 12 = 0 x(x – 4) –3 (x – 4) = 0 (x – 3) (x – 4) = 0 x = 3, 4 II. 2y² – 11y + 15 = 0 2y² – 6y – 5y + 15 = 0 2y(x – 3) – 3 (y – 3) = 0 (2y – 5) (y – 3) = 0 y = 5/2, 3 x≥y 51

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8. (d); I. 2x² + 11x + 15 = 0 2x² + 6x + 5x + 15 = 0 2x (x + 3) +5 (x + 3) = 0 (2x + 5) (x + 3) = 0 x = –5/2, –3 II. 2y² + 9y + 10 = 0 2y² + 4y + 5y + 10 = 0 2y (y + 2) +5 (y + 2) =0 (2y + 5) (y + 2) = 0 Y = –5/2, –2 y≥x 9. (c); I. 3x² + 7x – 40 = 0 3x² + 15x – 8x – 40 = 0 3x (x + 5) – 8x – 40 = 0 (3x – 8) (x + 5) = 0 x = 8/3, –5 II. 5y² – 29y + 42 = 0 5y – 14y – 15 y + 42 =0 y(5y – 14) – 3 (5y – 14) = 0 (y – 3) (5y – 14) = 0 y = 3, 14/5 y>x 10. (e); I. 3x² – 23x + 42 = 0 3x² – 9x – 14x + 42 = 0 3x( x – 3) – 14(x – 3) =0 (3x – 14) (x – 3) = 0 x =3, 14/3 II. 3x² – 19y + 45 = 0 2y² – 10y – 9y + 45 = 0 2y (y – 5) – 9 (y – 5) =0 (2y – 9) (y – 5) = 0 y=9/2, 5 No relation can be established between 𝑥 𝑎𝑛𝑑 𝑦

52

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11. (a); I. 𝑥 2 − 5𝑥 + 4 = 0 𝑥 2 − 4𝑥 − 𝑥 + 4 = 0 𝑥(𝑥 − 4) − 1(𝑥 − 4) = 0 (𝑥 − 1)(𝑥 − 4) = 0 𝑥 = 1, 4 II.

𝑦 2 + 5𝑦 + 6 = 0 𝑦 2 + 3𝑦 + 2𝑦 + 6 = 0 𝑦(𝑦 + 3) + 2(𝑦 + 3) = 0 (𝑦 + 2)(𝑦 + 3) = 0 𝑦 = −2, −3 𝑥>𝑦

12. (d); I. 2𝑥 2 − 𝑥 − 15 = 0 2𝑥 2 − 6𝑥 + 5𝑥 − 15 = 0 2𝑥 (𝑥 − 3) + 5(𝑥 − 3) = 0 (2𝑥 + 5)(𝑥 − 3) = 0 −5 𝑥 = ,3 2

II.

3𝑦 2 − 23𝑦 + 42 = 0 3𝑦 2 − 14𝑦 − 9𝑦 + 42 = 0 𝑦(3𝑦 − 14) − 3(3𝑦 − 14) = 0 (𝑦 − 3)(3𝑦 − 14) = 0 14 𝑦 = 3, 3 𝑦≥𝑥

13. (a); I. 𝑥 2 − 15 + 54 = 0 𝑥 2 − 9𝑥 − 6𝑥 + 54 = 0 𝑥(𝑥 − 9) − 6(𝑥 − 9) = 0 (𝑥 − 6)(𝑥 − 9) = 0 𝑥 = 6, 9 II. 𝑦 2 + 15𝑦 − 54 = 0 𝑦 2 + 18𝑦 − 3𝑦 − 54 = 0 𝑦(𝑦 + 18) − 3(𝑦 + 18) = 0 (𝑦 + 18)(𝑦 − 3) = 0 𝑦 = −18, 3 𝑥>𝑦 53

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14. (c); I. 𝑥 2 + 14𝑥 + 40 = 0 𝑥 2 + 10𝑥 + 4𝑥 + 40 = 0 𝑥(𝑥 + 10) + 4(𝑥 + 10) = 0 (𝑥 + 4)(𝑥 + 10) = 0 𝑥 = −4, −10 II. 𝑦 2 − 5𝑦 − 24 = 0 𝑦 2 − 8𝑦 + 3𝑦 − 24 = 0 𝑦(𝑦 − 8) + 3(𝑦 − 8) = 0 (𝑦 + 3)(𝑦 − 8) = 0 𝑦 = −3, 8 𝑦>𝑥 15. (e); I. 𝑥 2 − 225 = 0 𝑥 2 = 225 𝑥 = ±15 II.

𝑥 2 + 𝑦 2 = 306 225 + 𝑦 2 = 306 𝑦 2 = 306 − 225 𝑦 2 = 81 𝑦 = ±9 No relation can be established between x and y.

16. (c); I. 2x² – 7x + 6 = 0 2x² – 4x – 3x + 6 = 0 2x (x –2) –3 (x – 2) = 0 (2x – 3) (x – 2) =0 3 𝑥 = 𝑜𝑟 2 II.

2

3y² – 19y + 28 = 0 3y² – 12y – 7y + 28 = 0 3y (y – 4) –7 (y – 4) = 0 (3y – 7) (y – 4) = 0 7 y= ,4 3

y>x 54

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17. (a); I. x² – 13x + 36 =0 x² – 9x – 4x + 36 =0 x (x – 9) – 4 (x – 9) =0 (x – 4) (x – 9) = 0 x = 4, 9 II.

3y² – 19y + 30 = 0 3y² – 10y – 9y + 30 = 0 y(3y – 10) – 3 (3y – 10)=0 (y – 3) (3y – 10) = 0 10 y = 3, 3

x>y 18. (a); I. x³ = 120 + 96 = 216 x=6 II. y² – 25 = 0 y² = 25 y = ±5 x>y 19. (e); I. x² – 2x – 48 = 0 x² – 8x + 6x – 48 = 0 x(x – 8) +6 (x – 8) =0 (x + 6) (x – 8) = 0 x = 8, –6 II. y² – 15y + 54 = 0 y² – 9y – 6y + 54 = 0 y(y – 9) –6 (y – 9) = 0 (y – 6) (y – 9) = 0 y = 6, 9 No relation can be established between x and y 20. (c); I. 2x² + 25x + 72 = 0 2x² + 16x + 9x + 72 = 0 2x (x + 8) +9 (x + 8) = 0 55

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(2x + 9) (x + 8) = 0 9 x=− ,–8 II.

2

3y² + 22y + 40 = 0 3y² + 10 y + 12y + 40 = 0 y (3y + 10) + 4 (3y + 10) = 0 (y + 4) (3y + 10) = 0 10 y = – 4, – y>x

56

3

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Chapter

13 What is data interpretation: When data is organized into tables and charts it is done with the purpose of making it meaningful. The objective of data interpretation is to assess whether a student can understand bars and charts and Answer some questions based on them. This act of organizing and interpreting data to get meaningful information under a given set of conditions is Data interpretataion. About data interpretation: This is the calculation intensive portion, it consists of a myrid of graph. charts and tables and analyze data. The key to crack this area is to quickly Identify the key pieces of information that you will require to work on. Basic key that will help you to solve this topic: • Calculation • Square • cube • table • BODMAS • Percentage • Profit and loss • Ratio and proportion • Average Types of Data Interpretation: • Data table • Line graph • Pie charts • Bar graph • Mixed graph • Line with pie chart • Table with Line, etc. • Triangular graph • Case study (Puzzle) • Venn Diagram 2

• Table with Bar • Radar graph

• Table format

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Approach for data interpretation: • First you look carefully at the table or graph and the direction. Note the years to which, the data refers to and the units. Sometimes the figures may be given in thousands. While the Answer may be millions Resulting in mistakes. • The level of approximation that can be done is assessed from the choices. If the answer is wide, time should not be wasted in working out exact figures. If the choice ‘none of the above exists, a close approximation may be required’. • Read the question carefully, it will give an indication as to which row and column should be seen. A carefull reading of the question will reveal exactly what is to be done and the units in which the answer is required. • There may be one or two very large question requiring calculations. Attempt these at the last. • Revise bar charts, table and line graphs before attempting D.I. question remember that the D.I. section is a scoring one and also time saving. Data Table: The numbers in the bracket give the maximum marks in each subject. The following table gives the percentage of marks obtained by seven students in six different subjects in an examination. Study it and answer the question given below it. Subject (Maximum Marks)

Students

3

Maths (150)

Chemistry (130)

Physics (120)

Geography (100)

History (60)

Computer Science (40)

Golu

90

50

90

60

70

80

Mithi

100

80

80

40

80

70

Suraj

90

60

70

70

90

70

Gapplu

80

65

80

80

60

60

Mahi

80

65

85

95

50

90

Khushi

70

75

65

85

40

60

Sheetal

65

35

50

77

80

80

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Line Graph: Answer the question based on the given line graph Following line graph shows the ratio of export to import of company A and company B over the year 2 1.75 1.5

Company

A

Company

B

1.25 1 .75 .5 .25 0 Years

2005

2006

2007

2008

2009

2010

1. In how many of the given years were the exports more than the imports for company A? (a) 2 (b) 3 (c) 4 (d) 5 2. If the imports of company A in 2007 were increased by 40%. What would be the ratio of exports to the increased imports? (a) 1.50 (b) 1.25 (c) 1.75 (d) 1.35 3. If the exports of company B in 2008 was Rs 237 Crore, what was the amount of imports in that years ? (a) 189.6 (b) 188.8 (c) 184.6 (d) 182.8 4

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4. In 2005, the export of company A was double that of company B. If the imports of company A during the year was 180 crore. What was the approximate amount of imports of company B during that year? (a) 190 crore (b) 210 crore (c) 225 crore (d) 200 crore 5. In which year were the exports of company A minimum proportionate to its imports. (a) 2008 and 2009 (b) 2009 (c) 2008 (d) 2005 Solution 1: required ratio, more than 1 for the years; 2005, 2006 and 2007, ie; for 3 years Solution 2: In 2007, for company A the ratio of exports to import = 175 : 100 Let exports of company A = 175 x Þ and imports of company A = 100x New imports of company = 140% of 100x = 140x 175x = 1.25 There for requied ratio = 140x Solution 3: Let import of company B in 2008 = x 237 237 Þ 189.6 crore = 1.25 Þ x = Then, 1.25 x Solution 4: In 2005 for company A Þ Export = 1.75 × 180 crore in 2005 for company B export =

1.75 ´ 180 2

Ex = .75 Im 157.5 Import of company B = = 210 crore .75 = 157.5 crore,

But

Solution 5: 2008 and 2009, A option. Pie Charts or Circle Graphs: Distribution of candidate who were enrolled in MBA and the candidate (out of those enrolled) who passed the exam in different institutes 5

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X 16%

V 12% 8%

P 22% Q

T S 17%

X 12%

V 15% 15%

R 10%

9%

Q

T S 16%

Total number of candidates enrolled = 8550

P 18% 17%

R 13%

Total number of candidates who passed the exam = 5700

1. What percentage of candidates passed the exam from institute T out of the total no. of candidate enrollled from the same institute? (a) 50% (b) 52.5% (c) 75% (d) 80% 2. Which institute has the highest % of candidates passed to the candidates enrolled? (a) Q (b) R (c) V (d) T 3. The no. of candidates passed from institute S and P together exceeds the no. of candidates enrolled from institutes T and R together by? (a) 228 (b) 279 (c) 399 (d) 407 4. What is % of candidates passed to the candidate enrolled for institutes Q and R together ? (a) 68% (b) 80% (c) 74% (d) 65% 5. What is the ratio of candidates passed to the candidates enrolled from institute P ? (a) 9 : 11 (b) 14 : 17 (c) 6 : 11 (d) 9 : 17 6

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Solution 1: Required % =

9%of 5700 ´ 100 = 75% 8%of8550

Solution 2: The % of candidates passed to candidates enrolled can be determined for each institute as under.

17%of 5700 For Q Þ 15%of 8550 ´ 100 = 75.56% ; For R Þ For T

Þ

For V Þ

13%of 5700 ´ 100 = 86.67% 10%of 8550 9%of 5700 ´ 100 = 75% ; 8%of 8550 15%of 5700 ´ 100 = 83.33% 12%of 8550

So the highest of these is 86.67% corresponding to institutes R. Solution 3: Required Difference = [(16% + 18%) of 5700] – [(8% + 10%) of 8550] = 1938 – 1539 = 399

30%of 5700 Solution 4: 25%of 8550 ´ 100 = 80% Solution 5:

7

18%of 5700 6 = = 6 : 11 22%of 8550 11

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Bar Graph: Percentage profit earned by two companies X & Y over the given years. Pr ofit% = 70

Profit%

60 50 40

Company X Company Y 55 50 45 45 40 35

Income Expenditure ´ 100 Expenditure 60 50

65 50 50

55

30 20

2006

2007

2008

2009

2010

2011

1. The income of two company X and Y in 2010 were in the rato of 3 : 4. respectively. What is the respective ratio of their expenditures in 2010? (a) 7 : 22 (b) 14 : 19 (c) 15 : 22 (d)27 : 35 2. If the expenditure of company Y in 2007 was Rs 220 crore, what was its income in 2007? (a) Rs. 312 crore (b) Rs. 297 crore (c) Rs. 283 crore (d)Rs. 275 crore 3. If the expenditures of company X and Y in 2006 were equal and the total income of the two companies in 2006 was Rs 342 crore, what was the total profit of the two company together in 2006? (a) Rs. 240 crore (b) Rs. 171 crore (c) Rs. 120 crore (d)Rs. 102 crore 4. The expanditure of company X in the year 2008 was Rs 200 crore and the income of company X in 2008 was the same as its expenditure in 2011. The income of company X in 2011 was? (a) Rs. 465 crore (b) Rs. 385 crore (c) Rs. 335 crore (d)Rs. 295 crore 5. If the income of two company were equal in 2009, then what was the ratio of expenditure of company X to that of company Y in 2009? (a) 6 : 5 (b) 5 : 6 (c) 11 : 6 (d) 16 : 15 8

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Solution 1:

165%of E1 3 = 150%of E 2 4

Þ Therefore,

E 1 15 = = 15 : 22 E 2 22

Solution 2: 220 + 35% of 220 Þ 220 + 77 = 297 crore Solution 3: Let the expenditure of each companies x and y in 2006 be 100 Then we have 100 + 40 + 100 + 45 = 342 crore Þ Required profit =

342 285

×85 = 102 crore

Solution 4: Expenditure of Company X in year 2008 is 200 crore, Then income of company X in year 2008 = 200 + 55% of 200 = 310 crore Hence, the expenditure of company X in year 2011 is 310 crore, Therefore, the required income of X in year 2011 = 310 + 50% of 310 = 465 crore Solution 5:

150 160 X= Y 100 100

x 160 16 be the incomes of two companies in 2009; y = 150 = 15 = 16 : 15

9

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Practice Exercise Based on new Pattern Direction (1 – 5): Table given below shows total number of students in two sections of five different schools, percentage distribution of students in these different sections (X & Y) and also ratio between boys to girls in each section. Read the table carefully and answer the questions:

Schools

Total students

LPS CMS DPS SKD Loyola

450 360 420 300 250

Percentage Ratio between of students boys: girls in Section ‘X’ in section ‘X’ 48% 2:1 55% 7:2 60% 5:4 55% 1:2 50% 3:2

Ratio between boys : girls in Section ‘Y’ 4:5 1:2 11 : 10 2:1 2:3

1. Total number of boys in section ‘X’ from CMS & Loyola together are what percent of total girls in section ‘Y’ from DPS & SKD together? (a) 183.2 % (b) 180.2 % (c) 188.2 % (d) 189.2 % (e) 185.2% 2. Find the ratio between total number of girls in section ‘Y’ from CMS & SKD together to total number of boys in section ‘X’ from DPS & SKD together? (a) 65 : 51 (b) 51 : 65 (c) 51 : 67 (d) 51 : 62 (e) 51 : 49 3. Out of total number of boys in section ‘X’ from D.P.S school ratio between number of boys who got first, second and third division is 1 : 2 : 4. Find total number of boys who got first and third division together in section ‘X’ from D.P.S school? (a) 40 (b) 60 (c) 80 (d) 100 (e) 120 10

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4. Find the difference between average number of girls in section ‘X’ from SKD & Loyola together and average number of boys in section ‘Y’ from LPS & CMS together? (a) 5 (b) 4 (c) 3 (d) 2 (e) 1 5. Total girls in section ‘Y’ from CMS & LPS together are how much more than total boys in section ‘X’ from both DPS & SKD together? (a)33 (b) 43 (c)53 (d) 63 (e) 45 Direction (6 – 10): Table given below shows total number of applicants, who have applied for CDS exam in the five years. Percentage of applicants appeared in exam, percentage of applicants qualified in tier I and tier II. Line graph shows percentage of applicants got selection in CDS exam. Read the data carefully and answer the questions: Years

Total Applicants

2013 2014 2015 2016 2017

2400 3200 4800 7500 8400

Appeared applicants in tier I 85% 75% 90% 80% 60%

% of applicants % of students qualified qualified in in tier I tier II 25% 40% 30% 25% 20% 25% 25% 20% 1 12 % 40% 2

Note – All applicants qualified in tier I appeared for tier II.

(Percentage)

50 40 30 20 2013 11

2014

2015

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6.

Number of applicants, who have got the final selection in the year 2017, is how many percent less than number of applicants, who have got the final selection in the year 2014? 2

2

(a) 22 %

(b) 24 %

(d) 18 %

(e) 21 %

9 2 9

9 2

2

(c) 28 % 9

9

7. Find the difference between number of applicants, who have got final selection in the years 2013 & 2016 together and number of applicants, who have got final selection in the year 2014 & 2015 together? (a) 5 (b) 8 (c) 3 (d) 6 (e) 11 8. Find the ratio between applicants qualified in tier II but did not get final selection in the year 2017 to applicants qualified in tier I but did not qualify for tier II in the year 2014? (a) 7 : 20 (b) 7 : 22 (c) 7 : 19 (d) 7 : 18 (e)7 : 16 9.

Total applicants appeared for tier I in the year 2017 is what percent more than total applicants appeared for tier II in the year 2015? 1

(b) 438 %

(d) 483 %

(e) 487 %

3 1 3

10.

1

(a) 498 %

3 1

1

(c) 485 % 3

3

Find the total number of applicants, who did not qualify in tier I in the year 2014, 2016 & 2017 together? (a) 10390 (b) 10590 (c) 10390 (d) 10290 (e) 10190

Direction (11 – 15): Given below pie graph shows total number of tickets of ‘Sanju’ movie sold by five multiplex on opening days, while bar graph shows percentage distribution of three different type of class ticket sold by each store. Read given data and answer the questions: 12

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Total sold tickets - 12500 A 12% E 36%

D 10%

Percentage

Jubali

B 18%

C 24%

Silver

Gold

100 90 80 70 60 50 40 30 20 10 0 A

B

C Multiplex

D

E

11. Total Jubali class ticket sold by multiplex C & E together is what percent more than total Gold class ticket sold by multiplex A & C together? 1 1 1 (a) 105 % (b) 102 % (c) 100 % 1

7

(d) 97 % 7

13

7 1

7

(e) 107 % 7

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12. Find difference between total Silver class ticket sold by multiplex A, C & E together and total silver gold class ticket sold by multiplex B, C & E together? (a) 400 (b) 200 (c) 650 (d) 600 (e) 450 13. Find sum of average number of Jubli class ticket sold by multiplex B & C and average number of Gold class ticket sold by multiplex C & E? (a) 2075 (b) 2175 (c) 2275 (d) 2375 (e) 2205 14. Total Silver class tickets sold by multiplex A & D is what percent less than total Gold class ticket sold by B & E? 4 4 4 (a) 42 % (b) 40 % (c) 38 % 9 4

(d) 44 % 9

9 4

9

(e) 36 % 9

15. The ratio of price of one Gold ticket in multiplex D, C & A is 10 : 6 : 7, and all three multiplex earned Rs. 40000 from the selling of all Gold Class tickets. Find difference between price of one Gold ticket in Multiplex A and D? (a) 10 Rs. (b) 15 Rs. (c) 5 Rs. (d) 20 Rs. (e) 25 Rs. Direction (16 – 20): Given below table shows total number of five different brands of mobiles sold by a store on Sunday. Also given percentage of 2G mobiles and ratio between 3G and 4G mobile phone sold out of total number of sold mobiles. Read the table carefully and answer the following questions: Brand Samsung Lava MI Huwai Oppo 14

Number of total sold mobiles 750 840 1050 960 640

Percentage of 2G mobiles 20% 25% 30% 15% 35%

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2

16. Total number of Samsung mobiles sold by store on Saturday is 𝑟𝑑 of 3 the number of same brand of mobile sold on Sunday. If total number of 3G Samsung mobile sold on Saturday is 40 less than that of total number of 3G Samsung mobile sold on Sunday, then find total number of 4G Samsung mobile sold on Sunday is what percent more/less than total number of 4G Samsung mobile sold on Saturday (on Saturday store sold only 3G and 4G mobile)? (a) 25% (b) 20% (c) 10% (d) 15% (e) 5% 17. Total number of MI mobile sold on Monday is equal to average number of Huwai & Oppo mobile sold on Sunday. If ratio between number of MI 2G, 3G and 4G mobiles sold on Monday is 4 : 5 : 7, then find the ratio between total number of MI 3G mobile sold on Monday to total number of Lava 3G mobile sold on Sunday? (a) 25 : 29 (b) 25 : 27 (c) 25 : 36 (d) 25 : 31 (e) 25 : 33 18. If ratio of total number of Oppo 4G mobiles sold by store on Sunday, Monday and Tuesday is 6 : 4 : 7, then find total number of Oppo 4G mobile sold by store on Monday & Tuesday together is what percent less/more than the total number of Samsung & Lava 4G mobile sold by store on Sunday together? 5 5 5 (a) 18 % (b) 16 % (c) 24 % 9 5

(d) 20 % 9

9 5

9

(e) 12 % 9

19. Find the difference between average numbers of Samsung & Lava 2G mobile and average numbers of Huwai & Oppo 3G mobile sold by store on Sunday? (a) 12 (b) 16 (c) 10 (d) 8 (e) 4 20. If total number of Vivo mobile sold by store on Sunday is 56 more than total number of 4G Lava & Huwai mobile sold by store on Sunday, then find total number of Oppo mobile phone sold by store on Sunday is 15

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what percent more/less than total number of Vivo mobile sold by store on Sunday ? 1 1 1 (a) 35 % (b) 38 % (c) 33 % 3 1

(d) 39 % 3

3 1

3

(e) 41 % 3

Direction (21- 25): Table given below shows total number of tickets which were estimated to be sold in five stadiums. Out of total tickets sold, percentage of A1 tickets sold, and ratio between A2 and A3 is given. Study the data carefully and answer the following questions. Stadiums Tickets which can be sold A1 tickets sold (in %) A2 : A3 Nehru 1800 15% 9:8 Patel 2600 25% 4:5 Eden 2200 20% 5:3 Kaloor 2000 20% 7:5 Raipur 2700 30% 3:4 Note: - 200 tickets remain unsold in each stadium Tickets which can be sold = A1 tickets + A2 tickets + A3 tickets + Unsold tickets 21. Total A1 and A3 tickets sold in Patel stadium is what percent more than total number of A1 and A2 tickets sold in Kaloor stadium? 2 2 (a) 16 % (b) 50% (c) 66 % 3

(d) 25%

1

3

(e) 33 % 3

22. Total A2 tickets sold in Nehru stadium is how much less than A3 tickets sold in Raipur stadium? (a) 220 (b) 240 (c) 260 (d) 280 (e) 300 23. Find the ratio between total A3 tickets sold in Eden and Kaloor stadium together to total A2 tickets sold in Patel and Eden stadium? (a) 3 : 2 (b) 2 : 3 (c) 20 :23 (d) 23 : 20 (e) 8 : 9 16

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24. Find the average number of total A1 tickets sold by all five stadiums together? (a) 514 (b) 490 (c) 470 (d) 450 (e) 430 25. A1 and A2 tickets sold in Nehru stadium is what percent more/less than A1 and A2 tickets sold in Raipur stadium? (a) 36% (b) 64% (c) 56.25% (d) 43.75% (e) 28% Direction (26-30): Given below bar graph shows percentage distribution of weekly income of four different Bollywood movies. Table shows total income of these four movies and percentage of income earning by these movies from foreign. Read the data carefully and answer the questions.

100

First week

Second week

Third week

80

60 40 20 0 Movies A B C D

A B Total income (in crores) 210 180 200 240

C D % of income from foreign 26 ⅔% 25% 20% 30%

26. What is the ratio between income of movie B in second week to income of movie C in third week? (a) 13 : 5 (b) 13 : 3 (c) 27 : 5 (d) 27 : 10 (e) 27 : 13 17

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27. Out of these four movies, which movie beard highest percentage drop in its income in second week with respect to first week? (a) A (b) B (c) C (d) D (e) None of these 28. What is the ratio of income of movie A from foreign to the income of movie C from foreign? (a) 8 : 5 (b) 8 : 7 (c) 7 : 5 (d) 28 : 27 (e) 7 : 8 29. If ratio of foreign income of B of given three weeks is 1 : 2 : 2 respectively. Then find the income earn by movie B in second week in India? (a) 36 crores (b) 18 crores (c) 26 crores (d) 30 crores (e) 32 crores 30. Income of movie D in second week and income of movie C in third week together is approximately what percent more or less then total income of movies A and B of second week? (a) 55 (b) 78 (c) 62 (d) 44 (e) 38 Directions (31-35): Study the following graph carefully and answer the question. The following bar graph shows the budget allocation (in crore) for education, sport and Health care from year 2018-22.

18

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Education

250

190

Budget (in crore)

200 160 150 100

125 105

130 125

Sport

Health care

200 170 150 130

220 220

190 170

160

50 0 2018

2019

2020 2021 2022 Year 31. What percent is the average budget of sport during 2018 to 2022 than that of average budget of Education throughout all the years? (approximately) (a) 84% (b) 88% (c) 87% (d) 90% (e) 94% 32. In 2022, budget allocated for sport for male to female are in the ratio of 3 : 2 and it proposed that budget allocated for males to be increased by 25% in 2023 compared to previous year for sport then, what is the percentage increase in sport budget allocation for 2023 with no change in sport budget allocated for female in 2023 as compared to previous year? (a) 20% (b) 18% (c) 15% (d) 12% (e) 10% 33. Find the ratio between total budget allocated for Education from year 2018 to 2020 together to total budget allocated for Health care from year 2020 to 2022 together ? (a) 17 : 29 (b) 18 : 29 (c) 21 : 29 (d) 27 : 29 (e) 17 : 27 19

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34. If in 2019 budget allocated for only three sports i.e. Cricket, Badminton and Hockey in the ratio of 5 : 4 : 4 respectively and budget allocated for Education in 2022 for Urban and rural area in the ratio of 5 : 6 respectively. Then find budget allocated for Education for rural area in 2022 how much more than budget allocated for Cricket & Badminton together in 2019 ? (a) 30 cr (b) 35 cr (c) 40 cr (d) 45 cr (e) 25 cr 35. Find the sum of average of budget allocated for Health care throughout all the year 2018-22 and average of budget allocated for Education throughout all the year 2018-2022 ? (a) 336 (b) 338 (c) 340 (d) 342 (e) 348 Direction (36 – 40): Given below table shows the percentage of number of cars sold by four different motor companies, percentage of number of total returned cars out of total number of sold cars in two successive years(2016, 2017) . Read the graph carefully and answer the following questions. Note; (I)

Total number of manufactured cars = number of Unsold cars + number of Sold cars (II) Total actual number of cars sold = Total number of sold cars – number of returned cars

TATA HYUNDAI MARUTISUZUKI HONDA

20

Percentage Percentage of Percentage of Percentage of of number number of number of number of of cars sold cars sold in cars returned cars returned in 2016 2017 in 2016 in 2017 80% 75% 15% 10% 65% 70% 20% 15% 80%

85%

12.5%

17.5%

75%

70%

15%

22.5%

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36. The ratio between total number of cars manufactured by TATA and HONDA in the year 2016 is 6 : 7 and the total actual number of cars sold by HONDA is 4590 units more than that of TATA in the same year. If total number of cars returned to HYUNDAI in the year 2017 is 1920 units less than total number of cars returned to TATA in 2016 and total number of cars returned to MARUTI SUZUKI in the year 2017 is 4830 units more than total number of cars returned to HONDA in 2016, then find the difference between total number of manufactured cars by HYUNDAI and MARUTI SUZUKI in the year 2017? (a) 36,000 (b) 42,000 (c) 44,000 (d) 40,000 (e) 32,000 37. Total number of cars manufactured by TATA in the year 2017 is 40% more than that of total number of cars manufactured by MARUTI SUZUKI in the year 2016 and total number of cars returned to TATA in 2017 is 120 units more than total number of cars returned to MARUTI SUZUKI in the year 2016. If total number of cars returned to HONDA in the year 2017 is 2010 units more than total number of cars returned to MARUTI SUZUKI in the year 2016, then find total number of cars manufactured by HONDA in the year 2017? (a) 24,000 (b) 20,000 (c) 28,000 (d) 18,000 (e) 16,000 38. The ratio between total number of cars manufactured by HYUNDAI, MARUTI SUZUKI & HONDA in the year 2016 is 2 : 3 : 4 and average number of cars returned to these three companies in the same year is 4040 units. If total number of cars manufactured by HYUNDAI, MARUTI SUZUKI & HONDA in the year 2017 is increased by 20%, 25% and 12.5% respectively over that of the previous year, then find the average number of cars manufactured by HYUNDAI, MARUTI SUZUKI & HONDA in the year 2017? (a) 42,400 (b) 42,800 (c) 42,600 (d) 42,000 (e) 41,600 21

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39. The ratio between total number of cars manufactured by HYUNDAI & HONDA in the year 2016 is 8 : 9 and the ratio of number of cars manufactured by HYUNDAI & HONDA in the year 2016 to 2017 is 2 : 3 and 3 : 5 respectively. If total actual number of cars sold by HYUNDAI & HONDA together in the year 2016 is 39590 units then find total number of cars manufactured by HONDA in the year 2017 is what percentage more/less than total number of cars manufactured by HYUNDAI in the same year? (a) 20% (b) 15% (c) 10% (d) 25% (e) 35% 40. The ratio between total number of cars manufactured by TATA, HYUNDAI & HONDA in the year 2017 is 7 : 6 : 8 and total actual number of cars sold by these companies in the year 2017 is 75810 units, then find total number of cars manufactured by TATA, HYUNDAI & HONDA together in the year 2017? (a) 1,24,000 (b) 1,25,000 (c) 1,20,000 (d) 1,26,000 (e) 1,36,000 Direction (41 – 45): Pie-chart shows the distribution of total bikes manufactured by five different companies. Also Table shows percentage ‘150 CC’ and ‘200 CC’ & total bike sold by these five companies. Study the data carefully and solve the following questions

Yamaha 22%

Hero 15% Bajaj 25%

TVS 18% Honda 20% 22

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Companies Hero Bajaj Honda TVS Yamaha

Total sold bikes — — 80% — 75%

150CC Sold bikes 200CC Sold bikes 35% 40% — 75% 70%

65% 60% — — —

Note All companies manufactured only two types of bike ‘150 CC’ & ‘200 CC’. Some values are missing you have to calculate it according to questions. Total bikes sold of any company= Total 150 CC bikes sold + Total 200 CC bikes sold 41. Total ‘200 CC’ Honda bikes sold by store is 25% more than total ‘150 CC’ bikes sold by same store. If difference between ‘200 CC’ & ‘150 CC’ bike sold by Honda is 640 then find total ‘200 CC’ bike sold by Yamaha? (a) 1785 (b) 1782 (c) 1780 (d) 1787 (e) 1791 42. If Hero & Bajaj sold 80% & 90% of total manufactured bikes respectively and difference between total ‘200 CC’ bikes sold by these two companies is 4104, then find total ‘150 CC’ bike sold by Bajaj is what percent more than total ‘150 CC’ bike sold by Hero? (a) 88

10

%

(b) 92

(d) 114 %

(e) 96

21 2 7

10 21 10 21

%

2

(c) 94 % 7

%

43. If Tvs sold 80% of total manufactured bikes and difference between total ‘150 CC’ bikes sold by Tvs and Yamaha is 450, then find bike manufactured by Hero & Bajaj together? (a) 28000 (b) 26000 (c) 32000 (d) 24000 (e) 36000 23

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44. If Hero, Bajaj & Tvs sold 75%, 80% & 90% of total manufactured bikes respectively and total number of ‘150 CC’ bikes sold by these three companies is 23124, then find the total bike sold by Honda? (a) 15568 (b) 16164 (c) 17162 (d) 15360 (e) 17172 45. If total 84000 bikes manufactured by all five companies and ratio between ‘150 CC’ to ‘200 CC’ bike manufactured by Honda is 3 : 2, then find total ‘200 CC’ bike sold by Yamaha is what percent less than total ‘150 CC’ bike manufactured by Honda? (a) 56.75% (b) 54.75% (c) 62.75% (d) 59.75% (e) 58.75% Directions (46-50): Given below are two pie-charts which shows the percentage distribution of employees in Adda247 who travel to their office in Gurgoan by two different means i.e. by metro and by cab on different days of week. First pie chart shows data for Metro and second pie chart shows data for Cab

Saturday 15%

Monday 20%

Friday (50/3)% Thursday Wed. 10% 15%

24

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Tuesday (70/3)%

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Saturday Monday 10% 25% Friday 20% Thursday (20/3)%

Wed. 25%

Tuesday (40/3)%

Note: 1. Ratio of total employees travelling to Gurgaon by metro to by cab is 5 : 4. 2. If difference of persons travelling by metro and by cab on Saturday is 420. 46. If number of employees travelling on Sunday by cab decreases by ‘Y’ with respect to employees travelling on Friday by cab and number of employees travelling on Sunday by metro is twice than that of travelling on Friday by metro . Then find value of ‘Y’?( Given that total employees on Sunday is 75% of total employees on wed.) (a) 1280 (b) None of these (c) 1385 (d) 1415 (e) 1255 47. What is the difference of number of employees travelling by metro on Wednesday and Thursday together and number of employees travelling by cab on same days together? (a) 50 (b) 20 (c) None of these (d) 40 (e) 60 48. If total fare per person travelling by metro & cab are Rs.120, then ratio of total amount spent on Thursday by all employees travelling by cab to total amount spent by all employees on same day travelling by metro? (given that ratio of fare per person travelling by metro to by cab is 5 : 7) (a) 21 : 23 (b) 56 : 73 (c) None of these (d) 56 : 75 (e) 53 : 73 25

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49. If total fare on Monday by all employees travelling by metro is Rs. 48000. Then find the total fare on same day by all employees travelling by cab? (if per person fare for each employee is same) (a) Rs. 48000 (b) Rs. 42000 (c) Rs. 56000 (d)None of these (e) Rs. 26000 50. Total number of employees travelling by metro on Friday and Saturday together is approximately what percent more or less than number of employees travelling by cab on same days together? (a) 48% (b)28% (c) 42% (d) 38% (e) 32% Directions (51-55): Table given below shows number of students appeared in preliminary examination of an exam ‘ASK’ in six different cities and percentage of students failed in preliminary, mains and in only reasoning. Those students who cleared the Preliminary examination can give the Mains examination. In Mains examination of ASK, there are only two subjects i.e. Reasoning and Quant. Student has to clear both the subjects to pass in Mains examination of ‘ASK’.

25% 40%

Students Failed in Mains examination (in %) 87.5% 75%

Students Failed in only Reasoning (in %) 25% 37.5%

1,20,000

35%

80%

16 %

D E

1,60,000 1,25,000

15% 28%

75% 70%

25% 40%

F

72,000

33 %

68.75%

25%

Students appeared in Preliminary exam

Students Failed in Preliminary exam (in %)

A B

96,000 80,000

C

26

1 3

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51. Total number of students who failed only in Quant in City ‘A’ is three times of total number of students passed the mains exams in city ‘B’ while total number of students who failed in both subjects in City ‘B’ is 3000 less then total number of students who failed is both subjects in city ‘A’. Find number of students who failed only in Quant in city ‘A’ is what percent more than number of students who failed in Reasoning in city ‘B’? (a) 250% (b) 50% (c) 150% (d) 200% (e) 100% 52. In city D, total number of students who failed in Reasoning is ‘x’ more than total number of students who failed in only Quant. If total number of students who passed in mains exam in city D is ‘x’ then find the ratio between total number of students who failed in Reasoning to total number of students who failed in only Quant in city ‘D’. (a) 1 : 3 (b) 3 : 1 (c) 1 : 1 (d) 1 : 2 (e) 2 : 1 53. Total number of students who failed in Quant in city E is how much more than total number of students who failed in Quant in city F? (a) 3000 (b) 6000 (c) 9000 (d) 12000 (e) 15000 54. Ratio between total number of students who failed in Quant in city ‘B’ to total number of students who failed in at most one subject in city ‘C’ is 30 : 91. Find what percent of students failed in both subject in city ‘C’ in mains exam? (a) 25% (b) 15.5% (c) 17.5% (d) 19.5% (e) 30% 55. Total number of students who failed in at most one subjects in city F is 24000 less than total number of students who failed in at least one subject in city A. Find total number of students who failed in both subjects in city ‘F’ is how much less than total number of students who failed in only reasoning in city ‘A’. (a) 3000 (b) 6000 (c) 9000 (d) 12000 (e) 15000 27

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Direction (56-60): - A school has four hostels in which there are two sections one for boys and one for girls. Line chart given below shows increase/decrease in total number of students in 2013 in these hostels as compare to 2012(previous year). Table given below shows ratio between number of boys to number of girls in these hostels in 2013. Study the data carefully and answer the following questions 300

Hostel

Boys : Girls

Aravali

2:3

150

Nilgiri

4:1

100

Udaygiri

5:3

50

Shivalik

4:7

250 200

2…

Aravali Nilgiri Udaygiri Shivalik 56. In Aravali, number of girls in 2013 is 20% more than that in 2012. If change in the number of girls in Aravali is 25% greater than change in number of boys in Aravali then find the Number of boys in Aravali in 2013. (a) 900

(b) 600

(d) 750

(e) 800

(c) 500

57. Number of students is increases and decreases in Aravali and Nilgiri in 2013 respectively as compare to 2012. Number of girls who left Aravali joins Nilgiri (No other change in Number of girls) such that ratio between increase and decreases in Number of boys in Aravali and Nilgiri in 2013 as compare to 2012 is 16 : 15, then find the change in Number of boys in Nilgiri?

28

(a) 350

(b) 320

(d) 420

(e) 400 Adda247 Publications

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58. In 2012, Number of boys in Udaygiri and Shivalik are equal and Number of girls in Shivalik is 540 more than the Number of girls in Udaygiri. If in 2013, Number of boys in both hostels are equal then find the Number of total student in Shivalik, If In 2013 strength of both hostels increases as compare to previous year (a) 1650 (b) 1200 (c) 960 (d) 1500 (e) 900 59. In 2012 and in 2013 ratio of total strength of Udaygiri to that of Nilgiri is 4 : 5 and 5 : 8 respectively. Find the strength of Nilgiri in 2012 if strength of Udaygiri decreases (a) 1600 (b) 1200 (c) 1250 (d) 1400 (e) 1750 60. In2013, strength of all hostels increases as compare to 2012. If total strength of hostels in 2012 is 1030, then find the Number of boys in Aravali in 2013 is how much more than the Number of boys in Udaygiri in 2013, given that ratio of strength of Aravali, Nilgiri, Udaygiri and Shivalik is 2 : 3 : 2 : 3 respectively (a) 80 (b) 100 (c) 120 (d) 90 (e) 10

29

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Solutions 1. (a); Total boys in section ‘X” from CMS & Loyola = 360 ×

55 100

7

50

9

100

× + 250 ×

×

3 5

= 154 + 75 = 229 Total girls in section ‘Y’ from DPS & SKD = 420 ×

(100 –60) 100

×

= 80 + 45 = 125

10 21

+ 300 ×

Required percentage =

229 125

(100–55) 100

×

1 3

× 100 = 183.2%

2. (b); Total girls in section ‘Y’ from CMS & SKD = 360 ×

(100–55) 100

2

(100 –55)

3

100

× + 300 ×

×

1 3

= 108 + 45 = 153 Total boys in section ‘X’ from DPS & SKD = 420 ×

60 100

5

55

9

100

× + 300 ×

= 140 + 55 = 195 Required ratio =

153 195

×

1 3

= 51 : 65

3. (d); Total number of boys in section ‘X’ from D.P.S school = 420 ×

60 100

5

× = 140 9

5

Required number of students = 140 × = 100 7

4. (e); Average number of girls in section ‘X’ from SKD & Loyola 55

=

2

50

2

300×100×3+250×100×5 2

=

110+50 2

= 80

Average number of boys in section ‘Y’ from LPS & CMS = =

(100–48) 4 (100–55) 1 ×9+360× 100 ×3 100

450×

104+54 2

2

= 79

Required difference = 80 – 79 = 1 30

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5. (b); Total girls in section ‘Y’ from CMS & LPS together (100–48) (100–55) 5 2 = 450 × × + 360 × × = 130 + 108 = 238 100 9 100 3 Total boys in section ‘X’ from DPS & SKD 60 5 55 1 = 420 × × + 300 × × 100 9 100 3 = 140 + 55 = 195 Required difference = 238 – 195 = 43 6. (a); Applicants got final selection in the year 2017= 8400 × 40

25

60 100

1

× × 8

× = 63 100 Applicants got final selection in the year 2014 75 30 25 45 = 3200 × × × × = 81 100

100

100

100 100 81−63

Required percentage = =

18 81

2

81

× 100 2

× 1100 = × 100 = 22 % 9

9

7. (d); Total applicants got final selection in the year’s 2013 & 2016 together 85 25 40 25 80 25 20 30 = 2400 × × × × + 7500 × × × × 100 100 100 100 100 100 100 100 = 51 + 90 = 141 Total applicants got final selection in the year 2014 & 2015 together 75 30 25 45 90 20 25 25 = 3200 × × × × + 4800 × × × × 100 100 100 100 100 100 100 100 = 81 + 54 = 135 Required difference = 141 – 135 = 6 8. (a); Applicants qualified in tie II but did not get final selection in the years 2017 60 1 40 (100−25) = 8400 × × × × = 189 100 8 100 100 Applicants qualified in tier but did not qualify for tier II in the year 2014 (100−25) 75 30 = 3200 × × × = 540 100

100 189

Required ratio = 31

540

100

= 7 : 20 Adda247 Publications

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9. (d); Total appeared applicants for tier I in the year 2017 = 8400 ×

60 100

= 5040

Total appeared applicants for tier II in the year 2015 = 4800 ×

90 100

×

20 100

= 864

Required percentage = =

4176 864

1

5040−864 864

× 100

× 100 = 483 % 3

10. (b); Total number of applications did not qualify in tier I in the year 2014, 2016 and 2017 = 3200 ×

75 100

×

(100−30) 100

+ 7500 ×

80 100

×

(100−25) 100

+ 8400 ×

60 100

×

(100−12.5) 100

= 1680 + 4500 + 4410 = 10590 11. (e); Total Jubali class ticket sold by multiplex C & E = 12500 ×

24 100

×

35 100

+ 12500 ×

36 100

×

25 100

= 1050 + 1125 = 2175 Total gold class ticket sold by multiplex A & C = 12500 ×

12 100

×

30 100

+ 12500 ×

Required percentage =

24

100 2175−1050 1050

×

20 100

= 450 + 600 = 1050 1

× 100 = 107 % 7

12. (d); Total silver class ticket sold by multiplex A, C and E = 12500 ×

12 100

×

50 100

+ 12500 ×

36 100

×

35 100

+ 12500 ×

24 100

×

45 100

= 750 + 1350 + 1575 = 3675 Total gold class ticket sold by multiplex B, C & E = 12500 ×

18 100

×

30 100

+ 12500 ×

24 100

×

20 100

+ 12500 ×

36 100

×

40 100

= 675 + 600 + 1800 = 3075 Required difference = 3675 – 3075 = 600 32

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13. (b); Average number of jubali ticket sold by Multiplex B & C 12500×

18 40 24 35 × +12500× × 100 100 100 100 900+1050

12500×

24 20 36 40 × +12500× × 100 100 100 100

= = 975 2 2 Average number of gold class ticket sold by multiplex C & E 600+1800

= = 2 2 Required sum = 975 + 1200 = 2175

= 1200

14. (d); Total silver class ticket sold by multiplex A & D 12 50 10 50 = 12500 × × + 12500 × × = 750 + 625 = 1375 100 100 100 100 Total Gold glass ticket sold by B & E 18 30 36 40 = 12500 × × + 12500 × × = 675 + 1800 = 2475 100

100

Required percentage =

100 2475−1375 2475

100

× 100 =

1100 2475

4

× 100 = 44 % 9

15. (b); Let price of one Gold ticket in multiplex D, C and A be Rs.10x, Rs6x and Rs7x respectively. ATQ, Total earning = 40000 10 10 24 20 12 30 12500× × ×10x+12500× × × 6x + 12500 × × × 7x 100 100 100 100 100 100 1250x + 3600x + 3150x = 40000 40000 x= ⇒ x = 5 Rs 8000 Required difference = 5 × 10 – 5 × 7 = 15 Rs. 16. (b); Total Samsung mobile sold by store on Saturday 2 = 750 × = 500 3 Total Samsung 3G mobile sold on Saturday 80 2 = 750 × × – 40 = 240 – 40 = 200 100 5 Total Samsung 4G mobile sold on Saturday = 500 – 200 = 300 Total Samsung 4G mobile sold on Sunday 80 3 = 750 × × = 360 100

5

Required percentage = 33

360–300 300

× 100 =

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17. (b); Total MI mobile sold on Monday =

960+640 2

=

1600 2

= 800

MI 3G mobile sold on Monday 5

= 800 × (4+5+7) = 250 Required ratio =

250

=

75 3 840×100×7

250 270

= 25 : 27

18. (d); Total Oppo 4G mobile sold on Monday and Tuesday =

640×65 100

3

(4+7)

4

6

× ×

= 572

Total Samsung & Lava 4G mobile sold by store on Sunday =

750×80 100

3

75

5

100

× + 840 ×

= 360 + 360 = 720

Required percentage = =

148 720

×

4 7

720 –572 720

5

× 100

× 100 = 20 % 9

19. (d); Average number of Samsung & Lava 2G mobile sold by store on Sunday 20

=

25

(750×100+840×100) 2

=

150+210 2

= 180

Average number of Huwai & Oppo 3G mobile sold by store on Sunday 85

=

1

65

1

960×100×3+640×100×4 2

=

272+104 2

= 188

Required difference = 188 – 180 = 8 20. (c); Total Vivo mobile sold by store on Sunday = (840 ×

75 100

4

85

7

100

× + 960 ×

2

× ) + 56 3

= (360 + 544) + 56 = 960 Required percentage = = 34

320 960

1

960–640 960

× 100

× 100 = 33 % 3

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21. (e); Total A1 and A3 tickets sold in Patel stadium = 2400 ×

25 100

+ 2400 ×

75 100

×

5 9

= 600 + 1,000 = 1600

Total number of A1 and A2 tickets sold in Kaloor stadium = 1800 ×

20 100

+ 1800 ×

Required % =

1600−1200 1200

80 100

×

7 12

= 360 + 840 = 1200

× 100 =

400

1

1200

× 100 = 33 % 3

22. (d); A2 tickets sold in Nehru stadium = 1600 ×

85 100

×

9 17

= 720

A3 tickets sold in Raipur stadium = 2500 ×

70 100

4

× = 1,000 7

Required difference = 1000 − 720 = 280 23. (b); A3 tickets sold in Eden and Kaloor stadium together = 2000 ×

80 100

3

80

8

100

× + 1800 ×

×

5 12

= 600 + 600 = 1200

A2 tickets sold in Patel and Eden stadium together = 2400 ×

75 100

4

× + 2000 ×

Required ratio =

9 1200 1800

=

2

80 100

5

× = 800 + 1000 = 1800 8

3

24. (c); Total number of A1 tickets sold by all five stadiums together = 1600 ×

15 100

+ 2400 ×

25 100

+ 2000 ×

20 100

+ 1800 ×

20 100

+ 2500 ×

30 100

= 240 + 600 + 400 + 360 + 750 = 2350 Required average =

2350 5

= 470

25. (a); A1 and A2 tickets sold in Nehru stadium = 1600 ×

15 100

+ 1600 ×

85 100

×

9 17

= 960

A1 and A2 tickets sold in Raipur stadium = 2500 ×

30 100

+ 2500 ×

Required % = 35

1500−960 1500

70 100

3

× = 1500 7

× 100 = 36%

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26. (d); Income of movie B in second week =

30 100

× 180 = 54 crores

Income of movie C in third week =

10 100

× 200 = 20 crores

Required ratio = 27 : 10 27. (d); Income of movie A in first week =

50 100

× 210 = 105 crores

Income of movie A in second week =

30 100

× 210 = 63 crores

% drop =

42 105

× 100 = 40%

Similarly, Income of movies B in first week = 180 ×

40

= 72 crore

100

Income of movies B in second week = 180 ×

30 100 18

% drop =

=54 crore × 100 = 25%

72

Income of movie C in first week = 200 ×

60 100

= 120 crores

Income of movie C in second week = 200 ×

30

= crores

100 60

% drop =

120

× 100 = 50%

Income of movie D in week 1 = = 168 crore income of movie D in week 2 = = 24 crore % drop =

144 168

100 10 100

× 240 × 240

5

× 100 = 85 %

Hence D is the answer. 36

70

7

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28. (c); Foreign income of movie A =

80 300 20

× 210 = 56 crores

Foreign income of movie C = × 200 = 40 crores 100 Required ratio = 7 : 5 29. (a); Total income of B from foreign =

25 100 2

× 180 = 45 crores

Foreign income in second week = × 45 crores = 18 crores 5 Income earn by movie B in second week in India 30 = × 180 – 18 = 36 crores 100

30. (c); Second week income of movie D 10 = × 240 = 24 crore 100

10

Third week income of movie C = × 200 = 20 crore 100 Second week income of movie A 30 = × 210 = 63 crore 100 Second week income of movie B 30 = × 180 = 54 crore 100

Required % =

117–44 117

× 100 =

73 117

31. (e); Average budget for sport 125+130+150+170+160 735 = = = 147 5 5 Average budget for Education 105+125+130+200+220 780 = = = 156 5

Required% =

147 156

5

× 100 = 94.23 ≈ 94%

32. (c); Budget allocated for sport for male in 2022 3 = 160 × = 96 𝑐𝑟. 5 Budget allocated for sport for male in 2023 5 = 96 × = 120 cr 4

Required increase = 33. (b); Required ratio = 37

24 160

× 100 = 15%

105+125+130 170+190+220

=

360 580

= 18 ∶ 29

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34. (a); Budget allocated for Cricket and Badminton together in 2019 (5+4)

= 130 × (5+4+4) = 90 cr Budget allocated for Rural Education in 2022 6 = 220 × = 120 cr 11 Required difference = 120 – 90 = 30 cr 35. (d); Average of budget allocated for Health care in the year 2018-22 160+190+170+190+220 930 = = = 186 5 5 Average of budget allocated for Education in year 2018-22 105+125+130+200+220 780 = = = 156 5 5 Required sum = 186 + 156 = 342 36. (e); Let, total number of manufactured cars by TATA & HONDA in the year 2016 be 6x and 7x units respectively. ATQ, (100−15) (100−15) 75 80 7𝑥 × × − 6𝑥 × × 100 100 100 100 = 4590 75 85 80 85 7𝑥 × × − 6𝑥 × × = 4590 357𝑥

100 100 102𝑥

−

100

= 4590

80 25 1785𝑥−1632𝑥

=

100

400 4590×400

= 4590

𝑥= ⇒ x = 12,000 units 153 Total number of cars returned to HYUNDAI in 2017 = (12000 × 80 15 6) × × − 1920 100 100 = 8640 – 1920 = 6720 Total number of cars returned to MARUTI SUZUKI in 2017 75 15 = (12000 × 7) × × + 4830 = 9450 + 4830 = 14,280 units 100 100 Let total number of cars manufactured by HYUNDAI in 2017 be x units 70 15 𝑠𝑜, 𝑥 × × = 6720 𝑥= 38

100 100 6720×100×100 70×15

⇒ x = 64,000 units Adda247 Publications

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Let total number of cars manufactured by MARUTI SUZUKI in 2017 be y units so, 85 17.5 𝑦× × = 14280 100 100 14280×100×100

𝑦= ⇒ y =96,000 units 85×17.5 Required difference = 96000 – 64000 = 32,000 units 37. (c); Let total number of cars manufactured by MARUTI SUZUKI in the year 2016 be 100x units So, total number of cars manufactured by TATA in 2017 be 140x units ATQ, 75 10 80 12.5 140𝑥 × × − 100𝑥 × × = 120 100

21𝑥

100

100

100

− 10𝑥 = 120 ⇒ x = 240 Total number of cars returned to HONDA in 2017 = (240 × 100) × 80 12.5 × + 2010 = 2400 + 2010 = 4410 units 100 100 Let total number of cars manufactured by HONDA in year 2017 be x units ATQ, 70 22.5 𝑥× × = 4410 2

𝑥=

100 100 4410×100×100 70×22.5

⇒ x = 28,000 units

38. (c); Let total number of cars manufactured by HYUNDAI, MARUTI SUZUKI & HONDA in 2016 be 2y, 3y and 4y respectively. ATQ, 65

20

80

12.5

75

15

2𝑦×100×100+3𝑦×100× 100 +4𝑦×100×100 13𝑦

3𝑦

9𝑦

3

= 4040

+ + = 12120 50 10 20 101y = 1212000 ⇒ y = 12,000 Required average = = 39

(12000×2)×

120 125 112.5 +(12000×3)×100+(12000×4)× 100 100

28800+45000+54000 3

3

=

127800 3

= 42,600

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39. (d); Let total number of cars manufactured by HYUNDAI & HONDA in the year 2016 be 8x and 9x units respectively. ATQ, (100−20) (100−15) 65 75 8𝑥 × × + 9𝑥 × × = 39590 104𝑥

100 459𝑥

+

25 3959𝑥

80

100

100

100

= 39590

= 39590 ⇒ x = 4,000 units Total number of cars manufactured by HONDA in 2017 36000 = × 5 = 60,000 3 Total number of cars manufactured by HYNDUAI in 2017 32000 = × 3 = 48,000 400

2

Required percentage =

60000−48000 48000

× 100 =

12000 48000

× 100 = 25%

40. (d); Let total number of cars manufactured by TATA, HYUNDAI & HONDA in the year 2017 is 7y , 6y & 8y units respectively ATQ – 75 100−10 70 100−15 70 100−22.5 7𝑦 × × + 6𝑦 × × + 8𝑦 × × = 75810 189𝑦

100 357𝑦

+

+

100 217𝑦

40 100 50 945𝑦+714𝑦+868𝑦

100

100

100

100

= 75810

= 75810 2527y = 75810 × 200 ⇒ y = 6,000 Total number of cars manufactured by TATA, HYUNDAI & HONDA together in the year 2017 = 7 × 6000 + 6 × 6000 + 8 × 6000 = 42000 + 36000 + 48000 = 1,26,000 200

41. (b); Let total bike manufactured by all five companies = 100x So, total bike manufactured by Honda = 20x Ratio between total ‘200 CC’ & ‘150 CC bikes sold by Honda = 5 : 4 80 5 80 4 20x × × − 20𝑥 × × = 640 80𝑥

100 64𝑥

9

100

9

− = 640 ⇒ x = 360 9 Total ‘200 CC’ bike sold by Yamaha 22 75 (100−70) = 36000 × × × = 1782 9

100

40

100

100 Adda247 Publications

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42. (d); Let total bike manufactured by all five companies = 100x Total ‘150 CC’ bike sold by Bajaj 25

= 100x ×

90

×

100

100

×

40

= 9x

100

Total ‘150 CC’ bike sold by Hero = 100x×

15 100

×

90 100

×

35

= 4.2x

100 9𝑥−4.2𝑥

Required percentage =

4.2𝑥

2

× 100 = 114 % 7

43. (d); Let total bike manufactured by all five companies = 100x ATQ – 22x ×

75 100

×

70 100

− 18𝑥 ×

80

×

100

75 100

= 450

11.55x − 10.8x = 450 0.75x = 450 ⇒ x = 600 Total bike manufactured by Hero & Bajaj = 60000 ×

(15+25)

= 24000

100

44. (d); Let total bike manufactured by all five companies = 100x 15x ×

75 100

×

35 100

+ 25𝑥 ×

80 100

×

40 100

+ 18𝑥 ×

90 100

×

75 100

= 23124

3.9375x + 8x + 12.15x = 23124 x = 960 Total bike sold by Honda = 96000 ×

20 100

×

80 100

= 15360

45. (e); Total ‘200 CC’ bike manufactured by Honda = 84000 ×

20 100

×

3 5

= 10080

Total ‘150 CC’ bike sold by Yamaha = 84000 ×

22 100

×

75 100

×

(100−70)

Required percentage = 41

= 4158

100 10080−4158 10080

× 100 = 58. 75%

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Solutions (46-50): Let total employees travelling by metro & by Cab be 5𝑥 & 4𝑥 respectevely. 5𝑥 × x 75𝑥 100

−

15

100 40𝑥 100

– 4𝑥 × x

10 100

= 420

= 420 ⇒ X=1200

∴ Total employees travelling by metro = 1200 x 5 = 6000 & total employees travelling by cab = 1200 x 4 = 4800 46. (c); Total employees travelling on Friday by metro = Total employees travelling on Sunday by cab = ATQ, 960 – y + 2 x 1000 =

75

[

15

100 100

× 6000 +

25 100

50

300 20

100

x 6000 = 1000

x 4800–y=960 – y

× 4800]

2960 – y = .75 x 2100 Y = 2960 - 1575 = 1385 47. (b); No. of employees travelling by metro on Wednesday & Thursday together. =

15 100

× 6000 +

10 100

× 6000

= 900 + 600 = 1500 No. of employees travelling by Cab on same days together. =

25 100

× 4800 +

20

× 4800

300

= 1200 + 320 = 1520 Required difference = 1520 – 1500 = 20 48. (d); Fare per person travelling by metro =

5 12

× 120 = Rs. 50

Fare per person travelling by cab . 7 = × 120 = 𝑅𝑠 70 12

Require ratio =

42

20 ×4800×70 300 10 ×6000×50 100

= 56 : 75

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49. (a); Total employees travelling by metro on Monday 20 = × 6000 = 1200 100

48000

∴ Per person fare = = 𝑅𝑠. 40 1200 ∴ Total fare on Monday of all employees travelling by cab 25 = × 4800 × 40 = 𝑅𝑠. 48000 100

50. (e); Total No. of employees travelling by metro on Friday & Saturday together 50 15 = × 6000 + × 6000 300 100 = 1000 + 900 = 1900 Total no. of employees travelling by cab on same days together 20 10 = × 4800 + × 4800 100 100 = 960 + 480 = 1440 1900−1440 Required Percentage = × 100 1440 approximately 32% 51. (b); In city A, Number of students appeared in mains exam 75 = 96000 × = 72000 100 Number of students passed in mains exam 12.5 = 72000 × = 9000 100 In city B, Number of students appeared in mains exam 60 = 80,000 × = 48000 100 Number of students passed in mains exam 25 = 48000 × = 12000 100 Total number of students who failed only in quant in city A = 3 × 12000 = 36000 Total number of students who failed in both subjects in city ‘A’ 25 = 72000 – 9000 – 36000 – 72000 × = 9000 100 Total number of students who failed in both exam in city ‘B’ = 9000 – 3000= 6000 43

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Total number of students who failed in Reasoning = 48000 × 37.5 + 6000 100 = 18000 + 6000 = 24000 36000−24000 Required % = × 1000 = 50% 24000 Vein diagram for failed students:-

52. (e); In city D, Number of students appeared in mains exam 85 = 1,60,000 × = 1,36,000 100 Number of students passed in mains exam 25 = 1,36,000 × = 34,000 = 𝑥 100 Total number of students who failed in reasoning = 34,000 + x Total number of students who failed in Quant 75 = 1,36,000 × − 34,000 = 68,000 100 Let Total number of students who failed in only Quant = a And Let Let Total number of students who failed in both subjects = b 44

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ATQ, 𝑎 + 𝑏 = 68,000 And 34,000 + 𝑏 = 𝑎 + 34,000 So 𝑎 = 𝑏 = 34,000 Total number of students who failed in Reasoning = 34,000 + 34,000 = 68,000 Total number of students who failed in only Quant = 34,000 68,000 2 Required Ratio = = 34,000

1

53. (b); Total number of students who failed in Quant in city number of students who failed in Mains exam – Total students who failed in only Reasoning 72 70−40 = 1,25,000 × × = 27,000 100 100 Total number of students who failed in Quant in city number of students who failed in Mains exam – Total students who failed in only Reasoning 2 68.75−25 = 72,000 × × = 21,000 3 100 Required difference = 27,000 − 21,000 = 6,000

‘E’ = Total number of

‘F’ = Total number of

54. (e); Total number of students who failed in Quant in city ‘B’ 60 75−37.5 = 80,000 × × = 18,000 100 100 Total number of students who failed in at most one subject in city 18,000 ‘C’ = × 91 = 54,600 30 In city ‘C’ Total number of students who failed in at most one subject in city ‘C’ = Total number of students who failed in Reasoning only + Total 45

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number of students who failed in only Quant + Total number of students who passed in both subjects Total number of students who passed in both subjects 65 20 = 1,20,000 × × = 15,600 100 100 Total number of students who failed in only Reasoning 65 50 = 1,20,000 × × = 13,000 100 300 Total number of students who failed in only Quant = 54600 − 15,600 − 13,000 = 26,000 Total number of students who failed in Quant = 120000 ×

65 100

×[

80 100

−

50 300

] = 49,400

Total number of students who failed in both exams = 49,400 − 26,000 = 23,400 23,400 Required % = × 100 = 30% 78000

55. (c); Total number of students who failed in atleast one subject in city ‘A’ 75 87.5 = 96,000 × × = 63,000 100 100 Total number of students who failed in at most one subject in city ‘F’ = 63,000 − 24,000 = 39,000 Total number of students who failed in only Quant in city ‘F’ 2 31.25+25 = 39,000 − 72,000 × × = 12,000 3 100 Total number of students who failed in both subjects 2 68.75−25 = 72,000 × × − 12,000 = 21,000 − 12,000 = 9,000 3 100 Total number of students who failed in only reasoning in city ‘A’ 75 25 = 96,000 × × = 18,000 100

46

100

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Required difference = 18,000 − 9,000 = 9,000

56. (b); Let Number of girls and boys in Aravali in 2012 be x and y respectively. In 2013 Number of girls = 1.2 x Change in girls = 0.2x 0.2𝑥 Change in boys is = × 100 = 0.16x 125 Total increment = 0.2x + 0.16x 0.36x = 270 270×100 x= ⇒ x = 750 36 Number of boys in Aravali in 2013 2 = × 1.2 × 750 = 600 3

57. (c); Let Number of Girls in Aravali be x. Let Number of girls who left Aravali and joins Nilgiri is ‘a’ Total strength of Aravali increases which means change in number of boys is = 270 + a 47

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Total strength of Nilgiri decreases, which means change in number of boys in Nilgiri is = 250 + a 270+𝑎 250+𝑎

=

16 15

⇒ a = 50

Number of boys increases in Nilgiri = 250 + 50 = 300 58. (a); Let Number of boys in Udaygiri in Shivalik in 2012 = x Let Number of Girls in Udaygiri in 2012 = y So, Number of Girls in Shivalik = y + 540 In 2013, Let Number of boys in Udaygiri and Shivalik = 20a ⇒ Number of Girls in Shivalik = 35a And, Number of Girls in Udaygiri = 12a ATQ x + y + 150 = 32a … (i) x + (y + 540) + 300 = 55a … (ii) Solving (i) and (ii) a = 30 Shivalik strength = 55a = 55 × 30 = 1650 59. (e); Let in 2012 Total strength of Udaygiri and Nilgiri be 4x and 5x respectively And total strength of Udaygiri and Nilgiri in 2013 be 5y and 8y respectively ATQ, 4x – 5y = 150 … (i) Two cases formed Either 8y – 5x = 250 ... (ii) Or 5x – 8y = 250 ... (iii) On solving (i) and (iii) it does give-ve integer value So, on solving (i) and (ii) x = 350 ⇒ y = 250 Strength of Nilgiri in 2012 = 1750 48

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60. (d); Total strength in 2013 = 1030 + 970 = 2000 Number of boys in Aravali in 2013 2000 2 = × 2 × = 160 10 5 Number of boys in Udaygiri in 2013 2000 5 = × 2 × = 250 10 8 Required Difference = 250 – 160 = 90

49

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1

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Chapter

14

Arithmetic Data Interpretation

BEST APPROACH TO SOLVE THE QUESTIONS

Number of mobiles

INTRODUCTION: DI stands for DATA INTERPRETATION. Interpretation means representing a data in a sensitive way after analyzing it well. In all the banking exam or particularly quant section of bank, DI plays vital role. The basic DI consists of pie table, bar graph, numerical value table or graph. Look at this example: 900 800 700 600 500 400 300 200 100 0 Mon

Tue

Wed

Thu

Fri

Sat

Instead of telling ‘number of mobiles sold’ for each day it is represented in the graph. So that it is more visible and interested for a reader in this form. Common form of DI are ratios, percentages or average. ARITHMETIC DI Arithmetic DI means representing arithmetic equations and conditions in bar graphs, line graph, pie graph or in any other form of DI. The questions on arithmetic DI also cover major portion in mains of any banking exam. So 2

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it is necessary to understand the actual concept. Easy way to understand this is through an example.

Total Road construct = 3600 meters E 46° D 90°

A 72°

B 80° C 72°

In this pie chart, portion of ‘Road construct’ is given for 5 different 72 companies. For eg. Company A constructs × 3600 = 720 meters road. 360

Now to make question more complicated different individual for each team may be given. Let us consider through a bar graph. 32

No. of Labour in 1 team

28

24 20 16 12 8

4 0 A

3

B C Name of teams Adda247 Publications

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E

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Here we have considered number of individuals in each team. Now they may ask you questions like “How many meters 1 member of team A constructs”. It is definitely easy one. We just had to do 720 = 40 meters. 18

or to make it more complicated they may ask you “How many day will 8 members of team C will take to construct 3600 meters road?” Do it yourself. HINT: Calculate road constructed by an individual of team C in 1 day as we have calculated previously. Then calculate how many days 8 men will take to construct given road. A key point to solve any arithmetic DI is that to understand what had been given in DI. Then consider these as an individual separate questions from arithmetic, also these questions are way easier. Now let us consider another example: Series 1

25%

20%

20% 15%

15% 12%

10%

10%

5%

5% 0% A

B

C

D

E

Look carefully, here a % graph is given and nothing have been mentioned. Examiner may mention it “% of distance covered by these (A, B, C, D & E) individuals in 1 hour and total distance is 100 km” OR “ % of work done by individuals in given time” OR “ Profit/loss % earned on selling these items for shopkeeper” 4

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“ Interest rate given by a bank to different individuals” or many other things. Point is arithmetic DI is not as hard as student consider it in exams, it is always easier than actual arithmetic questions that are asked in exam. Before solving a question, we must exactly know the information he had given. For eg. Consider that this graph shows profit % for selling these items. In question, if examiner ask to calculate actual selling price, then he must mention the CP or some other clue in exam. Example: Profit earned on selling item C is Rs. 160, what is amount of profit earned for item D, if both items have same selling price? Solution: This is a typical DI question. Now look, if we know that 20% means Rs 160 we can calculate actual CP of item C. (Profit % is always calculated on CP). 100 CP OF ITEM C = 160 × = 800. 20

It means SP of item C is Rs. 960 Also SP of item D is Rs. 960. And then we can calculate CP and amount of profit of item D.

5

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Practice Exercise Based on new Pattern Direction (1 - 4): Given below bar graph shows number of hours taken by six person to complete a task individually. Read the data carefully and answer the questions: 50 40 30 20 10 0 Neeraj

Aniket Saurabh

Gopal

Veer

Sameer

1. Neeraj, Saurabh and Gopal start working together but due to bad health 1 1 the efficiency of Neeraj and Gopal decreases by 12 % and 33 % 2 3 respectively. Then find in how many hours task will be completed by all three? 1 1 1 (a) 11 hours (b) 10 hours (c) 12 hours 1

4

(d) 9 hours 4

4 1

4

(e) 13 hours 4

2. Veer and Sameer started working together on another task, where Sameer work with 25% less efficiency. Veer and Sameer work for y hours and remaining work complete by Aniket in (y + 1) hours, if ratio of work done by Veer and Sameer together and by Aniket alone is 2 : 1, then in how many hours Neeraj will complete same task alone? 1 1 1 (a) 15 hours (b) 13 hours (c) 17 hours 2 1

(d) 11 hours 2

6

1

2

2

(e) 9 hours 2

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3. If Divyaraj can do 50 % more work in one hour as Neeraj can do in one hour, while Manish can do 25% less work in one hour as Aniket can do in one hour. Saurabh started working alone and after some time he left and remaining work completed by Divyaraj & Manish together in 11.75 hours more than time for which Saurabh worked. Then find total time in which work is completed? 1 1 1 (a) 14 hours (b) 12 hours (c) 10 hours 4 1

1

4

4

(d) 16 hours (e) 8 hours 4 4 4. Neeraj, Aniket, Veer and Sameer worked together on a work in first hour, while Saurabh & Gopal together destroyed the same work in second hour. If this work continues till the total work completed. Find how many hours required to complete the whole work? 270 270 270 (a) 22 hours (b) 20 hours (c) 18 hours (d) 24

401 270

401

hours

(e) 16

401 270 401

401

hours

Direction (5 - 9): The given bar-graph shows the number of marbels of different colors which are contained by two different bags X and Y. Read the data carefully and answered the following questions. X

7

Y

6 5 4 3 2 1 0 Yellow 7

Red

Black

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5. If two marbels are picked at random from bag Y then what is the probability that both marbels are blue in color? (a) (d)

21 190 11 95

(b) (e)

19 190 10

(c)

63 190

99

6. If two marbels are picked at random from bag X then what is the probability that both are yellow in color? (a) (d)

3 53 5 51

(b) (e)

4 51 8

(c)

7 59

59

7. If two marbels are picked at random from bag Y then what is the probability that one is red and one is black in color? (a) (d)

1 19 8 95

(b) (e)

6 95 13

(c)

7 95

190

8. If two marbels are picked at random from bag X then what is probability that either both are black or both are blue in color? (a) (d)

1 15 3 37

(b) (e)

4 105 1

(c)

9 115

17

9. If three marbels are picked at random from bag Y then what is the probability that one is black, one is blue and one is yellow in color? (a) (d)

17 95 13 95

(b) (e)

49 285 47

(c)

53 381

360

Directions (10-14): Given below pie chart (I) shows percentage distribution of total filled quantity of water in seven tanks. Line graph shows time (minutes) taken by seven pipe A, B, C, D, E, F and G in filling tanks P, Q, R, S, T, U and V respectively. Give the answer of the question according to given data : 8

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Total quantity of seven tanks = 1500 ℓ V 12%

P 15%

U 18%

Q 12%

Time (in minutes)

T 15%

R 8% S 20%

65 60 55 50 45 40 35 30 25 20 A

B

C D Pipes

E

F

G

10. A man has a tank which has quantity of water equal to the quantity of water in P and T together. He opened pipe B and D together for filling his tank. After X minutes man closed both pipes and opened pipe C and E for filling remaining tank. If tank filled in next (x + 15) minutes then find how many liter of water filled by B and D together ? (a) 160 l (b)180 l (c)150l (d)120l (e) 144 l 9

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11. For filling tank S, man opened pipe D which fills the tank with 20% more efficiently. After X minutes man close the pipe D and opened pipe C and G together. Pipe C fills the tank with 25% more efficiently. If pipe C and G fills remaining tank in 15 minutes then find the value of X ? (a) 12 minutes (b) 10 minutes (c) 15 minutes (d) 8 minutes (e) 18 minutes 12. Quantity of another tank K is 120% of total quantity of all seven tank. For filling the bigger tank man opened pipes alternatively in such a way that three pipe B, D and G together for first minute, pipe A alone for second minute and pipe F and C together for third minutes. Find the total time required in filling the tank alternatively? (a) 151 7 minutes (b)141 7 minutes (c)151 minutes 9

9

(d)150 minutes

(e) 170 minutes

13. A man have a tank M, which has equal quantity of water as tank S and U together . For filling tank M man opened pipe A, C and G together for some minutes man thought tank will be fully filled, but it takes 7

17 19

minutes more because of a leak. Find in what time leak will empty the fully filled tank? (a) 3hr (b) 3 2 hr (c)2 1 hr (d) 3 1 hr 6

3 1

6

(e) 4 hr 3

14. A man have a tank of 570 liter. Man opened three pipes B, D and C together for filling the tank but pipe C fills the tank at 175% of its initial Efficiency. There is a leak at a height of 2/3 from the bottom of tank ,which empty the tank with half of the efficiency of what B, D and C filling the tank. Man noticed the leak after 32 minutes and closed all pipes and leak. Find how many liter of tank remain unfilled? (a) 90l (b) 76l (c) 78l (d) 85l (e) 96l 10

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Directions (15 − 𝟏𝟗): Given below the table shows Investment of five person, time and share of profit. Some data are missing, calculate that according to the question and answer the questions – Person Investment (Rs) Time (months) Profit share (Rs.) P — 12 65700 Q 32000 — — R — — — S 24000 — — T — — 24300 15. P, Q and S invested in a business together, investment of P is 75% of investment of S and all three invested for same time. If Q and S gets extra 12% and 15% of total profit respectively and remaining profit is distributed according to their share, then find the total profit. (a) 270000 Rs. (b) 370000 Rs. (c) 375000 Rs. (d) 275000 Rs. (e) 325000 Rs. 16. Person Q and R invested in the ratio of 4 : 3 and ratio between time period of Investment Q to R is 5 : 3. If both person agree that 65% of the total profit should be divided equally and remaining profit is to be divided into ratio of their capital. If Q gets 10010 Rs. more to R. then find the total profit share of R ? (a) 27765 Rs. (b) 28190 Rs. (c) 42497 Rs. (d) 34307 Rs. (e) 32695 Rs. 17. Q started a business with his investment, after some month S came to joined with him and invest his amount in business. At the end of 3 year, the ratio of profit of Q to S is 3 : 2. Find after how many month S joined the business ? (a) 4 months (b) 6 months (c) 3 months (d) 5 months (e) 6 months 11

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18. If T received Rs. 24300 as profit out of the total profit of Rs. 40500 which T and R earned at the end of one year. If T invested Rs. 81000 for 9 months, whereas R invested his amount for the whole year, what was the amount invested by R ? (a) 30500 Rs. (b) 40500 Rs. (c) 35500 Rs. (d) 32500 Rs. (e) 40050 Rs. 19. If Q and R invested into the ratio of 8 : 9 and R and S Invested into the ratio of 3 : 2. At the end of the year if they all got a total profit of 37030 Rs. then find the share of profit of Q, R and S individually ? (a) 14490 Rs, 9660 Rs, 12880 Rs. (b) 9660 Rs, 17710 Rs, 11005 Rs. (c) 12880 Rs, 14490 Rs, 9660 Rs. (d) 12434 Rs, 13594 Rs, 11006 Rs. (e) None of these Direction (20- 23): Data about investments of different persons is given below. Study the data and solve the questions carefully. → Investment of Neeraj is 50% more than investment of Aman while investment of Neeraj is 25% less than that of Sandeep.

Total Investment = 40,000

Rakesh, 30%

Saurabh, 40%

Sandeep, 30%

20. Neeraj and Saurabh started a business together. After 8 months of business they invested Rs 1000 per month for every month. If annual profit is Rs 8000 then find Saurabh’s profit out of total profit? (a) Rs 2950 (b) Rs 4000 (c) Rs 5050 (d) Rs 4450 (e) Rs 4850 12

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21. Aman and Rakesh started a business together. Rakesh left the business ‘x’ months before the completion of year while Aman increased his investment by Rs 2000 after 8 months and then after 2 months more he increased his investment by Rs 2000 again. If ratio between profit share of Aman and Rakesh is 7 : 8 then Rakesh worked for how many months? (a) 8 months (b) 4 months (c) 6 months (d) 2 months (e) 10 months 22. Sandeep and Neeraj started a business together. After 8 months, Neeraj is replaced by another partner, Satish whose investment is Rs 4000 more than investment of Saurabh. If Satish worked for 3 months, then find the profit share of Sandeep if annual profit is Rs 11,500. (a) Rs 2500 (b) Rs 3000 (c) Rs 4500 (d) Rs 5000 (e) Rs 6000 23. Aman invested his amount in a scheme which after 20% p.a. at C.I. for 2 years while Saurabh invested his amount in a car whose value will depreciate at 20% p.a. every year. Find the total value of amount Aman and Saurabh will have after 2 years. (a) Rs. 31,680 (b) Rs. 18,880 (c) Rs. 26,880 (d) Rs. 16,320 (e) Rs. 22,880 Direction (24 – 25): Given below table shows quantity of four different liquid in four different vessels. Read the data carefully and answer the questions. Vessels P Q R S 13

Milk (ml) 240 160 — 160

Water (ml) — 80 144 —

Orange Juice Mango juice (ml) (ml) 144 — — 210 192 — — 120

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24. In vessel P quantity of water is 50% of quantity of milk in same vessel, while quantity of mango juice in vessel P is 6.25% less than quantity of Orange juice in vessel R. In vessel S quantity of water is 50% more than quantity of milk in same vessel and quantity of Orange juice in vessel S is 40 ml less than quantity of water in same vessel. If 114 ml mixture from vessel P and 180 ml from vessel S taken out and mixed in vessel T, then find percentage of orange juice in vessel T? 25 25 25 (a) 29 % (b) 25 % (c) 27 % (d) 23

147 25

147

%

(e) 21

147 25

147

147

% 1

25. Quantity of milk in vessel R is 33 % less than quantity of water in 3 same vessel and quantity of orange juice in vessel Q is 25% less than quantity of milk in same vessel. For making 1086 ml mixture solution mixture of vessel Q & R mixed together, if ratio of quantity of total mango juice in vessel R to total quantity of four liquid in vessel R is 7 : 43, then find the ratio between total mango juice in resulting mixture to total given mixture? 49 49 49 (a) (b) (c) (d)

161 49

(e)

121

151 49

181

111

Direction (26-30): - Table given below shows details of 3 types of item of a garment shop MRP, discount% (on MRP) and certain offer. Shopkeeper have shoes, Jeans & T-Shirts in 2 types of each. S₁, S₂; J₁, J₂; and T₁, T₂ represents the types of two types of shoes, jeans and T-shirts respectively. Items

Types MRP (Rs) Discount Shop offer

Shoes S₁ 1200 0% 

S₂ 1647 1 11 % 9 

Jeans J₁ 1200 0% 

J₂ — 25% 

T-shirts T₁ T₂ 400 600 10% 0%  

Note: Shop offer: Buy 2 get 2 free  → Offer given  → Offer not given 14

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26. If a person wants to buy T-shirts, what is minimum average price/tshirts he could get? (a) Rs. 400 (b) Rs. 360 (c) Rs. 300 (d) Rs.600 (e) Rs. 500 27. If cost price of J₂ for shopkeeper is Rs. 1000. Find MRP of J₂ if he earns a profit of 20% on selling a single piece. (a) Rs. 1600 (b) Rs. 1800 (c) Rs. 1500 (d) Rs. 1700 (e) None of these 2

28. Find the cost price of S₂ if he earns a profit of 14 % on C.P. 7 (a) Rs. 1464 (b) Rs. 1098 (c) Rs. 1281 (d) Rs. 1379 (e) None of these 29. A boy buys 5 pieces of T₂, 4 pieces of J₁, & 1 piece of S₁ including offer. Find the amount paid by him (in Rs.) (a) 6000 (b) 7200 (c) 4800 (d) 3600 (e) 5400 30. If a person buys 4 pieces of T₂, 2 pieces of T₁ & 4 pieces of J₁ including offer. Find overall discount% obtained by him. (a) 40% (b) 44% (c) 42% (d) 46% (e) 48% Directions (31-35): Given below line graph shows the distance travelled by car A in given time intervals and table shows ratio of average speed of car A to average speed of car B in same time intervals. Both car started at 12:00 Noon. 90 80

70 60 50 40

30 12:00 – 1:00 1:00 – 2:00 15

2:00 – 3:00

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3:00 – 4:00

4:00 – 5:00

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Time Internal 12:00 – 1:00 1:00 – 2:00 2:00 – 3:00 3:00 – 4:00 4:00 – 5:00

Ratio 6:5 14 : 11 6:7 4:3 4:7

31. What is average speed of car A during whole journey? (a) 63 km/hr (b) 65 km/hr (c) 62 km/hr (d) 70 km/hr (e) 68 km/hr 32. If both cars started from same point at same time, what is difference between distance travelled by them till 3: 00 PM. (a) 20 km (b) 15 km (c) 24 km (d) 16 km (e) 12 km 33. Average speed of B between 2: 00 PM to 4: 00 PM is what % more or less than average speed of A in same time interval? 1

(a) 7 % 7

5

(d) 35 % 7

2

(b) 14 % 7

4

(c) 28 % 7

(e) None of these

34. If B have to cover a distance of 540 km till 9 : 00 PM, even after having a rest of 45 minutes between 5 : 00 to 5 : 45 PM. What should be his average speed for 4 hours between 5:00 PM to 9:00 PM. 1

(a) 60 km/hr

(b) 59 km/hr

(d) 58 ¾ km/hr

(e) 56 ⅔ km/hr

7

4

(c) 58 km/hr 7

35. If A consumes 1liter petrol for every 15 km, and if A is 25% more economic than B. What is the ratio of petrol consumed by them till2:00PM? (a) 91 : 92 (b) 93 : 97 (c) 104 : 105 (d) 106 : 107 (e) 103 : 106 16

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Directions (36-40): The following line graph shows principal amount(in 10000) submitted by 5 persons in a scheme which offers simple interest and table shows the rate of interest (ROI) per annum obtained by these persons, time (in years) for which they invested their amount and third column shows ratio of amount obtained after they withdraw their money to that of interest obtained. Some figures shown are missing. Calculate them according to question. 12 11 10 9 8 7 6 5 4 3 A

B Persons A B C D E

C ROI — 16% — 8% 15%

D Time 5 — 6 5 —

E

A:I 8:3 41 : 16 — — 29 : 9

36. What is the time period in months for which B invested his money? (a) 60 (b) 48 (c) 64 (d) 72 (e) None of these 37. If the ratio of amount obtained by A to that of C is 4: 3. Find the ROI obtained by C. (a) 8% (b) 12% (c) 8.5% (d) 10% (e) 12.5% 17

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38. If a new person Vikas deposit half of the money invested by A, for 3 more years than C, calculate his ROI, if SI obtained by him is same as SI obtained by D. 1

1

1

(a) 11 %

(b) 9

(d) 14 %

(e) None of these

9 2 7

11

%

(c) 12 % 3

39. What is the ratio of time period for which E invested to that of B? (a) 1 : 3 (b) 7 : 2 (c) 5 : 4 (d) 4 : 5 (e) 3 : 4 40. If after drawing his whole money, D invested 50% of this amount in CI for 2 years at ROI/annum of 10% and remaining of the money he kept with him. Calculate total money with him after 2 years. (a) Rs. 169400 (b) Rs. 84700 (c) Rs. 154700 (d) Rs. 167400 (e) Can’t be determined Directions (41-44): Bar-graph shows the percentage distribution of distance covered in upstream and downstream by Rahul on different dates. Upstream distance=6000 km

Downstream distance=2000 km

32

28 24 20 16

12 8 4 0 1st Aug. 18

2nd Aug.

3rd Aug.

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4th Aug.

5th Aug.

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Table given below shows the speed of current in km/hr on different dates. Date 1st Aug. 2nd Aug. 3rd Aug. 4th Aug. 5th Aug.

speed of current (in km/hr) 10 — 4 — 3 1

41. If time taken by Rahul to swim upstream on 1st August. is 61 hr. more 3 than time taken by him to swim downstream on same date, then find the speed of Rahul in still water? (a) 16 km/hr (b)30 km/hr (c) 25 km/hr (d) 20 km/hr (e) None of these 42. If speed of Rahul in still water on 2nd Aug. is 8 km/hr and speed of Rahul in still water on 3rd Aug. is 25% more than his speed on 2nd Aug. 2 and time taken by him to travel upstream on 2 nd August is 166 % more 3 than time taken by him to travel downstream on same date, then find the seven times of the speed of current on 2nd Aug.? (a) 8 km/hr (b) 6 km/hr (c) 4 km/hr (d)None of these (e) 2 km/hr 43. Speed of current on 4th Aug. is 50% of downstream speed of Rahul on 5th August, then time to cover 128 km upstream on 4th Aug. is ? [Given that time of Rahul to cover upstream distance on 5th August is 80 hr and assuming that speed of Rahul in still water is same on both days] 1 2 (a) 12 hr (b) 6 hr (c) 10 hr 3 3 (d) 21 hr (e)None of these 44. If ratio of speed of current on 2nd August to 4th August is 2 : 3 & speed of Rahul on both days are same and time in upstream on 4th August. is 24 times the time taken in downstream on 2th August then find the speed of Rahul in still water if speed of current on 3rd August is 50% of speed of current on 2nd Aug.(approximately)? (a) 21 km/h (b) 23km/h (c) None of these (d) 25km/h (e) 20km/h 19

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Directions (45-48): The following bar-graph shows the ratio of speed of three different trains Rajdhani exp., Shatabadi exp. and Duranto exp. on five different days of a week. And line graph shows the number of coaches attached to Rajdhani express on different days of week. Length of each train is the sum of the length of all coaches and length of engine. 7

Rajdhani Exp

Shatabadi Exp

Duranto Exp

6 5 4 3

2 1 0 Monday

Tuesday

Wednesday

Friday

Sunday

30 25 20

15 10 Monday

Tuesday Wednesday

Friday

Sunday

Length of each coaches for every train = 15 m And length of engine for every train = 20 m 20

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45. On Tuesday, Rajdhani exp. crosses a tunnel of some length in 12.5 sec while Duranto express having length 25% more than that of Rajdhani exp. crosses the same tunnel on the same day in 11.9 sec. Then find the time taken by Rajdhani exp. and Shatabadi exp. on the same day to cross each other if they are running in opposite direction and the length of Shatabadi exp. is 10 m less than that of Rajdhani exp. (a) 8 sec (b) 9 sec (c) 10sec (d) 11 sec (e) 12 sec 46. On Sunday, Duranto exp. having length 110 m less than that of Rajdhani 8

exp. crosses Rajdhani exp. travelling in opposite direction in 8 sec 9

then find the approximate time taken by Shatabadi exp. to cross Duranto exp. on the same day when both are running in the same direction and length of the Shatabadi exp. is 20% less than that of Rajdhani exp. (a) 79 sec (b) 81 sec (c) 75 sec (d) 84 sec (e) 72 sec 47. On Friday, the speed of Rajdhani exp. was 90 km/hr and on the same day, Shatabadi exp. started from station P and going to Q while Duranto started from station Q and going to P. If speed of Shatabadi is increased by 10% after first hour and 20% after second hour whereas the speed of Duranto is increased by 20% after first hour and 25% after second hour then they meet in 3 hours. Find the distance between P and Q. (a) 947 km (b) 952 km (c) 955 km (d) 957 km (e) 960 km 48. Duranto exp. having length 28% more than that of Rajdhani exp. on Monday crosses a platform of same length in 12.8 sec. Then find the difference between average speed of shatabadi and Rajdhani and the speed of Duranto exp. on Monday.(in km/hr) (a) 32.8 (b) 59.4 (c) 42.6 (d) 54.2 (e) 58.8 21

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Directions (49-53): Pie chart given below shows length of six different trains and table given below shows ratio between speed of six trains on three different days. Study the data carefully and answer the following questions. Total length = 1600m F, 8%

A, 16%

E, 22% B, 18% D, 12% C, 24%

Train A B C D E F

Speed on Monday 2 3 4 4 6 4

: : : : : : :

Speed on Tuesday 3 4 6 4 9 5

: : : : : : :

Speed on Wednesday 2 5 5 7 5 3

49. On Wednesday, train ‘B’ crosses train ‘D’ coming from opposite direction in 6 seconds. If speed of train ‘B’ on Monday is 97.2 km/hour then in how much time train ‘F’ can cross train ‘D’ on Monday if train ‘D’ is coming from opposite direction and speed of train ‘F’ on Monday is 20 m/sec. (2 Marks) (a) 6 seconds (b) 8 seconds (c) 10 seconds (d) 12 seconds (e) 14 seconds 22

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50. Train ‘C’ start from Delhi on Monday at 7:00 p.m and reach Kanpur on next day at 3:00 pm. In return journey on Tuesday, train ‘C’ start from Kanpur at 6 : 00 pm and reach Jaipur which is 180 km ahead of Delhi at 5 : 48 pm on Wednesday. Find the time taken by train ‘C’ to cross a pole on Monday? (2 Marks) (a) 8 seconds (b) 16 seconds (c) 19.2 seconds (d) 14.4 seconds (e) 28.8 seconds 51. On Monday, train ‘A’ takes 2.5 hours more to cover 900 km distance than train ‘C’. If train ‘A’ can cross a platform of length 128 in 12.8 seconds on Tuesday then find in how much time (in seconds) train ‘C’ can cross two poles 66 m apart from each other on Tuesday? (2 Marks) (a) 12 seconds (b) 16 seconds (c) 20 seconds (d) 24 seconds (e) 30 seconds 52. Ratio between speed of train ‘E’ to train ‘F” on Monday is 3 : 2. On Tuesday train ‘E’ cross train ‘F’ running in same direction in 24 seconds then find the time in which train ‘E’ can overtakes train ‘F’ on Wednesday? (2 Marks) (a) 48 seconds (b) 24 seconds (c) 12 seconds (d) 36 seconds (e) 60 seconds 53. Ratio between time taken by train ‘B’ to train ‘D’ to cross a pole on Monday is 1 : 1. The time taken by train ‘B’ to cross a pole on Wednesday is what percent more/less than time taken by train ‘D’ to cross a pole on Monday? (2 Marks) (a) 30% (b) 40% (c) 50% (d) 60% (e) 70% Direction (54-58): Table given below shows profit percentage earned on selling two different items X and Y and discount percentage offered by five retailers on these items. Mark price of each article sold by each retailer is same while cost price of article for each retailor may vary. Study the data carefully & answer the following question 23

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Items→ X Y Retailor Profit % Discount % Profit % Discount % ↓ A 20% – – 15% B – 26.5% 20% 32.5% C 25% – 60% 24% D 20% 34% – 37% E 35% 46% – 28% 54. On article ‘X’, 28% discount is offered by ‘A’. If selling price of article ‘Y’ sold by ‘A’ is Rs 312 more than selling price of article ‘X’ sold by ‘A’ then find profit percent earned by ‘A’ on selling article ‘Y’ given that average of cost price of both article for ‘A’ is Rs 1520. (2 Marks) (a) 18.5% (b) 22.5% (c) 27.5% (d) 32.5% (e) 37.5% 55. Discount % on article ‘X’ offered by ‘C’ is 15% while profit % of article 1 ‘Y’ sold by ‘E’ is 33 %. If difference between cost price of article ‘Y’ sold 3 by ‘B’ and ‘E’ together is Rs 216 more than cost price of article ‘X’ sold by ‘C’ and ‘E’ together then find the cost price of article ‘Y’ sold by ‘C’? (2 Marks) (a) Rs. 4940 (b) Rs. 3420 (c) Rs. 3800 (d) Rs. 4180 (e) Rs. 4560 56. Find the ratio between cost price of article ‘X’ for ‘B’ to cost price of article ‘Y’ for ‘D’. If profit % of article ‘X’ sold by ‘B’ and profit % of article ‘Y’ sold by ‘D’ is 68% and 20% respectively. (2 Marks) (a) 2 : 3 (b) 5 : 6 (c) 1 : 2 (d) 5 : 8 (e) 5 : 7 57. Selling price of article ‘X’ sold by ‘E’ is same as cost price of article ‘Y’ for ‘E’. Find the profit % earned by ‘E’ on selling article ‘Y’? (1 Mark) 2 (a) 15% (b) 16 % (c) 25% 1

(d) 33 % 3

24

3 2

(e) 41 % 3

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58. Seller A marked article ‘X’ 100% above its cost price. If on selling both article he earns total 25% profit, then find selling price of article ‘X’ sold by ‘A’ is what percent less than cost price of article ‘Y’ for ‘A’. (2 Marks) 1

(a) 10%

(b) 9

(d) 20%

(e) 15%

11

1

%

(c) 8 % 3

Directions (59-63): Line graph given below shows the distance between Delhi to five different cities in kilometer and Table given below shows the speed of five different cars in km/hr

Distance from Delhi (in km)

3500 3000 2500 2000 1500 1000 500 0 City A Cars P Q R S T

City B

City C

City D

City E

Speed (in kmph) 40 — 60 — 75

NOTE: - Some data is missing you have to calculate according to question. 25

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59. Time taken by car ‘P’ to travel from city ‘E’ to Delhi and then Delhi to city ‘B’, is equal to the time taken by car ‘R’ to travel from Delhi to city ‘A’ and then city ‘A’ to city ‘B’. Find the distance between city ‘A’ and city ‘B’. (a) 5650 km (b) 5750 km (c) 5450 km (d) 5550 km (e) 5320 km 60. Find the approximate time car ‘T’ takes to reach city ‘E’ from city ‘A’ if city ‘A’ and city ‘E’ is north and east direction of Delhi respectively. (a) 24 hours (b) 27 hours (c) 20 hours (d) 36 hours (e) 42 hours 61. Car Q and Car S start from Delhi for city B and city C respectively and they reached in equal time. If Car Q and Car S starts from city B and city D respectively at same time and move towards each other, then time taken by car Q to cross car S is what percent of the time taken by car Q to reach city B from Delhi. Distance between city B and city D is 1500 km. (a) 25% (b) 20% (c) 30% (d) 40% (e) 50% 62. A thief runs in a car S from Delhi to city E and after 6 hours of running, a policeman started to catch him in a car R. Due to this, thief increases the speed of his car by 100%. By this, the policeman is able to catch him 3 at th of the distance of city E from Delhi. Find the initial speed of car ‘S’. 5

(a) 15 km/hr (d) 25 km/hr

(b) 27 km/hr (e) 40 km/hr

(c) 20 km/hr

63. Car P and Car Q start from Delhi for city A. Car Q first reaches at city A and meets car P in between the way, 200 km from city ‘A’. Find after how much time they will meet second time after first time meeting if they continue their to and fro motion. (a) 24 hours (b) 15 hours (c) 16 hours (d) 25 hours (e) 20 hours 26

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Direction (64-68): - Line chart given below shows time taken by five different persons to complete a work ‘M’ alone. Ratio of efficiency of all five persons remain same throughout any work. Study the data carefully and answer the following questions. 70

60 50 40 30

20 10 0

Abhi

Bhavya

Neeraj

Satish

Veer

64. All five starts working together to complete work ‘X’. ‘Veer’ left after 8 days. Work done by ‘Bhavya’ is same as work done by ‘Neeraj’ while ‘Abhi’ and ‘Neeraj’ worked for same time. ‘Satish’ worked for ‘y’ days. If ‘Bhavya’, ‘Neeraj’ and ‘Satish’ together can complete work ‘X’ in 24 days then find the value of ‘y’ if Bhavya worked for starting 10 days. (a) 7 days (b) 9 days (c) 11 days (d) 13 days (e) 15 days 65. Abhi and Neeraj together can complete work ‘Z’ in (A + 42) days while Bhavya and Satish together can complete work ‘Z’ in (A + 15) days. All start the work Z such that ratio between work done by Abhi, Bhavya and Veer is 1 : 2 : 3, while ratio between days, Neeraj, Satish and Veer worked is 2 : 2 : 1. Find how many days ‘Bhavya’ worked. (a) 10 days (b) 15 days (c) 20 days (d) 30 days (e) 40 days 27

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66. All five persons started together to complete work ‘Y’. Veer worked for starting 6 days and left the work. After 3 days more both Bhavya and Satish left too. Remaining 40% work should be completed by Abhi and Neeraj together but ‘Abhi’ left after ‘x’ days. Remaining work is completed by ‘Neeraj’ in ‘z’ days. If ‘z – x = 3’, then number of days for which ‘Neeraj’ worked is what percent more than number of days for which ‘Abhi’ worked. 1 2 (a) 33 % (b) 50% (c) 66 % 3 3 (d) 75% (e) 100% 67. Abhi, Bhavya and Neeraj together starts to do work ‘M’. After 7 days ‘Neeraj’ left and after 3 days more ‘Abhi’ and ‘Bhavya’ left. Remaining work is completed by Satish and Veer working alternatively in ‘y’ days. If ‘y’ is integer then find ‘Veer’ worked for how many days? (a) 3 days (b) 4 days (c) 5 days (d) 6 days (e) Cannot be determined 68. Abhi, Bhavya and Satish starts working together to complete work ‘M’. After 5 days, Bhavya and Satish replaced by Neeraj and Veer. After 5 more days Abhi left the work. After 1 more day Veer left too. Neeraj worked for total ‘x’ days. In other case Abhi and Bhavya starts working together to complete ‘M’. After 4 days both are replaced by Veer. Veer worked for 5 days and replaced by Satish who worked for 8 days. Remaining work is completed by Neeraj in ‘y’ days. Find (y − x)2 ? (a) 25 (b) 36 (c) 49 (d) 64 (e) 81

28

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Solutions

Solutions (1-4): Total work = 5040 units (LCM of days taken by all) Efficiency of Neeraj = Efficiency of Aniket =

5040 35 5040

= 144 units/hour = 168 units/hour

30 5040

Efficiency of Saurabh = Efficiency of Gopal = Efficiency of Veer =

45 5040

16 5040 24

Efficiency of Sameer =

= 112 units/hour

= 315 units/hour

= 210 units/hour

5040 18

= 280 units/hour

1. (a); New efficiency of Neeraj = 144 ×

7 8

= 126 units/hour New efficiency of Gopal = 315 ×

2 3

= 210 units/hour Required time =

5040 (126+112+210)

1

= 11 hours

2. (c); New efficiency of Sameer = 280 ×

4 3 4

= 210 units/hour ATQ – (210+210 )(y) 168(y+1)

=

2 1

420y = 336y + 336 420y – 336y = 336 y = 4 hour Total work = 420 × 4 + 168 × 5 = 2520 units Neeraj will complete alone in = 29

2520 144

1

= 17 hours

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

3. (d); Divyaraj work in one hour = 144 × 1.5 = 216 units Manish work in one hour = 168 ×

75 100

= 126 units Let Saurabh work for n hours and Divyaraj & Manish work for (n + 11.5) hours ATQ – N × 112 + (216 + 126)(n + 11.75) = 5040 112n + 342n = 5040 – 4018.5 n=

1021.5 454

n = 2.25 hours Total time = (2.25 + 2.25 + 11.75 ) 1 = 16 hours 4

4. (d); Total work completed by Neeraj, Aniket, Veer and Sameer together in one hour = (144 + 168 + 210 + 280) = 802 units In Second hour total work destroyed by Saurabh & Gopal together = −(315 + 112) = − (427) Total work in 2 hours = 802 – 427 = 375 units Total work in 24 hours = 375 ×

24 2

= 4500 unit On 25th hour remaining work by Neeraj, Aniket, Veer and Sameer together =

5040 −4500 802

=

270 401

hours

Required time = 24

270 401

hours

5. (a); Total marbels in bag Y = 20 7C

Required probability = 20 2 = C2

30

7×6 20×19

=

21 190

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

6. (d); Total marbels in bag X = 18 6C

Required probability = 18 2 = C2

6×5 18×17

=

5 51

7. (c); Total marbels in bag y = 20 Required probability =

2C × 7C 1 1 20C 2

=

7 95

8. (e); Total marbels in bag X = 18 3C

4C

Required probability = 18 2 + 18 2 =

3 9×17

+

4×3 18×17

=

9 153

=

C2

1

C2

17

9. (b); Total marbels in bag Y = 20 Required probability =

7C × 7C × 4C 1 1 1 20C 3

=

49 285

10. (b); Quantity of tank P and T together = 1500 ×

(15+15) 100

= 450 ℓ

Efficiency of pipe B and D Pipe B = Pipe D =

12 100

1500×

40 20 1500×

100

40

= 4.5 ℓ⁄m = 7.5 ℓ⁄m

Efficiency of pipe C and E Pipe C = Pipe E 

8 100

1500× 30

1500  45

= 4 ℓ⁄m

15 100  5l / m

According to question = 12X + 9(X + 15) = 450 = 21X = 450 – 135 =X=

315 21

= 15 minutes

Liter of water filled by pipe B and D together = 15 (4.5 + 7.5) = 15 × 12 = 180ℓ 31

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

11. (c); Quantity of tank S 20 = 1500 × = 300ℓ 100 New efficiency of Pipe D 300 6 = × = 9 ℓ⁄m 40 5 Efficiency of pipe C and G 8

12

1500×100

1500×100

= × 125 + 30 30 = 5 + 6 = 11 l/ m According to question = 9X + 11 × 15 = 300 = 9X = 300 – 165 = 9X = 135 135 X= = 15 minutes 9

12. (a); Given, total quantity of all seven tank = 1500ℓ Quantity of bigger tank K 120 = 1500 × = 1800ℓtr 100 Efficiency of pipe B, D and G together 1500×

12 100

20 100

1500×

12 100

1500×

= + + 40 40 = 4.5 + 7.5 + 6 = 18 ℓ/m

30 15 100

1500×

Efficiency of pipe A = 25 Efficiency of Pipe F and C 1500×

18 100

= 9 ℓ⁄m

8 100

1500×

= + 60 30 = 4.5 + 4 = 8.5 ℓ/m According to question Tank filled in three minutes alternatively = (18 + 9 + 8.5) = 35.5 litre In (50 × 3) minutes = 35.5 × 50 = 1775 litre Next by (B + D + G) = 18 ℓtr/m After 151 minutes remain quantity of bigger tank = 1800 – (1775 + 18) = 7 ℓtr 7 Tank K, total filled in = 151 + 7

9

= 151 minutes 9

32

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13. (d); Total quantity of tank M = (total quantity of tank S + U together) 20 18 = 1500 × + 1500 × 100 100 = 300 + 270 = 570ℓ Efficiency of pipe A, C and G together 15

1500×100

8

1500×100

12

1500×100

= + + 25 30 30 = 9 + 4 + 6 = 19 ℓ/m (A + C + G) filled tank M together in 570 = = 30 minutes 19 But according to question — 17 tank 7 minutes more due to leak 19 leak empty quantity of water in 30 minutes 17 = (A + C + G) filled in 7 minutes = 19 × 7

17 19

= 19 ×

19 150 19 150

= 150 ℓtr

Efficiency of leak = = 3 ℓ⁄m 30 Leak empty tank M in 570 = = 190 minutes 3 1

= 3 hr 6

14. (b); Given, total quantity of tank = 570 ℓ Efficiency of pipe B, D and E together 1500×

12 100

20 100

1500×

8 100

1500×

175

= + + × 40 40 30 100 = 4.5 + 7.5 + 7 = 19 ℓ/m Leak at a height of 2/3 from the bottom of tank 2 = 570 × = 380 ℓtr 3 380 ℓ filled without leak 380 = = 20 minutes 19 Given, man notice after 32 minutes Tank filled in 32 minutes = (20 + 12) minutes 1 = 380 + 12 × (19 – 19 × ) = 380 + 114 = 494 2 Remain = 570 – 494 = 76 ℓ 33

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15. (b); Q

S

P 75

32000 : 24000 : 16 : 12 : Let total profit = 100X

24000 × 100 9

Q extra profit share = 100X × S extra profit share = 100X ×

12 100 15 100

= 12X = 15X

Remaining profit = 100X – (12X + 15X) = 73X 73X

Share of P = (16+12+9) × 9 = 65700 =

657X

= 65700

37 65700×37

X=

657

= 3700

Total profit = 3700 × 100 = 370000 Rs. 16. (e); Ratio of investment by Q & R = 4 : 3 ⇒ 4x : 3x Ratio of time period = 5 : 3 ⇒ 5y : 3y Ratio of profit share between Q & R = 4x × 5y ∶ 3x × 3y = 20xy ∶ 9xy = 20a ∶ 9a [say] ATQ, 20a – 9a = 100% a=

10010 11

= 910

Now 35% profit = 29a = 29 × 910 65% profit = 29 × 910 × 65 2

% profit =

29×910×13 2×7

65 35

= 24505

R’s total profit share = 24505 + 9a = 24505 + 8190 = 32695 34

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17. (a); Lets S joined the business after X months According to question 32000×36 3 36 = ⇒ x = = 4 months 24000×(36 –X)

2

9

18. (b); Profit share of R = 30500 – 24300 = 16200 Rs. Lets R invested X Rs. for 12 months ATQ 81000×9 24300 81000×3 3 = = = X×12 16200 4X 2 X = 40500 Rs. 19. (c); Q : R = 8 : 9 R:S=3:2 Q:R:S=8:9:6 Total profit = 37030 Q’s share 8 × 37030 = 12880 23 R’s share 9 × 37030 = 14490 23 S’s share 6 × 37030 = 9660 23

20. (c); Investment of Saurabh = 40,000 ×

40

100 30

= 16,000 75

Investment of Neeraj = 40,000 × × = 9,000 100 100 Ratio of profit share of Neeraj to Saurabh

Saurabh’s profit = 35

8000 59+101

× 101 =

8000 160

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

21. (a); Investment of Rakesh = 40,000 ×

30 100

= 12,000

Investment of Aman 30 75 100 = 40,000 × × × = 6,000 100

100

150

Ratio of profit share of Aman to Rakesh

Aman

:

Rakesh

6000  8  8000  2  10, 000  2 : 12000(12  x) : 12000(12  x)

84000 7

:

(12  x)

ATQ, 7 7 = ⇒ x = 4 months 12−x 8 Rakesh worked for (12-x) = 12 -4 = 8 months 22. (e); Investment of Sandeep = 40000 × Investment of Neeraj = 40000 × Investment of Satish = 40000 ×

30

= 12000

100 30 75

100 40

100

×

100

= 9000

+ 4000 = 20000

Ratio of profit share of Sandeep, Neeraj and Satish

11500

Sandeep’s profit share = (12+6+5) × 12 = Rs 6000 23. (b); Investment of Aman 30 75 100 = 40,000 × × × = 6,000 100

100

150

Investment of Saurabh = 40,000 ×

40 100

= 16,000

Total amount they have after 2 years 120 120 80 80 = 6000 × × + 16000 × × 100

100

= 8,640 + 10,240 = 18,880 36

100

100

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24. (b); Total quantity of in vessel P = 240 × Quantity of Mango juice in vessel P 93.75 = 192 × = 180 ml.

50 100

100

Quantity of water in vessel S = 160 ×

= 120 ml

150 100

= 240 ml Quantity of Orange juice in vessel S = 240 – 40 = 200 ml Ratio of (Milk : Water : Orange juice : Mango juice) in vessel P = 240 : 120 : 144 : 180 = 20 : 10 : 12 : 15 Ratio of (milk : Water : Orange juice : Mango juice) in vessel S = 160 : 240 : 200 : 120 = 4 : 6 : 5 : 3 Total orange juice in vessel T 12 5 = 114 × (20+10+12+15) + 180 × (4+6+5+3) = 24 + 50 = 74 ml 74 Required % = (114+180) × 100 =

74

294

× 100 = 25

25

147

%

25. (c); Quantity of milk in vessel R = 144 ×

2 3

= 96 ml

Quantity of Orange juice in vessel Q 3 = 160 × = 120 ml 4 Total solution in vessel Q = (160 + 80 + 120 + 210) = 570 ml Total solution in vessel R = 1086 – 570 = 516 ml Let mango juice and total solution in vessel R be 8n and 43 n respectively ATQ— 96 + 144 + 192 + 7n = 43n 36n = 432 n = 12 Total mango juice in vessel R = 12 × 7 = 84 ml 210+84 294 49 Required ratio = = = 1086

37

1086

181

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26. (c); Minimum average is obtained when he will buy 2T₂ t- shirts. Price for 2 shirts = 600 × 2 = Rs. 1200 He will get 4 t-shirts in this price 1200 Price/t-shirt = = Rs. 300 4 NOTE: If he buys T₁ t-shirt, price will be 90 400 × = Rs. 360 100

27. (a); 20% profit means he is selling it in 1000 (1 +

20 100

) = Rs. 1200

This amount is obtained after 25% discount 4 Hence MRP = 1200 × = Rs. 1600 3

1

28. (c); SP of S₂ = 1647 [1–

11 9 100

8

] = 1647 [ ] 9

= 183 × 8 = Rs. 1464 ATQ, CP [1 +

14

2 7

100

] = Rs. 1464

CP = 1464 ×

7 8

= Rs. 1281

29. (e); Price paid for 5 pieces of T₂ is actually price paid for 2 pieces of T₂ + 1 piece of T₂. As he will get 2 – t-shirts (T₂) free after buying two = 2 × 600 + 600 = 1800 Similarly price for 4 pieces of J₁ = 2 × 1200 1 piece of S₁ = 1 × 1200 = Rs. 1200 Total = Rs. 1800 + Rs. 2400 + Rs. 1200 = Rs. 5400 30. (d); Actual price of 4 pieces of T₂ = 4 × 600 = 2400 Rs. Actual price of 2 pieces of T₁ = 2 × 400 = 800 rs. Actual price of 4 pieces of J₁ = 4 × 1200 = Rs 4800 Total actual price is Rs (2400 + 800 + 4800) = Rs 8000 Price paid for 4 pieces of T₂ = 2 × 600 = Rs. 1200 38

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For 2 pieces of T₁ = 400 × (

90 100

) × 2 = Rs. 720

For 4 pieces of J₁ = 2 × 1200 = Rs. 2400 Discount amount obtained = (8000 – 4320) Rs. = 3680 3680 D% = × 100 = 46% 8000

31. (c); Average speed = =

60+70+60+80+40 5

Total distance

=

Total time 310 5

= 62 km/hr

32. (b); Distance travelled by A till 3 PM is = 60 + 70 + 60 = 190 km Now note that if A is travelling a distance of 60 km in 1 hour, then its speed is also 60 km/hr. Also, we know that ratio of distance travelled by two objects in same time is same as ratio of their speed. D S ∴ A= A DB

SB

Distance travelled by B between 12:00 to 1:00 PM 60 6 = ⇒ DB =50 km DB

5

DB

11

DB

7

Similarly distance travelled by B between 1:00 PM – 2:00 PM 70 14 = ⇒ DB = 55 km Distance travelled between 2:00 – 3:00 PM 60 6 = = DB = 70 km Required difference = 190 – [50 + 55 + 70] = 15 km 33. (a); Average speed of A between 2:00 to 4:00 pm 60+80 = = 70 km/hr 2 Average speed of B between 2:00 to 3:00 pm 7 = × 60=70 km/hr 6

Similarly between 3:00 to 4:00 pm 3 = 80 × = 60 km/hr 4

39

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Average speed of B between 2:00 to 4:00 pm 60+70 = = 65 km/hr 2

Required answer = ( =

1 14

70–65 70

1

) × 100

× 100 = 7 % 7

34. (d); Distance covered by B till 5:00 PM is 5

11

6

14

(60 × ) + (70 ×

7

3

7

6

4

4

) + (60 × ) + (80 × ) + (40 × )

= 50 + 55 + 70 + 60 + 70 = 305 km Remaining distance = 540 – 305 = 235 km Required average =

235 4

= 58¾ km/hr

35. (c); A consumes 1 litre/15 km B consume 15 × 4

100 125

15 × = 1 litre/12 km 5

Distance travelled by A till 2:00 PM = 60 + 70 = 130 km Petrol consumed =

130 15

=

26 3

litre

Distance travelled by B till 2:00 PM = 50 + 55 = 105 km Petrol consumed = Required Ratio =

105

=

12 26 35 3

∶

35 4

litre

4

⇒ 104: 105 36. (b); Let the amount obtained by B is 41x and S.I. obtained is 16x. Therefore, principal is 41x – 16x = 25x = Rs75000 ⇒ x = 3000/– Therefore, SI obtained 16x = 16 × 3000 = 4800/Using formulae PRT 100

= 48000 ⇒ T =

48000×100 16×75000

= 4 years = 48 months 40

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37. (d); If we assume amount of A 8x and SI obtained 3x, then we can calculate principal of A (5x) = Rs. 80000 x = 16000 and amount obtained by A = 16000 × 8 = 128000 3 ∴ amount obtained by C = × 128000 = Rs. 96000 4 Let ROI obtained by C is m%/annum and from line graph, we know that Principal submitted by him is Rs 60000. Then, 60000(m)×6 = Rs. (96000 – 60000) = 36000 100 m = 10% 38. (a); Time for which Vikas invested is 6 + 3 = 9 years. 80000 Sum deposit by him = = Rs. 40000 SI obtained by D = ∴ ROI of Vikas = R=

100 9

1

2 100000×8×5

100 40000×R×9 100

= Rs. 40000

= 40000

= 11 % 9

39. (e); Let amount obtained by E is 29y & SI obtained is 9y ∴ 29y – 9y = 20y = 50000 [∵ 50000 is principal] ⇒ y = 2500 Interest obtained = 2500 × 9 = Rs. 22500 50000×15%×T ∴ = 22500 100 22500

T= = 3 years. 500×15 Similarly, For B principal is Rs. 75000 (41–16) = 25 units = Rs. 75000 75000 ⇒ 16 units (SI) = × 16 = Rs. 48000 25 ∴ time for which B invested 75000×16×T = 48000 100 48×100

T= = 4% 75×16 Required ratio = 3 : 4 41

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40. (c); Amount obtained by D after drawing his whole money 100000×8×5 = 100000 + = Rs. 140000 100 Now he invested 50% of this = 70000 Amount obtained after two years from this 110 2

= 70000 × ( ) = Rs. 84700 100 Total amount with him = Rs. 84700 + Rs.70000 = Rs. 154700 41. (d); Upstream distance travelled on 1st Aug. 12 = × 6000 = 720km. 100 Downstream distance travelled on 1st Aug. 16 = × 3000 = 320 km 100 Let speed of Rahul in still water be x km/hr. ATQ. 720 320 184 = = + x−10 x+10 3 On solving x = 20km/hr. ∴ speed of Rahul in still water = 20Km/hr. 42. (a); Speed of Rahul in still water on 3rd Aug. 125 = × 8 = 10Km/hr. 100 ATQ. 8 ×600 100

12 ×2000 100

8

= × 3 8+x Solving we get x = 8/7 km/hr. ∴ seven times Speed of Current = 8 Km/hr 8−x

43. (c); Let speed of Rahul in still water be x Km/hr. 32 ×6000 100

∴ 80 = ⇒ x = 27Km/hr. x−3 Downstream speed on 5 August =27 + 3 = 30 km/h Speed of current on 4th Aug. 50 = (27+3) × = 15 Km/hr. 100 Time to cover 128 Km up stream on 4th Aug. 128 2 = = 10 hr. 27−15

42

3

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44. (e); Let speed of current on 2nd Aug. & 4th Aug. be 2x km/hr & 3xkm/hr. Let speed Rahul in still water be y km/hr. ATQ. If speed of current on 3 August is 4 km/h then 2x = 8 Therefore 3x = 12 Now, 12 ×2000 100

24×

y+8

= 2x ×

28 ×6000 100

y−12

7y + 56 = 24y − 288 y = 20.22 km/h Approximately 20km/h 45. (c); On Tuesday, Length of Rajdhani exp → 24 × 15 + 20 = 380 m Let the speed of Rajdhani exp on Tuesday be 8 x m/s Speed of shatabadi exp = 7x m/s Speed of duranto exp = 10 x m/s Let the length of tunnel be y m ATQ, 380+y = 12.5 8x

⇒ 380 + y = 100x ⇒ 100x – y = 380

… (i)

And, length of Duranto exp = 380 × = 475 m 475+y 10x

100

= 11.9

⇒ 475 +y = 119x ⇒ 119x – y = 475 From (i) and (ii) x = 5 m/s Required time = 43

125

… (ii)

380+370 15×5

=

750 75

= 10 sec

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46. (a); Length of Rajdhani on Sunday = 29 × 15 + 20 = 455m Length of Duranto exp = 345 m Let the speed of Rajdhani, shatabadi and Duranto exp be 6x m/s, 5x m/s and 4x m/s respectively. ATQ, 455+345 80 = ⇒ x = 9 m/s 10x

9

Length of Shatabadi = 455× Required time =

364+345 9

80 100

= 364 m

≈ 79 sec

47. (d); Speed of Shatabadi exp on Friday = 150 km/hr Speed of Duranto exp on Friday = 120 km/hr Distance travelled by Shatabadi exp in 3 hours = 150 + 150 × 1.10 + 150 × 1.1 × 1.2 = 513 km Distance travelled by Duranto exp in 3 hours = 120 + 120 × 1.2 + 120 × 1.2 × 1.25 = 444 km Total distance between P and Q = 513 + 444 = 957 km 48. (b); On Monday, Length of Rajdhani exp. = 15 × 17 + 20 = 275 m Length of Duranto exp. = 275 × 1.28 = 352 m Let the speed of Rajdhani, shatabadi and Duranto exp. on Monday be 3x m/s, 4x m/s, 5x m/s respectively. ATQ, 352+352 = 12.8 ⇒ 704 = 64x ⇒ x = 11 5x

Required difference = 5x − = 59.4 km/hr 44

7x 2

3

18

2

5

= ×11×

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49. (b); Length of train B =

18

× 1600 = 288m

100 12

Length of Train D = Length of Train F =

100 8 100

× 1600 = 192m × 1600 = 128m

Speed of train B on Monday = 97.2 × = 27 m/sec Speed of train ‘B’ on Wednesday 27 = × 5 = 45 m/sec

5 18

3

ATQ, 288 + 192 = (45 +y)×6 where y is the speed of train ‘D’ on Wednesday ⇒ y = 80 – 45 = 35 m/sec Speed of train ‘D’ on Monday =

35 7

×4

= 20 m/sec 192+128 Time required to cross train F = 20+20

= 8 seconds

50. (e); Let speed of train ‘C’ on Monday, Tuesday and Wednesday be 4x, 6x and 5x respectively. Train ‘C’ travel 5 hours on Monday and 15 hours on Tuesday. ∴ Total distance = 5 × 4x + 15 × 6x = 110x On the same day i.e, Tuesday, train ‘C’ start from Kanpur. It travels 6 hours on Tuesday and 17.8 hours on Wednesday. ∴ total distance travel = 6 × 6x + 17.8 × 5x = 36x + 89x = 125x ATQ, 125x = 110x + 180 ⇒ 15x = 180 ⇒ x = 12 speed of train ‘C’ on Monday 40 = 12 × 4 = 48 km/hour = m/sec Length of train ‘C’ = Required time = 45

384 40

24 100

3

× 1600 = 384

× 3 = 28.8 sec Adda247 Publications

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

51. (a); Let, speed of train ‘A’ and train ‘C’ on Monday be ‘4x’ and ‘4y’ respectively ATQ, 900 900 2.5 = − 4x

4y 1 1

2.5 = 225 [ − ] x

y

xy = 90 (y – x) length of train ‘A’ =

16 100

× 1600 = 256

speed of train ‘A’ on Tuesday = =

384 12.8

256+128 12.8

= 30 m/sec

⇒ Speed of train ‘A’ on Monday 30 = × 2 = 20 m/sec = 72 km/hr 3

⇒ 4x = 72 ⇒ x = 18 xy = 90(y – x) y = 5(y – 18) ⇒ y = 22.5 Speed of train ‘C’ on Monday = 4y = 4 × 22.5 = 90 km/hr 90 Speed of train ‘C’ on Tuesday = × 6 4 = 135 km/hr = 37.5 m/sec 24 Length of train ‘C’ = × 1600 = 384 Required time =

100 384+66 37.5

52. (a); Length of train ‘E’ = Length of train ‘F’ =

22 100 8 100

= 12 seconds × 1600 = 352 × 1600 = 128

Let speed of train ‘E’ and train ‘F’ on Monday be 6x and 4y respectively. 6x 3 x 1 ⇒ = ⇒ = 4y

2

y

1

Let speed of train ‘E’ on Tuesday = 9x So speed of train ‘F’ on Tuesday = 5y = 5x ATQ, 46

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352+128

9x – 5x = = 20 ⇒ 4x = 20 24 ⇒x=5 Speed of train ‘E’ on Wednesday = 5 × 5 = 25 m/sec Speed of train ‘F’ on Wednesday = 3 × 5 = 15 m/sec 352+128 480 Required time = = 25−15

= 48 seconds

10

53. (b); Let, speed of train ‘B’ on Monday, Tuesday & Wednesday be 3x, 4x & 5x respectively. And speed of train ‘D’ on Monday, Tuesday & Wednesday be 4y. 4y & 7y respectively. 18 Length of train ‘B’ = × 1600 = 288 Length of train ‘D’ = ATQ, 288 3x 192 4y

=

1 1

y

1

x

2

⇒ =

⇒

3 2

×

4y 3x

=

100 12

100

× 1600 = 192

1 1

⇒ x = 2y

Time taken by train ‘B’ on Wednesday to cross pole = Time taken by train ‘D’ on Monday to cross a pole = Required % =

96 57.6 ( − )×100 x x 96 x

=

38.4 96

288

5x 192 4y

=

= 96

57.6 x

x

× 100 = 40%

54. (c); Let, marked price of both articles be 600x 72 S.P. of article X = 600x × = 432x S.P. of article Y = 600x ×

100 85 100

= 510x

ATQ, 510x − 432x = 312 ⇒ 78x = 312 ⇒x=4 47

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Cost price of article ′X ′ =

432×4 120

× 100 = 1440

Cost price of article ‘Y’ = 1520 × 2 – 1440 = 3040 – 1440 = 1600 Profit % earned on selling article Y =

510×4−1600 1600

× 100 =

440 1600

× 100 = 27.5%

55. (e); Let M.P. of each article sold by each seller be 800x Cost price of article Y sold by B =

800x×67.5 120

= 450x Cost price of article Y sold by E =

800x×72 4×100

× 3 = 432x

Cost price of article X sold by C =

800x×85 125

= 544x Cost price of article X sold by E =

800x×54 135

= 320x ATQ, (450x + 432x) – 544x – 320x = 216 882x – 864x = 216 ⇒x=

216 18

= 12

M.P. of each article = 800 × 12 = 9600 Cost price of article Y sold by C =

9600×76 160

= Rs 4560 56. (b); Let M.P. of each article be 400x Cost price of article X sold by B = Cost price of article Y sold by D = Required ratio = 48

175x 210x

=

5

400x×73.5 168 400x×63 120

= 175x

= 210x

6

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57. (d); Let M.P. of each article = 400x S.P. of article X sold by E =

400x×54

= 216x

100

Cost price of article Y sold by E = 216x 400x×72 Selling price of article Y sold by E = 100

= 288x 288x−216x Profit % = × 100 =

72x 216x

216x

1

× 100 = 33 % 3

58. (b); Let, M.P. of each article be 400x C.P. of article X sold by A =

400x 2

= 200x

S.P of article X sold by A = 200x × = 240x SP of article Y sold of A = 400x × = 340x Let, CP of article Y sold by A = y ATQ, 125 (200x + y) × = 240x + 340x (200x + y) =

100 580x

Required % =

120 100

85 100

× 4 ⇒ y = 264x

5 264x−240x 264x

× 100 = 9

1 11

%

59. (b); Distance travel by car P = 1500 + 3000 = 4500 km Total Time taken =

4500 40

= 112.5 hour

Time taken by car R from Delhi to City A =

1000 60

=

50 3

hours

Time taken from city A to city B = 112.5– Distance from between City A to city B = 49

287.5 3

50 3

=

287.5 3

× 60 = 5750 km Adda247 Publications

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

60. (a);

Distance between city A and city E = √10002 + 15002 = √1000000 + 2250000 = √3250000 = 500√13km Approximate time taken by car ‘T’ 500√13

=

75

≈ 24 hours

61. (c); Let speeds of car Q and car S be x and y respectively. ATQ— ⇒

3000 x

=

2000

⇒

y

x y

=

3 2

Let speed of car Q and car S be 3a and 2a respectively Distance between city B and city D = 1500 km Time taken to cross each other =

1500 5a

=

300 a

Time taken by car Q to reach city B from Delhi = Required% = 62. (d);

1500×3 5

300×100 1000

3000 3a

=

1000 a

= 30%

= 900 km

Time taken by car R to cover this distance =

900 60

= 15 hour

Let initial speed of car S = x km/hr So, ATQ 6x + 15(2x) = 900 6x + 30x = 900 36x = 900 ⇒ x = 25 km/hr 50

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63. (e); Distance between Delhi and city A = 1000 km Distance covered by Car Q before first meeting = 1200 km Distance covered by Car P before first meeting = 800 km Speed of car P = 40 km/hr 800 ⇒ Time for first meeting = = 20hr Speed of car Q =

40

1200

= 60km/hr

20

When car P reaches city ‘A’ distance covered by car ‘Q’ 200 = × 60 = 300km 40

Time taken by car ‘Q’ to reach Delhi = Distance covered by car ‘P’ in =

25 3

× 40 =

1000 3

25 3

500 60

=

25 3

hr

hour

km

Distance between car ‘Q’ and car ‘P’ 1000 2000 = 1000 − = 3

3

Time to meet = Total time =

2000 3

60+40 200 25 40

+

3

= +

20 3 20 3

hour = 20 hours

64. (d); Total work = 24 × (4a + 2a + 4a) = 240a Bhavya worked for10 days ⇒ Work done by ‘Bhavya’ = 40a 40a ⇒ Time taken by ‘Neeraj’ = = 20 days 2a

⇒ ‘Neeraj’ and ‘Abhi’ worked for 20 days And ‘Veer’ worked for 8 days 20 × 3a + 10 × 4a + 20 × 2a + y × 4a + 8 × 6a = 240a ⇒ 118a + y × 4a = 240a 240a−188a 52a ⇒y= ⇒ y= = 13 days 4a

4a

65. (b); 5a × (A + 42) = 8a × (A + 15) ⇒ 5A + 210 = 8A + 120 ⇒ A = 30 Total work = 5a (30 + 42) = 5a(72) = 360a 51

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Let, Abhi, Bhavya and Veer worked for x, y and z days ATQ, 3x : 4y : 6z = 1 : 2 : 3 Ratio between working days of Abhi, Bhavya and Veer ⇒x:y:z=2:3:3 Ratio between working days of Neeraj, Satish and Veer =2:2:1 Let working days of Abhi, Bhavya, Neeraj, Satish and Veer be 2m, 3m, 6m and 6m and 3m days respectively ATQ, 3a × 2m + 4a × 3m + 2a × 6m + 4a × 6m + 6a × 3m = 360a ⇒ 72am = 360a ⇒ m = 5 Bhavya worked for 5 × 3 = 15 days 66. (b); Veer, Bhavya and Satish worked for 6 days, 9 days and 9 days respectively. Total 60% of work completed by them ⇒ 60% of work = 4a × 9 + 4a × 9 + 6a × 6 = 108a 108a ⇒ Total work = × 5= 180a 3

‘Abhi’ worked for (9 + x) days & ‘Neeraj’ worked for (9+ x + z) days and completed 40% of work ⇒ 72a = 3a (9 + x) + 2a(9 + x+ z) 72 = 27 + 3x + 18 + 2x + 2z 27 = 5x +2z … (i) And, z–x=3 … (ii) On solving (i) & (ii) z = 6, x= 3 ‘Abhi’ worked for (9 +3) = 12 days ‘Neeraj’ worked for (9 + x+ z) = 9 + 3 + 6 = 18 18−12 Required % = × 100 = 52

6 12

12

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67. (b); There are two possibilities First – Satish worked first Second – Veer worked first When satish worked first remaining work → 120a – 10(3a+4a) – 7(2a) = 36a Satish and Veer worked in 3 days = 30a 4th day satish’s worked = 4a Veer’s worked for = 1/3 day Now ‘y’ cannot be in fraction 2nd caseWhen Veer worked firstVeer and Satish worked for first 3 day = 30a Remaining work = 6/6= 1 day So Veer worked for 4 days. 68. (d); In first case ATQ, 10 5 x + + + 40

⇒

x 60

30

+

53 60

60

5 30

+

=1 ⇒

In second case ATQ, 4 4 y 8 + + + + 40

y

30

1

60

30

6 20 x 60

5 20

=1 =

7 60

⇒ x=7

=1 ⇒

y 60

3

+ =1 4

⇒ = ⇒ y = 15 60 4 2 (y − x) = (15 − 7)2 = 82 = 64

53

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1

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Chapter

15

Data Sufficiency

BEST APPROACH TO SOLVE THE QUESTIONS Data Sufficiency These questions formerly asked in only MAINS Examination but now-adays these questions were seen in PRELIMINARY Examinations too. So, student can’t deny the importance of DATA SUFFICIENY Questions. In these questions some data are given in form of statements and one question is asked. Student must choose that statement/statements which is/are sufficient to solve the given question. They consist of two types. In first type, two statements are given while in second type, three statements are given, and student is expected to choose the statement/statements among the given statements which is/are required to solve the question. Five questions of each type are given below to practice this topic. Example1. What is the length of train ‘A’? A. Train A takes 8 second to cross a pole B. Train A cross train B coming from opposite side in 8 second, speed of train B is 25 m/sec and length of train B is half of train A. C. Train A takes 10 sec to cross two pole 100 m apart. (a) Only A and B together (b) Only A and C together (c) Any two of them (d) Either B alone or A and C together (e) All statements are required 2

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Solution: 1. (c); Let length of train A = L Speed of train A = x L

From A = = 8 From B = From C =

x L+0.5L

x+25 L+100 x

=8 = 10

From any of these two-statement question be solved. Example 2: In how many days B and C together can complete work? A. Per day efficiency of A, B and C is in the ratio 3 : 2 : 4 B. A and B together can complete the work in 7.2 days C. Time taken by A, B and C alone to complete the work is in the ratio 4:6:3. (a) Only A and B together (b) Either A and B or B and C (c) Any two of them (d) Either B alone or A and C together (e) All statements are required Solution: 2. (b); From A per day efficiency = 3 : 2 : 4 ⇒ ratio of time taken by A, B and C alone to complete work = 4 : 6 : 3 same as (C) From B) time taken by A and B together = 7.2 days Either B and A or B and C questions can be solved Example3: What is the base radius of the cone? A. Total surface area of the cone is 462 cm² 3

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B. Volume of the given cone is

1078√3 3

𝑐𝑚3 and height = 7√3 cm

C. Ratio of height and radius of the given cone is √3 : 1 (a) Only A and B together (b) Only A and C together (c) Any two of them (d) Either B alone or A and C together (e) All statements are required Solution: 3. (d) From A) T.S.A = πrℓ + πr² = 462 cm2 From B) V =

1078√3 3

= πr 2

h 3

And h = 7√3 cm From C) h : r = √3 ∶ 1 Either B alone or A and C together, question can be solved. Example 4: Find the share of Satish in profit after one year among Satish, veer and Yogesh A. Investment of Veer, Satish and Yogesh is in the ratio 6: 8: 9 and share of Yogesh in profit is Rs.900 B. Satish and Veer invested Rs.8000 and Rs.6000 for 9 months and one year respectively. Difference b/w profit earned by Satish and Veer is Zero. C. Satish, Veer and Yogesh invested for 9 months, 1 year and 8 months respectively and share of Veer in profit is 900. (a) Either A and B or B and C (b) Either A and B or A and C (c) Any two of them (d) Either B alone or A and C together (e) Either A and C or B and C 4

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Solution: 4. (e); Veer : Satish : Yogesh From A and C) 6×12 : 8×9 : 72 : 72 : 72 1 : 1 : 1 Share of Veer = Share of Satish Rs 900 Satish : Veer From (B and C) 8000×9 : 6000×12 1 : 1 Share of Veer share of Satish = Rs 900 Either A and C or B and C question can be solved.

9×8

Example 5: What will be the probability of choosing two fresh eggs simultaneously from tray A given that there are only two trays i.e. tray A and tray B? A. Tray B contains 30 Eggs in which fresh and rotten eggs in the ratio 7 : 3. Ratio of fresh eggs in Tray A and Tray B is 2 : 3 B. Tray A contains 20 Eggs in total out of which 14 are fresh and rest are rotten. C. Tray B contains 30 Eggs and Tray A contains 20 Eggs respectively. Ratio of Rotten eggs and fresh eggs in Tray A is 7 : 3 (a) Either A or B (b) Either B or C (c) Any one of them (d) A alone (e) All statements are required Solution: 5. (b); From A) Tray B = 30 Rotten eggs in Tray B = 9 eggs & fresh eggs in Tray B = 21 eggs Tray A, Fresh eggs = 14 eggs From B) Tray A = 20 eggs Fresh eggs in Tray A = 14 Rotten eggs in Tray A = 6 From C) Tray B = 30 eggs Tray A = 20 Eggs Rotten eggs in Tray A =6 eggs Fresh eggs in Tray A = 14 eggs From Either B or C question can be solved. 5

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Practice Exercise Based on new Pattern

Directions (1-5): Given below in each question there two statements (I) and (II). You have to determine, which statement is sufficient to give the answer of question. Also there are five alternatives given, you have choose one alternative as your answer of the questions: 1. What will perimeter of smaller rectangle? I. Ratio between length of smaller and larger rectangle is 4 : 5 and breadth of both rectangle is equal. Difference between Perimeter of both rectangle is 8 cm. II. Breadth of both rectangle is equal to side of square, whose area is 196 cm2 . (a) Only statement I is sufficient (b) Only statement II is sufficient (c) Statement I and II both together sufficient (d) Either statement I or Statement II alone sufficient (e) Neither statement I or statement II sufficient 2. How many Students in college? I. Ratio between girls to boy is 9 : 11. II. Out of total girls in the college 20% are belongs to below eighteen years age group. Total girls belongs to below eighteen years age group are 9% of total students in college (a) Only statement I is sufficient (b) Only statement II is sufficient (c) Statement I and II both together sufficient (d) Either statement I or Statement II alone sufficient (e) Neither statement I or statement II sufficient 3. What was profit shopkeeper made on article? I. Shopkeeper sold article on 5% discount in the Rs. of 3800. II. If shopkeeper sold article on marked price, he would made a profit of 25%. 6

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(a) Only statement I is sufficient (b) Only statement II is sufficient (c) Statement I and II both together sufficient (d) Either statement I or Statement II alone sufficient (e) either statement I or statement II sufficient 4. What is speed of boat? I. Speed of boat in still water is two times more that speed of current. II. Boat takes equal time to cover a distance downstream to 50% of that distance upstream. (a) Only statement I is sufficient (b) Only statement II is sufficient (c) Statement I and II both (d) Either statement I or Statement II alone sufficient (e) Neither statement I or statement II sufficient 5. Find the amount invested at the rate of 10%? I. Total amount of Rs. 4500 invested in two different parts at the rate of 20% p.a. and 10% p.a. for two years. Simple Interest obtained from both parts are equal. II. A man invested an amount in two schemes A and B in the ratio of 2 : 1 respectively. Scheme A offered simple interest at the rate of 10% p.a. and Scheme B offered compound interest at the rate of 20% p.a. and man got a total interest of Rs. 1260 after two years from both scheme. Amount invested on SI is same as amount invested at the rate of 10% in statement I? (a) Only statement I is sufficient (b) Only statement II is sufficient (c) Statement I and II both (d) Either statement I or Statement II alone sufficient (e) Neither statement I or statement II sufficient 7

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Directions (6-10): The following questions are accompanied by two statements (I) and (II). You have to determine which statements(s) is/are sufficient/necessary to answer the questions. (a) Statement (I) alone is sufficient to answer the question but statement (II) alone is not sufficient to answer the questions. (b) Statement (II) alone is sufficient to answer the question but statement (I) alone is not sufficient to answer the question. (c) Both the statements taken together are necessary to answer the questions, but neither of the statements alone is sufficient to answer the question. (d) Either statement (I) or statement (II) by itself is sufficient to answer the question. (e) Statements (I) and (II) taken together are not sufficient to answer the question. 6. Ratio between length of two trains is 4 : 3. What will be difference between lengths of both trains? I. Speed of larger trains and smaller train is 72 km/hr and 90 km/hr respectively. Both trains cross each other in

28 3

𝑠𝑒𝑐, when running

in opposite direction. II. Speed of smaller train is 90 km/hr and it can cross a pole in 7.2 sec. 7. There are three men P, Q and R. Find the difference between time taken by P & Q together to complete a task and time taken by Q & R together to complete the same task? I. ‘R’ takes twice as much time as ‘Q’ and thrice as much time as ‘P’ takes alone. II. If they all three works together work will be completed in 4 days. 8.

Satish sold an article to Ayush at 20% profit. If Ayush purchased article from Satish in Rs. 1440, then find the profit percentage of Veer if Satish bought this article form Veer? I. Veer sold the article on Rs. 240 more than its cost price to Satish. II. If Veer sold article to Ayush on same price as Satish sold to Ayush, then he made overall profit of 50%. 8

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9. If x : y = 11 : 9 and y : z = 3 : 4, then find (x + y) – 1.5z = ? I. Average of all three is two more than average of x and y. 1 1 II. Sum of 9 % of x and 11 % of y is equal to (36)0.5. 11

9

10. Ratio between length and breadth of rectangle ‘X’ is 7 : 4. Find area of a square ‘Y’? I. Length of rectangle ‘X’ is two times of radius of circle, whose area is 616 cm2. II. Perimeter of rectangle ‘X’ is 20 cm more than perimeter of square ‘Y’. Directions (11-14): The following questions are accompanied by three statements (A) or (I) and (B) or (II), You have to determine which statement(s) is/are sufficient/necessary to answer the questions. 11. What is the minimum passing percentage in a test? I. Raman scored 25% marks in the test and Sunil scored 288 marks which is 128 more than that of Raman. II. Raman scored 64 marks less than the minimum passing marks. (a) Both I and II together are not sufficient (b) Both I and II together are needed. (c) Only I alone is sufficient. (d) Only II alone is sufficient. (e) Either I or II alone is sufficient 12. Whose body weight is second highest among the five boys Arun, Vinay, Suraj, Raju and Pratap? I. Average weight of Arun, Suraj and Vinay is 68 kg and average weight of Raju and Pratap is 72 kg. Also Suraj is 78 kg. Raju is 68 kg and Vinay is 46 kg. II. Average weight of Arun, Suraj, Vinay and Raju is 68 kg and also Suraj is 78 kg. Raju is 68 kg and Vinay is 46 kg. All of them have different weights. (a) Both I and II together are not sufficient (b) Both I and II together are needed. (c) Only I alone is sufficient. (d) Only II alone is sufficient. (e) Either I or II alone is sufficient 9

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13. What is the population of the city A? I. The ratio of the population of males and females in city A is 27 : 23 and the difference between their population is 100000. II. The population of city A is 80% of that of city B. The difference between populations of city A and city B is 312500. (a) Both I and II together are not sufficient (b) Both I and II together are needed. (c) Only I alone is sufficient. (d) Only II alone is sufficient. (e) Either I or II alone is sufficient 14. How many students did participate in Singing? I. The students who participated in dancing were 150% more than that who participated in Singing. II. 150 students participated in dancing. (a) Both I and II together are not sufficient (b) Both I and II together are needed. (c) Only I alone is sufficient. (d) Only II alone is sufficient. (e) Either I or II alone is sufficient Directions (15-19): The following questions are accompanied by two statements A and B. You have to determine which statements(s) is/are sufficient/necessary to answer the questions. (a) Statement A alone is sufficient to answer the question but statement B alone is not sufficient to answer the questions. (b) Statement B alone is sufficient to answer the question but statement A alone is not sufficient to answer the question. (c) Both the statements taken together are necessary to answer the questions, but neither of the statements alone is sufficient to answer the question. (d) Either statement A or statement B by itself is sufficient to answer the question. (e) Statements A and B taken together are not sufficient to answer the question. 10

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15. An equilateral triangle is inscribed in a circle. What will be the difference between the area of circle and area of triangle? (A) Radius of circle is given (B) Sum of the perimeter of triangle and circle is given. 16. Is ‘n’ even or odd, if n is a natural number. (A) 2n + 1 is divisible by 3 (B) 4n − 1 is divisible by 3 17. Five positive naturals numbers are given which may or may not be equal the sum of these numbers. (A) If numbers are arranged is ascending order then median is 18 while sum of first and last number and second and fourth number is equal (B) Median of these numbers is 18 while mode of these numbers is 12 and 24. 18. ‘X’ mark up an article 50% above its cost price. Find the cost price of article. (A) ‘X’ gave Rs 60 discount on mark price and earn 20% profit. (B) If ‘X’ gave two successive discounts of 10% each then he will earn Rs 43 as profit. 19. Amit tells truth 2 times out of x times while Ankush tells truth 5 times out of ‘y’ times. Find the product of ‘x’ and ‘y’ (A) Probability that Amit and Ankush contradict with each other on facts is 50% (B) Probability that both tells truth is 20% Directions (20-24): The following questions are accompanied by two statements (A) and (B). You have to determine which statements(s) is/are sufficient/necessary to answer the questions. 11

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20. O is center, find area of shaded region

A. Length of AO is given. B. AC is given in multiple of radius of semicircle. (a) Only A (b) Only B (c) Either Only A or Only B (d) A and B together (e) A and B together are not sufficient 21. In a box three type of balls are there, Black, Red and White. If no. of white balls is given then find out the probability of getting one white ball. A. Probability of getting one Red ball is given. B. Probability of getting one black ball is given. (a) Only A (b) Only B (c) Either Only A or Only B (d) A and B together (e) A and B together are not sufficient 22. What is the volume of the sphere? A. Surface area of hemisphere is equal to the total surface area of the cylinder having radius and height in ratio 3 : 4. B. When we cut sphere into two hemi-sphere then total surface area is equal to the area of a circle whose radius is 21 cm. (a) Only A (b) Only B (c) Either Only A or Only B (d) A and B together (e) A and B together are not sufficient 12

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23. PR is diameter of circle. Find, ∠QPO – ∠SRO = ?

A. ∠SPO = 40° B. ∠PQR + ∠QRO = 120° (a) Only A (b) Only B (c) Either Only A or Only B (d) A and B together (e) A and B together are not sufficient 24. A shopkeeper gets a loss of 70 Rs. when he sold an article at 20% discount on M.P. Find cost price of Article. A. % of mark up above cost price is equal to % discount given on M.P. B. When no discount is given, article sold at profit of 350 Rs. (a) Only A (b) Only B (c) Either Only A or Only B (d) A and B together (e) A and B together are not sufficient Directions (25-29): The following questions are accompanied by three statements (I), (II), and (III). You have to determine which statement(s) is/are sufficient /necessary to answer the questions 25. What is distance between A and B ? I. Two persons Amit and Abhi started simultaneously from A to B with their speed in ratio 4 : 5. 13

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II. Abhi reached reached Q one hour earlier than Amit. III. Difference between speed of Amit and Abhi is 20 km/hr. (a) Only I and II. (b) Only II and III (c) All I, II and III (d) Cannot be answered even including all three statement (e) None of these 26. What is the area of rectangle ? I. If ratio of length and breadth of the rectangle is 3 : 2. II. Circumference of a circle is 440 m and breadth of rectangle is 1/7 th of diameter of circle. III. If length is 50% more than breadth. (a) Only III (b) Only I and II or II and III. (c) Only II (d) All I, II and III (e) None of these 27. How many students failed in class 11th ? I. 400 Students passed in class 11th. II. No. of students failed in class 11th is 20% of those failed in class 12th. III. Ratio of student appeared to that of failed in class 11th is 5 : 3. (a) Only I and III (b) Only II (c) Only I and II (d) All I, II and III (e) Cannot be answered even including all three statement 28. What is the rate of interest on some amount? I. S.I. accrued in two years on same amount at same rate of interest is Rs. 44000. II. The amount after some years on S.I. is Rs. 154000. III. Difference between the C.I. and S.I. earned in two years on the same amount invested is Rs. 120. 14

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(a) Only I and III (b) Only III (c) Only II and III (d) Cannot be answered even including all statement (e) None of these 29. What is the sum of two number? I. The bigger no. is 6 more than the smaller no. II. 40% of smaller no. is equal to 30% of bigger no. III. The ratio b/w half of the bigger no. & one-third of smaller no. is 2 : 1. (a) Only II & III (b) Only I & II (c) Any two of the three statement (d) All statement is required (e) I and II or I and III Directions (30-33): In the following questions three statements either A, B and C or I, II and III are given. You have to use your knowledge of mathematics to answer which statement(s) is/are sufficient to answer the question. 30. What is the cost price of wrist watch? A. Shopkeeper gives 20% discount on a speaker. B. The marked price of wrist watch is 25% more than that of speaker. C. The shopkeeper earns a profit of 10% after selling the speaker. (a) Any two of them are sufficient (b) All the three statements are required (c) Only C is sufficient. (d) Even using all statements, answer cannot be found (e) Only A is sufficient 31. What is the total quantity of milk in final mixture of milk and water after adding some milk ? A. 68 liters of initial mixture has ratio of milk and water as 11:6. B. A certain amount of milk is added to this mixture to make the ratio of water to milk as 6:13. C. Selling the mixture at a certain rate, 35 % profit is obtained. 15

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(a) Any two of them is sufficient (b) A and B together are sufficient (c) All the three statements are required (d) Answer cannot be found even using all the three statements (e) Only B and C are required 32. The speed of a train A and time taken by it to cross a tunnel is known. Find length of tunnel. A. Another train B is running in opposite direction to A with a speed 40% more than A. B. Train B crosses a platform X and a pole in 24 sec and 8 sec respectively before crossing the train A and tunnel. C. The ratio of length of train A and platform X is 3 : 5. (a) Only A and B are sufficient. (b) Only B and C are sufficient (c) Even using all the three statements answer cannot be found. (d) All the three statements are required. (e) Only B is sufficient 33. In how many ways 2 green balls can be chosen from box ‘A’ which contains red, green and black balls? A. Ratio of total number of balls in boxes A and B is 2 : 3 and box A contains 5 green balls. B. Total balls in box B is 18 while ratio of red and black balls in box A is 3:4. C. Box ‘B’ contains 5 red 7 black and 6 green balls only while ratio of red and green balls in box ‘A’ is 3:5. (a) Statement A and either B or C are sufficient (b) Any two statements are sufficient (c) All the three statements are required (d) Statement B and either A or C are sufficient (e) None of these Directions (34-38): The following questions are accompanied by three statements (I), (II), and (III). You have to determine which statements(s) is/are sufficient/necessary to answer the questions.

16

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34. Making a new t-shirt, consists of two parts first is cutting of cloth and second is stitching the cloth. In how many days Nitin and Vishal together can make 100 t-shirts. I. Nitin can cut cloth for 20 t-shirts in a day and he is 20% slow as that of Vishal in cutting clothes and 25% faster than Vishal for stitching the cloth. II. Time taken by Nitin alone to make 60 t-shirts is 9 days. III. The ratio of days taken by Vishal to cut cloth for 50 t-shirts and to stich 100 t-shirts is 4 : 25. (a) Only I and III together or only II and III together (b) All I, II and III together (c) Any two of the three (d) Only I and III together (e) Either from I & II or I & III 35. Ratio of age of Aashish & Kullu is 4 : 5 and that of Nikhil and Yash is 7 : 8. What is their average age after 4 years. I. Age difference of Nikhil and Aashish is 18 years. II. Ratio of 12 years ago age of Aashish & Nikhil is 2 : 5 and Kullu & Yash is 1 : 2. III. 12 years later Aashish will be 150% of his present age. (a) Any two of I, II and III are required. (b) Only I and II are required (c) Only II and III are required (d) All I, II and III together are required (e) II alone or from I and III together are required 36. Three athletes practice running on a circular track of 400 m. who runs fastest among them? I. When A starts running clockwise and B starts running 1 anticlockwise, they meet 2nd time at a distance of 57 meter from 6

7

starting point in clockwise direction after 22 seconds. 7 II. B runs 5 m/s faster than C. A is not twice or more fast than B and neither B is twice or more fast than A or C. III. When A, C and B, all runs in same direction, A and C only meet at starting point, while ratio of speed of B to C is 3 : 2. 17

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(a) Any one of them (b) Only I and II together are sufficient (c) Any two of the three together are sufficient (d) None of the above (e) All the three statements are not sufficient. 37. A man has 16 balls with him, in 3 colors i.e. Red, green and blue. Three balls are drawn at random, what is the probability that all are blue. 1 I. The probability of drawing a blue ball is greater than and less than 8

5

. Also, the number of green balls he has is odd and number of red balls he has is even. 3 1 II. Probability of drawing red ball is which is lesser than that of 8 16 green balls. III. If a man lost one ball, probability of drawing a blue ball is ⅓. (a) Only I and III or only II and III (b) All I, II and III together (c) Any one of them is sufficient (d) Only I and III together sufficient (e) None of the above 16

38. A man bought two bats and 6 identical balls, he sold all of them in a day, calculate his overall profit %. I. He sold one bat at a price of Rs. 600 and other at Rs. 420 and profit is 20% on both bats. Each ball is sold at 12% profit. II. Profit earned from 6 balls is Rs. 36 and profit earned from both bats is Rs. 170. III. Profit earned on each bat is 20% while cost price of each ball is Rs. 50. (a) Any one of them (b) Only I and II together are sufficient (c) Any two of the three together are sufficient (d) None of the above (e) All three together are sufficient Directions (39-42): The following questions are accompanied by three statements A, B and C. You have to determine which statement(s) is/are necessary/sufficient to answer the question. 18

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Cracker Book for Bank (IBPS | SBI | RRB PO | Clerk) Mains Exams

39. Find the area of the shaded region?

A. B. C.

Radius of bigger circle is given. Diagonal of square ABCD is given. Difference between area of bigger circle and area of smaller circle is given. (a) Either A or B alone is sufficient to answer the question (b) Either B or C alone is sufficient to answer the question (c) Either A or C alone is sufficient to answer the question (d) Any of A, B or C alone is sufficient to answer the question (e) Either only A or B and C together is sufficient to answer the question 40. Find the surface area of cone? A. Ratio between height and radius of cone is 8 : 7. B. Radius of cone is half of the radius of hemisphere, which volume is 19404 cm³. C. Radius of cone is 75% of radius of cylinder, which total surface area of 2640 cm³ and ratio between height and radius of cylinder is 8 : 7. (a) Either A and B or B and C are sufficient to answer the question (b) Either A and B or A and C are sufficient to answer the question (c) Either A and C or B and C are sufficient to answer the question (d) A, B and C together are sufficient to answer the question (e) Either only A or B and C together are sufficient to answer the question 41. What will be 26 times of 26th term of Arithmetic progression (AP)? A. If eight times of the 8th term of that A.P. is equal to 18th times the 18th term of that A.P. 19

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B. If first term of AP is a prime number. C. If difference between two terms in A.P. is an odd number. (a) Only statement ‘A’ alone is sufficient to answer the question (b) Only statement ‘B’ alone is sufficient to answer the question (c) Only statement ‘C’ alone is sufficient to answer the question (d) Any of two statements are sufficient to answer the question (e) All three together are sufficient to answer the question 42. What is cost price of the article? A. On selling the article at 119 Rs. shopkeeper earn profit % equal to cost price of article. 2 B. If 14 % discount is given on the mark price, then its selling price 7 will be 102 Rs. C. If article sold on profit percent half of the mark price, then profit is 41.65 Rs. (a) Either A and B or B and C are sufficient to answer the question (b) Either A and B or A and C are sufficient to answer the question (c) Either A and C or B and C are sufficient to answer the question (d) A, B and C together are sufficient to answer the question (e) Either only A or B and C together are sufficient to answer the question 43. A joker has three identical boxes having same number of balls but are of different colors. In each box ‘x’ balls are red, 7 balls are blue and ‘y’ balls are green. Find (𝒙 + 𝒚)? A. Probability of choosing one ball which is either red or blue from 𝟑 any one of the box, is . 𝟒 B. Number of ways to select two balls from any one of the box is 6. C. Probability of choosing one ball which is either red or green from 𝟗 any one of the box is . 𝟏𝟔 (a) Only statement ‘A’ alone is sufficient to answer the question (b) Only statement ‘B’ alone is sufficient to answer the question (c) Only statement ‘C’ alone is sufficient to answer the question (d) Any of two statements are sufficient to answer the question (e) Both A and C are required together 20

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44. Is (bc − ab) even integer or odd integer if a, b and c are integers? A. a, b and c are in arithmetic progression. B. Set {a,b,c} is co-prime and a,b and c are consecutive integers. C. a, b and c all three are prime numbers. (a) Either statement ‘A’ or ‘B’ alone is sufficient to answer the question (b) Either statement ‘B’ or ‘C’ alone is sufficient to answer the question (c) Either statement ‘C’ or ‘A’ alone is sufficient to answer the question (d) Any of two statements are sufficient to answer the question (e) Any one of them is sufficient to answer the question 45. Find the area of a rhombus? A. The ratio of the diagonals of the rhombus is 3 : 4. B. The perimeter of the rhombus is equal to that of a square whose diagonal length is 20√2m. C. The difference between the squares of the diagonals of rhombus is 448 m2. (a) Statement ‘A’ and ‘B’ together are sufficient to answer the question (b) Statement ‘B’ and ‘C’ together are sufficient to answer the question (c) Statement ‘C’ and ‘A’ together are sufficient to answer the question (d) Any of two statements are sufficient to answer the question (e) Any one of them is sufficient to answer the question 46. What will be the sum of the ages of father and the son after five years? A. Father’s present age is twice son’s present age B. After ten years the ratio of father’s age to the son’s age will become 12 : 7. C. Five years ago the difference between the father’s age and son’s age was equal to the son’s present age. (a) Statement ‘A’ and ‘B’ together are sufficient to answer the question (b) Statement ‘B’ and ‘C’ together are sufficient to answer the question (c) Statement ‘C’ and ‘A’ together are sufficient to answer the question (d) Any of two statements are sufficient to answer the question (e) Either ‘A’ and ‘B’ together or ‘B’ and ‘C’ together 47. In how much time Bhavya can row 64km in a stream to reach a destination ‘A’ and return back? 21

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A. Time taken by Bhavya to reach destination ‘A’ is twice than that of to return back. B. Ratio between speed of boat in still water to speed of boat in downstream is 3 : 4 C. Upstream speed of boat is 4kmph less than downstream speed. (a) Either statement ‘A’ and ‘B’ together or ‘B’ and ‘C’ together are sufficient (b) Either statement ‘A’ and ‘C’ together or ‘B’ and ‘C’ together are sufficient (c) Either statement ‘A’ and ‘B’ together or ‘A’ and ‘C’ together are sufficient (d) Any of two statements are sufficient to answer the question (e) All three are required to solve the question 48. A, B and C together can complete work ‘X’ in 5 days. Find in how many day ‘C’ alone can complete work ‘X’? A. A and B together can complete work ‘X’ in 15 days if both worked with half of their efficiencies. B. B, who is 25% more efficient than ‘C’, takes 8 days more than ‘A’ to complete work alone C. ‘A’ is 25% less efficient than ‘B’. (a) Either statement ‘A’ alone or ‘B’ alone is sufficient to answer the question (b) Either statement ‘B’ alone or ‘C’ alone is sufficient to answer the question (c) Either statement ‘C’ alone or ‘A’ alone is sufficient to answer the question (d) Any of two statements are sufficient to answer the question (e) Any one of them is sufficient to answer the question 22

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49. Find the sum of ‘x’ and ‘y’ ? A. H.C.F and L.C.M of ‘x’ and ‘y’ is 21 and 3003 respectively. B. Difference between x and y is 42 C. Both ‘x’ and ‘y’ are multiple of 21. (a) Either statement ‘A’ and ‘B’ together or ‘B’ and ‘C’ together are sufficient (b) Either statement ‘A’ and ‘C’ together or ‘B’ and ‘C’ together are sufficient (c) Either statement ‘A’ and ‘B’ together or ‘A’ and ‘C’ together are sufficient (d) Any of two statements are sufficient to answer the question (e) All three are required to solve the question 50. Find the area of shaded region if radius of each circle is same?

A. Area of circle is given. B. Perimeter of triangle is given. C. Area of triangle is approximately 310% more than area of a circle. (a) Either statement ‘A’ or ‘B’ alone is sufficient to answer the question (b) Either statement ‘B’ or ‘C’ alone is sufficient to answer the question (c) Either statement ‘C’ or ‘A’ alone is sufficient to answer the question (d) Any of two statements are sufficient to answer the question (e) Any one of them is sufficient to answer the question 23

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Solutions

1. (c); From I. Lets length of larger rectangle and smaller rectangle be 5x and 4x respectively. 2(5x+b) – 2(4x+b) = 8 Form II. Breadth of rectangle = side of square Side of square = 14 cm From I & II we get 10x – 8x = 8 x = 4cm perimeter of smaller rectangle = 2 (4×4+14) = 60 cm So, I and II both together sufficient to give answer 2. (e); From I, Lets number of girls and boys be 9x and 11x respectively From II, Let total students 100x Girls below eighteen years group 9 = 100x × = 9x Given

9𝑥 20

100

× 100 = 45x

Boys : girl = (100x–45x) : 45x = 11 : 9 From I and II we get same equation but we can’t get the answer. So, Neither I nor II both sufficient to give answer of question. 3. (c); Form I, Selling price of article = 3800 Rs. 3800 Marked price of article = × 100 = 4000 95

From II 4000 Cost price = × 100 = 3200 125

24

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From I & II Profit of shopkeeper = 3800 – 3200 = 600 Rs. 4. (e); From I, Let speed of boat in still water x km/hr and speed of current y km/hr x = 3y …(1) From II, ′𝑑′ Let bot cover ‘d’ distance downstream and distance in upstream d 3y+y

=

d 2

3y−y

2

⇒

1 4y

=

1 4y

So, From I & II both not sufficient to answer of the question– 5. (d); From I, Let amount invested on 20% is x Rs and on 10% is (4500–x) Rs. 𝑥 × 20 × 2 (4500– 𝑥 ) × 2 × 10 = 100 100 40x = 90000 – 20x 60x = 90000 x = 1500 Rs. amount invested on 10% = (4500–1500) = 3000 Rs. From II, Lets man invested Rs 3x Equivalent CI of two years on 20% = 20 + 20 +

20×20 100

= 44%

ATQ – 20 44 2𝑥 × + 𝑥× = 1260 100 100 108x = 126000 x = 1500 Rs. Amount invested on 10% = 2 × 1500 = 3000 𝑅𝑠. So, Either statement I alone or statement II alone sufficient is to give answer of question 25

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6. (d); Let length of two trains be 4x meter and 3x meter From I, 5

(4𝑥+3𝑥)3

(90 + 72) × = 18 28 21x = 1260 x = 60 meters Required difference = 60 × 4 − 60 × 3 = 60 meters From II, 5 3𝑥 90 × = 18 7.2 3x = 180 meters 180 Length of larger train = 4x = × 4 = 240 3

Required difference = 60 meters So, either I or II alone sufficient to give answer of question. 7. (c); From I and II together Let P, Q and R takes 2x days, 3x days and 6x days respectively So efficiency of P, Q and R is 3x, 2x and x unit/day Total work = 4 × (3𝑥 + 2𝑥 + 𝑥 ) = 24x So, we can determine required difference with I and II together 8. (d); Given, Cost price of article for Satish 1440 = × 100 = 1200 Rs. 120

From I, Cost price of article for Veer = 1200 – 240 = 960 Rs. 240 Profit percentage of Veer = × 100 = 25% 960

From II, 100 Cost price of article for Veer = 1440 × = 960 Veer profit percentage 1200−960 = × 100 = 25%

150

960

So, either I or II alone sufficient to give answer of question. 26

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9. (d); x : y : z = 11 : 9 : 12 Let x, y and z be 11a, 9a and 12a respectively From I, 11a + 9a + 12a 11a + 9a − =2 3 2 32a − 10a = 2 3 𝑎=3 So, (x + y) – 1.5z = (11 × 3 + 9 × 3)–1.5×3×12 = 6 From II, 11𝑎 ×

1 11

1

+ 9𝑎 × = (62)0.5 9

2𝑎 = 6 𝑎=3 So, we can determine (x + y) – 1.5z from II also So, Either statement I or Statement II alone sufficient 10. (c); Let length and breadth of rectangle be 7x and 4x respectively From I, Given, πr2 = 616 616×7 r2 = 22 r = 14 cm length of rectangle = 14 × 2 = 28 𝑐𝑚 28 breadth of rectangle = × 4 = 16 𝑐𝑚 7

From I and II together, 2( l + b) – 4a = 20 2 (28 + 16) – 4a = 20 4a = 88 – 20 a = 17 cm Area of Square = (17)2 = 289 cm2 So, Statement I and II both together sufficient 27

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11. (b); If the max marks of exam = x x Raman = 4 x ⇒ = 288 − 128 = 160 4 x = 640 ∴ Minimum passing marks = 160 + 64 = 224 224 Required %= × 100 = 35% 640

12. (c); From I, A + S+ V = 3 × 68 = 204 kg R + P = 144 kg A ⇒ 204 – 46 – 78 = 80 kg P ⇒ 144 – 68 = 76 kg S = 78 kg 13. (e); From I 27x − 23x = 100000 x = 25000 Population of city A = 50x = 1250000 From II Population of city B = x 4x Population of city A = 5 4x x− = 312500 5 x = 1562500 4 Population of city A = × 1562500 5 = 1250000 14. (b); From I and II Students participating in dance = 150 Students who participate in singing 150×100 = = 60 250

28

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15. (d); Form A → Let radius of circle = r Then side of equilateral ∆ = √3𝑟 Area of equilateral triangle and circle can be find out and required difference can also be find out easily. From B → Let radius of circle =r ⇒ side of equilateral triangle = √3 r Sum of peremeter of triangle and circle is given. By this value of ‘r’ can be find out and after this required difference can be find out easily. Hence, either only A or only B is sufficient to answer the question 16. (a); From A → 2𝑛 + 1 is divisible by 3 ⇒ n =1, 3, 5, 7…. ⇒ n = odd From B) → 4𝑛 − 1 is divisible by 3 ⇒ n =1, 2, 3……. ⇒ ‘n’ can be even as well as odd Hence, Only A is sufficient to answer the question 17. (b); From A → Let numbers are a, b, c, d, e in ascending order ⇒ Median = 18 = c And a+e=b+d From B → let number are a, b, c, d, e is ascending order. Median = 18 = c Mode = 12 and 24 This means two number are 12 and two number are 24 ⇒ Numbers are 12, 12, 18, 24, 24 Required sum 12 + 12 + 18 + 24 + 24 = 90 Hence, Only B is sufficient to answer the question. 18. (d); Let CP = 200x ⇒ MP = 300x From A → SP = 300x – 60 120 200𝑥 × = 300𝑥 − 60 100 ⇒ 60𝑥 = 60 29

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⇒ CP = 200x = 200 From B → Two successive discount 10 × 10 = 10 + 10 − = 19% 100 81 S.P =200x + 43 = 300𝑥 × 100

200x + 43 = 243x x=1 CP = 200x = 200 Hence, either alone A or alone B is sufficient to answer the question. 19. (b); Probability that Amit tells truth = Probability that Amit tells lie =

2

𝑥 𝑥−2 𝑥

Probability that Ankush tells truth= Probability that Ankush tells lie = 2

𝑦−5

𝑥 2

𝑦 5

𝑥

𝑦

From A→ ×

+

From B → × =

1

𝑥−2 𝑥

5

1

𝑦

2

× =

5

𝑦 𝑦−5 𝑦

5

⇒ 𝑥𝑦 = 50 Hence only B is sufficient to answer the question. 20. (d); Given AO + OC = OB From A = AO → find From B → AC = x (AO) When ‘x’ multiple From A & B together area of triangle and Area of semi circle find out. Required area =

π(AO)2

S=

− √S(S – AO). (S – OC). (S – AC)

2 AO+OC+AC 3

∴ A & B together sufficient to answer the question 30

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21. (d); Given no. of white ball Let → a x From A let probability →

y

Let no. of red ball → px, total balls → py 𝑠 From B → Let probability = t Let no. of black ball = qs, total balls = qt From A & B px + a + qs = qt = py we know the values of x, y, s, t and a so we can find the value of p and q a a So probability of white ball found = or 𝑞𝑡

py

∴ A & B together sufficient to answer the question 22. (b); Only B is sufficient to answer the question When we cut sphere into hemisphere total surface area of two hemisphere Total surface area of two hemisphere 3πr² + 3πr² = π × 21 × 21 r = find out So, volume of sphere can be find out. 23. (d); Given ∠PQR = ∠PSR = 90° {Angle of diameter} A → ∠SRO = 180°– 90°(∠PSR) – 40°(∠ SPO) ∠SRO = 50° B→ ∠PQR + ∠QRO = 120° ∠QRO = 120°– 90 = 30° ∠QPO = 60° From B & A together, required difference can be find out. 24. (c); From A→ Discount % = 20% = Mark up% If cost price is 100x then Markup price 120x and selling price is → 96x So ATQ, 100x – 96x = 70 31

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70

C. P = 100x = × 100x = 1750 4x From B→ Let mark up price is → 100x Then selling price is → 80x ATQ, 100x – 80x = 350 + 70 20x = 420 100x = 2100 80x = 1680 C.P. → 1680 + 70 = 1750 So Either A or B alone required. Solutions (25-29) 25. (c); From I, II & III Let speed of Amit and Abhi be 4x and 5x km/hr respectively. 5x – 4x = 20 ∴ x = 20 km/hr Let distance be d km d 80

–

d

=1

100 80×100

∴d=

20

= 400 km

26. (b); From I and II Let length and breadth be 3x and 2x 2πr = 440 [r → radius of circle] r = 70 m ∴ breadth = 10 m & length = 15 m ∴ Area = 10 × 15 = 150 m² Statement I and III are same. 27. (a); From I Passed = 400 From III Let appeared & Passed student be 5x and 3x respectively 2x = 400 ⇒ x = 200 32

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∴ failed = appeared – passed = 1000 – 400 = 600 28. (a); From I PRT(2)

= 44000 PR = 2200000 From II PRT P+ = 15400 100 From III 100

Difference = PR2 1002

PR2

1002

= 120

From I and III R can be found. 29. (e); Let the smaller no. is x & bigger no. is y. From I y=x+6 From II, 40 30 ×𝑥 = ×𝑦 100

100

From III, 𝑦 2 =2 𝑥 1 3 ⇒ 3y = 4x ∴ from I and II or I and III 30. (d); From statement A, Let MP of speaker = Rs. 100 ∴ SP = 80 rupees From B, MP of wrist watch = 1.25 M.P of speaker = Rs. 125 (From st. A) From C, 33

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C. P. of speaker = 80 ×

100 110

=

800 11

(from st. A)

Here, there is no information about S.P. and Profit of watch. So, answer cannot be found. 31. (b); From A, 11 Milk = 68 × = 44 ℓ 17 6 Water = 68 × = 24 ℓ 17 From B, let x ℓ milk is added to the mixture. 24 6 = ⇒x=8ℓ 44+x

13

∴ Total milk in final mixture = 44 + 8 = 52 ℓ 32. (d); From st. A, Speed of train A = P m/sec (Given) 140 14P Speed of train B = of P = From st. B+A, Length of platform 14P 14P 24 × – × 8

100

10

10 10 14P

= 16 × = 22.4 P 10 From C, Length of train A 3 = × 22.4P (with help of st. A and B together) 5

3

∴ Length of tunnel = Pt(given) – × 22.4P 5

33. (e); With help of statement A only, required number of ways can be found out. 34. (e); From statements (I) Nitin can cut cloth for 20 t-shirts in a day 5 ∴ Vishal can cut cloth for 20 × = 25 t-shirts in a day 4 And if Vishal can stich 4x t-shirts in a day, Nitin will stich 5x tshirts. It can’t be solved further. 34

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From statement (II) We only can obtain time taken by Nitin to build 100 t shirts which is15 days Nothing more can be obtained from (II) alone. From statement (III) Question can’t be solved from this statement alone. So, by using (I) & (II) Nitin will take 3 days to cut cloth for 60 t-shirts, Hence 6 days to stich them. Therefore, Nitin stiches 10 t-shirts per day And Vishal stiches 8t-shirts per day. (From ratio obtained in (I)) Total time required for Nitin alone to make 100 t-shirts 100 100 = + = 15 days 20

10

25

8

Total time required for Vishal alone to make 100 t-shirts 100 100 = + = 4 + 12.5 = 16.5 days Hence time can be calculated for both. Using statement (I) and (III) Days required by Vishal to cut cloth for 50 t-shirts is 2 days ∴ Days required by Vishal to stich 50 t-shirts in 12.5 days Hence days required for both Nitin and Vishal can be calculate, [first we will calculate, days required individually and then for both of them] Hence question can be solved through I & II or I & III. 35. (e); From Question, Let age of Aashish and Kullu is 4x and 5x and that of Nikhil & Yash is 7y and 8y. From (I) statement 7y – 4x = 18 …eqn. (A) It can’t be solved further. From (II) statement 35

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4𝑥–12 7𝑦–12

=

2 5

⇒ 20x – 60 = 14y – 24 ⇒ 10x – 30 = 7y – 12 ⇒ 10x – 7y = 18 …eq. (B) 5x–12 1 = 8y–12

2

⇒ 10x – 24 = 8y – 12 ⇒ 5x – 12 = 4y – 6 5x – 4y = 6 …eqn. (C) Solving eqn. (B) & eqn. (C) We will get x=y=6 Hence their age can be calculated. Also their average age after 4 years. Therefore, statement (II) alone is efficient to solve question. From (III) statement 4𝑥+12 3 = ⇒ x = 6, present age of Aashish is 24 years as that of Kullu 4𝑥 2 is 30 years. Nothing more can be calculated Now combining statement (III) & (I) We will get x = y = 6. Hence question can be solved. Therefore, this question can be solved by statement (II) alone or combining (I) and (III). 36. (b);

36

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Let us draw a circular track of 400 meters with 4 point U, V, W and X at anti-clockwise distance of 100m, 200 m and 300 meter. From statement (i) 160 They meet second time in seconds that is somewhere between 7 U and X. Therefore their sum of speed [as both were travelling in opposite direction] is 800×7 = = 35 ms¯¹ 160 As they meet between U and X. There are two chances, either A have only covered a distance of 57 1 1 meter or total distance of (400 + 57 ) meter, hence, we can’t 7 7 determine who is more faster. From statement (II) It is clearly given that B is faster than C, but we can’t determine who is faster between A and C. From statement III A and C only meet at starting points means either their speed is coprime of each other or one of their speed is integral multiple of other and ratio of speed of B and C is given. Now combining statement (i) and statement (ii) It is given that B doesn’t run faster at twice or more speed. Hence 1 chances that A only travelled (400 + 57 ) meter distance, hence 7 faster than B, and it is given that B runs faster than C in statement (ii). Hence question can be answered from statement (i) and (ii) We can’t answer question from (i) and (iii) or (ii) or (iii), as we don’t know who is faster among A and C. 37. (c); From statement …(I) Probability is greater than ⅛ i.e.

2 16

and less than

5 16

. Therefore, he

could either have 3 balls or 4 balls. Now sum of green balls and red balls is odd. [Odd + Even = Odd]. 16 is even. Therefore, blue balls must be odd. [Odd + Odd = Even]. Hence there are 3 blue balls. 37

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(II) Probability of drawing red ball is

3 8

Therefore, he has 6 red balls. 3 Probability of drawing green ball = + 8

1 16

=

7 16

He has 7 green balls. ∴ No. of blue balls he has = 16 – (6 + 7) = 3 Hence question can be answered. (III) Let he have x balls. If one ball is lost, probability of drawing a blue ball is

Which gives us x = 3 Hence question can be answered. Question can be answered from either statement alone. 38. (c); From (I) Total SP of bats = 1020 5 Therefore, CP of bats = 1020 × = Rs. 850 6

Profit % on ball = 12% And more can’t be calculated. From (II) Profit on 1 ball = Rs.6 Profit on selling both bats = Rs. 170 From (III) Cost price of each ball = Rs.50 Profit earned on each bat = 20% From (I) and (II) Total SP of bats = Rs. 1020 And CP of bats = Rs. 850 Total profit of 36 on balls is 12% of total CP of balls. 38

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Hence profit on each ball is 12% which is

36 6

= Rs. 6

Hence CP = Rs. 50/ ball And question can be answered. From (I) and (III) Question can be answered, as we can calculate both CP and SP of both balls & bat. From (II) and (III) Total profit earned is Rs. 206 [from (II)] We can say that Rs. 170 is 20% of total CP of but, hence SP and CP of both bats can be calculated. CP of balls is given in statement (III), Hence question can be answered from (II) and (III). Hence, we can answer question from any two statements. 39. (d);

By joining the centers of all four circles, a square is formed which contain 4 quadrant of circles which together makes a complete circle. Area of shaded region = Area of square(EFGH) - Area of bigger circle – Area of smaller circle Let radius of smaller circle and bigger circle be ‘r’ and ‘R’ respectively Then relation between ‘r’ and ‘R’ is 𝑟 = 𝑅(√2 − 1) A → Radius of bigger circle is given 39

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By this side of square, radius of smaller circle can be find out. So, area of shaded region can be find out B → Diagonal of square is given By this, side of square then radius of bigger circle and then radius of smaller circle can be find out After that shaded region can be find out C → by using this we can find out the radius of the both circles and side of square. 𝜋𝑅2 − 𝜋𝑟 2 = 𝑔𝑖𝑣𝑒𝑛 Relation between R and r is:𝑟 = 𝑅(√2 − 1) Hence, Any of A, B or C alone is sufficient to answer the question 40. (b); From A, Given 𝑟 ∶ ℎ = 7 ∶ 8 From B Radius of cone =

Radius of hemisphere 2 2

Volume of hemisphere = 𝜋𝑟 3 2

3

3

𝜋𝑟 = 19404 19404 × 3 × 7 𝑟3 = 22 × 2 3 𝑟 = 9261 𝑟 = 21 cm 3

From A & B together — 21 Radius of cone = cm 2 10.5

Height of cone = × 8 = 12 cm 7 So from A and B we can determine the surface area of cone. From A and C, Given, radius of cone : height of cone = 7 : 8 Ratio between height and radius of cylinder = 8 : 7 2πr (r + h) = 2640 cm³ 2π7x (8x + 7x) = 2640 40

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x = 2 cm Radius of cone = 7 × 2 × Height of cone =

10.5 7

75 100

= 10.5 cm

× 8 = 12 cm

So, from A and B or from A and C, we can determine the surface area of cone 41. (a); From A, 8𝑡8 = 18𝑡18 We have to find 26𝑡26 8(𝑎 + 7𝑑 ) = 18(𝑎 + 17𝑑) 8𝑎 + 56𝑑 = 18𝑎 + 306𝑑 10𝑎 + 250𝑑 = 0 (𝑎 + 25𝑑) = 0 So 26th term of = (𝑎 + 25𝑑 ) = 0 26 times = 26 × 0 = 0 From B, No result determines From C No result determines So, only statement A sufficient to give answer of question. 42. (e); From A: Let cost price = profit percent = x So, 𝑥×(100+𝑥)

= 119 Rs. x² + 100x = 11900 x² + 100x – 11900 = 0 x = 70 Rs. From B: Selling price = 102 Rs. 102 M.P. ⇒ × 7 = 119 Rs. 100

6

From C: If M.P. ⇒ 119 Rs. 41

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Let cost price = x Rs. From B and C together: 𝑥 119 × = 41.65 Rs. 100

2

x = 70 Rs. Either only A or B and C together. 43. (c); From A, ATQ, 𝑥+7 3 = ⇒ 3y – x = 7 𝑥+𝑦+7

4

𝑥+𝑦+7

16

From B, As all boxes are identical and all colored balls are identical then number of ways to choose balls is 6 ⇒ 𝑅𝑅, 𝐺𝐺, 𝐵𝐵, 𝑅𝐺, 𝑅𝐵, 𝐺𝐵 Where GG = Green | Green By this value of x and y cannot be find out From C ATQ, 𝑥+𝑦 9 = ⇒x+y=9 So, only statement ‘C’ is sufficient to give answer of question. 44. (a); From A, a, b and c are in A.P Let common difference = ‘d’ and ⇒a = b−d And c = b + d (bc − ab) = b2 + bd − b2 + bd = 2bd ⇒ (bc − ab) is divisble by 2 From B, Set {a, b, c} is coprime ⇒ all three have ′1′ as common multiple And they are consecutive numbers So , either b is even or a and c are both even. In both casesb(c-a)= even 42

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From C All three are prime numbers but we can’t say whether in these prime numbers 2 is taken or not. And if 2 is taken then b = 2 or not. SO, in some cases (bc − ab) is divisble by 2 and in some cases (bc − ab) is not divisble by 2. Correct conclusion cannot be inferred from statement ‘C’ alone So, either statement ‘A’ or ‘B’ alone is sufficient to give answer of question. 45. (d); From A, diagonal = 3x, 4x From B, 4 a = 80, a = 20 ,𝑑12 + 𝑑22 = 1600 From C, (4𝑥 )2 − (3𝑥 )2 = 448 x = 8, diagonal = 24, 32 1

Area = × 24 × 32 = 384 𝑚2 2

So, using any two of the three statement we can find area of rhombus 46. (e); Let age of father = 𝑥 Age of son = 𝑦 From A, 𝑥 = 2𝑦 From B, 𝑥+10 𝑦+10

=

12 7

From C, 𝑥 − 𝑦 = 𝑦, 𝑥 = 2𝑦 So, Either A and B together or B and C together are sufficient to answer the question 43

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47. (b); Distance = 64 km From A Ratio between time to reach destination A and to return back = 2 : 1 1 Upstream to downstream Speed ratio = ∶ 1 ⇒ 1 : 2 2 From this we can find out the ratio between speed of boat in still water to speed of stream =3:1 From B Ratio of speed of boat in still water to speed of boat in downstream =3:4 From this we can find out the ratio between speed of boat in still water to speed of stream =3:1 From C Difference between upstream speed to downstream speed = 4 km/hr By using ‘A’ and ‘C’ together or ‘B’ and ‘C’ together, answer can be find out. 48. (a); Let A, B and C can complete the work ‘X’ in a, b and c days Respectively 1 1 1 1 ⇒ + + = a

b

From A 1 1 + = 2a

2b

c

5

1 15

From B 4 b= c 5 and b=a+8 From C 5 a=b 4 So, either statement ‘A’ alone or ‘B’ alone is sufficient to solve the question. 44

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49. (c); From A, 𝑥 × 𝑦 = 21 × 3003 = 21 × 21 × 11 × 13 L.C.M is 21 ⇒ Minimum value of ‘x’ or ‘y’ should be 21 From B, (𝑥 − 𝑦) 𝑜𝑟 (𝑦 − 𝑥 ) = 42 From C Both x and y are multiple of 21 ⇒ L.C.M of x and y is 21. Hence, either statement ‘A’ and ‘B’ together or ‘A’ and ‘C’ together are sufficient 50. (a); In the figure Triangle ABC is an equilateral triangle And relation between radius or circle (r) and side of triangle (a) is given below 𝑟 1 = 𝑎

2(1+√3)

From A, Area is given ⇒ radius is given ⇒ side of triangle can be find out ⇒ Area of shaded region can be find out From B, Perimeter of triangle is given ⇒ radius of circle can be find out ⇒ Area of shaded region can be find out From C It is a basic condition of this figure and anything cannot be concluded from this statement Hence, either statement ‘A’ or ‘B’ alone is sufficient.

45

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1

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Chapter

16

Caselets

BEST APPROACH TO SOLVE THE QUESTIONS A Caselet is another way of providing data, more often, explicitly. Unlike various graphical methods of data representation, a caselet requires to be solved completely before arriving at the answers to the questions that follow it. Usually, in a caselet, data is provided through statements that imply the logic or relation of various components of data. Some caselets are based on Set Theory and concepts of Set theory are applied to them. All other caselets are based on basic mathematical understanding. A sample example of a caselet is as follows: ‘PO’ exam in Lucknow is 60% more than candidates who are appearing for same exam in Agra. Number of candidates appearing for ‘PO’ exam in Allahabad is 48% of total candidates appearing for ‘PO’ exam. Ratio between total candidates appearing for ‘PO’ exam and ‘Clerk’ exam is 5 : 9. Difference between candidates appearing for ‘PO’ exam in Lucknow and Allahabad is 400. Ratio between candidates appearing for ‘PO’ exam to ‘Clerk’ exam in Agra and Lucknow is 5 : 12 and 8 ; 9 respectively. Let total candidates appeared for ‘PO’ exam and ‘Clerk’ exam is 500x and 900x Number of candidates appearing for ‘PO’ exam in Allahabad 48 = × 500x = 240x 100

Remaining candidates appeared for ‘PO’ exam in Agra and Lucknow together = 500x – 240x = 260x Let total candidates appeared for ‘PO’ exam in Agra = y Then total candidates appeared for ‘PO’ exam in Lucknow = 1.6y ATQ, y + 1.6y = 260x ⇒ y = 100x 2

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total candidates appeared for ‘Clerk’ exam in Agra = 100x ×

12 5

= 240x 9

total candidates appeared for ‘Clerk’ exam in Lucknow = 160x × = 180x 8

total candidates appeared for ‘Clerk’ exam in Allahabad = 900x – 240 – 180x = 480x EXAM PO CLERK

AGRA 100x 240x

LUCKNOW ALLAHABAD 160x 240x 180x 480x

AGRA 500 1200

LUCKNOW ALLAHABAD 800 1200 900 2400

Now ATQ 240x – 160x = 400 400 ⇒𝑥 = =5 80

EXAM PO CLERK

Practice Exercise Based on new Pattern

Directions (1-5): Study the given passage carefully and answer the questions. Rahul, Sandy and Sati invested in ratio 2 : 3 : 4. After 4 months Sandy added Rs. 1500 more in his investment and Rahul withdrew Rs. 800 from his investment. After six months more Sati invested half of the investment done by Rahul in first four months and Sandy invested 50% more than the investment done by Sati in first 10 months. Rahul invested same as investment done by Sandy in first four months. Ratio of profit of Sati to total profit at the end of year is given as 125 : 376. 1. Profit of Sandy is approximately what percent of total profit? (a) 64% (b) 48% (c) 72% (d) 68% (e) 42% 3

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2. What is the difference between profit share of Rahul and Sandy if total profit is Rs.37,600? (a) 12,000 (b) 16,400 (c) 18,500 (d) 22,900 (e) 20,000 3. Veer have 250% more than initial investment of Sati for a year. Find total interest earned by him if he invested his amount in a scheme which offers 20% p.a. for 2 years? (a) Rs. 1400 (b) Rs. 1500 (c) Rs. 1540 (d) Rs.1600 (e) Rs.1640 4. What is the average of profit share of Sandy and Sati out of total profit of Rs. 37,600? (a) 18,220 (b) 18,250 (c) 16,420 (d) None of these (e) 12,490 5. If initial investment of Bhavya is one-third of initial investment of Rahul, Sandy and Sati together then find the difference between initial investment of Bhavya and Rahul. (a) Rs.1000 (b) None of these (c) Rs.750 (d) Rs.500 (e) Rs.250 Directions (6-10): There are 2000people live in a village. Out of 2000 people some people are watching ‘3 idiots’ ‘Ravan’ and ‘PK’ on Sunday and some are not watching any movie. Number of people did not watch any movie is equal to number people watching ‘PK’ only. Number of people who are watching ‘Ravan’ is 720. Ratio between number of people watching ‘3 Idiots’ and ‘Ravan’ only to number of people watching ‘Ravan’ and ‘PK’ only is 2 : 3. Number of people watching ‘3 Idiots ’ and ‘PK’ both is half of number of people watching only ‘PK’. Number of people watching ‘3 idiots’ only is 50% more than number of people watching ‘Ravan’ only. Number of people watching all the three movies is 4% of the total number of people in the village. Number of people watching ‘Ravan’ only is same as number of people watching ‘Ravan’ and ‘PK’ only. 4

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6. Number of people who did not watch any movie are what percent less than number of people who watching only Ravan, only PK and only 3 Idiots together? (approx.) (a) 55% (b) 50% (c) 65% (d) 62% (e) 70% 7. Find the ratio between people who watching only Ravan and only PK together to people who watching only Ravan & 3 idiots and only PK & 3 idiots together ? (a) 16 : 5 (b) 16 : 7 (c) 16 : 13 (d) 16 : 11 (e) 76 : 43 8. Out of total people watching 3 idiots only, ratio between male to female is 4 : 1 and that of people watching Ravan only is 5 : 3. Find total number of males , who watching 3 idiots and Ravan only ? (a) 436 (b) 428 (c) 440 (d) 438 (e) 442 9. People watching only Ravan & PK are what percent more than people watching only Raven & 3 idiots? (a) 50 % (b) 45% (c) 55% (d) 40% (e) 60% 10. People who are watching only 3 Idiots, only PK and only 3 Idiots and PK both is how much more than people who are watching Ravan? (a) 252 (b) 192 (c) 672 (d) 488 (e) 272 Directions (11-15): Given below is the data about students appeared in two exams i.e., A and B in six different years i.e., 2011 to 2016. Total students appeared in both exam in 2016 is 8000 while in 2013 it is 5800. Average number of students appeared in exam B in 2011 and 2013 is 3100 and is in the ratio 18 : 13. Students appeared in exam A in year 2015 is 1 33 % more than students appeared in exam B in same year. Total students 3

appeared in 2016 is 25% more than total students appeared in 2011. 5

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Students appeared in exam A in 2016 is 62

26 27

% more than students

appeared in exam B is 2015. Ratio of total students appeared in 2016 & 2014 is 16 : 13. Total number of students appeared in exam A in all six years is 21,100. Students appeared in exam B in 2011 is same as student appeared in exam A in 2015.Students appeared in exam A in 2012 is 700 more than that of students appeared in same exam in 2014. Students appeared in exam B in 2014 is 1200 less than that of in same exam in 2012. 11. In which year total students appeared in both exam is 3rd highest? (a) 2012 (b) 2014 (c) 2016 (d) 2011 (e) Other than the given options 12. What is the respective ratio between students appeared in exam A in year 2011, 2012 and 2014 together to the students appeared in exam B in year 2013, 2014 and 2016 together? (a) 95 : 97 (b) 99 : 97 (c) 98 : 97 (d) 99 : 95 (e) None of these 13. Find the difference between average number of students appeared in exam A and average of students appeared in exam B in starting four years? (a) 250 (b) 225 (c) 215 (d) 200 (e) None of these 14. Students appeared in exam A in 2013 is how much less than students appeared in exam B in 2012? (a) 1400 (b) 1000 (c) 1100 (d) 1200 (e) 1300 15. Total number of students appeared in both exam in 2012 is what percent more than total number of students appeared in both exam in 2011? (a) 25.25% (b) 28.25% (c) 31.25% (d) 34.25% (e) 37.25% 6

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Directions (16-20): There are 1000 students in a college. Out of 1000 students some appeared in exams ‘X’, ‘Y’ and ‘Z’ while some not. Number of student not appeared in any exam is equal to number of students appeared in exam ‘Z’ only. Number of students appeared in exam ‘Y’ is 360. Ratio of number of students appeared in exam ‘X’ and ‘Y’ only to number of students appeared in exam ‘Y’ and ‘Z’ only is 2 : 3. Number of student appeared in exam ‘X’ and ‘Z’ both is half of number of students appeared in only exam ‘Z’. Number of students appeared in exam ‘X’ only is 50% more than number of students appeared in ‘Y’ only. Number of students appeared in all the three exam is 4% of the total number of students in the college. Number of students appeared in ‘Y’ exam only is same as number of students appeared in ‘Y’ and ‘Z’ only. 16. How many students appeared in at least two exams? (a) 240 (b) 260 (c) 300 (d) 360 (e) 500 17. How many students appeared in two exams only? (a) 280 (b) 220 (c) 340 (d) 300 (e) 260 18. How many students appeared in at most two exams? (a) 240 (b) 260 (c) 300 (d) 500 (e) 960 19. How many students not appeared in exam Y? (a) 440 (b) 360 (d) 640 (e) None of these

(c) 540

20. How many students appeared in exam X or in exam Z? (a) 240 (b) 360 (c) 500 (d) 680 (e) 760 Directions (21-25): Number of songs recorded by Sonu Nigam in 2008 is 300. And number of songs recorded by Sandeep and Arijit Singh in 2010 are in ratio 5 : 4. Total number of songs recorded in 2009 is 250% more than songs recorded by Arijit Singh in 2009. Total songs recorded by Sandeep in 7

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4

all three years is 650. Average of songs recorded in 2010 is th of songs 3 recorded by Arijit Singh in 2008. Songs recorded by Arijit Singh in 2009 is 2th of songs recorded by Sonu Nigam in 2008. Total songs recorded in 2008 5 is 6 times of songs recorded by Arijit Singh in 2009 and songs recorded by Sandeep in 2008 is 25% less than songs recorded by Arijit Singh in same year. Ratio of songs recorded by Sonu Nigam and Sandeep in 2009 is 2 : 3. 21. Number of songs recorded by Arijit Singh in all years is what percent more/less than total songs recorded in 2008 ? 2 7 2 (a) 12 % (b) 17 % (c) 21 % 4 2

(d) 27 % 5

9

3

(e) None of these

22. What is the difference of average of total number of songs recorded by Sandeep in 2008 and 2009 and average number of songs recorded by Sonu Nigam and Arijit Singh in 2010? (a) 155 (b) 165 (c) None of these (d) 150 (e) 145 23. What is the ratio of total number of songs recorded in 2010 to total number songs recorded in 2011, if total number of songs recorded in 2011 is 50% more than total number of songs recorded in 2009 ? (a) 32 : 17 (b) None of these (c) 21 : 19 (d) 33 : 19 (e) 32 : 21 24. Average number of songs recorded by Sonu Nigam is approximately what percent more/less than average number of songs recorded in 2009? (a) 104% (b) None of these (c) 124% (d) 114% (e) 110% 25. Average number of songs recorded by all in all three years and is how much more than average number of songs recorded by Sandeep and Arijit Singh in 2009? (a) None of these (b) 525 (c) 650 (d) 550 (e) 450 8

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Directions (26-30): Study the following passage & answer the questions that follows. There are three schools in a town named A, B & C. In each school students knows only Hindi, Only English and both language. Total number of student in school A is 1600. The no. of student who know only Hindi in school C is 40% of the total students in school A. And total students who know only single language in school A are equal to student who knows both language in same School and ratio of student who know only Hindi to only English in school A is 2 : 3. Student who know both language in all three schools is 50% more than student who knows only English in school B. Student who knows only Hindi in all schools is equal to the total students in school A except student who knows only English. Ratio of students who know only English in school B to C is 1 : 4 and average of student who know only English in all the school is 460. Student who know both language in school C is half of student who know only Hindi in school 3

A. Total student in school B is th of total student from all the school who 4

knows only Hindi.

26. Total student in school A who know only one language is what percent more/less than total student in school B who knows both language? (a) 50% (b) 60% (c) 70% (d) None of these (e) 45% 27. What is ratio of total student who knows only English in school B and C together to total student in school B ? (a) 3 : 2 (b) 7 : 5 (c) None of these (d) 5 : 3 (e) 15 : 14 28. What is sum of average of student who knows only English in all the school and

3th 4

of student who knows only single language in school B

and C together. (a) 1735 (d) 1525 9

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29. 40% of total student in school B is what percent more/less than half of student who know only Hindi in school A ? (a) 140% (b) 110% (c) 90% (d) None of these (e) 130% 30. What is difference of average of total student in all three schools and average of student who knows both language in school B and C ? (a) 920 (b) 840 (c) 890 (d) 990 (e) None of theses Directions (31-35): Data regarding investment of three different persons in three different schemes is given below. Study the data carefully and answer the following questions. → Out of total amount invested by ‘Rahul’ in all the three schemes, 25% is invested in scheme ‘X’. Remaining amount is invested in scheme ‘Y’ and ‘Z’ equally. → ‘Veer’s’ investment in scheme ‘X’ is 37.5% less than that of ‘Anurag’s’ in same scheme. Total amount invested by ‘Rahul’, ‘Veer’ and ‘Anurag’ in all the three schemes is in the ratio of 4 : 5 : 6. Ratio between amount invested by ‘Veer’ in scheme ‘Y’ to in ‘Z’ is 3 : 2. Amount invested by ‘Veer’ in scheme ‘Y’ is 80% more than that in scheme ‘X’. → Amount invested by Anurag in scheme ‘Y’ and ‘Z’ together is Rs. 20,000 more than the amount invested by ‘Rahul’ in both the same schemes together. Amount invested by Anurag in scheme ‘Y’ is 200% more than that in scheme ‘Z’. 31. Rate of interest on scheme ‘X’, ‘Y’ and ‘Z” is 10%, 20% and 30% p.a. at Simple interest. Find total interest earned by Rahul after 2 years. (a) Rs. 44,000 (b) Rs. 41,000 (c) Rs. 34,000 (d) Rs. 30,000 (e) Rs. 24,000 10

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32. Veer and Anurag both invested in scheme ‘X’. After 8 months Veer withdraw his total amount while Anurag withdraw his total amount after 12 months. If total profit of both is Rs. 7,650 then find the profit share of Veer. (a) Rs. 2,250 (b) Rs. 3,150 (c) Rs. 4050 (d) Rs. 4950 (e) Rs. 6750 33. Total amount invested in scheme ‘Y’ by all three together is what percent more than total amount invested by all three in scheme ‘Z’ all together? (a) 50.25% (b) 56.75% (c) 62.75% (d) 68.75% (e) 72.25% 34. Find the ratio between amount invested by Veer on scheme ‘X’ to amount invested by Rahul in scheme ‘Z’? (a) 4 : 3 (b) 1 : 1 (c) 2 : 3 (d) 5 : 8 (e) 5 : 6 35. Scheme ‘Z’ offers 20% p.a. at Compound interest. Find interest earned by Veer is what percent more than interest earned by Anurag in that scheme? (a) 75% (b) 50% (c) 100% (d) 150% (e) 200% Direction (36-40): Data given below about five vessels M, N, O, P & Q contains mixture of different liquid read the data carefully and answer the questions. Vessel M contains mixture of milk and water, vessel N contains mixture of water and orange juice, vessel O contains mixture of orange juice and Vodka, vessel P contains mixture of Vodka and apple juice and vessel Q contains mixture of apple juice and Scotch. Mixture of milk and water in vessel M in the ratio of 7 : 2 and quantity of water in vessel N equal to quantity of milk in vessel M. Mixture of orange juice and vodka into the vessel O in the ratio of 5 : 3 and total quantity of mixture in vessel O is 30 liter more than total quantity of mixture in vessel N. 11

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2

Total quantity of mixture in vessel Q in 66 % of total quantity of mixture in 3

vessel M. Total quantity of mixture in vessel O is 50% more than total quantity of mixture in vessel P. Ratio of orange juice in vessel N and O is 7 : 15, while ratio of vodka in vessel O and P is 3 : 2. Total quantity of Scotch in vessel Q is 30 liter and total quantity of water in vessel M is 40 liter. Quantity of orange juice in vessel N is 70 liters. 36. If 72 𝑙 of mixture from vessel M and 36 mixture taken out from vessel N and mixed in another vessel G, then what will be ratio of milk, `water and orange juice in vessel G? (a) 15 : 5 : 8 (b) 14 : 10 : 3 (c) 14 : 7 : 3 (d) 15 : 8 : 3 (e) 14 : 10: 7 37. 36 liter of mixture from vessel Q taken out and mixed in a vessel C, which contains 21 liter scotch, what quantity of apple juice should mixed in vessel C to obtain ratio of apple juice and scotch 2 : 1: (a) 36 (b) 33 (c) 38 (d) 37 (e) 32 38. If vessel N and O mixed together in a big vessel Z and 22.5 liter of mixture taken out from vessel Z, find quantity of Vodka in 22.5 liter of mixture? (a) 4.5 liter (b) 9 liter (c) 2.25 liter (d) 1.5 liter (e) 3.6 liter 39. If 64 liter of mixture from vessel O, 48 liter of mixture from P and 60 liter of mixture from vessel Q taken out and mixed together in vessel A, find the ratio of orange juice, vodka, apple juice and scotch in vessel A? (a) 40 : 42 : 75 : 25 (b) 40 : 25 : 75 : 52 (c) 40 : 52 : 75 : 15 (d) 30 : 52 : 75 : 15 (e) 40 : 42 : 75 : 15 12

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Directions (40-42): Given below is the data about players participated for state level and National level championship in six different years i.e., 2001 to 2006. Total players participated for both Championship in 2006 is 6000 while in 2003 is 4800. Total number of players participated for national level championship in 2001 and 2003 is 3600 and is in ratio of 11 : 7. Player participated for state championship in 2005 is 25% more than player participated for national Championship in same year. Total players participated in 2006 is 20% more than total players participated in 2001. 1 Players participated for state Championship in 2006 is 62 % more than 2 players participated for national championship in 2005. Ratio of total players participated in 2006 to 2004 is 15 : 13. Total number of players participated for state championship in all six years is 20000. Players participated for national championship in 2001 is same as players participated for state championship in 2005. Players participated for state championship in 2002 is 800 more than that of players participated for same championship in 2004. Players participated for National level championship in 2004 is 1200 less than that of for same Championship in 2002.? 40. What is the respective ratio between players participated for state level championship in year 2001, 2002 and 2004 together to the players participated for national level championship in year 2003, 2004 and 2005 together? (a) 1145:493 (b) 1154:441 (c) 1154:439 (d) 439:1154 (e) 1105:439 41. Find the difference between average number of players participated for state level championship and average of players participated for national level championship for starting four years? (a) 1900 (b) 1920 (c) 1820 (d) 1780 (e) 1940 42. Players participated for state level championship in 2003 is how much more than players participated for national level championship in 2002? (a) 930 (b) 940 (c) 950 (d) 960 (e) 970 13

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Direction (43-47): Given below statistical data of top five wheat producer states of India in the three successive years. Read the data carefully and answer the question given below. Year 2015– Uttar Pradesh and Punjab together produced 15860 thousand ton, while Punjab production was 1180 ton less than that of Uttar Pradesh. Madhya Pradesh produced 25% less than Uttar Pradesh, Haryana produced 690 thousand ton less than that of Madhya Pradesh. Production of Rajasthan was 70% of total production of Haryana Year 2016– Production of Punjab decreased by 20% compare to previous year, while production of Haryana increased by 800 thousand ton compare 1

to previous year. Madhya Pradesh produced 33 % more than previous 3

year, while Rajasthan produced 2390 thousand ton more wheat than previous year. All five states produced total 36612 thousand ton wheat in 2016. Year 2017– Uttar Pradesh and Haryana together produced 16520 thousand ton, while Uttar Pradesh production was 2840 ton more than that of Haryana. Punjab produced 628 thousand ton more wheat than previous year and Madhya Pradesh produced 360 thousand ton more wheat than previous year. All five states produced total 38200 thousand ton wheat in 2017. 43. Total wheat produced by Uttar Pradesh is how much more than total wheat produced by Madhya Pradesh in the all three given years? (a) 3550 thousand ton (b) 3750 thousand ton (c) 3650 thousand ton (d) 3250 thousand ton (e) 3150 thousand ton 44. Total wheat produced by Madhya Pradesh in the year 2016 & 2017 together is what percent more than total wheat produced by Haryana in the year 2015 & 2016 together? (a) 40 (d) 42 14

38

%

(b) 44

%

(e) 48

61 38 61

38 61 38 61

%

(c) 36

38 61

%

%

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45. Find the difference between average quantity of wheat produced by Uttar Pradesh, Madhya Pradesh and Haryana in the year 2015 and average quantity of wheat produced by Madhya Pradesh, Haryana and Rajasthan in the year 2017? (a) 360 thousand ton (b) 400 thousand ton (c) 420 thousand ton (d) 450 thousand ton (e) 470 thousand ton 46. Find the ratio between total wheat produced by Rajasthan & Punjab together in the year 2017 to total wheat produced by Haryana & Rajasthan together in the year 2016? (a) 160 : 161 (b) 150 : 161 (c) 170 : 171 (d) 160 : 169 (e) 160 : 171 47. Find the total wheat produced by Rajasthan and Uttar Pradesh in the year 2016 and 2017 together? (a) 30700 thousand ton (b) 30800 thousand ton (c) 31700 thousand ton (d) 32400 thousand ton (e) 33400 thousand ton Direction (48–51): Given below data is about the number of candidates who got final selection into UPSC exam from five different states (Uttar Pradesh, Kerala, Maharashtra, Madhya Pradesh and Bihar) in three different successive years. Read the data carefully and answer the questions: 2012– Total number of Candidates selected from Uttar Pradesh are 20% more than total number of Candidates selected from Kerala and total number of selected Candidates from these two states are 440. Average number of Candidates selected from Uttar Pradesh, Kerala and Maharashtra is 220 and total number of Candidates selected from Madhya Pradesh is 70 less than that of from Maharashtra. Total number of Candidates selected from these five states in the year 2012 is 930. 2013– Total number of Candidates selected from Uttar Pradesh is 33 1 % more than that of total number of Candidates selected from same state 3 in previous year, while total number of Candidates selected from Kerala is 30 less than average number of Candidates selected from Kerala & Maharashtra in the year 2012. Total number of Candidates selected from 15

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Maharashtra is 55 more than that of total number of Candidates selected from same state in previous year and total number of candidates selected from Madhya Pradesh is equal to total number of candidates selected from Kerala. Total number of Candidates selected from these five states in the year 2013 is 1130. 2014– Total number of candidates selected from Uttar Pradesh and Maharashtra is 40 more and 25 less than total number of candidates selected from Uttar Pradesh and Maharashtra in previous year respectively. Average number of candidates selected from Uttar Pradesh, Kerala and Maharashtra is 280, while average number of candidates selected from Madhya Pradesh & Bihar is 195. Total number of selected candidates from Bihar is 20 more than that of Madhya Pradesh. 48. Total number of candidates selected from Kerala & Maharashtra together in the year 2014 is what percent more/less than total number of candidates selected from Uttar Pradesh in the year 2012 & 2014 together? (a) 10% (b) 20% (c) 15% (d) 25% (e) 5% 49. If total number of candidates selected from Karnataka in the year 2014 is 50% more than total number of candidates selected from Madhya Pradesh in the year 2013, then find average number of candidates selected from Karnataka, Uttar Pradesh, Madhya Pradesh & Bihar in the year 2014? (a) 225 (b) 255 (c) 205 (d) 215 (e) 200 50. Find the ratio between total number of candidates selected from Kerala to total number of selected candidates from Bihar in the given three years? (a) 63 : 50 (b) 71 : 50 (c) 61 : 50 (d) 67 : 50 (e) 69 : 50 16

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51. Total number of candidates selected from Maharashtra, Madhya Pradesh and Bihar together in the year 2013 is what percent more/less than total number of candidates selected from Madhya Pradesh in the year 2012 & 2013 together? 10 10 10 (a) 92 % (b) 90 % (c) 94 % (d) 91

11 10 11

%

(e) 95

11 10 11

11

%

Directions (52-56): Neeraj have some toys which are in the form of different structures. These are cylindrical, conical, spherical. Other than solid conical structure, all two are of both types i.e., hollow as well as solid. → Volume of a conical toy is three times of the volume of a solid cylindrical toy while radius of a solid spherical toy is half than that the radius of a conical toy. Outer radius of hollow cylindrical toys is same as radius of solid spherical toy while average of outer radius and inner radius of hollow cylindrical toys is equal to radius of solid cylindrical toy. Height of cylindrical, conical and hollow cylindrical toys is same i.e, 14c.m → Number of solid spherical toys is 20% of total number of toys Neeraj have. Number of hollow spherical toys is 150% more than number of conical toys. Ratio between number of solid cylindrical toys to number of conical toys is 3 : 2. Total number of hollow cylindrical toys is 40% of total number of toys Neeraj have and also ‘20’ more than the total number of solid spherical toys Neeraj have. → Volume of a hollow spherical toy is 33,957 cm² whose inner radius is half of its outer radius. Volume of a hollow spherical toy is 5.25 time of volume of conical toy. 52. Find the total space taken by all solid spherical toys? (in cm³) (a) 97020 (b) 48510 (c) 72765 (d) 14553 (e) 24255 53. Find the number of conical toys Neeraj have? (a) 40 (b) 20 (d) 12 (e) 8 17

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54. Find the curved surface area of one hollow cylindrical toy? (in cm²) (a) 616 (b) 1232 (c) 924 (d) 462 (e) 1386 55. Find the ratio between outer radius of hollow spherical toy to radius of solid cylindrical toy? (a) 4 : 1 (b) 3 : 2 (c) 3 : 1 (d) 4 : 3 (e) 2 : 1 56. Volume of one hollow cylindrical toy is how much more then volume of one cylindrical toy?(in cm3 ) (a) 4312 (b) 3234 (c) 2696 (d) 2156 (e) 1078 Direction (57-61): - Data about speed and length of six different trains is given below. Study the data carefully and answer the following question. Speed of six trains A, B, C, D, E and F are a, b, c, d, e and f respectively while time taken by six trains to cross a pole is 𝑃𝐴 , 𝑃𝑏 , 𝑃𝑐 , 𝑃𝑑 , 𝑃𝑒 and 𝑃𝑓 respectively. Length of six trains A, B, C, D, E and F are 𝑙𝑎 , 𝑙𝑏 , 𝑙𝑐 , 𝑙𝑑 , 𝑙𝑒 and 𝑙𝑓 respectively → Train C can cross a pole in double time in which train E can cross a pole. Time taken by train D to cross a platform of 180 m length is twice than that time taken by same train to cross a pole. Train F can cross a platform double of its length in 36 seconds. Speed of train ‘E’ is 20% more than speed of train ‘C’. Train ‘A’ can cross a platform of 100 m length in same time in which train ‘F’ can cross a pole. → Train ‘B’ can cross two poles 60 m apart from each other in 24 seconds. Train ‘B’ can cross train ‘C’ in 10 seconds if they move towards each other. Time taken by train ‘D’ to cross a platform of 300 m length is same time in which train ‘C’ can cross a pole. Given that a:b→5:3 ⇒d:f→4:3 𝑃𝐴 ∶ 𝑃𝑏 → 2 ∶ 5 ⇒ 𝑙𝑑 ∶ 𝑙𝑒 → 1 ∶ 2 18

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57. Train ‘C’ and Train ‘D’ enter in a tunnel from opposite sides at same time. Find the length of tunnel if both trains completely take exit from the tunnel at same time? (a) 900 m (b) 1200 m (c) 1500 m (d) 1800 m (e) 2100 m 58. Find the time in which train ‘B’ can cross a platform of same as length as of train ‘F’? (a) 50 seconds (b) 52 seconds (c) 54 seconds (d) 56 seconds (e) 58 seconds 59. Train ‘A’ starts from station Rewari and move towards Gurgaon. After 4 hours Train ‘F’ starts from station Rewari and move towards Gurgaon. After how much time Train ‘F’ will overtake train ‘A’ if length of both trains is taken as negligible? (a) 5 hours after start of Train ‘F’ (b) 5 hours after start of train ‘A’ (c) 9 hours after start of train ‘F’ (d) 9 hours after start of train ‘A’ (e) Both (a) and (d) 60. Time taken by train ‘A’ to cross a platform of length 250 m length is what percent less than time taken by train ‘D’ to overtake train ‘F’ when they move in same direction? (a) 37.5% (b) 62.5% (c) 50% 2

(d) 166 % 3

2

(e) 266 % 3

61. Train ‘C’ starts from Gurgaon station while train ‘D’ starts from Delhi station at same time. Both starts to move towards each other. After how much time trains will completely cross each other if distance between Delhi and Gurgaon is 30km? (a) 208 seconds (b) 218 seconds (c) 223 seconds (d) 228 seconds (e) 238 seconds 19

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Directions (62-66) Satish, Abhi and Bhavya are three persons who each invested some amount in three different schemes (P, Q and R). Data tells about amount invested and time of investment by them. Study the data carefully & answer the following questions. → Amount invested by Satish in scheme ‘P’ is half than that amount invested by Bhavya in same scheme. Abhi invested Rs 80,000 in scheme ‘P’ and time for which he invested in scheme ‘P’ is 5 months more than time for which Satish invested in same scheme. Ratio of amount invested by Satish and Bhavya in Scheme ‘R’ is same as ratio of amount invested by Satish and Bhavya in scheme ‘P’. Abhi invested 20,000 less in scheme ‘R’ than that of amount invested by Bhavya in scheme ‘R’. → Time of investment of Abhi and Bhavya is same in scheme ‘Q’. Amount invested by Abhi in scheme ‘Q’ is 96% of amount invested by Satish in scheme ‘Q’. Bhavya invested Rs 1,00,000 more than Abhi in scheme ‘Q’. Ratio of amount invested by Satish in scheme ‘P’ to scheme ‘Q’ is 3 : 5. Ratio of amount invested by Bhavya in scheme ‘P’ to scheme ‘Q’ is 15 : 22. → Out of total profit earned from scheme ‘Q’ Bhavya got 50%. Ratio of profit share of Satish and Abhi is 3 : 4 in scheme ‘R’ while time of investment of Satish and Abhi is scheme ‘R’ is 4 : 3. Bhavya invested for 10 month in scheme ‘R’. Satish got 25% of total profit both in scheme ‘R’ as well as in scheme ‘P’. Time of investment of Satish in scheme ‘P’ is 2 months less that of in scheme ‘R’. Satish invested for 16 months in scheme ‘Q’ 62. Abhi and Bhavya both invested same amount they invested in scheme ‘Q’ in two different schemes i.e, S1 and S2 respectively. S1 and S2 offers 20% p.a at C.I and 25% p.a at S.I respectively. Find the difference between interest earned by both after 2 years? (a) 36,800 (b) 41,900 (c) 47,000 (d) 52,100 (e) 57,200 20

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63. If Bhavya earned Rs. 28,980 profits from scheme ‘R’ which is 125% more than profit earned by him from scheme ‘P’, then find total profit earned by Satish from scheme ‘P’ and ‘R’ together? (a) 33,948 (b) 26,588 (c) 30,268 (d) 22,908 (e) 37,628 64. Find interest earned by Satish if scheme ‘Q’ offers S.I. at the rate of 15% p.a? (a) Rs26500 (b) Rs29000 (c) Rs28000 (d) Rs25000 (e) Rs27500 65. Amount invested by Bhavya in scheme ‘R’ is what percent more than amount invested by Satish in scheme ‘P’? (a) 140% (b) 100% (c) 144% (d) 50% (e) 20% 66. If Abhi doubles his investment in scheme ‘P’ than before, then find what percent of decrement is seen in Bhavya’s profit percentage? 4 (a) 20% (b) 40% (c) 28 % 6

(d) 42 % 7

21

1

7

(e) 57 % 7

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Solutions

Solutions (1-5) Let investment of Rahul, Sandy and Sati be 2x, 3x and 4x respectively. Ratio of profit Rahul : Sandy : Sati 2x × 4 : 3x × 4 : 4x × 10 +(2x – 800)×6 +(3x+1500)× 6 +(5x × 2) +(5x–800)× 2 (9x +1500) × 2 30x – 6400 : 48x + 12000 : 50x ATQ,

50x

125

= ⇒ x = 250 128x+5600 376 Ratio of profit share of Rahul, Sandy and Sati is 1100 ∶ 24000 ∶ 12500 → 11 ∶ 240 ∶ 125 240

1. (a); Required percentage = × 100 376 = 63.829% ≃ 64% 2. (d); Required difference 240−11 = × 37600 = 22,900 376

3. (c); Investment of Veer 350 = 4 × 250 × = 3500 100 Interest earned by Veer = 3500 [1 +

20 2

100

] − 3500 = 1540

4. (b); Required average =

240+125 2

×

37600 376

=18,250

2x+3x+4x

5. (e); Investment of Bhavya = = 3x 3 = 3 × 250 = Rs.750 Required profit = 750 - 2× 250 = 750 – 500 = 250 22

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Solutions (6–10): Total people = 2000 Let total people watching only ‘PK’ = a Total number of people watching ‘Ravan’ = 720 Ratio between number of people watching only ‘3 Idiiots’ and ‘Ravan’ to number of people watching only ‘Ravan’ and ‘PK’ = 2 : 3 Number of people watching ‘3 Idiot’ and ‘PK’ = Number of people watching all three movies 4 = 2000 × = 80

𝑎 2

100

Number of people watching ‘Ravan’ only = number of people watching ‘Ravan’ and ‘PK’ = 3x 2 Number of people = watching ‘3 Idiots’ and ‘Ravan’ only = × 3x = 2x 3

Number of people watching ‘3 Idiots’ only = Number of people watching ‘Ravan’ only ×

150 100

3

= 3𝑥 × = 4.5x 2

Now, 3x + 3x + 2x + 80 = 720 640 8x = 720 – 80 ⇒ x = ⇒ x = 80 8

23

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Now, a 12.5x + a + + a = 2000 - 80 5a 2

2

= (1920 – 1000) ⇒ a = 368

6. (d); People watching only Ravan, only PK and only 3 Idiots = 240 + 360 + 368 = 1000 968 −368 Required percentage = × 100 = 62% 968

7. (e); Required ratio =

240+368 160+184

= 76 : 43

8. (d); Total number of males , who watching 3 idiots and Ravan only 4 5 = 360 × + 240 = 438 5

8

9. (a); Required percentage =

240 −160 160

× 100 =

80 160

× 100 = 50%

10. (b); Required difference = 360 + 184 + 368 – 240 – 160 – 80 – 240 = 912 – 720 = 192 Solutions (11-15): Total students appeared in 2016 = 8000 Total students appeared in 2013 = 5800 Total students appeared in exam B is 2011 & 2013 = 6200 24

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Total students appeared in exam B in 2011 =

6200 31 6200

× 18 = 3600

Total students appeared in exam B in 2013 = × 13 = 2600 31 Total students appeared in exam A in 2013 = 5800 – 2600 = 3200 8000 Total students appeared 2011 = × 100 = 6400 125 Total students appeared in exam A in 2011 = 6400 – 3600 = 2800 8000 Total students appeared in 2014 = × 13= 6500 16 Students appeared in exam B in 2011 = Students appeared in exam A in 2015 = 3600 3600 Students appeared in exam B in 2015 = × 3 = 2700 4 1700

] × 2700 = 4400 Students appear in exam A in 2016 = [1 + 2700 Students appear in exam B in 2016 = 8000 – 4400 = 3600 Let, student appeared in exam A in 2014 = x student appeared in exam A in 2012 = x + 700 ⇒ x + x + 700 + 2800 + 3200 + 3600 + 4400 = 21,100 2x = 6400 ⇒ x = 3200 Students appeared in exam A in 2014 = 3200 Students appeared in exam A in 2012 = 3200 + 700 = 3900 Students appeared in exam B in 2014 = 6500 – 3200 = 3300 Students appeared in exam B in 2012 = 3300 + 1200 = 4500 2011 2012 2013 2014 2015 2016 Total

A 2800 3900 3200 3200 3600 4400 21,100

B 3600 4500 2600 3300 2700 3600 20,300

Total 6400 8400 5800 6500 6300 8000

11. (b); According to table its in 2014. 12. (d); Required ratio = 25

2800 + 3200 + 3900 2600 + 3300 + 3600

=

9900 9500

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=

99 95

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13. (b); Average students appeared in exam A in starting four years 2800+3900+3200+3200 = = 3275 4 Average students appeared in exam B in starting four years 3600+4500+2600+3300 = = 3500 4 Required difference = 225. 14. (e); Required difference = 4500 – 3200 = 1300 15. (c); Required % =

8400 –6400 6400

× 100 =

2000 6400

× 100 = 31.25%

Solutions (16-20): Total students = 1000 Let, students appear in exam Z only = a Total students appeared in exam Y = 360 Ratio of number of students appeared in exam X and Y only to students appeared in exam Y and Z only = 2 : 3 Students appeared in exam X and Z both = a/2 4 Number of students appeared in all three exams = × 1000 = 40 100 Number of students appeared in Y exam only = No. of students appeared in Y and Z only = 3x 2 Number of students appeared in exam X and Y only = × 3x = 2x 3

Now, 2x + 3x + 3x + 40 = 360 ⇒ x = 40 a and, 12.5x + a + + a = 1000 2

26

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5a 2

= 500 ⇒ a = 200

16 (c); Students appeared in atleast two exams = 80 + 60 + 40 + 120 = 300 17. (e); Students appeared in two exams only = 80 + 60 + 120 = 260 18. (e); Students appeared in atmost two exams = 180 + 120 + 200 + 60 + 80 + 120 + 200 = 960 19. (d); Student not appeared in exam Y = 1000 – 360 = 640 20. (d); Students appeared in exam X or in exam Z = 180 + 60 + 40 + 80 + 200 + 120 = 680 Solutions (21-25): No. of songs recorded by Sonu Nigam in 2008 = 300 2 No. of songs recorded by Arijit Singh in 2009 = × 300 = 120 5 Total number of songs recorded in 2008 = 6 × 120 = 720 Let number of songs recorded by Arijit Singh in 2008 be 𝑥. 75 ∴𝑥+ 𝑥 = (720– 300) 100

∴ 𝑥 = 240 ∴ No. of songs recorded by Arijit Singh and Sandeep in 2008 are 240 & 180 respectively 27

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Total no. of songs recorded in 2009 =

350 100

× 120 = 420 2

No. of songs recorded by Sonu Nigam in 2009 = × (420– 120) = 120 5

∴ No. of songs recorded by Sandeep in 2009 = 180 4 Total no. of songs recorded in 2010 = 3 × × 240 = 960 3

No. of songs recorded by Sandeep in 2010 = 650 – 180 – 180 = 290 290 No. of songs recorded by Arijit Singh in 2010 = × 4 = 232. 5

No. of songs recorded by Sonu Nigam in 2010 = 960 – 290 – 232 = 438 Years Singers Sonu Nigam Arijit Singh Sandeep Total 21. (b); Required percentage =

2008

2009

2010

300 240 180 720

120 120 180 420

438 232 290 960

720–(240+120+232) 720

22. (a); Required difference = ( 23. (e); Required ratio =

438+2321

960 420 ×

150 100

2

)–(

7

× 100 = 17 % 9

180+180 2

) = 335 – 180 = 155

= 32 : 21

24. (a); Average number of songs recorded by Sonu Nigam 300+120+438 858 = = = 286 3

3

Average number of songs recorded in 2009. = Required percentage =

286–140 140

420 3

= 140

× 100 ≃ 104%

25. (d); Required difference =( 28

720+420+960

120+180

3

2

)–(

) = 700 – 150 = 550

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Solutions (26-30): Total student in school A = 1600 40 Student who knows only Hindi in C = × 1600 = 640 100 Let student who knows only Hindi & only English in school A be 2𝑥 and 3𝑥 respectively. ∴ Total student who knows both language in school A = 5𝑥 ATQ, 10𝑥 = 1600 ⇒ 𝑥 = 160 ∴ Student who knows only Hindi in A = 320 & only English in A = 480 Both language in A = 800 Student who knows only Hindi in B = 800 + 320 – 640 – 320 = 160 Total student who knows Only English = 3 × 460 = 1380 1 Student who knows only English in B = (1380– 480) = 180 5 Student who knows only English in C = 900 – 180 = 720 320 Student who knows both language in C = = 160 2

2

Total student in school B = 1120 × = 840 4 ∴ Student who knows both language in B = 840 – 160 – 180 = 500 School Only Hindi Only English Both Hindi and English Total A B C Total

320 160 640 1120

480 180 720 1380

26. (b); Required percentage = 27. (e); Required ratio =

(320+480)–500 500

180+720 840

1380

800 500 160 1460

=

900 840

× 100 =

1600 840 1520

300 500

× 100 = 60%

= 15 : 14

3

28. (a); Required sum = + [160 + 180 + 640 + 720] 3 4 = 460 + 1275 = 1735 29

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29. (b); Required percentage = 30. (d); Required difference= (

40 1 ×840– ×320 100 2 1 ×320 2

1600+840+1520 3

× 100 = )–(

336–160

500+160 2

160

× 100 = 110%

) = 1320 – 330 = 990

Solutions (31-35): Total amount invested by ‘Rahul’, ‘Veer’ and ‘Anurag’ is in the ratio 4 : 5 : 6. Let total amount invested by Rahul, Veer and Anurag in all the three schemes be 16x, 20x and 24x 25 Amount invested by Rahul in scheme ‘X” = × 16𝑥 = 4𝑥 100

16𝑥−4𝑥

Amount invested by Rahul in scheme ‘Y’ or ‘Z’ = = 6𝑥 2 Let Amount invested by Veer in scheme Y and Z be 3y and 2y respectively 3𝑦 5𝑦 Then amount invested by Veer in scheme X is = × 100 = 180

3

5𝑦

Ratio between amount invested by Veer in scheme ‘X’, ‘Y’ and ‘Z’ = ∶ 3𝑦 ∶ 3 2𝑦 → 5 ∶ 9 ∶ 6 Amount invested by Veer in scheme ‘X’, ‘Y’ and ‘Z’ is 5x, 9x and 6x respectively. 5𝑥 Amount invested by Anurag in scheme ‘X’ = × 8 = 8𝑥 5 Amount invested by Anurag in scheme ‘Y’ and ‘Z’ together = 24𝑥–8𝑥 = 16𝑥 Amount invested by Rahul in scheme ‘Y’ and ‘Z’ together = 12𝑥 ATQ, 16𝑥 − 12𝑥 = 20,000 ⇒ 𝑥 = 5,000 Total amount invested by Anurag in scheme ‘Y’ and ‘Z’ together = 80,000 Let amount invested by Anurag in scheme ‘Z’ = ‘a’ Amount invested by Anurag in scheme ‘Y’ = 3𝑎 ⇒ 𝑎 + 3𝑎 = 80,000 ⇒ 𝑎 = 20,000 X Rahul 20,000 Veer 25,000 Anurag 40,000 30

Y 30,000 45,000 60,000

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31. (c); Required interest 20,000×10×2 30,000×20×2 30,000×30×2 = + + 100 100 100 = 4,000 + 12,000 + 18,000 = Rs. 34,000 32. (a); Ratio between profit share of Veer and Anurag Veer : Anurag → 25,000 × 8 ∶ 40,000 × 12 → 5 ∶ 12 5 Profit share of Veer = × 7650 = Rs 2,250 17

33. (d); Total amount invested in scheme ‘Y’ = 30,000 + 45,000 + 60,000 = 1,35,000 Total amount invested in scheme ‘Z’ = 30,000 + 30,000 + 20,000 = 80,000 1,35,000−80,000 55,000 Required % = × 100 = × 100 = 68.75% 80,000

34. (e); Required Ratio =

25,000 30,000

80,000

=

5 6

35. (b); Interest earned by Veer = 30,000 × [1 +

20 2

Interest earned by Anurag = 20,000 × [1 + Required % =

13,200−8,800 8,800

] − 30,000 = 13,200

100 20 2 100

] − 20,000 = 8,800

× 100 = 50%

Solutions (36 – 40) Vessel M Water ⇒ 40 liters 40 So milk in M ⇒ × 7 = 140 liters 2 Total quantity in vessel M = 180 liters Total quantity quantity in vessel Q 180×2 = = 120 liters 3 Scotch in vessel Q = 30 liters Apple juice in vessel Q = 90 liters 31

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In vessel N Total quantity = 70 + 140 = 210 liter Water = 140 liters Orange juice = 70 liters In Vessel O Total quantity = 210 + 40 = 240 liters Orange juice in vessel O : Orange juice in vessel N = 7 : 15 70 Orange juice in vessel O = × 15 = 150 liters 7 Vodoka in vessel = 240 – 150 = 90 liters In Vessel P 2 Total quantity = 66 % of total quantity of vessel P 3

2

= × 240 = 160 3

2

Vodka = of vodka in vessel O 2

3

90 × = 60 liters 3 Apple juice = 160 – 60 = 100 liters Milk Water Orange juice Vodka Apple Juice Scotch

M 140 40 — — — —

N — 140 70 — — —

O — — 150 90 — —

P — — — 60 100 —

Q — — — — 90 30

36. (b); In vessel G— 7 Milk in 72 liter of mixture = 72 × = 56 liter 9

Water in 72 liter of mixture

2

= 72 × = 16 liter 2

9

Water in 36 liter of mixture =36 × =24 ltr. 3

1

Orange juice in 36 liter of mixture = 36 × = 12 liter 3 Ratio of milk : water : Orange juice in vessel G = 56 : (16 + 24) : 12 = 14 : 10 : 3 32

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3

37. (b); Apple juice in vessel C = 36 × = 27 liter 1

4

Scotch in vessel C = 36 × + 21 = 9 + 21 = 30 liter 4

Let 𝑥 liter of apple juice mixed = 𝑥 = 60 – 27 ⇒ 𝑥 = 33 liter

27+𝑥 30

=

2 1

38. (a); Ratio of water, Orange juice and Vodka vessel Z = 140 : (70 + 150) : 90 = 14 : 22 : 9 Vodka in 22.5 liter of mixture 9 9 = 22.5 × (14+22+9) = 22.5 × = 4.5 liter 45

39. (e); Ratio of orange juice, Vodka, apple juice & scotch in vessel A 5 3 3 5 3 1 =[(64 × ) : (64 × + 48 × ) : (48 × + 60 × ) ∶ 60 × ] 8

8

8

8

4

= [ 40 : (24 + 18) : (30 + 45) : 15) = 40 : 42 : 75 : 15

4

Solutions (40-42): Total players participated in 2006 = 6000 Total players participated in 2003 = 4800 Total players participated for national level Championship in 2001 & 2003 = 3600 Total players participated for National level championship in 2001 11 = 3600 × = 2200 18 Total players participated for national Level championship in 2003 7 = 3600 × = 1400 18 Total player participated for state level championship in 2003 = 4800 – 1400 = 3400 Total player participated in 2001 100 = 6000 × = 5000 120 Total player participated in 2001 for state level championship = 5000 – 2200 = 2800 Total player participated in 2004 6000 = × 13 = 5200 15

33

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Players participated for national championship in 2001 = Players participated for state championship in 2005 = 2200 2200 Player participated for national championship in 2005 = × 4 = 1760 5 Player participated for state level championship in 2006 5 13 = [1 + ] × 1760 = × 1760 = 220 × 13 = 2860 8 8 Player participated for national championship in 2006 = 6000–2860 = 3140 Let players participated for state level championship in 2004 = X Players participated for state level championship in 2002 = X + 800 ⇒ X + X + 800 + 3400 + 2800 + 2200 + 2860 =20000 2X = 20000 – 12060 7940 X= ⇒ X = 3970 2 Player participated for state championship level in 2004 = 3970 Player participated for state level championship in 2002= 3970+800 = 4770 Players participated for national level championship in 2004 = 5200 – 3970 = 1230 Players participated for national level championship in 2002 = 1230 + 1200 = 2430 Years 2001 2002 2003 2004 2005 2006

state level championship 2800 4770 3400 3970 2200 2860

40. (c); Required ratio =

2800+4770+3970 1400+1230+1760

National Level Championship 2200 2430 1400 1230 1760 3140 =

11540 4390

Total 5000 7200 4800 5200 3960 6000

= 1154 : 439

41. (b); Required difference =(

2800+4770+3400+3970 4

)–(

2200+2430+1400+1230 4

)= 3735 – 1815 = 1920

42. (e); Required difference = 3400 – 2430 = 970 34

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Solutions (43-47): 2015 → Production of wheat in Punjab =

15860−1180 2

= 7340 thousands ton

Production of wheat in Uttar Pradesh = 734 + 1180 = 8520 thousands ton Production of wheat in Madhya Pradesh= 8520 ×

75 100

= 6390 thousands ton

Production of wheat in Haryana = 6390 – 690 = 5700 thousands ton Production of wheat in Rajasthan = 5700 ×

70 100

= 3990 thousands ton

2016 → Production of wheat in Punjab = 7340 ×

80 100

= 5872 thousands ton

Production of wheat in Haryana = 5700 + 800 = 6500 thousands ton 4

Production of wheat in Madya Pradesh= 6390 × = 8520 thousands ton 3

Production of wheat in Rajasthan = 3990 + 2390 = 6380 thousands ton Production of wheat in Uttar Pradesh = 36612 – (5872 + 6500 + 8520 + 6380) = 9340 thousands ton 2017 → Production of wheat in Haryana =

16520−2840 2

= 6840 thousands ton

Production of wheat in uttar Pradesh = 6840 + 2840 = 9680 thousands ton Production of Wheat in Punjab = 5872 + 628 = 6500 thousands ton Production of wheat in Madhya Pradesh= 8520 + 360 = 8880 thousands ton Production of wheat in Rajasthan = 38200 – (6840 – 9680 + 6500 + 8880) = 6300 thousands ton Production of wheat (thousands ton) 2015 2016 2017 Uttar Pradesh 8520 9340 9680 Punjab 7340 5872 6500 Madhya Pradesh 6390 8520 8880 Haryana 5700 6500 6840 Rajasthan 3990 6380 6300 States

35

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43. (b); Total wheat produced by Uttar Pradesh in all three years = 8520 + 9340 + 9680 = 27540 thousands ton Total wheat produced by Madhya Pradesh in all three years = 6390 + 8520 + 8880 = 23790 thousands ton Required difference = 27540 – 23790 = 3750 thousands ton 44. (d); Total wheat produced by Madhya Pradesh in year 2016 & 2017 together = 8520 + 8880 = 17400 thousands ton Total wheat produced by Haryana in the year 2015 & 2016 together = 5700 + 6500 = 12200 thousands ton Required% =

17400 –12200 12200

× 100 =

5200 12200

× 100 = 42

38 61

%

45. (e); Average quantity of wheat produced by Uttar Pradesh, Madhya Pradesh and Haryana in the year 2015 =

8520+6390+5700 3

=

20610 3

= 6870 thousands ton

Average quantity of wheat produced by Madhya Pradesh, Haryana and Rajasthan in the year 2017 =

6840+6300+8880 3

=

22020 3

= 7340 thousands ton

Required difference = 7340 – 6870 = 470 thousands ton 46. (a); Required ratio =

6300+6500 6500+6380

=

12800 12880

= 160 : 161

47. (c); Total Wheat produced by Rajasthan and Uttar Pradesh in the year 2016 & 2017 together = (9340 + 6380) + (9680 + 6300) = 31700 thousands ton 36

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Solutions (48-51): 2012 → Let total number of candidates selected from Kerala be 5𝑥 and from Uttar Pradesh be 6𝑥 ATQ— 5𝑥 + 6𝑥 = 440 ⇒ 𝑥 = 40 Total number of selected candidate from Uttar Pradesh = 240 Total number of selected candidate from Kerala = 200 Total number of candidate selected from Maharashtra = 220 × 3 – (240 + 200) = 660 – 440 = 220 Total number of candidates selected from Madhya Pradesh = 220 – 70= 150 Total number of candidates selected from Bihar = 930 – (240 + 200 + 220 + 150) = 930 – 810 = 120 2013 → 4 Total number of candidates selected from Uttar Pradesh = 240 × = 320 3 Total number of candidates selected from Kerala (200+220)

= – 30 = 210 – 30 = 180 2 Total number of candidates selected from Maharashtra = 220 + 55 = 275 Total number of candidates selected from Madhya Pradesh = 180 Total number of candidates selected from Bihar = 1130 – (320 + 180 + 275 + 180) = 1130 – (955) = 175 2014 → Total number of candidates selected from Uttar Pradesh = 320 + 40 = 360 Total number of candidate selected from Maharashtra = 275 – 25 = 250 Total number of candidate selected from Kerala = 280 × 3 – (360 + 250) = 840 – 610 = 230 Total number of candidate selected from Madhya Pradesh and Bihar = 195 × 2 = 390 390–20 Total number of candidate selected from Madhya Pradesh = = 185 2 Total number of candidate selected from Bihar = 185 + 20 = 205 37

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48. (b); Total number of candidates selected from Kerala & Maharashtra together in 2014 = (250 + 230) = 480 Total number of candidates selected from Uttar Pradesh in 2012 & 2014 together = 240 + 360 = 600 600 –480 Required percentage = ×100 =20% 600

49. (b); Total number of candidate selected from Karnataka in the year 2014 150 = 180 × = 270 100

Required average =

270+360+185+205 4

(200+180+230)

50. (c); Required ratio = (120+175+205) =

610 500

=

1020 4

= 255

= 61 : 50

51. (b); Total number of candidates selected from Maharashtra, Madhya Pradesh and Bihar together in 2013 = (275 + 180 + 175) =630 Total number of candidates selected from Madhya Pradesh in the year 2012 & 2013 together = (150 + 180) = 330 630 –330 300 1000 10 Required percentage = × 100 = × 100 = = 90 % 330

330

11

11

Solution (52-56): Volume of hollow spherical toy = 33,957𝑐𝑚2 Let Outer radius of hollow spherical toy = R 𝑅 Inner radius of hollow spherical toy = 2

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ATQ, 4 3

𝑅 3

𝜋 (𝑅3 − ( ) ) = 33,957 3

2

⇒ R = 9261 ⇒ R = 21 Outer radius of hollow spherical toy = 21cm Inner radius of hollow spherical toy = 10.5 cm Volume of conical toy 33957 14 = = 6468 = 𝜋(𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑐𝑜𝑛𝑒)2 × 5.25 3 ⇒ Radius of cone = 21 cm 6468 Volume of solid cylindrical toy = = 2156 3 = π (radius of cylinder)² × (height of cylinder) ⇒ Radius of Cylinder = 7cm 21 Radius of Solid Spherical toy = = 10.5𝑐𝑚 2 Outer Radius of hollow cylindrical toy = 10.5𝑐𝑚 Inner radius of hollow cylindrical toy = 7 × 2 − 10.5 = 3.5 cm

Let total number of toys = 100x ATQ, Number of solid spherical toys = 20x Number of hollow cylindrical toys = 40x ATQ, 40𝑥 − 20𝑥 = 20 ⇒ 𝑥 = 1 Let Number pf conical toys = 2𝑦 Number of hollow spherical toys = 5𝑦 Number of solid cylindrical toys = 3𝑦 ATQ, 2𝑦 + 5𝑦 + 3𝑦 = 100 − 40 − 20 = 40 ⇒ 𝑦 = 4 39

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Toy Conical Solid Cylindrical Solid Spherical Hollow Cylindrical Hollow Spherical

Number of Toys 8 12 20 40 20

52. (a); Space taken by one solid spherical toy = Volume of one solid spherical toy 4 = 𝜋(10.5)3 = 4851𝑐𝑚3 3 Total space taken by solid spherical toys = 20 × 4851 = 97020𝑐𝑚2 53. (e); Number of conical toys Neeraj have = 8 54. (b); Curved surface area of one hollow cylindrical toy = 2𝜋 × (3.5 + 10.5) × 14 = 1232 cm2 55. (c); Required Ratio =

21 7

=

3 1

56. (d); Volume of one hollow cylindrical toy = 𝜋 × 14 × (10.52 − 3.52 ) = 4312 cm3 Required difference = 4312 − 2156 = 2156𝑐𝑚3 Solution (57-61): Time taken by train ‘D’ to cross a platform of 180 m length is twice than that time taken by same train to cross a pole. ℓ +180 2ℓ ⇒ d = d ⇒ ℓd = 180m d d But, ℓd ∶ ℓe = 1 ∶ 2 ⇒ ℓe = 360m Train ‘F’ can cross a platform double of its length in 36 seconds ℓ +2ℓ ℓ ⇒ f f = 36 ⇒ f = 12 seconds f f ⇒ Train ‘F’ can cross a pole in 12 seconds Train ‘A’ can cross a platform of 100m length in same time in which train ‘F’ can cross a pole i.e, 12 seconds ℓ +100 ⇒ A = 12 a

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Train ‘B’ can cross two poles 60 m apart from each other in 24 seconds ℓ +60 ⇒ 24 = B b But a : b → 5 : 3 And, PA : PB → 2 ∶ 5 ℓa a ℓb b

2

ℓa

5

a

⇒ =

×

b ℓb

⇒

ℓa ℓb

2

2

3

3

= ⇒ ℓa = ℓb

And, 2(ℓA +100) a

=

(ℓb +60) b 15

5 3ℓa 3 2

⇒ 2(ℓa + 100) = ( 3

+ 60)

6ℓa + 600 = ℓa + 300 ⇒ 300 = ℓa 2 2 ⇒ ℓa = 200m And, ℓb = 300m 200+100 300 a= = = 25 m/sec 25

12

12

b = × 3 = 15 m/sec 5 Train ‘C’ can cross a pole in double time in which train ‘E’ can cross a pole. ℓ 2ℓ ⇒ C= e c e But, e = 1.2c ℓ 2ℓ 5 ⇒ c = e ⇒ ℓc = ℓe c

1.2c

3

But ℓe = 360 m ⇒ ℓc = 600m Train ‘B’ can cross train ‘C’ in 10 seconds if they move in opposite direction ℓ +ℓ 300+600 ⇒ b c=b+c⇒ = 15 + c ⇒ 90 – 15 = C ⇒ c = 75 m/sec 10 10 Time train by train ‘D’ to cross a platform of 300m is same time in which train ‘C’ can cross a pole. ℓd +300 ℓ 180+300 600 480 ⇒ = c⇒ = ⇒d= = 60 m/sec d c d 75 8 d:f→4:3 60 ⇒ f = × 3 = 45 m/sec ℓf f

4

= 12 ⇒ ℓf = 540 m 41

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But e = 1.2c ⇒ e = 1.2 × 75 = 90 m/sec Train

Speed (in m/sec)

Length (in m)

A B C D E F

25 15 75 60 90 45

200 300 600 180 360 540

57. (c); Let length of tunnel is ‘x’ meter Time taken by train ‘C’ to cross tunnel is same as time taken by train ‘D’ to cross tunnel 180+𝑥 600+𝑥 = ⇒ 900 + 5𝑥 = 2400 + 4𝑥 60 75 𝑥 = 1500 𝑚 58. (d); Required time =

300+540 15

=

840 15

= 56 seconds

59. (e); Speed of train ‘A’ = 25m/sec = 90km /hr Distance covered by train ‘A’ in 4 hours = 90 × 4 = 360 km Let Train ‘F’ will overtake train ‘A’ in ‘x’ time 360×5 Then, 𝑥 = (45−25)×18 =5 hours

Train ‘F’ meet train ‘A’ after 5 hours start of train ‘F’ or after 9 hours start of train ‘A’

60. (b); Time taken by train ‘A’ to cross a platform of 250m length 200+250 = = 18 25 Time taken by train ‘D’ to overtake train ‘F’ 180+540 = = 48sec 60−45

Required % =

48−18 48

61. (d); Required time = 42

× 100 = 62.5%

30,000+600+180

= 228 seconds

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Solutions (62-66): Ratio of amount invested by Satish and Bhavya in scheme ‘P’ and in ‘R’ is same that is 1 : 2 Let Bhavya and Satish invested 2z and z in scheme ‘R’ ⇒ Amount invested by Abhi in scheme ‘R’ is (2z – 20,000) Ratio of profit share of Satish and Abhi in scheme ‘R’ is 3 : 4 while time of investment of Satish and Abhi in scheme ‘R’ is 4 : 3. 3 𝑧×4 ⇒ = ⇒ 18z – 1,80,000 = 16z 4

(2𝑧−20,000)×3

z = 90,000 Scheme R

Amount invested Amount invested Amount invested by by Satish by Abhi Bhavya 90,000 1,60,000 1,80,000

In scheme R, Bhavya invested for 10 months while Satish got 25% of profit out of total profit If ratio between Satish’s and Abhi’s profit share is 3 : 4, then Bhavya’s profit 3 share is × 100 − 7 = 5 25 ⇒ Ratio of profit share of Satish, Abhi and Bhavya is 3 : 4 : 5. Let Satish invested for ‘y’ month 1,80,000×10 5 = ⇒ y = 12 month 90,000×𝑦

3

12

Abhi invested for = × 3 = 9 month 4 Time of investment of Satish in scheme ‘P’ is 2 months less than that of in scheme ‘Q’ ⇒ Satish invested for 10 months in scheme ‘P’. Time of investment of Abi in scheme ‘P’ is 5 months more than time for Satish invested ⇒ Abhi invested for 15 months in scheme ‘P’ Let amount invested by Satish in scheme ‘Q’ is = 5b 5𝑏 ⇒ Amount invested by Satish in scheme ‘P’ is = × 3 = 3𝑏 5

96

And, Amount invested by Abhi in scheme ‘Q’ is = 5𝑏 × = 4.8𝑏 100 Amount invested by Bhavya in scheme ‘Q’ is = 4.8b + 1,00,000 43

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Amount invested by Bhavya in scheme ‘P’ is = 3b × 2 = 6b ATQ,

6𝑏

4.8𝑏 + 1,00,000

b= 25,000 Scheme P Q R

=

15 22

⇒ 132b = 72b + 15,00, 000

Amount Invested Satish Abhi 75,000 80,000 1,25,000 1,20,000 90,000 1,60,000

Bhavya 1,50,000 2,20,000 1,80,000

Let, Bhavya invested for ‘y’ month in scheme ‘P’. ATQ, 1 = 4

75,000×10

75,000×10+80,000×15+1,50,000×𝑦

⇒ y = 7 months

Time of investment of Abhi and Bhavya is same in scheme ‘Q’ and ratio between profit sharing of Bhavya to total profit in scheme ‘Q’ is 1 : 2. Let, Abhi and Bhavya invested for ‘5a’ months and Satish invested for ‘x’ month. ATQ, 1 = 1

2,20,000×5𝑎

1,25,000×16+1,20,000×5𝑎

⇒ 1,25,000 × 16 = 1,00,000 × 5a ⇒ 5a= 20 months So Bhavya and Veer invested in scheme ‘Q’ for 20 months

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120

120

62. (e); Interest earned by Abhi = 1,20,000 × × − 1,20,000 = 52,800 100 100 Interest earned by Bhavya 25 = 2,20,000 × 2 × = 1,10,000 100 Required difference = 1,10,000 –52,800 = 57,200 63. (b); Ratio of profit sharing between Satish, Abhi and Bhavya in scheme ‘P’ and scheme ‘R’ is 5 : 8 : 7 and 3 : 4 : 5 respectively. 28,980 Profit earned by Satish from scheme ‘R’ = × 3 = 17,388 Profit earned by Bhavya from scheme ‘P’ =

5 28980

225 12,880

× 100 = 12,880

Profit earned by Satish from scheme ‘P’ = × 5 = 9,200 7 Total profit earned by Satish from scheme ‘P’ and ‘R’ together = 17,388 + 9,200 = 26,588 64. (d); Time of investment of Satish =

20 5

× 4 = 16 months

Interest earned by Satish = 1,25,000 × 65. (a); Required % =

1,80,000−75,000 75,000

× 100 =

15 100

105 75

×

16 12

= 25,000

× 100 = 140%

66 (c); Ratio of profit sharing between Satihs, Abhi and Bhavya in scheme ‘P’ is 5 : 8 : 7 Let total profit = 20 7 Bhavya’s profit = × 100 20 = 35% When Abhi doubles his investment then ratio of profit sharing between Satish, Abhi and Bhavya in scheme ‘P’ is 5 : 16 : 7 Let total profit = 28 7 Bhavya’s profit = × 100 = 25% 28

% Decrement in profit %

45

35−25

4

35

7

×100 = 28 %

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