Imo Level1 Solution Class 10 Set 3

CLASS 10 SET-3

HINTS AND SOLUTIONS Also, in each step, the element that reaches the encircled position gets replaced by a new element. 10. (C) : Order of persons from the tallest to the shortest is, M, T, D, R, N So, M is the tallest among all. 11. (B) :

1. (C) : The positions of flats are I

Flats facing North

A

C

B

E

Flats facing South

F G J D H So, G's flat is to the left of F's flat. 2. (D) : From figure (1) to (2) : The element '

' rotates

45° clockwise and decreased by 1 and element '

'

rotates 135° anticlockwise and increased by 1. 3. (C) : The series is, aabc/bbca/ccab/aabc 4. (A) : We have, [{(17 × 13) – (5 ÷ 5)} + (23 – 6)] After decoding sign, we get [{(17 – 13) + (5 × 5)} ÷ (23 + 6)] = (4 + 25) ÷ 29 = 29 ÷ 29 = 1 5. (D) :

12. (B) : Opposite faces of cube are (Yellow, Green), (Red, White), (Blue, Brown) 13. (D) 14. (A) : Mars

Star Moon

15. (C) :

Mirror

6. (A) : According to question, the new arrangement of alphabets are A B C D E F

G H I

Z D B C Z D F

Q R S Z

D V

K L

G Z D J

Q R S T U V W X P

J

M N

K L

O P

M Z

D

Y Z

W X Y

Therefore code for STANDING is RSZMCZMF. 7. (C) :

8. (B) :

Distance he rode northwards initially = AB = CD = 2 km. 9. (A) : The elements move in the sequences in alternate steps. Class-10 | Level-1 | Set-3

and

16. (D) : We have,

=

14587

×

14587 1250 23

=

14587 × 23

=

116696 10000

54 × 2 23 54 × 2 4 14587 \ The decimal expansion of the rational number 1250 will terminate after four decimal place. 17. (D) : Let the length of the shortest side (AB) = x According to question, Perimeter = 5x ⇒ AB + BC + AC = 5x ⇒ x + BC + AC = 5x ⇒ BC + AC = 4x …(i) 1 Area of DABC = × BC × AB = 15x 2 1 × x × BC = 15x ⇒ 2 ⇒ BC = 30 From (i), AC = 4x – BC = 4x – 30 In DABC, by Pythagoras theorem, AC2 = AB2 + BC2 ⇒ (4x – 30)2 = x2 + (30)2 ⇒ 16x2 + 900 – 240x = x2 + 900 ⇒ 15x2 – 240x = 0  ⇒  15x (x – 16) = 0 ⇒ x = 0  or  x = 16 But x cannot be equal to 0. Hence x = AB = 16 and AC = 4x – 30 = 64 – 30 = 34  1

18. (B) : Frequency of the class (18-24) is maximum i.e., 35 so this is the modal class. f − f1 Mode = l + ×h 2 f − f1 − f 2 l = lower limit of the modal class = 18 f = frequency of the modal class = 35 h = width of the modal class = 6 f1 = frequency of the class preceding the modal class = 25 f2 = frequency of the class following the modal class = 18 \ Mode = 18 +

35 − 25 ×6 2 × 35 − 25 − 18

= 18 +

10 20 182 × 6 = 18 + = 20.22 = 27 9 9

19. (B) : First term = a, let common difference = d Sum of first n terms = Sn According to question, n Sn = 0  ⇒  (2a + (n – 1)d) = 0 2 −2a ⇒ 2a + (n – 1)d = 0  ⇒  d = n −1 Now, sum of next m terms = sum of (m + n) terms = Sm + n m+n Sm + n = (2a + (m + n – 1)d) 2 m+n  −2a  =  2a + (m + n − 1) ×  n −1 2 =

m + n  2a (n − 1) + (m + n − 1) × (−2a )    n −1 2 



∠FDG = ∠ADB (Common)

\ DDFG ~ DDAB (AA similarity) DF FG =  DA AB In trapezium ABCD ⇒

A

B

F

E G

D

C

…(i)

AF BE = (Q AB || EF || CD) FD EC  BE 3  AF 3 = (given)  =  Q  EC 4 DF 4 Adding 1 on both sides, we get

AF 3 AF + DF 7 AD 7 +1= + 1 ⇒ = ⇒ = DF 4 DF DF 4 4 DF 4 =  ⇒ …(ii) AD 7  2

EC + BE 4 + 3 BC 7   ⇒  = = BE BE 3 3 BE 3 =  ⇒ …(v) BC 7 EG 3 = From (iv) and (v), CD 7 3 3 ⇒ EG = CD = × 2AB ( CD = 2AB (given)) 7 7 6 ⇒ EG = AB …(vi) 7 Adding (iii) and (vi), we get 4 6 AB + AB 7 7 FE 10 10 = ⇒ FE = AB  ⇒ AB 7 7 FG + EG =

21. (C) : P(2, –1), Q(3, 4), R(–2, 3) and S(–3, –2)

m+n = (2an – 2a –2am – 2an + 2a) 2(n − 1) m+n = × (– 2am) 2(n − 1) − am(m + n) = (n − 1) 20. (D) : In DDFG and DDAB, ∠DFG = ∠DAB (Corresponding angles)

DF FG 4 4 …(iii) = = ⇒ FG = AB  AD AB 7 7 In DBEG and BCD, ∠BEG = ∠BCD (Corresponding angles) ∠GBE = ∠DBC (Common) \ DBEG ~ DBCD (AA similarity) BE EG  …(iv) = BC CD BE 3 EC 4 Now, =   ⇒  = EC 4 BE 3 Adding 1 on both sides, we get From (i) & (ii),

PQ =

(3 − 2)2 + (4 + 1)2 = 1 + 52 = 26

QR =

(−2 − 3)2 + (3 − 4)2 = (−5)2 + (−1)2 = 26

RS =

(−3 + 2)2 + (−2 − 3)2 = (−1)2 + (−5)2 = 26



SP =

(−3 − 2)2 + (−2 + 1)2 = (−5)2 + (−1)2 = 26



PR =

(−2 − 2)2 + (3 + 1)2 = (−4) 2 + (4) 2 = 4 2



QS =

(−3 − 3)2 + (−2 − 4)2 = (−6) 2 + (−6) 2 = 6 2

\ PQ = QR = RS = SP = 26 But diagonal PR ≠ diagonal QS Hence, PQRS is a rhombus. 22. (D) : Let CD = x cm then DB = (125 – x) cm Now, in DADC, by Pythagoras theorem, AC2 = CD2 + AD2  ⇒  (75)2 = x2 + AD2 ⇒ AD2 = 5625 – x2 In DADB, by Pythagoras theorem, AB2 = AD2 + DB2 ⇒ (100)2 = AD2 + (125 – x)2 ⇒ AD2 = 10000 – 15625 – x2 + 250x ⇒ AD2 = –5625 – x2 + 250x From (i) and (ii), we get 5625 – x2 = –5625 – x2 + 250x ⇒ 11250 = 250x  ⇒  x = 45

…(i)

…(ii)

 Class-10 | Level-1 | Set-3

Now, from (i) AD2 = 5625 – (45)2  ⇒  AD2 = 5625 – 2025 ⇒ AD =

3600   ⇒  AD = 60 cm

23. (A) : Let f (x) = Ax3 + x2 + Bx + C When f (x) divided by (1 – 2x) leaves the remainder

21 . 4

 1  21 f  = 2 4

\

3

Now, l =

2

21 1 1 1 ⇒ A   +   + B   + C = 2 2 2 4

16 + 196 = 212 = 2 53 = 2 × 7.28 = 14.56 cm 26. (D) : Since the system of equations 3x + y = 1 and (2k – 1)x + (k – 1)y = 2k + 1 is inconsistent. a b c \ 1 = 1 ≠ 1 a2 b2 c2

A B 21 1 20 + +C = − = =5 8 2 4 4 4 ⇒ A + 4B + 8C = 40 Also, when f (x) divided by x leaves the remainder \ f (0) = 18 ⇒ A(0)3 + (0)2 + B(0) + C = 18 ⇒ C = 18 Now, (x – 2) is the factor of f (x). \ f (2) = 0 ⇒ A(2)3 + (2)2 + B(2) + C = 0 ⇒ 8A + 4 + 2B + C = 0 ⇒ 8A + 2B + 18 = – 4 ⇒ 8A + 2B = –22 ⇒ 4A + B = –11 Putting value of C in (i), we get A + 4B + 8 × 18 = 40 ⇒ A + 4B = 40 – 144 ⇒ A + 4B = –104 ⇒ A = –104 – 4B Putting value of A in (ii), we get 4(–104 – 4B) + B = –11 ⇒ –416 – 16B + B = –11  ⇒  –15B = 405 ⇒ B = –27 Now, from (iii) A = –104 – 4 × (–27) = 4 ⇒

…(i) 18.



=

…(ii)

a c

=

−b  ac

…(iii)

(Using (i) and (ii))

25. (B) : Internal radius of sphere (r) = 2 cm External radius of sphere (R) = 4 cm Radius of the base of cone (r1) = 4 cm Let height of the cone be h. According to question, Class-10 | Level-1 | Set-3

3 1 1 = ≠ (2k − 1) (k − 1) 2k + 1

From first two, we get 3(k – 1) = 2k – 1 ⇒ 3k – 3 = 2k – 1  ⇒  k = 2 3 1 1 1 = ≠ ⇒1=1≠ Also, 4 −1 2 −1 4 +1 5 27. (B) : Let the least number of years be n. According to question, n

n

20  6  2P < P 1 + ⇒2<   5  100  Now, for checking n = 1, 2, 3 …, we get the least number for which the inequality holds is 4. 28. (A) : Probability of selecting a blue marble P(B) =

α+β α β ( α )2 + ( β )2 = + = αβ β α αβ −b a

⇒



24. (C) : a and b are the zeroes of the polynomial ax2 + bx + c. −b  …(i) Sum of roots = a + b = a c Product of roots = ab =  …(ii) a Now,

r12 + h 2 = 42 + 142

⇒ l =

A 1 B 21 + + +C = 8 4 2 4

⇒

Volume of hollow sphere = Volume of cone 4 1 ⇒ π( R3 − r 3 ) = π(r1 )2 × h 3 3 4 1 ⇒ π(43 − 23 ) = π(4)2 × h 3 3 4(64 − 8) 56 =h ⇒h= ⇒ = 14 cm 4×4 4 Let the slant height of cone be l.

1 3

4 9 Now, probability of selecting white marble P(W) Probability of selecting a green marble P(G) =

1 4 7 2 = 1 –  +  = 1 – = 3 9 9 9 2 Therefore number of white marbles = ×54 = 12 9 29. (C) : We have equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 Since, the roots of the equation are equal. \ D = b2 – 4ac = 0 ⇒ [–2(a2 – bc)]2 – 4(c2 – ab) (b2 – ac) = 0 ⇒ 4(a4 + b2c2 – 2a2bc) – 4(c2b2 – ac3 – ab3 + a2bc) = 0 ⇒ 4a4 + 4b2c2 – 8a2bc – 4b2c2 + 4ac3 + 4ab3 – 4a2bc = 0 ⇒ 4a4 + 4ac3 + 4ab3 – 12a2bc = 0 ⇒ 4a(a3 + c3 + b3 – 3abc) = 0 ⇒ 4a = 0  ⇒  a = 0 or (a3 + b3 + c3 – 3abc) = 0 ⇒ a3 + b3 + c3 = 3abc 

30. (B) : Let the diameters of two circles be 3x and 4x 3x and 2x. So, the radius of two circles are 2  3

Radius of bigger circle =



A + B + C = 180° °  105  ⇒ A + C +   = 180° 2

30 = 15 cm 2

According to question,

°

 3x  p   + p(2x)2 = p(15)2  2

105  360° − 105° ⇒ A + C = 180° −  =  2  2

  9 x2 ⇒ p  + 4 x 2  = 225 p   4

255  ⇒ A + C =    2  Putting value of B in (ii), we get

2

°

9 x 2 + 16 x 2 = 225  ⇒  25x2 = 225 × 4 4 225 × 4 ⇒ x2 = = 36  ⇒  x = 6 25 Therefore diameters of the two circles are 3 × 6 = 18 cm and 4 × 6 = 24 cm

°

⇒





=2

=



°

°

15 ⇒ C – A =    2 Adding (iii) and (iv), we get

12 13 2 sin θ cos θ

°

cos 2 θ − sin 2 θ sin θ cos θ 2

cos θ cos θ sin 2 θ − cos 2 cos 2 θ 2 tan θ

°

[Dividing both numerator and denominator by cos2q]

2

1 − tan 2 θ  12  2   13 

24 24 169 13 = = = × 2 169 − 144 13 25  12  1−   169  13  =

33. (D) : tan(A + B – C) = 1 ⇒ tan(A + B – C) = tan45° ⇒ A + B – C = 45° sec(B + C – A) = 2 ⇒ sec(B + C – A) = sec 60° ⇒ B + C – A = 60° Adding (i) and (ii), we get

(Given) …(i) (Given) …(ii) °

 105   1 =  52  2B = 105° ⇒ B =   2   2 Now by the angle sum property of a triangle, we have  4

°

°

 270   135   1 = =  67  ⇒ C =   4   2   2 Now, from (iv)

°

°

°

°

°

…(iv)

 255   15   270  +  =  2C =    2  2  2 

24 × 13 312 = 25 25 32. (A) : Area of the quadrant of circle of radius 42 cm 90° 1 22 = × p(42)2 = × × 42 × 42 = 1386 cm2 360° 4 7 1 Area of the DCEF = × base × height 2 1       = × 6 × 6 = 18 cm2 2 Therefore, area of the shaded region = area of quadrant BCD + area of DCEF = (1386 + 18) cm2 = 1404 cm2

 105    + C – A = 60° 2 

105  120° − 105°  15  ⇒ C – A = 60° −  = =   2  2 2

31. (D) : Given tanq = We have,

…(iii)

°

°

 135   15   120  −  =  A=  = 60°   2  2  2 

34. (C) : Let the height of the light house AB = h m and CB = x m, then BD = (100 – x) m In DABC, AB h tan 45° =   ⇒  1 = BC x ⇒ x = h …(i) In DABD, AB tan60° = BD h ⇒ 3 =   ⇒  (Q x = h) 3 (100 – h) = h 100 − x ⇒ 100 3 – 3 h = h  ⇒  h + 3 h = 100 3 (100 3 ) ( 3 − 1) 100 3 ⇒ h =   ⇒  h = × (1 + 3 ) ( 3 + 1) ( 3 − 1) ⇒ h = 100 3 ( 3 − 1) = 50 3 ( 3 − 1) m. 2 35. (C) : Since perpendicular from the centre of a circle to a chord of the circle, bisect the chord. OQ ^ AB  ⇒  AQ = QB In DAOQ by Pythagoras theorem AO2 = OQ2 + AQ2  ⇒ 52 = 32 + AQ2 ⇒ AQ2 = 25 – 9 = 16  ⇒  AQ = 16 = 4 cm. Length of tangents drawn from an external point to a circle are equal. \ QB = BP = 4 cm Also OP ^ BC So, BP = PC \ BC = BP + PC = (4 + 4) cm = 8 cm  Class-10 | Level-1 | Set-3

36. (A) : Since production of mobile phones in 3rd year = 6000 ⇒ a3 = 6000 ⇒ a + 2d = 6000 ...(i) Also production of mobile phones in 5th year = 6500 ⇒ a5 = 6500 ⇒ a + 4d = 6500 ...(ii) Subtracting (i) from (ii), we get 2d = 500 ⇒ d = 250 From (i), a + 2 × 250 = 6000 ⇒ a = 5500 Hence, production in 8th year, a8 = a + 7d = 5500 + 7 × 250 = 5500 + 1750 = 7250 units 37. (B) : Let lamp post is at point A, so AB = 3.9 m Height of Sudha, CD = 120 cm = 1.2 m Let the length of the shadow, DE = x m Distance BD walked by Sudha in 3 seconds = 3 × 1.5 m = 4.5 m In DABE and DCDE ∠ABE = ∠CDE = 90° ∠BEA = ∠DEC (Common) \ DABE ~ DCDE (AA similarity) AB BE 3.9 4.5 + x \   ⇒  = = CD DE x 1.2 ⇒ 3.9x = 1.2 (4.5 + x)  ⇒  3.9x = 5.4 + 1.2 x 5.4 ⇒ 3.9x – 1.2x = 5.4  ⇒  x = =2 2.7 Hence, the length of the shadow is 2 m. 38. (C) : Let the amount of sugar at which 8% profit is obtained  = x kg. According to question, 14 8 18 x× + (1000 − x) × = × 1000 100 100 100 ⇒ 8x + 18000 – 18x = 14000 ⇒ 10x = 4000  ⇒  x = 400 Therefore amount of sugar which is sold at 18% profit  = 1000 – 400 = 600 kg 39. (B) : Largest size of the tile = H.C.F. of 378 cm and 525 cm = 21 cm 40. (A) : Let the number of candidates applied for the examination be x. Number of eligible candidates = 95% of x Number of eligible candidates of other categories  = 15% of (95% of x) = 4275. 15 × 95 × x 4275 × 10000 = 4275 ⇒ x = = 30,000 ⇒ 100 × 100 15 × 95 Therefore number of candidates applied for the examination are 30,000. 41. (A) : Let the present age of son be x and father be y. Two years ago son's age was x – 2 and father's age was y – 2 Two years later son's age will be x + 2 and father's age will be y + 2 According to question, (y – 2) = 5(x – 2)  ⇒  y = 5x – 8 …(i) and y + 2 = 3(x + 2) + 8 …(ii) Class-10 | Level-1 | Set-3

From (i) putting value of y in (ii), we get 5x – 8 + 2 = 3x + 6 + 8 ⇒ 5x – 3x = 20  ⇒  2x = 20 ⇒ x = 10 Now, from (i) y = 5x – 8 = 50 – 8 = 42 Therefore present age of son is 10 years and present age of father is 42 years. 42. (B) : Marked price of VCR = ` 12,000 Selling price of the VCR = 95% of 90% of 85% of 12000 95 90 85 = × × × 12000 = ` 8721. 100 100 100 43. (B) : Let the two trains meet x hours after 7 a.m. Distance travelled by train at station A in x hours = x × 20 Distance travelled by train at station B in (x – 1) hours = (x – 1) × 25 According to question, x × 20 + (x – 1)25 = 110 ⇒ 20x + 25x – 25 = 110 135 =3 ⇒ 45x = 135 ⇒ x = 45 Therefore, the time at which two trains will meet is 3 hours after 7 a.m. i.e., 10 a.m. 44. (D) : Let the principal be `P and rate of interest be r % Difference of compound interest and simple interest for 2 years 2  P×r×2  r  =  P 1 + − P −  100   100    2 Pr   r2 2r  =  P 1 + + − P −    10000 100   100 2 Pr 2 Pr Pr 2 Pr 2 + −P− = 10000 100 100 10000 Difference of compound interest and simple interest for 3 years 3  P×r ×3   r  =  P 1 + − P −  100   100   3  3Pr   r 3r 3r 2  =  P 1 + + + − P −   100   1000000 100 10000  = P +



 3Pr  3Pr 3Pr 2 Pr 3 = P + + + − P −  1000000 100 10000  100 3 2 Pr 3Pr 3Pr 3Pr = + + − 1000000 100 10000 100

3Pr 2 Pr 3 Pr 2  r  + = + 3  1000000 10000 10000  100  Now, according to question

=

Pr 2  300 + r    10000  100  25 = 8 Pr 2 10000 300 + r 25 = ⇒ 100 8 2500 100 1 ⇒ r = − 300 = = 12 % 8 8 2  5

(x1 ≠ –3 as x1 lies on positive x-axis) \ x1 = 1 \ Co-ordinates of R are (1, 0) Now, AP = (PQ – QA) = 3QA – QA (\ PQ = 3QA) ⇒ AP = 2QA So, AQ : AP = 1 : 2 Since A lies on the y-axis let co-ordinates of A be (0, y) Using section formula, we have 1× 7 + 2 ×1 m y + m2 y1   ⇒  y = y= 1 2 1+ 2 m1 + m2 9 ⇒ y = = 3 3 \ Coordinates of A are (0, 3).

5 3 45. (C) : Volume of earth dug out =  4 × ×  m3 = 15 m3 .  2 2 Area over which earth is spread 5  =  31 × 10 − 4 ×  m 2 = 300 m 2 .  2  Volume   15  = × 100  cm = 5 cm. \ Rise in level =   Area   300  46. (D) : Let O be the centre S of the circle. Join OS and OQ. O Radius of the circle is T perpendicular to the R 62° tangent. U 25° Q \ ∠OQR = 90° ∠OQS = 90° – 62°     = 28° P Since, OQ = OS (radii of circle) ∠OQS = ∠OSQ = 28°  (Angle opposite to equal sides are equal) In DOQS, by angle sum property ∠QOS + ∠OSQ + ∠OQS = 180° ⇒ ∠QOS = 180° – 28° – 28° = 124° Now, angle subtended by a chord of circle at centre is double to the angle subtended by the chord at the remaining part of the circle. 1 1 ∠SOQ = × 124° = 62° \ ∠SUQ = ∠STQ = 2 2 Also, ST = SQ (given) ⇒ ∠STQ = ∠SQT = 62°  (Angle opposite to equal sides are equal) Now, in DSTQ, by angle sum property ∠STQ + ∠SQT + ∠TSQ = 180° ⇒ 62° + 62° + ∠TSQ = 180° ⇒ ∠TSQ = 56° In DQUP, ∠UPQ + ∠PQU = ∠QUS (Exterior angle property) ⇒ ∠UPQ + 25° = 62° ⇒ ∠UPQ = 37° 47. (D) : Length of PQ =

(2 + 1)2 + (7 − 1)2 = 32 + 62

       = 9 + 36 = 45 = 3 5 Area of PQRS = PQ × QR = 15 ⇒ 15 = 3 5 × QR ⇒ QR = 5 Since R lies on positive x-axis let its co-ordinates be (x1, 0) Length of QR = 5 = ( x1 + 1)2 + (0 − 1)2 ⇒ 5 = x12 + 2 x1 + 2 Squaring both sides, we get 5 = x12 + 2x1 + 2  ⇒  x12 + 2x1 – 3 = 0 ⇒ x12 + 3x1 – x1 – 3 = 0 ⇒ x1(x1 + 3) – 1 (x1 + 3) = 0 ⇒ (x1 + 3) (x1 – 1) = 0 ⇒ (x1 – 1) = 0  or  (x1 + 3) = 0  6

48. (D) : Putting t =

3×

1

1 in the given equation, we get 2 1

4 2 + 156(4) 2 = 216(4

⇒ (2)

6×

2×

1 2)

1 1 2× 2 + 156( 2) 2 = 216 × 4

⇒ 23 + 156 × 2 = 216 × 4 ⇒ 8 + 312 = 864 ⇒ 320 = 864 1 Which is false, so t = does not satisfy the given equation. 2 Hence, Statement I is false. If (x + a) exactly divides ax2 + 2a2x + b3 then x = –a satisfy the equation. \ a(–a)2 + 2a2(–a) + b3 = 0 ⇒ a3 – 2a3 + b3 = 0 ⇒ b3 – a3 = 0 ⇒ (b – a) (b2 + a2 + ab) = 0 ⇒ b – a = 0  or  a2 + b2 + ab = 0 \ a2 + b2 + ab = 0 (Q a ≠ b) Hence, Statement II is true. 49. (B) : Step-3 is incorrect as the correct step is at O, construct ∠A'OB = 30° and let OB intersect the circle at B.  (Hence ∠AOB = 150°) 50. (C) : (a) Length of tangents from external point to the circle are equal. \ PA = PB = 7 cm and CQ = CA = 2.5 cm Now, PC = PA – AC = (7 – 2.5) cm = 4.5 cm (b) Let r be the radius of the circle. Now, in DPOT, by Pythagoras theorem, PO2 = PT2 + TO2 ⇒ (PA + AO)2 = PT2 + TO2 ⇒ (2 + r)2 = 62 + r2 (Q AO = TO = radius of the circle) ⇒ 4 + r2 + 4r = 36 + r2 ⇒ 4r = 32 ⇒ r = 8 cm (c) ∠APB = 60°  ⇒  ∠APO = 30° Now, in DOAP, OA sin 30° = ⇒ 1 OP = 3 3 ⇒ OP = 6 3 cm. OP 2

JJJ

 Class-10 | Level-1 | Set-3