Imo Level1 Solution Class 10 Set 6

CLASS 10 SET-6

HINTS AND SOLUTIONS 1. (A) : The pattern followed is : –4 –3 –2

–1

10. (B) : We have, D E T +1

20 , 20 , 19 , 16 , 17 , 13 , 14 , 11 , 10 , 10 –1 –2 –3 –4 So, the missing terms are 10 and 10. 2. (D) : From figure (i) to (ii) : Out of the two parallel lines, the smaller line converted to a pin and the longer line converted to an arrow, both pointing towards the square boundary. The third line is converted to a triangle. 3. (A) : Rita’s mother’s son is Rita’s brother. Nikhil is the son of Rita’s brother. So, Rita is Nikhil’s aunt. 4. (C) : Total number of cubes = 5 × 10 + 5 × 4 + 5 × 3 + 5 × 2 + 5 = 100 5. (D) : No two persons are of the same age. 6. (B) :

7. (B) : The rule followed is, (15 – 5) × (2 + 6) = 80 (9 – 4) × (7 + 6) = 65 Similarly, (13 – 11) × (16 + 8) = 48 Hence, the missing number is 48. B

30 m

20 m A (Starting position)

D –1

T

T S Similarly, S +1

T

–1

U A

+1

D –1

B U

–1

T

E

T S B

+1

C

C

+1

T

–1

D E

+1

F –1

S

R

+1

S M

–1

L C

+1

D

I

O

–1

J E

+1

N N

–1

F R

–1

M +1

Q

I J

N

+1

O T

+1

U B

–1

E

+1

A

F

11. (D) : All the elements get vertically inverted in each step. Also in the first step; first and second elements interchange their positions, in second step; second and third element interchange their positions, in third step; third and fourth element interchange their positions. These three steps are repeated to continue the series.

13. (C) : The squares formed are : AEI, BFJ, CHL, DGK, EI, FJ, GK, HL, I, J, K, L, IJKL, EIFJHLGK, ABCDEFGHIJKL Therefore total number of squares = 15

C

35 m G

S

+1

+1

12. (D) : Let K denotes Kamala’s current position and K' denotes Kamala’s original position. S denotes Sujata’s position. 10th 14th 17th (Left) (Right) K K' S Total number of girls = 40 Sujata is at 17th position from left. Therefore from right Sujata’s position = (40 – 17 + 1) = 24th

Water level

8. (D) :

E

–1

(Final position) F

B

14. (C) : According to the given information, we have

15 m D 15 m E

Required distance, AF = AG + GF ⇒ AF = (30 + 15) m = 45 m Also, F lies to the East of A. Hence, Rasik is 45 metres away and in East from his starting position.

Hence B and C are not facing the centre. 15. (B) :

9. (D) :

Class-10 | Level-1 | Set-6

 1

16. (D) : We have, a =

3+ 3+

 a + b + a − b a − b + a + b   2a + b + x 2a − b + y  , , ⇒   =    2 2 2 2  2a + b + x 2a − b + y  ⇒ ( a, a ) =  ,   2 2 2a + b + x 2a − b + y = a and =a ⇒ 2 2 ⇒ 2a + b + x = 2a  and 2a – b + y = 2a ⇒ x = –b and y = b Hence, coordinates of the fourth vertex of the parallelogram are (–b, b).

3 +....∞

⇒ a = 3 + a On squaring, we get a2 = 3 + a ⇒ a2 – a – 3 = 0 ⇒ a =

1 ± 12 + 4 × 3 1 ± 13 ⇒a= 2 2

⇒ a =

1 − 13 1 + 13  or a = 2 2

⇒ a =

1 − 3.605 1 + 3.605   or a = 2 2

4.605 2.605  or a = − 2 2 ⇒ a = 2.302  or  a = –1.302 Hence, 2 < a < 3  or  –2 < a < –1 ⇒ a =

21 17. (C) : We have, f(x) = 2x2 + 5x + k and a2 + b2 + ab = 4 −5 Sum of zeros = a + b =  …(1) 2 Product of zeros = a·b =

k  2

Now, (a + b)2 = a2 + b2 + 2ab ⇒ (a + b)2 – ab = a2 + b2 + ab Putting values from (1) and (2) in (3), we get

…(2) …(3)

20. (D) : The given system of equations will have infinite number of solutions, if 2 3 9 = = p + q ( 2 p − q ) 3 ( p + q + 1) 2 3 3 ⇒ = = p + q ( 2 p − q ) ( p + q + 1) From 1st and 3rd 2 3 = p + q ( p + q + 1) ⇒ 2p + 2q + 2 = 3p + 3q ⇒ p + q = 2 From 2nd and 3rd 3 3 = (2 p − q ) ( p + q + 1) ⇒ p + q + 1 = 2p – q ⇒ –p + 2q = –1 Adding (1) and (2), we get

2

k 21  −5    − = 2 2 4 ⇒

3q = 1  ⇒  q =

4×2 25 21 k = k  ⇒  k = 2 − =   ⇒  4 4 4 2

18. (B) : Area of sector =

θ × πr 2 360°

56° × π × r2 360° 4.4 × 360 × 7 =9 ⇒ r2 = 56 × 22 ⇒ 4.4 =

⇒ r = 9 = 3 cm Hence, the radius of the circle is 3 cm. 19. (A) : Let A(a + b, a – b), B(2a + b, 2a – b), C(a – b, a + b) and D(x, y) be the vertices of a parallelogram ABCD. (a + b, a – b)A B(2a + b, 2a – b)

O (x, y)D C(a – b, a + b) Since, the diagonals of a parallelogram bisect each other. \ Coordinates of the mid-point of AC  = Coordinates of the mid-point of BD

 2

Putting q =

…(1)

…(2)

1 3

1 in (1), we get 3

1 5 = 3 3 21. (A) : We have, OA = r, OO' = h

p = 2−

r2  9 In DOO'B', by Pythagoras theorem (OB')2 = (OO')2 + (O'B')2 2 and OB ′ = h +

O'

A'

h

A r2 2 r O = h 2 + (O ′ B ′ ) 9 r2 r 2 ⇒ (O ′ B ′ ) = ⇒ O ′B ′ =   9 3 1  r 2 r 2  \ Volume of the frustum = π r 2 + +  h 3  9 3  ⇒ h2 +



=

B'

B

1 13 2 13 2 πh r = πr h 3 9 27

22. (B) : We have,

(

) 23 (sin

4 sin 4 30° + cos 4 60° −

2

)

1 60° − cos 2 45° + tan 2 60° 2

Class-10 | Level-1 | Set-6

2   1  4  1  4  2   3   1  2  1 2 = 4   +    −    + 2 3  −      2 2 3 2 2        1 1 2 3 1 1     = 4 +  −  −  + ×3  16 16  3  4 2  2  2  2 3− 2 3 = 4  −  +  16  3  4  2 1 1 3 11 = − + = 2 6 2 6 23. (C) : We have given, 5x2 – 4x + 2 + k(4x2 – 2x – 1) = 0 ⇒ 5x2 – 4x + 2 + 4kx2 – 2kx – k = 0 ⇒ x2 (5 + 4k) –x (4 + 2k) + 2 – k = 0 Here, a = 5 + 4k, b = – (4 + 2k), c = 2 – k The equation will have real and equal roots if, D = 0 ⇒ b2 – 4ac = 0 ⇒ {– (4 + 2k)}2 – 4(5 + 4k)(2 – k) = 0 ⇒ 16 + 4k2 + 16k – 4(10 – 5k + 8k – 4k2) = 0 ⇒ 16 + 4k2 + 16k – 4(10 + 3k – 4k2) = 0 ⇒ 16 + 4k2 + 16k – 40 – 12k + 16k2 = 0 ⇒ 20k2 + 4k – 24 = 0  ⇒  5k2 + k – 6 = 0 ⇒ 5k2 + 6k – 5k – 6 = 0 ⇒ k(5k + 6) – 1(5k + 6) = 0 ⇒ (k – 1)(5k + 6) = 0 6 ⇒ k = 1, − 5 24. (C) : We have given BN2 = AN ⋅ NC A In DABN by Pythagoras theorem AB2 = AN2 + NB2 …(1) N In DBNC by Pythagoras theorem 2 2 2 …(2) BC = NC + NB  Adding (1) and (2), we get C B AB2 + BC2 = AN2 + NB2 + NC2 + NB2 ⇒ AB2 + BC2 = AN2 + NC2 + 2BN2 ⇒ AB2 + BC2 = AN2 + NC2 + 2AN ⋅ NC ⇒ AB2 + BC2 = (AN + NC)2 ⇒ AB2 + BC2 = AC2 \ In DABC, (proved above) AC2 = AB2 + BC2 Hence, DABC is a right angled triangle, right angled at B. \ ∠ B = 90°

( )

25. (B) : Let AB = AC = 13 cm Draw AD ⊥ BC. Let BC = 2x then BD = DC = x In DABC, by pythagoras theorem, AB2 = AD2 + BD2 ⇒ 132 = AD2 + x2 ⇒ 169 = AD2 + x2

B D ⇒ AD = 169 − x 2 Now, area of DABC = 60 cm2 1 2 1 × AD × BC = 60 ⇒ × 169 − x × 2 x = 60 ⇒ 2 2

Class-10 | Level-1 | Set-6

26. (D) : Since, a tangent at a point to a circle is perpendicular to the radius at the point of contact. \ ∠ACB = 90° In DABC, by Pythagoras theorem, AB2 = AC2 + CB2 ⇒ AB2 = 42 + 32 ⇒ AB2 = 16 + 9 ⇒ AB =

C 3

4 A

P

B

D

25 = 5 cm

Now, let AP = x then BP = 5 – x Since, line joining the centre of circles is perpendicular bisector of their common chord. \ AB ⊥ CD and CP = PD In DAPC, by Pythagoras theorem, AC2 = AP2 + PC2 ⇒ 42 = x2 + PC2 ⇒ 16 = x2 + PC2 In DPCB, by Pythagoras theorem, CB2 = PC2 + PB2 ⇒ 32 = PC2 + (5 – x)2 ⇒ 9 = PC2 + 25 + x2 – 10x Subtracting (ii) from (i), we get 7 = – 25 + 10x ⇒ 10x = 32 ⇒ x = 3.2 cm Now, from (i) 16 = (3.2)2 + PC2 ⇒ PC2 = 5.76 ⇒ PC = 5.76 ⇒ PC = 2.4 cm \ Length of CD = 2 × PC = 2 × 2.4 = 4.8 cm

...(i)

...(ii)

27. (D) : We have,

sec θcosec (90° − θ) − tan θ cot (90° − θ) + sin 2 55° + sin 2 35° tan 10° tan 20° tan 60° tan 70° tan 80°

A 13 cm

Squaring both sides, we get (169 – x2)x2 = 3600 ⇒ x4 – 169x2 + 3600 = 0 ⇒ x4 – 144x2 –25x2 + 3600 = 0 ⇒ x2(x2 – 144) – 25 (x2 – 144) = 0 ⇒ (x2 – 144) (x2 – 25) = 0 ⇒ x2 = 144 or x2 = 25 ⇒ x = 12 or x = 5 \ Base BC = 2x = 2 × 12 = 24 cm or BC = 2x = 2 × 5 = 10 cm

13 cm

C

=

sec 2 θ − tan 2 θ + sin 2 (90° − 35°) + sin 2 35° tan (90° − 80°) tan (90° − 70°) tan 60° tan 70° tan 80°

sec 2 θ − tan 2 θ + cos 2 35° + sin 2 35° cot 80° cot 70° tan 70° tan 80° tan 60° 1+1 sin 2 θ + cos 2 θ = 1 and  =   2  tan 60°  sec θ − tan 2 θ = 1  2 = 3 =

 3

18 (18 − 13) (18 − 13) (18 − 10)



18 × 5 × 5 × 8 = 60 cm 2

=

cm

cm

Area of DADC =

13

m

13

c 10

28. (D) : Let ABCD be the rhombus, 13 cm A B AC be the smaller diagonal of length 10 cm and BD be the longer diagonal. Perimeter of rhombus = 4 × side ⇒ 4 × side = 52 D C 13 cm 52 = 13 cm ⇒ Side = 4 13 + 13 + 10 36 Semiperimeter of DADC (s) = = = 18 cm 2 2

Area of rhombus ABCD = 2 × Area of DADC   = 2 × 60 = 120 cm2 1 Also, area of rhombus ABCD = × BD × AC 2 1 ⇒ 120 = × BD × 10 ⇒ BD = 24 cm 2 Hence, length of the longer diagonal = 24 cm 29. (D) : The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ......, 99. This forms an A.P. with first term a = 3 and common difference d = 6 an = a + (n – 1)d ⇒ 99 = 3 + (n – 1)6 96 = n – 1  ⇒  n = 17 ⇒ 6 17 (2 × 3 + (17 − 1) 6) \ Required sum, S17 = 2 17 (6 + 96) 2 17 = × 102 = 867 2 Hence, sum of all odd integers between 2 and 100 which are divisible by 3 is 867.

=

30. (A) : Let AB be the tower of height h and BD = y. A In DABC, AB tan 30° = BC  h 1 h ⇒ = 2x + y 3 45° 30° C B D ⇒ 2x + y = 3h 2x y ⇒ y = 3h − 2 x  …(1) In DABD, tan 45° =

h AB   ⇒  1 = y BD

⇒ y = h Putting value of y from (2) into (1), we get h = 3h − 2 x ⇒  4

3h − h = 2 x ⇒ h

(

)

3 − 1 = 2x

…(2)

⇒ h =

( 3 + 1) = x ( ( 3 − 1) × ( 3 + 1) 2x ×

31. (D) : We have, 1 − sin A = 1 + sin A =

(1 − sin A) (1 − sin A) (1 + sin A) (1 − sin A)

(1 − sin A)2

1 − sin A 1 sin A = − cos A cos A = sec A – tan A 2

2

)

3 +1

=

(1 − sin A) cos A

32. (C) : Tangents drawn from an external point to a circle are equal. \ AP = BP ⇒ ∠PAB = ∠PBA(angles opposite to equal sides are equal) In DABP, ∠APB + ∠PAB + ∠PBA = 180° (Angle sum property) ⇒ 30° + ∠PBA + ∠PBA = 180° ⇒ 2∠PBA = 180° – 30° = 150° 150° A C = 75° ⇒ ∠PBA = 2 O Also, CA || PB   (given) \ ∠CAB = ∠PBA = 75° 30°  (Alternate angles) P B Join OA and OB. Since, radius drawn from centre is perpendicular to the tangent at the point of contact. \ ∠OBP = 90° Now, ∠OBP = ∠OBA + ∠ABP ⇒ 90° = ∠OBA + 75° ⇒ ∠OBA = 90° – 75° = 15° Now, OA = OB (radii of circle) \ ∠OAB = ∠OBA = 15°(Angles opposite to equal sides are equal) In DOBA, by angle sum property ∠OBA + ∠OAB + ∠BOA = 180° ⇒ 15° + 15° + ∠BOA = 180° ⇒ ∠BOA = 180° – 30° = 150° Since, angle subtended by an arc at the centre is double the angle subtended by same arc at any point on the circle. 1 150° \ ∠BCA = ∠BOA = = 75° 2 2 In DABC, by angle sum property ∠ABC + ∠BCA + ∠CAB = 180° ⇒ 75° + 75° + ∠ABC = 180° ⇒ ∠ABC = 180° – 150° = 30° 33. (A) : Let number of correct answers be x and number of incorrect answers be y. According to question, x + y = 90 …(i) and 5x – 2y = 387 …(ii) Multiplying (i) by 2 and then adding to (ii), we get 7x = 567  ⇒  x = 81 Put x = 81 in (i) we get, y = 90 – 81 = 9 \ Number of questions attempted wrong = 9. Class-10 | Level-1 | Set-6

31 1 × 6 31x

41 = hr = 6 hr 50 min. 6 38. (B) : Number of bottles produced by 6 machines in 1 minute = 2700 Number of bottles produced by 1 machine in 1 minute 2700 = = 450 6 \ Number of bottles produced by 10 machines in 4 minutes = 450 × 10 × 4 = 18000 39. (B) : Let the age of daughter-in-law at the time of marriage be x years. Class-10 | Level-1 | Set-6

41. (A) : Number of English, Mathematics and Science books are 336, 240 and 96 respectively. Now, H.C.F. of (336, 240, 96) = 48 Number of stacks for English books = 336 ÷ 48 = 7 Number of stacks for Mathematics books = 240 ÷ 48 = 5 Number of stacks for Science books = 96 ÷ 48 = 2 Therefore, total number of stacks = 7 + 5 + 2 = 14 42. (D) : Let the value of certificates in the first year = `a Total value of the certificates after 8 years = `48000 Increase in value every year = `400 We have, a, a + 400, a + 800,… Therefore it forms an A.P. with first term a and common difference d = 400. According to question S8 = 48000 8 (2a + (8 − 1) × 400) 2 ⇒ 48000 = 4(2a + 2800) ⇒ 2a = 12000 – 2800 9200 = 4600 ⇒ a = 2 Therefore, value of the certificates purchased by Sonali in first year was `4600. ⇒ 48000 =

43. (C) : Let the depth of the drainlet be h m. P 10 m B A 240 m D

10 m

C

Q 200 m

= 41x ×

40. (B) : Let the quantity of spirit in mixture = 3x Let the quantity of water in mixture = 2x According to question 3x = 2x + 3  ⇒ 3x – 2x = 3 ⇒ x = 3 Therefore quantity of spirit in the mixture = 3 × 3 = 9 litres

180 m 10 m



Sum of ages of husband, wife and son, at the time of marriage of son = 42 × 3 = 126 After the child turned to 5 years, sum of ages of all family members = 126+ 18 + x + 6 + 5 = 155 + x 155 + x = 36 Average age of the family = 5 ⇒ 155 + x = 180 ⇒ x = 180 – 155 = 25 Therefore, age of daughter-in-law at the time of marriage was 25 years.

10 m

34. (A) : Average of the numbers 1 + 2 + 2 + 3 + 3 + 3 + 4 + 4 + 4 + 4 30 = =3 = 10 10 Total number of outcomes = 10 3 occurs 3 times, so favourable outcomes = 3 3 \ Required probability = 10 1 5 35. (C) : Let AB be the line segment and C A (2,3) C(x, y) B (3, 4) be the point dividing AB in the ratio 1 : 5. Let the coordinates of C be (x, y). Using section formula, we have m x + m2 x1 m y + m2 y1 x= 1 2 , y= 1 2 m1 + m2 m1 + m2 1× 3 + 5 × 2 1× 4 + 5 × 3 , y= ⇒x= 1+ 5 1+ 5 13 19 ⇒x= ,y= 6 6  13 19  Therefore coordinates of C are  ,  . 6 6 36. (C) : Let the C.P. of the article = ` x M.P. = x + 10% of x 11x x ×10 = x+ =` 10 100 Discount of 10% is given on M.P. 11x 11x −10% of \ S.P. = 10 10 11x 11x 10 99 x − × = = 10 10 100 100 Now, C.P. > S.P., therefore loss has incurred. 99 x x = Loss = C.P. – S.P. = x − 100 100 x × 100 = 1% Loss % = 100 × x 37. (C) : Let the speed of motorboat be 36x km/hr and the current of water be 5x km/hr. Speed downstream = (36x + 5x) = 41x km/hr Speed upstream = (36x – 5x) = 31x km/hr 31   Distance covered downstream =  41x ×  km  6 Time taken to cover same distance in upstream

R 260 m Area of rectangular plot ABCD = 240 × 180 = 43200 m2 Area of PQRS = 260 × 200 = 52000 m2 Area of the drainlet = 52000 – 43200 = 8800 m2 S

 5

Volume of earth dug out from the drainlet = Volume of earth spread over the rectangular plot ABCD 43200 × 25 ⇒ 8800 × h = 100 43200 × 25 = 1.227 ⇒ h = 8800 × 100 Therefore, depth of the drainlet is 1.227 m. 44. (B) : Let A’s investment for 1 year

1 1  8x =  x × 4 + x × 8  = 3  3 6

x × 12 = 3 x 4 1 12 x Let C’s investment for 1 year = x × 12 = 5 5 Total annual profit = ` 8470 8x 12 x 121x + 3x + = Total investment of A, B and C = 3 5 15 8x 3 Therefore A’s share in profit = 121x × 8470 15 8 x 15 × × 8470           = 3 121x 8 × 15 × 8470           = = ` 2800 3 × 121 45. (C) : Let each installment be ` x. Principal (P) = ` 5300 and Rate(R) = 12% P×R×T Interest for first year = 100 Let B’s investment for 1 year =

5300 × 12 × 1 = ` 636 100 Now, amount for second year at which interest will be paid = 5300 + 636 – x = 5936 – x According to question, 12   x = (5936 − x ) 1 +  100 

=

⇒ 100x = 664832 – 112x ⇒ 212x = 664832 664832 = 3136 ⇒ x = 212 Therefore, value of each installment is ` 3136. 46. (B) : Join O2P and O2S2. T1 ∠O1PS1 = 20° (given) S1 ∠O1PS1 = ∠O2PS2 O2 O1 P (vertically opposite angles) S2 \ ∠O2PS2 = 20° T2 In DO2PS2, ∠O2PS2 = ∠O2S2P = 20° (angles opposite to equal  sides are equal) Now, the radius of a circle is perpendicular to the tangent at the point of contact. \ ∠O2S2T2 = 90°  6

Now, ∠O2S2T2 = ∠O2S2P + ∠PS2T2 ⇒ 90° = 20° + ∠PS2T2 ⇒ ∠PS2T2 = 90° – 20° = 70° 47. (B) : Possible outcomes are (1, 1), (1, 4), (1, 5), (1, 6), (4, 1), (4, 4), (4, 5), (4, 6), (5, 1), (5, 4), (5, 5), (5, 6), (6, 1), (6, 4), (6, 5), (6, 6) \ Total number of outcomes = 16 (i) Outcomes for odd score are (1, 4), (1, 6), (4, 1), (4, 5), (5, 4), (5, 6), (6, 1) (6, 5) \ Number of favourable outcomes = 8 8 1 = Probability of getting odd score = 16 2 (ii) Outcomes for score to be atleast 7 are (1, 6), (4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 1), (6, 4) (6, 5), (6, 6) \ Number of favourable outcomes = 11 11 Probability of getting score atleast 7 = 16 (iii) Outcomes for score to be atmost 7 are (1, 1), (1, 4), (1, 5), (1, 6), (4, 1), (5, 1), (6, 1) \ Number of favourable outcomes = 7 7 Probability of getting score atmost 7 = 16 48. (C) : Produce BD to E, join AE and OD. ∠BEA = 90° (Angle in the semicircle) Since, radius is perpendicular to the tangent at the point of contact. \ ∠BDO = 90° In DBAE and DBOD, A O ∠BEA = ∠BDO = 90° E B ∠ABE = ∠OBD  (Common) D \ DBAE ∼ DBOD  (AA similarity) \

BE AE BA BE AE 32 = = ⇒ = = BD OD BO BD 10 16

So,

1 BE = 2 ⇒ BE = 2BD ⇒ BD = DE = BE 2 BD

…(1)

AE = 2 ⇒ AE = 20 cm 10 Now, in DOBD, by Pythagoras theorem OB2 = OD2 + BD2  ⇒  (16)2 = (10)2 + BD2 and

⇒ BD2 = 256 – 100 = 156  ⇒  BD = 156 cm Now from (1), DE = BD = 156 cm In DADE, by Pythagoras theorem AD2 = AE2 + DE2 ⇒ AD2 = (20)2 + \ AD =

(

156

)

2

= 400 + 156 = 556

556 = 2 139 cm Class-10 | Level-1 | Set-6

49. (D) 50. (C) : Since, AB || CD || XY \ ∠ACD = ∠AXY = q In DACI, CI CI cot θ = = AI b

(Corresponding angles)

⇒ CI = b cot q Similarly, in DBJD JD = b cot q \ CD = CI + IJ + JD = b cot q + a + b cot q ⇒ CD = a + 2 b cot q

We have, GH || XY \ ∠AGK = ∠AXY = q

(Corresponding angles)

In DAGK, GK GK = cot q = AK 3b ⇒ GK = 3 b cot q Similarly, in DBLH LH = 3 b cot q \ GH = GK + KL + LH ⇒ GH = 3 b cot q + a + 3 b cot q ⇒ GH = a + 6 b cot q

JJJ

Class-10 | Level-1 | Set-6

 7