Solutions S1. Ans.(c) Try Using option and divide 64329 With given option S2. Ans.(a) Let the fraction be x/y. According to the questions, x y of 4 7 + 4 7 = 15 14 x y 4 7 + 4 7 = 15 14 x y 4 7 = 1 2 x y = 7 8 S3. Ans.(b) Let the no. of boys = x and no. of girls = y According to the questions, x + y = 14 (ii) Solving equation (i) & equation (ii) x = 8 & y = 6 Total = 14 S4. Ans.(d) Let the numbers be LCM (14 and 4) = 28 According to the question, difference
3 4 28 3 4 28 = 21 6 = 15 This 15r 150 1r 10 Thus, the given number = 280 S5. Ans.(d) The sum of first 50 odd natural number = (50)² = 2500 S6. Ans.(d) Let the number be = 1000x + 100y + 10x + y = 10x(100 + 1) + y(100 + 1) = (101) (10x + y) Thus, the number is divisible by 101. S7. Ans.(b) Since, the difference is same. So, the number whose product is greater will be smaller & vice-versa. Here, largest is 5 3. S8. Ans.(a) Let the 3 rd no. be 200. First no. is 90% of the second number. S9. Ans.(d) x² + (x + 1)² + (x + 2)² = 110 3x² + 6x = 105 x² + 2x 35 = 0 (x + 7) (x - 5) = 0 x = -7, 5 Smallest number = 5 S10. Ans.(b) Let the maximum marks be x. 20 42 x + 10 = 100 100 x 1
11 = 22x 100 22% 11 1% 1 2 100% 50 Thus, the maximum marks = 50 S11. Ans.(b) E : S 5 : 3 New saving = 896 575 = 321 % increment = 7% S12. Ans.(c) According to the question, 5832 = x (1 10 100 ) 3 5832 = x ( 9 10 ) 3 x = 8000 Thus, the purchased price = Rs. 8000 S13. Ans.(c) Biology = 72% Mathematics = 44% Both = 72% + 44% 100% = 116% 100% = 16% Here, 16 % 40 1% 40 16 100% 40 16 100 = 250 S14. Ans.(c)
Let the numbers of sum scored be x and n number of wickets. x n = 12.4 x = 12.4 n According to the question, 12.4n + 26 = 12.2 n + 5 12.4n + 26 = 12.2n + 61 0.2n = 35 n = 175 S15. Ans.(b) Quantity Price I 120 Rs. 360 II 120 Rs. 240 240 Rs. 600 Selling price of 25 Rs. 60 Selling price of 240 Rs. 576 Now, Loss = 600 576 = 24 Loss %= 4% S16. Ans.(c) Here, 7r corresponds to 7 technician. So, 15r 15. Total no. of workers = 22 S17. Ans.(b) A + B = 20 A + B = 40 B + C 2 C + A 2 2 = 19 B + C = 38 = 21 C + A = 42 2(A + B + C) = 120 A + B + C = 60 A 60 38
A = 22 S18. Ans.(c) a + b + c = 1 and ab + bc + ca = 1 3 (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) 1 = a² + b² + c² + 2 1 3 a² + b² + c² = 1 3 Now, a² + b² + c² = ab + bc + ca = 1 3 Thus, a = b = c a : b : c = 1 : 1 : 1 S19. Ans.(b) Rs. 1 : 50p : 25p 8 : 5 : 3 value Rs. 8 : Rs. 2.50 : 0.75 Total value 11.25 According to the question, This, 11.25r Rs. 112.50 1r 10 No. of 50 paise coin = 50 S20. Ans.(b) Let the no. of boys = 100 Girls = 120 B : G Ratio 5 : 6 According to the question, 11r 66 1r 6 Girls 36, Boys 30 Ratio 30 = 3 : 4 40 S21. Ans.(a)
Milk : Water Vessel I 7 : 5 Vessel II 17 : 7 Mixture 5 : 3 Let the capacity of container = 48l. S22. Ans.(b) Share of Milk : Share of Water 4 9 1 11 3 2 3 2 5 20 : 5 20 5 5 33 : 17 S23. Ans.(a) Total cost = 80 13.50 + 120 16 Cost of 200 kg = 3000 Cost of 1 kg Rs. 15 Required profit = 20% So, S.P = 15(1.2) Then, selling price = Rs. 18/kg S24. Ans.(a) S.P = M.P (1 20 10 ) (1 ) (1 5 ) 100 100 100 = 2000 4 5 9 10 19 20 = 1368 S25. Ans.(c) Let the original price be Rs. 100 C.P Rs. 80 S.P 80 (1.4) = 112
Profit % = 12% S26. Ans.(d) C.P. of Rita = 16800 800 = 16000 M. R. P. 80 100 = 16000 M.R.P. = 20,000 Rs S27. Ans.(d) Rebate = 25% = 1 4 This rebate of Rs. 80 is on 1 shirt. Rebate of Rs. 400 is on 5 shirts. MP 4 : SP 3 4r 320 1r 80 3r 240 S28. Ans.(c) Let the CP be Rs. x. According to the questions, (0.75x) (1.32) = (1.15x) 60 0.99x = 1.15x 60 0.16x = 60 x = 375 S29. Ans.(a) Here, 165r 330 1r 2 Cost of A = Rs. 200
S30. Ans (c) Let price of each Article is Rs 1 1 ST TRANSACTION CP of 100 articles = 100 SP of 50 Articles= 60 [Profit % 20] SP of 50 Articles=70 [Profit % 40] Total SP = 130 2 ND TRANSACTION CP OF 100 ARTICLES = 100 SP OF 100 ARTICLES = 125[PROFIT % 25] DIFFERENC IN PROFIT = 130-125=5 5 UNIT = RS100 1 UNIT = RS 20 S31. Ans.(b) 95 (1.1) MP = 110 0.95 Thus, marked price = Rs. 110 S32. Ans.(a) Total cost price of Tea = 96 (0.8) + 140 (1.25) = 120 + 112 Cost price of 2 kg = 232 CP of 1 kg = 116 S.P of 1 kg = 174 Gain % = 58 116 100 = 50% S33. Ans.(c) C.P S.P Profit Profit in (Rs.) I 100 120 20%(on CP) 20 II 96 120 20%(on SP) 24 Now, difference of their profit = 4 Here, 4r Rs. 85
120 85 4 120 = 2550 S34. Ans.(d) Let the expenses be Rs. 120, 170 and 30. Total expense = 320. Now, increment in the respectively expenses increament = 410 320 = 90 % increament = 90 320 100 = 28 1 8 % S35. Ans.(b) 130% Rs. 1326 100% 1020 Thus, principal = Rs. 1020 Let the amount in t years = 1530 Now, Interest = 1530 1020 = 510 So, 510 = 1020 = 5 100 t t = 10 years S36. Ans.(b) Due debt 100 Installment = r t (t 1) 100t + 2 S37. Ans.(b) I = P r t 100 Let the money lent at 5% be x & 4% be (60,000 x) According to the questions = 6450 100 400 + 5 4 3 = 1500 Rs. 2
2560 = x 5 4 + (60,000 x) 100 100 x = 16,000 Money lent at 4% = 44,000 S38. Ans.(c) Borrowed money at the rate = 4% Lend money at the rate = 6 % = (3%, half yearly) 2 t = 2 1 = 2 years So, rate = 3 + 3 + 9 100 = 6.09% Difference of rate = 2.09% This 2.09% 104.50 1% 50 100% 5,000 Thus, borrowed money = Rs. 5000 S39. Ans.(b) 9261 = 8000 (1 + 5 100 )2t 2t 9261 8000 = (21 20 ) ( 21 20 ) 3 = ( 21 20 ) 2t t = 3 2 = 1.5 years S40. Ans.(a) Man takes 30 hours Man + son takes 20 hours Let the work be L.C.M (30, 20) = 60 units Time taken by son to finish alone = 60 = 60 hours 1
S41. Ans.(b) Tap p hours [q > p] Sink q hours Let the capacity of tank be pq. 1 hour work of both (q p) According to the question, pq r = (q p) 1 r = 1 p 1 q S42. Ans.(d) Seema 1 piece of work 6 hours Jaya 1 piece of work 8 hours Gelta 1 piece of work 40 3 hours Let the work be LCM (6, 8, 40 3 ) = 120 Time taken to work together = = 2 8 hours 11 120 (20+15+9) S43. Ans.(d) A 2 work 9 days 5 45 days 2 B + A 1 work 10 days A 1 Work Let the work be LCM ( 45, 10) = 90 2 Now, B finishes it in = 90 = 15 days 5
S44. Ans.(a) Time taken by A to finish the work = 63 3.50 Time taken by B to finish the same work = 75 Let work = 90 unit 2.50 = 18 days = 30 days Time taken to work together = 90 8 days Total cost = 90 8 = Rs. 67.50 (3.5 + 2.5) S45. Ans.(b) Man : Boy According to the question, 10(2) 8 = [3(2) + 4 (1)] d 160 = (10) d d = 16 days S46. Ans.(c) Let the capacity of 3 containers = 3x, 4x & 5x Milk : Water 4 1 1st 3x : 3x 5 5 3 1 2nd 4x : 4x 4 4 5 2 3rd 5x : 5x 7 7 Total milk 314x 35 106x 35 Ratio = 157 : 53 S47. Ans.(d) 2 : 1
Time taken to go P R = 16 hrs 40 min = 16 hrs 40 60 = 16 hrs 2 3 = 50 3 hours Time taken to go P Q + Q P = 12 hrs. Let PQ = d Let the speed of river = u Speed of boat = v d + v u d v+u = 12 (i) 2d v u = 50 3 d = 25 v u 3 Put equation (ii) in equation (i) (ii) 25 3 + d (v + u) = 12 d (v + u) = 11 3 Now, time to go from R P 2d v + u = 2 11 3 = 22 3 = 7 1 3 hrs. S48. Ans.(c) Distance of the journey = S t = 42 10 = 420 km New speed = 420 = 60 km/hrs 7 This, increment = (60 42) km/hr = 18 km/hr S49. Ans.(a) Total distance to travel = 450 km Let speed of car = V c 450 = 240 + V c x
240 T + 210 C = 8 2 3 (i) 180 T + 270 C = 9 (ii) = 45 km/hr. S50. Ans.(b) t = d S 2 (150 + 250)/1000 = 60 100 V c V c = 88 km/hr S51. Ans.(b) Volume of iron used in making material = πr²h = 22 7 (4² 3²) 20 = 22 7 7 1 20 = 440 cm³
S52. Ans.(c) For cuboid, V = A 1 A 2 A 3 = 12 15 20 = 60 cm³ S53. Ans.(c) l : b = 3 : 2 Let the length & breadth be 3x & 2x km So, perimeter = 2(5x) = 10x According to the question, 12 = 10x ( 8 60 ) x = 8 50 Now, Area = l b = 6x² km² = 6 8 50 8 1000 1000 m² 50 = 153600 m² S54. Ans.(d) r 25% h 25% Net increment / decrement = 25 + 25 25 25 100 = 6.25% Decrement by 6.25% S55. Ans.(a) r = 0.6 cm Volume of material removed = volume of cylinder of height 1.4 cm Volume of cone of height 1.4 cm
= πr 2 h 1 3 πr2 h = 2 3 πr2 h = 2 3 22 0.6 0.6 1.4 7 = 1.056 cm³ S56. Ans.(a) Area of park = 60 40 = 2400 m² Area of lawn = 2109 m² Area of crossroads = (2400 2109) m² = 291 m² Let the width be x m. 40 x + 60 x x² = 291 x = 3 m. S57. Ans.(a) Radius of each Circle = 70 cm Area of space enclosed = (140) 2 (4 90 360 πr2 ) = (140) 2 ( 22 70 70) 7 = 4200 cm²
S58. Ans.(d) 2πr = 8.8 cm 2 22 7 r = 8.8 r = 7 5 (2πr)h = 17.6 h = 2 Volume = πr²h = 22 7 7 5 7 5 2 = 12.32 m³ S59. Ans.(d) According to the questions, 2 3 πr3 = n πr 2 h 2 3 π (9)3 = n π 3 2 3 2 4 n = 54 S60. Ans.(d) Length of water column in 1 2 hr = (10 1000 1 2 ) = 5000 m Let the height raised be h m. According to the question, 80 40 h = h = 1 160 m = 5 8 cm 40 100 100 (10 1000 1 2 ) S61. Ans.(b) Diagonal of square = diameter of circle
2a = 2r a = r 2 Area of square = a² = 2r² Area of hexagon = 6 area of equal = 6 3 4 r2 = 3 3r2 2 Ratio = 2r2 2 3 3r 2 = 4 3 3 S62. Ans.(c) T.S.A = 2πrh + 2πr² C.S.A = 2πrh According to the question, C.S.A = 2 3 T.S.A C. S. A T. S. A = 2 3 2πr² = 77 r = 7/2 cm Now, C.S.A = 2πrh h = 77 2 2 2 π 7 = 7 Now, Volume = πr²h = 22 7 7 2 7 2 7 = 269.5 cm³
S63. Ans.(c) Area of base = 16 π cm 2 πr 2 2 = 16π R 2 = 4 Slant ht. = 6 cm Curved surface area = π(r 1 + r 2 ) l = π(4 + 2) 6 = 36π cm 2 S64. Ans.(b) r 1 : r2 2 : 1 SA 1 = V 2 4π(2r) 2 = 4 3 π(r)3 r = 12 Radius of first sphere = 24 S65. Ans.(a) r = 21 cm [cylinder] 2 h = 38 cm Volume of cylinder = n[volume of cone + Volume of hemisphere] π ( 21 2 2 ) 38 = n [ 1 2 3 π (7 2 ) 12 + 2 3] 3 π (7 2 ) n = 54 S66. Ans.(b) (x 2 + y 2 2xy) x 2 + y 2 [ (x y)(x2 + xy + y 2 ) 3xy] x y (x y)2 = (x 2 + y 2 ) 1 (x 2 + xy + y 2 3xy) 1 = (x 2 + y 2 )
S67. Ans.(a) a + b + c = 0 1(c + a) = (a + b)(b + c)(c + a) + (a + b) (b + c)(c + a)(a + b) + (b + c) (c + a)(a + b)(b + c) 2(a + b + c) = (a + b)(b + c)(c + a) = 0 S68. Ans.(a) x² + y² + 2x + 1 = 0 (x² + 2x + 1) + y² = 0 (x + 1)² + y² = 0 x = 1 y = 0 Now, x 31 + y 35 = ( 1) 31 + 0 = 1 S69. Ans.(b) x = 1 y xy = 1 x 2 + xy + y 2 x 2 xy + y 2 = x2 + y 2 + 1 x 2 + y 2 1 x2 + Now, x + y x + 1 x = 5 + 1 5 1 + 5 1 5 + 1 = ( 5 + 1)2 + ( 5 1) 2 ( 5 1)( 5 + 1) = 2( 5)2 + 1 4 x + 1 x = 6 2 = 3 x 2 + 1 x 2 = 9 2 = 7 Now, x 2 + xy + y 2 x 2 xy + y 2 = 7 + 1 7 1 = 8 6 = 4 3 1 x 2 + 1 x 2 + 1 x 2 1
S70. Ans.(c) x 1 x = 3 x + 1 x = ( 3) 2 + 4 = 7 Now, x 2 + 1 x 2 = (5) x 6 + 1 x 6 = (5)3 3 5 = 125 15 = 110 S71. Ans.(c) x 4 + 2x 3 + ax 2 + bx + 9 This is square of (x² + x + 3) ( x2 + x + 3 (a + b) 2 ) Let x² = a & (x + 3) = b = x 4 + x 2 + 9 + 2x 3 + 6x + 6x 2 = x 4 + 2x 3 + 7x 2 + 6x + 9 = a = 7, b = 6 S72. Ans.(c) a² + b² + c² = 16 x² + y² + z² = 25 We will use trick, a = 0, b = 0, c = 4 and x = y = 0, z = 5 Now, a+b+c x+y+z = 4 5 S73. Ans.(b) x + a 2 + 2c 2 + x + b2 + 2a 2 b + c c + a Do it from option Option (b) satisfy S74. Ans.(a) + x + c + 2b2 a + b = 0
a³ = 117 + b³ & a b = 3 a³ b³ = 117 (a b)³ = a³ b³ 3ab(a b) 27 = 117 3ab(3) 9ab = 117 27 9ab = 90 ab = 10 Now, (a b) (a² + ab + b²) = 117 3(a² + ab + b²) = 117 a² + ab + b² = 39 a² + b² = 39 10 = 29 Thus, (a + b)² = a² + 2ab + b² = 29 + 2(10) = 49 a + b = ±7 S75. Ans.(a) a + 1 a = 2 Put a = 1 Now, a 1000 + a 1000 = ( 1) 1000 + ( 1) 1000 = 1 + 1 = 2 S76. Ans.(b) Since s are similar, 2 ar( ABC) ar( DEF) = (BC EF ) 2 9 16 = (BC EF ) BC EF = 3 4 EF = 4 2.1 3 = 2.8 cm S77. Ans.(c)
AOB = 60 [ AOB is equilateral] So, APB = 1 2 60 = 30 S78. Ans.(b) AB = AC = chord X & Y are tangent on smaller circle. Thus, X & Y is bisecting AB & AC. So, XY is midpoint of BC. XY = 1 2BC S79. Ans.(b) 3a = 24 a = 8 AG = a 3 = 8 3 S80. Ans.(d)
PQ = (AB) 2 (11 6) 2 = 169 25 = 144 PQ = 12 S81. Ans.(b) Drawn AD BC. Since, it is isosceles. So, AD = Median In right angled ADB, AB 2 = AD 2 + BD 2 (12 5) 2 = (AD) 2 + (12) 2 (12 5) 2 (12) 2 = AD 2 AD = 576 = 24 Now, In OBD, x 2 = (24 x) 2 + (12) 2 x 2 (24 x) 2 = (12) 2 (x + 24 x)(x 24 + x) = 12 2 24(2x 24) = 144 (2x 24) = 6 2x = 30 x = 15
S82. And.(a) Let AB = AC = y Let BP = BQ = x (tangent from an external point) BQ = x. So, AQ = (y x) Now, AQ = AR = (y x) But AC = y So, RC = x Thus, BP = PC = x S83. Ans.(d) Line joining midpoint is to third side and half of third side. and thus, ADE ABC. So, ar( ADE) ar(abc) = ( x 2 2x ) = 1 4 ar( ADE) thus, ar( BCED) = 1 4 1 = 1 3
S84. Ans.(d) Since, O is circumcenter. So, BP = 3. R = abc 4 = 1 2 6 280 17 17 6 R = = 8.64 cm 4 2 6 280 S85. Ans.(b) From properties of lines, 1 C 1 C + 1 = 1 AA 1 BB 1 S86. Ans.(d) (1 + sec 22 + cot 68 )(1 cosec 22 + tan 68 ) Let θ = 22 [1 + sec θ + cot(90 θ)][1 cosec θ + tan(90 θ)]
[(1 + sec θ + tan θ)][1 cosec θ + cot θ] [1 + 1 cos θ + sin θ cos θ ] [1 1 sin θ + cos θ sin θ ] (cos θ + sin θ + 1) (sin θ + cos θ 1) [ ] cos θ sin θ [ (cos θ + sin θ)2 1 ] cos θ sin θ = 2 S87. Ans.(a) x sin 2 θ + y cos 3 θ = sin θ cos θ & x sin θ y cos θ = 0, x sin θ = y cos θ x sin θ (sin 2 θ) + ycos θ(cos 2 θ) = sin θ cos θ x sin θ (sin 2 θ + cos 2 θ) = sin θ cos θ [x sin θ = y cos θ] x sin θ = sin θ cos θ x = cos θ Thus, y = sin θ Now, x 2 + y 2 = 1 S88. Ans.(b) sec θ + tan θ = m [0 < θ < 90 ] ( 1 + sin θ cos θ ) = m Squaring both sides, 1 2 + sin 2 θ + 2 sin θ = m 2 cos 2 θ 1 + sin 2 θ + 2 sin θ = m 2 (1 sin 2 θ) 5 2 (1 + m 2 ) + 25 + (1 m 2 ) = 0 sin θ = S = (m2 1) (m 2 + 1) S89. Ans.(b)
(a 2 b 2 ) sin θ + 2ab cos θ = a 2 + b 2, tan θ = (a2 b 2 ) 2ab S90. Ans.(d) Let the speed of Vehicle = v Time taken = 10 min. So, d = 10 v tan 60 = h x x = h 3 h tan 45 = (x + 10v) x + 10v = h 10v = h h 3 10v = h (1 1 3 ) Now, time taken to cover x m t = x v h 10 3 = 3 h( 3 1)
10( 3 + 1) = ( 3 1)( 3 + 1) = 5( 3 + 1) = 13.66 min. S91. Ans.(d) 2y cos θ = x sin θ 2y cosec θ = x sec θ 2x sec θ y cosec θ = 3 2 x sec θ x sec θ = 3 2 x sec θ 3 2 = 3 x = 2 cos θ (i) y = sin θ (ii) Now, x 2 + 4y 2 = 4 cos 2 θ + 4 sin 2 θ = 4 S92. Ans.(a) Let the ht. of tower = h m. tan 60 = 100 x x = 100 3 (100 h) tan 45 = x x = 100 h 100 3 = 100 h
h = 100 100 1 ( 3 1) = 100 (1 ) = 100 3 3 3 = 100 (3 3) m. 3 S93. Ans.(b) tan α = x d (i) (h + x) tan β = d (h + x) tan β = tan α x x tan β = h tan α + x tan α x(tan β tan α) = h tan α h tan α x = (tan β tan α) h tan α thus ht. of tower = (tan β tan α) S94. Ans.(a) Let the velocity of the boat Let the time t = 10 min. tan 60 = h x x = h 3
h 1 tan 30 = (x + 10v) 3 = h h ( h 3 + 10v ) 3 = h 5 3 Time taken to cover x m = x v h = 3 h 5 3 = h 5 3 3 x = 5 min. S95. Ans.(b) cot θ + cosec θ 1 cot θ + cosec θ + 1 Put θ = 45 Option (b) satisfy. S96. Ans.(b) According to question, 36 825 1 825 36 Total Monthly income = 360 = 360 825 = Rs. 8250 36 + 10v = 3h10v = ( 3h h 3 ) 10v = h ( 2 3 ) v S97. Ans.(a) Saving = 54 % = 54 100 = 15% 360
S98. Ans.(c) Ratio 108 : 72 = 3 : 2 S99. Ans.(a) Average expenses on clothing & Rent = 126 2 = 63 = 63 825 36 = 1443.75 S100. Ans.(d) Ratio (Food + Clothing + Misc.) : (Saving + Rent) (108 + 36 + 72) : (54 + 90) 216 3 144 2 72 : 72 = 1 : 1