SSC CGL Quant Tier 2

2016 Solved Paper

Click here for 20 Online Tests for SSC CGL Tier 2

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1. Let 0 < x < 1, then the correct inequality is

(a) x < √x < x2 (b) √x < x < x2 (c) x2 < x < √x (d) √x < x2 < x

2. Three bells ring at interval of 36 seconds, 40 seconds and 48 seconds respectively. They start ringing together at a particular time. They will ring together after every (a) 6 minutes (b) 12 minutes

(c) 18 minutes (d) 24 minutes

3. If the sum of the digits of a three-digit number is subtracted from that number, then it will always be divisible by,

(a) 3 only (b) 9 only (c) both 3 and 9 (d) all of 3, 6 and 9

4. Which of the following is correct?

(a) 2/3 < 3/5 < 11/15 (b) 3/5 < 2/3 < 11/15 (c) 11/15 < 3/5 < 2/3 (d) 3/5 < 11/15 < 2/3

5. The greater of two numbers whose product is 900 and sum exceeds their difference by 30 is (a) 60 (b) 75 (c) 90 (d) 100

6. The smallest fraction, which should be added to the sum of 2 1/2, 3 1/3, 4 1/4 and 5 1/5 to make the result a whole number is (a) 13/60 (b) 1/4 (c) 17/60 (d) 43/60

7. Find the cube root of (-13824) Or

Find the value of 3 13824

(a) 38 (b) -38 (c) 24 (d) -24

8. The sum of three positive numbers is 18 and their product is 162. If the sum of two numbers is equal to the third, then the sum of squares of the numbers is (a) 120 (b) 126 (c) 132 (d) 138

9. The sum of three consecutive even numbers is 28 more than the average of these three numbers. Then the smallest of these three numbers is (a) 6 (b) 12 (c) 14 (d) 16

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10. In a division sum, the divisor ‘d’ is 10 times the quotient ‘q’ and 5 times the remainder ‘r’, if r =

46, the dividend will be (a) 5042 (b) 5328 (c) 5336 (d) 4276

11. A man can do a piece of work in 30 hours, if he works with his son then the same piece of work

finished in 20 hours, if the son works alone he can do the work in (a) 60 hours

(b) 50 hours (c) 25 hours (d) 10 hours

12. A water tap fills a tub in ‘p’ hours and a sink at the bottom empties it in ‘q’ hours. If p<q and

both tap and sink are open, the tank is filled in ‘r’ hours; then (a) 1/r = 1/p + 1/q (b) 1/r = 1/p – 1/q (c) r = p + q

(d) r = p – q

13. John does 1/2 piece of work in 3 hours, joe does 1/4 of the remaining work in 1 hour and George finishes remaining work in 5 hours, how long would it have taken the three working together to do work? (a) 2 1/7 (b) 3 1/7 (c) 3 8/11 (d) 2 8/11

14. A does 2/5 of a work in 9 days. Then B joined him and they together completed the remaining work in 6 days, B alone can finish the whole work in (a) 6 12/13 days (b) 8 2/11 days (c) 10 days (d) 18 days

15. The daily wages of A and B respectively are Rs. 3.50 and 2.50. when A finishes a certain work, he gets a total wage of Rs. 63. when B does the same work, he gets a total wage Rs.75. if both

of them do it together what is the cost of the work? (a) Rs. 67.50 (b) Rs. 27.50 (c) Rs. 60.50

(d) Rs. 70.50

16. A man does double the work done by a boy in the same time. The number of days that 3 men and 4 boys will take to finish a work which can be done by 10 men in 8 days is (a) 4

(b) 16 (c) 7 3/11 (d) 8 4/5

17. The marked price of an article is 30% higher than the cost price. If a trader sells the articles allowing 10% discount to customer, then the gain percent will be (a) 17 (b) 20 (c) 19

(d) 15

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18. A merchant marked the price of an article by increasing its production cost by 40%, now he

allows 20% discount and gets a profit of Rs.48 after selling it, the production cost is (a) Rs.320 (b) Rs.360 (c) Rs.400 (d) Rs.440

19. A watch dealer pays 10% customs duty on a watch which costs Rs.500 abroad. He desires to

make a profit of 20% after giving a discount of 25% to the buyer. The marked price should be (a) Rs.950

(b) Rs.800 (c) Rs.880 (d) Rs.660

20. A shopkeeper allows 20% discount on his advertised price and to make a profit of 25% on his

outlay. What is the advertised price (in Rs.) on which he gains Rs.6000? (a) 36000 (b) 37500 (c) 39000

(d) 42500

21. Rs.2420 were divided among A, B and C such that A: B = 5: 4 and B: C = 9: 10 then C gets (a) 680 (b) 800 (c) 900 (d) 950

22. 49 kg of blended tea contain Assam and Darjeeling tea in the ratio 5: 2. Then find the quantity of Darjeeling tea that is to be added to the mixture to make the ratio of Assam to Darjeeling tea 2: 1 is (a) 4.5 kg (b) 3.5 kg (c) 5 kg (d) 6 kg

23. Among 132 examinees of a certain school, the ratio of successful to unsuccessful students is 9: 2. Had 4 more students passed, then the ratio of successful to unsuccessful students will be (a) 14: 3

(b) 14: 5 (c) 28: 3 (d) 28: 5

24. In a regiment the ratio between the number of officers to that of soldiers was 3: 31 before battle. In a battle 6 officers and 22 soldiers were killed and the ratio became 1: 13, the number of officers in the regiment before battle was (a) 31 (b) 38

(c) 21 (d) 28

25. Three containers have their volumes in the ratio 3: 4: 5. They are full of mixtures of milk and

water. The mixtures contain milk and water in the ratio of (4: 1), (3: 1) and (5: 2) respectively. The contents of all these three containers are poured into a fourth container. The ratio of milk and water in the fourth container is (a) 4:1 (b) 151:48

(c) 157:53 (d) 5:2

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26. In what proportion must a grocer mix sugar at Rs.12 a kg and Rs.7 a kg as to make a mixture

worth Rs.8 a kg? (a) 7: 12 (b) 1: 4 (c) 2: 3 (d) 12: 7

27. Fifteen movie theatres average 600 customers per theatre per day. If six of the theatres close

down but the total theatre attendance stays the same, then the average daily attendance per theatre among the remaining theatres is

(a) 900 (b) 1000 (c) 1100 (d) 1200

28. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is (a) 31 kg (b) 32 kg

(c) 29.5 kg (d) 35 kg

29. The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is (a) 165 (b) 170 (c) 172 (d) 174

30. The average of 7 consecutive numbers is 20. The largest of these numbers is (a) 20 (b) 23 (c) 24 (d) 26

31. Mukesh has twice as much money as soham. Soham has 50% more money than Pankaj. If the average money with them is Rs.110, then Mukesh has

(a) 155 (b) 160 (c) 180 (d) 175

32. The average daily income of 7 men, 11 women and 2 boys is Rs.257.50. if the average daily

income of the men is Rs.10 more than that of women and the average daily income of the women is Rs.10 more than that of boys, the average daily income of a man is (a) Rs.277.5

(b) Rs.250 (c) Rs.265 (d) Rs.257

33. If the profit on selling an article for Rs.425 is the same as the loss on selling it for Rs.355, then the cost price of the article is (a) Rs.410 (b) Rs.380 (c) Rs.400

(d) Rs.390

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34. A & B jointly made a profit of Rs.1650 and they decided to share it such that 1/3 of A’s profit is

equal to 2/5 of B’s profit. Then profit of B is (a) Rs.700 (b) Rs.750 (c) Rs.850 (d) Rs.800

35. 4% of the selling price of an article is equal to 5% of its cost price. Again 20% of the selling

price is Rs.120 more than 22% of its cost price. The ratio of cost price and selling price is (a) 2:3

(b) 3:2 (c) 4:5 (d) 5:4

36. Due to 25% fall in the rate of eggs, one buy 2 dozen eggs more than before by investing Rs.162.

then the original rate per dozen of the eggs is (a) Rs.22 (b) Rs.24 (c) Rs.27

(d) Rs.30

37. Last year Mr. A bought two paintings. This year he sold them for Rs.20,000 each. On one, he made a 25% profit and on the other he had a 25% loss. Then his net profit or loss is (a) He lost more than Rs.2000 (b) He lost less than Rs.2000 (c) He earned more than Rs.2000 (d) He earned less than Rs.2000

38. A shopkeeper sells rice at 10% profit and uses weight 30% less than the actual measure. His gain percent is (a) 57 2/3 % (b) 57 1/7 % (c) 57 2/5 % (d) 57 3/7 %

39. What % of a day is 30 minutes? (a) 2.83 (b) 2.083

(c) 2.09 (d) 2.075

40. A businessman’s earning increase by 25% in one year but decreases by 4% in the next. Going

by this pattern, after 5 years, his total earnings would be Rs.72000. what is his present earnings? (a) Rs.10000 (b) Rs.80000 (c) Rs.40000

(d) Rs.54000

41. In an examination 73% of the candidates passed in quantitative aptitude test, 70% passed in General awareness and 64% passed in both. If 6300 failed in both subjects the total number of

examinees were (a) 60000 (b) 50000 (c) 30000 (d) 25000

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42. A man spends 75% of his income. His income increases by 20% and his expenditure also

increases by 10%. Find the percentage increases in his savings (a) 25% (b) 50% (c) 15% (d) 10%

43. On a river, Q is the mid-point between two points P and R on the same bank of the river. A boat

can go from P to Q and back in 12 hours, and from P to R in 16 hours 40 min. how long would it take to go from R to P?

(a) 3 1/3hr (b) 5 hr (c) 6 2/3hr (d) 7 1/3hr

44. A car can finish a certain journey in 10 hours at a speed of 42kmph. In order to cover the same distance in 7 hours, the speed of the car(km/h) must be increased by: (a) 12 (b) 15

(c) 18 (d) 24

45. A man travels 450km to his home partly by train and partly by car. He takes 8hrs 40mins if he travels 240km by train and rest by car. He takes 20 mins more if he travels 180km by train and the rest by car. The speed of the car in km/hr is (a) 45 (b) 50 (c) 60 (d) 48

46. A train ‘B’ speeding with 100kmph crosses another train C, running in the same direction, in 2 mins. If the length of the train B and C be 150m and 250m respectively. What is the speed of the train C (in kmph)? (a) 75 (b) 88 (c) 95 (d) 110

47. The compound interest on Rs.30,000 at 7% per annum for n years is Rs.4347. the value of n is (a) 3 (b) 2 (c) 4

(d) 5

48. If A borrowed Rs. P at x% and B borrowed Rs.Q (>P) at y% per annum at simple interest at the same time, then the amount of their debts will be equal after (a) 100(Q-P)/(Px-Qy) years

(b) 100(Px-Qy)/(Q-P) years (c) 100(Px-Qy)/(P-Q) years (d) 100(P-Q)/(Px-Qy) years

49. A man invested a sum of money at compound interest. It amounted to Rs.2420 in 2 years and to Rs.2662 in 3 years. Find the sum. (a) Rs.1000 (b) Rs.2000 (c) Rs.5082

(d) Rs.3000

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50. If a sum of money becomes 4000 in 2 years and 5500 in 4 years 6 months at the same rate of

simple interest per annum. Then the rate of simple interest is (a) 21 3/7 % (b) 21 2/7 % (c) 21 1/7 % (d) 21 5/7 %

51. A hollow cylindrical tube 20cm along is made of iron and its external and internal diameters are

8 cm and 6 cm respectively. The volume (in cubic cm) of iron used in making the tube is (take π=22/7)

(a) 1760 (b) 440 (c) 220 (d) 880

52. If the areas of three adjacent faces of a rectangular box which meet in a corner are 12cm2, 15cm2, and 20cm2 respectively. Then the volume of the box is (a) 3600cm3 (b) 300cm3

(c) 60cm3 (d) 180cm3

53. The ratio between the length and the breadth of a rectangular park is 3:2. If a man cycling along the boundary of the park at the speed of 12km/hr completes one round in 8 minutes, then the area of the park is (a) 153650 m2 (b) 135600m2 (c) 153600m2 (d) 156300m2

54. If the radius of a right circular cylinder open at both the ends, is decreased by 25% and the height of the cylinder is increased by 25%. Then the curved surface area of the cylinder thus formed. (a) remains unaltered (b) is increased by 25% (c) is increased by 6.25% (d) is decreased by 6.25%

55. A cylindrical pencil of diameter 1.2cm has one of its end sharpened into a conical shape of height 1.4cm. the volume of the material removed is (a) 1.056 cm3 (b) 4.224 cm3

(c) 10.56 cm3 (d) 42.24 cm3

56. A rectangular park 60m long and 40m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 m2 then

the width of the road is (a) 3m (b) 5m (c) 6m

(d) 2m

57. Four circles of equal radii are described about the four corners of a square so that each touches two of the other circles. If each side of the square is 140 cm then area of the space enclosed between the circumference of the circle is (taken π = 22/7)

(a) 4200 cm2 (b) 2100 cm2 (c) 7000 cm2 (d) 2800 cm2

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58. The amount of concrete required to build a concrete cylindrical pillar whose base has a perimeter

8.8 metre and curved surface area 17.6 sq.metre, is (Take π = 22/7) (a) 8.325 m3 (b) 9.725 m3 (c) 10.5 m3 (d) 12.32 m3

59. A hemispherical bowl of internal radius 9 cm, contains a liquid. This liquid is to be filled into

small cylindrical bottles of diameter 3 cm and height 4 cm. Then the number of bottles necessary to empty the bowl is

(a) 18 (b) 45 (c) 27 (d) 54

60. A rectangular water tank is 80m × 40 m. Water flows into it through a pipe of 40 sq.cm. at the opening at a speed of 10 km/hr. The water level will rise in the tank in half an hour is (a) 3/2 cm (b) 4/9 cm

(c) 5/9 cm (d) 5/8 cm

61. A square and a regular hexagon are drawn such that all the vertices of the square and the hexagon are on a circle of radius r cm. the ratio of area of the square and the hexagon is (a) 3:4 (b) 4:3√3 (c) √2:√3 (d) 1:√2

62. A solid cylinder has the total surface area 231 sq.cm. If its curved surface area is 2/3 of the total surface area, then the volume of the cylinder is (a) 154 cu.cm (b) 308 cu.cm (c) 269.5 cu.cm (d) 370 cu.cm

63. The lateral surface area of frustum of a right circular cone, if the area of its base is 16πcm2 and the diameter of circular upper surface is 4 cm and slant height 6 cm, will be

(a) 30 πcm2 (b) 48 πcm2 (c) 36 πcm2 (d) 60 πcm2

64. The diameter of a sphere is twice the diameter of another sphere. The surface area of the first

sphere is equal to the volume of the second sphere. The magnitude of the radius of the first sphere is (a) 12

(b) 24 (c) 16 (d) 48

65. A right circular cylinder having diameter 21 cm and height 38 cm is full of ice cream. The ice cream is to be filled in cones of height 12 cm and diameter 7 cm having a hemispherical shape on the top. The number of such cones to be filled with ice cream is (a) 54 (b) 44

(c) 36 (d) 24

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66. The simplified value of

[1 – (2xy/(x2+y2))] [(x3-y3)/(x-y) – 3xy] is (a) 1/(x2-y2) (b) 1/(x2+y2) (c) 1/(x-y) (d) 1/(x+y)

67. If a + b + c = 0 then the value of

[1/(a+b)(b+c)] + [1/(b+c)(c+a)] + [1/(c+a)(a+b)] (a) 0

(b) 1 (c) 3 (d) 2

68. If x2+y2+2x+1 = 0, then the value of x31 + y35 is

(a) -1 (b) 0 (c) 1 (d) 2

69. If x = (√5+1)/(√5-1) and y = (√5-1)/(√5+1) , then the value of (x2+xy+y2)/(x2-xy+y2)

(a) 3/4 (b) 4/3 (c) 3/5

(d) 5/3

70. If (x-(1/x))2 = 3, then the value of x6+(1/x6) equals (a) 90

(b) 100 (c) 110 (d) 120

71. If x4+2x3+ax2+bx+9 is a perfect square, where a and b are positive real numbers, then the

value of a and b are (a) a = 5, b = 6 (b) a = 6, b = 7 (c) a = 7, b = 6 (d) a = 7, b = 8

72. If a2 + b2 + c2 = 16, x2 + y2 + z2 = 25 and ax +by +cz = 20, then the value of (a+b+c)/(x+y+z) (a) 3/5 (b) 5/3 (c) 4/5 (d) 5/4

73. The value of x which satisfies the equation

((x+a2+2c2)/b+c) + ((x+b2+2a2 )/c+a) + ((x+c+2b2 )/a+b) = 0 is (a) (a2+b2+c2) (b) –(a2+b2+c2) (c) (a2+2b2+c2)

(d) –(a2+b2+2c2)

74. If a3 = 117+b3 and a = 3+b, then the value of a+b is (a) ±7 (b) ±49

(c) ±13 (d) 0

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75. If a + 1/a = -2 then the value of a1000 + a-1000 is

(a) 2 (b) 0 (c) 1 (d) 1/2

76. ABC is similar to DEF. If area of ABC is 9 sq.cm. and area of DEF is 16 sq.cm. and BC = 2.1 cm. Then the length of EF will be (a) 5.6 cm (b) 2.8 cm (c) 3.7 cm (d) 1.4 cm

77. A chord of a circle is equal to its radius. The angle subtended by this chord at a point on the circumference is (a) 80° (b) 60° (c) 30° (d) 90°

78. Let two chords AB and AC of the larger circle touch the smaller circle having same centre at X and Y. then XY =? (a) BC (b) (1/2)BC

(c) (1/3)BC (d) (1/4)BC

79. Let G be the centroid of the equilateral triangle ABC of perimeter 24 cm. then the length of AG is

(a) 2√3 cm (b) 8/√3 cm (c) 8√3 (d) 4√3

80. A and B are the centres of two circles with radii 11 cm and 6 cm respectively. A common tangent touches these circles at P and Q respectively. If AB = 13 cm, then the length of PQ is (a) 13 cm (b) 17 cm

(c) 8.5 cm (d) 12 cm

81. ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12√5 and BC = 24 cm then radius of circle is

(a) 10 cm (b) 15 cm (c) 12 cm (d) 14 cm

82. ABC is an isosceles triangle where AB = AC which is circumscribed about a circle. If P is the point where the circle touches the side BC, then which of the following is true? (a) BP = PC (b) BP > PC

(c) BP < PC (d) BP = (1/2)PC

83. If D and E are the mid points of AB and AC respectively of ABC, then the ratio of the areas of ADE and BCED is?

(a) 1:2 (b) 1:4 (c) 2:3 (d) 1:3

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84. O is the circumcentre of the isosceles ABC. Given that AB = AC = 17 cm and BC = 6 cm. The

radius of the circle is (a) 2.115 cm (b) 40/280 cm (c) 42/298 cm

(d) 42/280 cm

85. B1 is a point on the side AC of ABC and B1B is joined. A line is drawn through A parallel to B1B meeting BC at A1 and another line is drawn through C parallel to B1B meeting AB produced at C1. Then

(a) [(1/CC1) - (1/AA1)] = 1/BB1 (b) [(1/CC1) + (1/AA1)] = 1/BB1 (c) [(1/BB1) – (1/AA1)] = 2/CC1 (d) [(1/AA1) – (1/CC1)] = 2/BB1

86. The value of the expression (1+sec22°+cot68°)(1-cosec22°+tan68°) is

(a) o (b) 1 (c) -1

(d) 2

87. If Xsin3𝜃 + Ycos3𝜃 = sin 𝜃cos𝜃 and Xsin𝜃 - Ycos𝜃 = 0, then the value of x2+y2 equals (a) 1 (b) 1/2

(c) 3/2 (d) 2

88. If sec𝜃 + tan𝜃 = m (>1), then the value of sin 𝜃 is (0°< 𝜃<90°)

(a) 1-m2/1+m2 (b) m2-1/m2+1 (c) m2+1/m2-1 (d) 1+m2/1-m2

89. If (a2-b2)sin𝜃 + 2abcos 𝜃 = a2 + b2, then tan 𝜃 = (a) 2ab/a2-b2 (b) a2-b2/2ab (c) ab/a2-b2 (d) a2-b2/ab

90. A person from the top of a hill observes a vehicle moving towards him at a uniform speed. It takes 10 minutes for the angle of depression to change from 45° to 60°. After this the time required by the vehicle to reach the bottom of the hill is (a) 12 min 20 sec (b) 13 min (c) 13 min 40 sec (d) 14 min 24 sec

91. If 2ycos𝜃 = xsin𝜃 and 2xsec𝜃 – ycosec𝜃 = 3, then the value of x2+4y2 is (a) 1 (b) 2 (c) 3 (d) 4

92. From the top of a cliff 100-metre-high, the angles of depression of the top and bottom of a tower are 45° and 60° respectively. The height of the tower is (a) (100/3)(3-√3) metre

(b) (100/3)( √-1) metre (c) 100/3 (2√3-1) metre (d) 100/3 (√3 -√2) metre

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93. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height

h. At a point on the plane, the angle of elevation of the bottom of the flag staff is ⍺ and that of the top of the flag staff is 𝛽. Then the height of the tower is (a) h tan⍺

(b) [h tan⍺/(tan 𝛽 -tan⍺)]

(c) [h tan⍺/(tan⍺ - tan 𝛽)] (d) None of these

94. A man on the top of a tower, standing on the sea-shore, finds that a boat coming towards him takes 10 minutes for the angle of depression to change from 30° to 60°. How soon the boat

reach the sea-shore? (a) 5 minutes (b) 7 minutes (c) 10 minutes (d) 15 minutes

95. The expression of (cot𝜃 + cosec𝜃 - 1)/(cot𝜃 - cosec𝜃 + 1) is equal to (a) (1+cos𝜃)/sin𝜃

(b) (1-cos𝜃)/sin𝜃

(c) (cot𝜃 + 1)/cosec𝜃 (d) (cot𝜃 – 1)/cosec𝜃

Directions for Questions 96 to 100: The following pie-chart shows the monthly expenditure of a family on various items. If the family spends Rs.825 on clothing, answer the question

96. What is the total monthly income of the family ? (a) Rs.8025 (b) Rs.8250 (c) Rs.8520 (d) Rs.8052

97. What percent of the total income does the family save? (a) 15% (b) 50% (c) 20%

(d) 25%

Food30%

Rent25%

Savings15%

Clothing10%

Miscellaneous20%

Food Rent Savings Clothing Miscellaneous

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98. What is the ratio of expenses on food and miscellaneous?

(a) 3:4 (b) 2:3 (c) 3:2 (d) 2:5

99. What is the average of expenses on clothing and rent? (a) Rs. 1443.75 (b) Rs. 1344.57 (c) Rs. 1574.34

(d) Rs. 1734.45

100. The ratio of average of expenses on food, clothing and miscellaneous items to the average of expenses on savings and rent is (a) 3:2

(b) 1:3 (c) 2:1 (d) 1:1

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Answer keys:

1 (c) 2 (b) 3 (c) 4 (b) 5 (a) 6 (d) 7 (d) 8 (b) 9 (b) 10 (c)

11 (a) 12 (b) 13 (d) 14 (d) 15 (a) 16 (b) 17 (a) 18 (c) 19 (c) 20 (b)

21 (b) 22 (b) 23 (d) 24 (c) 25 (c) 26 (b) 27 (b) 28 (a) 29 (d) 30 (b)

31 (c) 32 (c) 33 (d) 34 (b) 35 (c) 36 (c) 37 (a) 38 (b) 39 (b) 40 (c)

41 (c) 42 (b) 43 (d) 44 (c) 45 (a) 46 (b) 47 (b) 48 (a) 49 (b) 50 (a)

51 (b) 52 (c) 53 (c) 54 (d) 55 (a) 56 (a) 57 (a) 58 (d) 59 (d) 60 (d)

61 (b) 62 (c) 63 (c) 64 (b) 65 (a) 66 (b) 67 (a) 68 (a) 69 (b) 70 (c)

71 (c) 72 (c) 73 (b) 74 (a) 75 (a) 76 (b) 77 (c) 78 (b) 79 (b) 80 (d)

81 (b) 82 (a) 83 (d) 84 (d) 85 (b) 86 (d) 87 (a) 88 (b) 89 (b) 90 (c)

91 (d) 92 (a) 93 (b) 94 (a) 95 (a) 96 (b) 97 (a) 98 (c) 99 (a) 100 (d)

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Solutions:

1. C

As 0 < x < 1, which means x is positive Now, x < 1 Multiplying both the sides by x

x2 < x ………… (i) Taking square root on both the sides in equation (i) x < √x …………….. (ii) From equation (i) and (ii)

x2 < x < √x ALTERNATE: Let x = 0.04, which lies between 0 to 1 x2 = (.04)2 = 0.0016 √x = √(0.04) = 0.2

From above, we can clearly see that x2 < x < √x and, it is true for all x belongs to (0, 1)

2. B L.C.M. of 36, 40 and 48 = 720 So, bells will ring together after every 720 seconds = 720/60 minutes = 12 minutes

3. C In the three-digit number let the unit’s place digit be z, ten’s place digit be y and hundredth’s place digit be x. Then, the three-digit number = (100x + 10y + z) Sum of the digits of three-digit number = (x + y + z) Difference = (100x + 10y + z) - (x + y + z) = 99x + 9y = 9(11x + y) From above, we can clearly see that the given difference is divisible by both 3 and 9.

4. B In 2/3, multiplying both numerator and denominator by 5 Then, 2/3 = 10/15 … (i) In 3/5, multiplying both numerator and denominator by 3 Then, 3/5 = 9/15 … (ii) Now, we can clearly see that 11/15 > 10/15 > 9/15,

which means 11/15 > 2/3 > 3/5 or we can say 3/5 < 2/3 < 11/15

5. A Let the two numbers be a and b, where a>b

a*b = 900 ……. (i) (a + b) – (a – b) = 30 2b = 30 b = 15 From equation (i)

a*15 = 900 a = 60 Hence, greater number = 60

6. D 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5 = 5/2 + 10/3 + 17/4 + 26/5 = (150 + 200 + 255 + 312)/60 = 917/60 = 15 + 17/60 To make this number a whole number, we need to add (1 – 17/60) to it = (60 – 17)/60 = 43/60

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7. D

-13824 = -24 * -24 * -24

Hence, 3 13824 = -24

8. B

Let the three positive numbers be x, y and z. x + y + z = 18 ………… (i) x*y*z = 162 …………. (ii) Also, x + y = z ……….. (iii)

From (i) and (iii) z + z = 18 2z = 18 z = 9 …………… (iv)

From (ii) and (iv) x*y = 18 Also, x + y = z = 9 From above, (x, y) = (6, 3) or (3, 6)

Sum of squares of the numbers = 32 + 62 + 92 = 9 + 36 + 81 = 126

9. B Let the three consecutive even numbers be x, x+2 and x+4.

Sum of three consecutive even numbers = (x) + (x+2) + (x+4) = 3x + 6 Average of three consecutive even numbers = (3x + 6)/3 = x + 2 Now, 3x + 6 = 28 + (x + 2) 2x = 24

x = 12 Hence, the smallest number is 12.

10. C

divisor (d) = 10*quotient(q) = 5*remainder(r) r = 46 Hence, d = 5 * 46 = 230 q = 46/2 = 23

We know that Dividend (D) = divisor (d) * quotient(q) + remainder(r) D = 230 * 23 + 46 = 5290 + 46 = 5336

11. A A man can finish the work by working alone in 30 hours Hence, work done by man in 1 hour = 1/30 Let his son can finish the work by working alone in x hours Hence, work done by his son in 1 hour = 1/x While working together, they can finish the work in 20 hours 20 * (1/30 + 1/x) = 1 1/30 + 1/x = 1/20 1/x = 1/20 – 1/30 = 10/600 = 1/60 x = 60 hours

12. B Let the volume of tub be 1 unit. Volume of tub filled by tap in 1 hour by working alone = 1/p Volume of tub emptied by sink in 1 hour by working alone = 1/q When both tap and sink are opened simultaneously for r hours, where p<q r * (1/p – 1/q) = 1 1/r = 1/p – 1/q

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13. D

Let the total work to be done be ‘w’ units Work done by John in 3 hours = w/2 Work done by John in 1 hour = w/6 Work done by Joe in 1 hour = 1/4 of the remaining work = 1/4 *(w - w/2) = 1/4 * w/2 = w/8 Work done by George in 5 hours = w – w/2 – w/8 = 3w/8

Work done by George in 1 hour = 3w/40 When all three work together, let work will get completed in x hours. x * (w/6 + w/8 + 3w/40) = w (20 + 15 + 9)/120 = 1/x

44/120 = 1/x x = 120/44 = 30/11 = 2 8/11

14. D Let the total work be 1 unit

Let B alone can finish the whole work in ‘b’ days Hence, work done by B in 1 day = 1/b Work done by A in 9 days = 2/5 Work done by A in 1 day = 2/45

When A did 2/5 of the work in 9 days, then remaining work = 1 – 2/5 = 3/5 Remaining work is completed by A and B by working together in 6 days. 6 * (2/45 + 1/b) = 3/5 2/45 + 1/b = 1/10 1/b = 1/10 – 2/45 = (45 – 20)/450 = 25/450 = 1/18 b = 18 days Hence B alone can finish the whole work in 18 days.

15. A Let the total work be 1 unit Number of days A takes to finish the work by working alone = 63/3.5 = 18 days Work done by A in 1 day = 1/18 Number of days B takes to finish the work by working alone = 75/2.5 = 30 days Work done by B in 1 day = 1/30 When both of them work together, work will be finished in x days x * (1/18 + 1/30) = 1 48/540 = 1/x x = 540/48 = 45/4 days Total cost of the work = 45/4 * (Daily wage of A + Daily wage of B) = 45/4 * (3.50 + 2.50) =

45/4 * 6 = 135/2 = Rs. 67.50

16. B Let work done by a boy in 1 day be x

Then, work done by a man in 1 day = 2x 10 men can finish the work in 8 days. Hence total work = 10 * 2x * 8 = 160x Let the number of days required by 3 men and 4 boys to finish the work be ‘n’ n * [(3 * 2x) + (4 * x)] = 160x n * 10x = 160x

n = 16 days

17. A Let the cost price of the article be ‘x’

Marked price = (1 + 30/100)x = 1.3x Selling price after 10% discount = 90/100 * 1.3x = 1.17x Gain percent = (1.17x – x)/x * 100 = 17%

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18. C

Let the production cost of the article be ‘x’ Marked price = (1 + 40/100)x = 1.4x Discount = 20% Selling price = 80/100 * 1.4x = 1.12x Profit = 1.12x – x = 48

0.12x = 48 x = 400 Hence, the production cost is Rs. 400.

19. C Final cost price of the watch to the dealer = 500 + (10/100)*500 = 550 Let the marked price be M Selling price = 75/100 * M = 0.75M Profit percent = (0.75M – 550)/550 * 100 = 20

0.75M – 550 = 110 0.75M = 660 M = 880 Hence, marked price should be Rs. 880

20. B

Let the cost price of his outlay be C and advertised price be A Selling price = 80/100 * A = 0.8A Gain = selling price – cost price = 6000 0.8A – C = 6000 ……………. (i) Profit percent = (0.8A – C)/C * 100 = 25 From equation (i) 6000/C = 25/100 C = 24000 ………….. (ii) From (i) and (ii); 0.8A -24000 = 6000 0.8A = 30000 A = 37500 Hence, the advertised price is Rs. 37500

21. B A: B = 5: 4 = 45: 36 B: C = 9: 10 = 36: 40

From above, A: B: C = 45: 36: 40 Share of C = 40/(45+36+40) * 2420 = 40/121 * 2420 = 800 Hence, C gets Rs. 800

22. B

In 49 kg of blended tea, Quantity of Assam tea = 5/7 * 49 = 35 kg Quantity of Darjeeling tea = 2/7 * 49 = 14 kg

Let the quantity of Darjeeling tea that is to be added to the mixture be ‘x’ kg 35/(14 + x) = 2/1 35 = 28 + 2x 2x = 7

x = 3.5 kg

23. D Successful students = 9/11 * 132 = 108 Unsuccessful students = 132 – 108 = 24

Had 4 more students passed, then Successful students = 108 + 4 = 112 Unsuccessful students = 24 – 4 = 20 Required ratio = 112: 20 = 28: 5

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24. C

Let the number of officers and the number of soldiers before battle be 3x and 31x respectively. After battle, Number of officers = 3x – 6 Number of soldiers = 31x – 22 Ratio = (3x-6)/(31x-22) = 1/13

39x – 78 = 31x – 22 8x = 56 x = 7 Hence, the number of officers in the regiment before battle = 3x = 3*7 = 21

25. C

Let the volume of first container be 3x, second container be 4x and that of third container be 5x In first container, Amount of milk = 3x * 4/5 = 12x/5

Amount of water = 3x * 1/5 = 3x/5 In second container, Amount of milk = 4x * 3/4 = 3x Amount of water = 4x * 1/4 = x

In third container, Amount of milk = 5x * 5/7 = 25x/7 Amount of water = 5x * 2/7 = 10x/7 When contents of all these three containers are poured into a fourth container. Amount of milk in fourth container = 12x/5 + 3x + 25x/7 = (84x + 105x + 125x)/35 = 314x/35 Amount of water in fourth container = 3x/5 + x + 10x/7 = (21x + 35x + 50x)/35 = 106x/35 Ratio of milk and water in fourth container = 314x/35: 106x/35 = 314: 106 = 157: 53

26. B Let ‘x’ kg of sugar at Rs. 12 per kg and ‘y’ kg of sugar at Rs. 7 per kg should be mixed to make a mixture worth Rs. 8 per kg 12x + 7y = 8(x + y) 12x + 7y = 8x + 8y 4x = y x/y = 1/4 Required ratio = x: y = 1: 4

27. B Total number of customers in 15 movie theatres per day = 600*15 = 9000

When six theatres close down and the total theatre attendance stays the same, then Total number of customers in 9 theatres = 9000 Average daily attendance per theatre among the remaining 9 theatres = 9000/9 = 1000

28. A Total weight of A, B and C = A + B + C = 45 * 3 = 135 …………… (i) Total weight of A and B = A + B = 40 * 2 = 80 ……………………. (ii) Also, B + C = 43 * 2 = 86 …………………………. (iii) From (i) and (ii)

80 + C = 135 C = 55 ……………. (iv) From (iii) and (iv) B + 55 = 86

B = 31 kg

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29. D

Total runs scored by the player in 40 innings = 40 * 50 = 2000 Let his highest score be H and lowest score be L. Then, H – L = 172 …………. (i) Total runs scored by the player in 38 innings (excluding the innings in which he scored the highest and lowest runs) = 38 * 48 = 1824

From above, H + L = 2000 – 1824 = 176 ……………….. (ii) From (i) and (ii) H = 174 and L = 2

Hence, the highest score of the player is 174 runs

30. B Let the 7 consecutive numbers be x, (x+1), (x+2), (x+3), (x+4), (x+5) and (x+6) x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) = 7 * 20

7x + 21 = 7 * 20 x + 3 = 20 x = 17 Largest of these numbers = x+6 = 17+6 =23

31. C

Let Pankaj has Rs.x Then, Soham has Rs.1.5x Mukesh has Rs.2*1.5x =Rs.3x Now, according to the question, (x+1.5x+3x)/3 = 110 => 5.5x=330 => x=60 then, Mukesh has 3*60 = Rs.180

32. C Total daily income = 257.50 × 20 = Rs.5150 Average Daily income of a boy = Rs.x Average Daily income of a woman = Rs.(x + 10) Average Daily income of a man = Rs.(x + 20) So, 7(x+20)+11(x+10)+2x = 5150

20x = 4900 x = Rs.245 Hence, average daily income of a man = 245+20=Rs.265

33. D

let the cost price of an article=Rs. x selling price of an article=Rs. 425 profit=selling price-cost price=425-x

in 2nd case, selling price of an article=Rs. 355 loss=cost price- selling price=x-355 according to the condition,

=> 425-x=x-355 => x+x=355+425 => 2x=780 => x=Rs. 390 cost price of an article=Rs. 390

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34. B

Let the profit share of A = x Profit share of B = 1650-x According to the condition, x/3=2(1650-x)/5 => 5x=9900-6x

=> x=900 then profit of B = 1650-900=Rs.750

35. C

Let the cost price of the article = x Selling price of the article = y According to the 1st condition, 4y/100=5x/100 => x/y=4/5

required ratio = 4:5

36. C Let the original rate per dozen = P

And initially number of dozens of eggs that can be bought with this rate = Q So, PQ= 162 ---------(i) Now according to the condition, 0.75P(Q+2)=162 --------(ii) On equating (i) and (ii), 0.75PQ+6P/4 = PQ 6P=PQ Q=6 So, P = 162/6 = 27 Original rate per dozen = Rs. 27

37. A For first painting SP = Rs. 20000, profit = 25% Let CP = Rs. X So, => 25/100 = (20000-x)/x => x=Rs. 16000 For 2nd painting SP= Rs. 20000, loss = 25%

Let CP = Rs. y So, => 25/100 = (y-20000)/y => y= Rs. 80000/3

total SP = 2*20000 = Rs. 40000 total CP = 16000 + 80000/3 = Rs. 42666.67 net loss = 42666.67-40000 = Rs.2666.67 which is more than Rs. 2000.

38. B

Let the actual price of per kg of wheat = Rs. 100 Then, selling price = 110% of 100 = ₹ 110 Now, shopkeeper sell 30% less than actual weight so cost price to shopkeeper will be = 70% of 100

= (70 x 100)/100 = Rs. 70 Hence, required profit gained by shopkeeper = [(110 - 70)/70] x 100 = (40/70) x 100 = 57 1/7 %

39. B

In a day, total minutes = 24*60 = 1440 Required percentage = 30*100/1440 = 2.083 %

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40. C

Let the present earning = Rs. x So, According to the question, x*1.25*0.96*1.25*0.96*1.25 =72000 => 1.8x = 72000 => x = Rs. 40000

41. C

Let the total candidates = 100x Candidate passed in qualitative aptitude test = 73x

Candidate passed in general awareness = 70x Candidate passed in both = 64x Candidates who passed in only quantitative aptitude = 73x-64x = 9x Candidate who passed in only general awareness = 70x -64x=6x So, candidates who failed in both subject = 100x-(64x+9x+6x) = 21x

But, 21x = 6300 So, 100x = 30000 Total number of candidates = 30000

42. B Let the initial income = 100x Expenditure = 75x Saving = 100x-75x = 25x Increased income = 1.2*100x = 120x Increased expenditure = 1.1*75x = 82.5x Saving2=120x-82.5x = 37.5x Percentage increase in savings = (37.5x-25x)*100/25x=50%

43. D

Let the speed of boat in still water = x kmph Let the speed of the stream = y kmph And distance between P to R= 2D km If river is flowing from R to P then, According to the condition,

D/(x+y)+D/(x-y) = 12 --------(i) And 2D/(x-y) = 16 hr 40 min = 16(2/3) ---------(ii) Replacing the value of equation (ii) in equation (i), 2D/(x+y)+ 16(2/3) = 24

=> 2D/(x+y)=24-16(2/3) => 2D/(x+y)=22/3 hrs Time taken to cover distance R to P=2D/(x+y) = 22/3 hrs = 7 1/3 hr If river is flowing from P to R then,

According to the condition, D/(x+y)+D/(x-y) = 12 --------(i) And 2D/(x+y) = 16 hr 40 min = 16(2/3) ---------(ii) Replacing the value of equation (ii) in equation (i), 2D/(x-y)+ 16(2/3) = 24 => 2D/(x-y)=24-16(2/3) => 2D/(x-y)=22/3 hrs Time taken to cover distance R to P=2D/(x-y) = 22/3 hrs = 7 1/3 hr

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44. C

With speed of 42kmph and time 10hrs Distance covered = 42×10=420km person wants to cover the distance in 7 hrs using formula speed=distance/time Required speed= 420/7= 60km/hrs so speed of the car should be increase by 60-42=18kmph in order to cover distance in 7 hrs.

45. A

Let the speed of train = x km/hr and speed of car= y km/hr Case 1

Distance travelled by train = 240 Km Distance travelled by car = 450-240 = 210 km Time taken by train = 240/x hr Time taken by car = 210/y hr Total time taken = 240/x+210/y hr

240/x+210/y = 8(2/3) ---------------(i) Case 2 Distance travelled by train = 180 Km Distance travelled by car = 450-180 = 270 km

Time taken by train = 180/x hr Time taken by car = 270/y hr Total time taken = 180/x+270/y hr 180/x+270/y = 9 180/x+270/y = 9 --------------(ii) on solving (i) and (ii), x=60,y=45 Speed of Car =45 km/hr Speed of train = 60 km/hr

46. B Let the speed of train C is ‘c’ kmph. As both trains are travelling in same directions, so relative speed = (100-c) kmph So, (0.150+0.250)/(100-c)=2/60 0.4*30=100-c c=88 kmph

47. B

Principal = Rs.30000 Interest = Rs. 4347 Compounded amount = 30000+4347 = Rs.34347 Now,

A=P(1+r/100)n => 34317=30000(1+7/100)n => 11449/10000=(1+7/100)n => (107/100)2=(1+7/100)n => n=2 years

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48. A

Let the amount be equal in T years. SI=PRT/100 Amount = P+SI Case 1: Amount of debt meet by A,

A1 = P + PxT/100 Case 2: Amount of debt meet by B, A2 = Q + QyT/100

It is given that A1=A2 So, => P + PxT/100= Q + QyT/100 => (PxT-QyT)/100 = Q-P => T(Px-Qy)=100(Q-P)

=> T=100(Q-P)/(Px-Qy) years

49. B Amount after 2 years will be the principal for the next year.

So, A=P(1+R/100) => 2662=2420(1+R/100) => (1+R/100)=1.1 => R=10% Now let the initial sum = Z So, 2420=Z(1+10/100)2 => Z=Rs.2000

50. A Let the initial sum =P Rate of interest =R % SI=PRT/100 Case 1: 4000-P=PR*2/100 ----------(i) Case 2: 5500-P=PR*4.5/100 ----------(ii) By (i)/(ii),

(4000-P)/(5500-P)=2/4.5 => 36000-9P=22000-4P => 5P=14000 => P=2800

Substituting the value of P in equation (i), 4000-2800=4000*R*2/100 => R=300/14 => R=21 3/7 %

51. B Volume of iron used to make a hollow cylindrical tube = π(r2

2-r12)h where r1 and r2 internal and

external radius of the hollow cylinder respectively. r1=6/2=3 cm

r2 = 8/2 =4 cm h=20 cm volume = 22*(42-32)*20/7 = 440 cubic cm

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52. C

If length, breadth and height of the rectangular box are l, b and h respectively. So, It is given that, lb=12,bh=15 and lh=20 now, on multiplying these three values,

(lbh)2=12*15*20 => (lbh)2 =3600 => lbh = 60 so, volume of the box is = 60 cm3

53. C

Speed of the man = 12 kmph = 12*5/18=10/3 m/sec Man completes one round in =8 min = 8*60=480 sec Area covered in one round will be the perimeter of the park.

So, perimeter of the park = 480*10/3=1600 m Now, Let the length and breadth of the park are 3x and 2x respectively. So, Perimeter = 2(L+B)

=> 1600=2(3x+2x) => x=160 so, area of the park = L*B=3x*2x=6x2 = 6*(160)2=153600 m2

54. D Let the radius of the initial cylinder be r and height be h. New radius = r – 25r/100 = 3r/4 New height = h + 25h/100 = 5h/4 So, the percentage change in the surface area can be calculated as: => % change = [2π(3r/4)(5h/4)-2πrh]*100/2πrh => % change = (-rh/16)*100/rh => % change = -6.25 % So, surface area is decreased by 6.25%.

55. A

Given : h=1.4 and r=0.6 Volume of material removed = volume of cylinder ABCD – volume of cone APB = πr2h – πr2h/3 = 2(πr2h)/3

= (2*22*0.6*0.6*1.4)/(3*7) = 1.056 cm3

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56. A

Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 x2 – 100x + 291 = 0 (x - 97)(x - 3) = 0 x = 3. Width of the road = 3 m

57. A

Radius of each circle = side/2 = 70 cm

The area of the circles in the square makes a complete circle so, Area of four sectors = 22*70*70/7 = 15400 cm2 Area of the square = 140*140=19600 cm2 So, area of enclosed space = 19600-15400 = 4200 cm2

58. D Perimeter of the base = 8.8 m Radius of the base = 8.8/(2π) =1.4 m Curved surface area of the cylinder = 2πrh =17.6 So h = 2 m

So amount of concrete required=volume of the cylindrical pillar= πr2h=12.32 m3

59. D R= 9cm and r = 3/2 cm, h = 4 cm

Let the number of bottles required = n Volume of hemispherical bowl = n * volume of one cylindrical bottle 2πR3/3 = n * πr2h n= 2R3/(3r2h) n=2*93/(3*1.52*4)

n=54

60. D Length of water column flown in 1 min = (10*1000/60)m = 500/3 m

Volume flown per minute = (500/3)*(40/1002) = 2/3 m3 Volume flown in half an hour = (2/3)*30 = 20 m3 So, rise in water level =20/(40*80) m = (1*100/160)cm = 5/8 cm

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61. B

Square: -

Diagonal of the square = diameter of the circle = 2r Side2 + Side2 = Diagonal2

2 x Side2 = 4r2

Side = 2r Area of the square = 2r2

Regular hexagon: -

If we join all the vertices with the centre, then we will get 6 equilateral triangles. Area of Hexagon = 6 x area of triangle = 6 x 3/4 x r x r = (33 r2/2) Ratio of area of square to area of Hexagon = 2r2/(33 r2/2) = 4/33

62. C Let, r be the radius of the cylinder and l be the height. Total surface area = 2πr2 + 2πrl = 2πr (r + l)

Curved surface area = 2πrl Curved surface area = 2/3 x Total surface area 2πrl = 2/3 x 2πr (r + l) 3l = 2r + 2l l = 2r … (1)

Cylinder has the total surface area 231 sq.cm So, 2πr2 + 2πrl = 231 From equation (1) 6πr2 = 231

πr2 = 231/6 … (2) Volume of the cylinder = πr2l = 7πr2

= 7 x 231/6 = 269.5 cu.cm

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63. C

Area of the base = πr2 = 16πcm2

Radius of the base = 4 Diameter of the base = 8 cm The given frustum can be drawn as:

In the above diagram,

Triangle ABC ≈ Triangle CED So, AC = CD = 6 cm Total slant height = 12 cm Lateral surface area of ADG = rl = 48

Lateral surface area of ACF = rl = 12 Lateral surface area of frustum = 48 - 12 = 36

64. B Let, radius of first sphere is 2R and that of another sphere is R.

Surface area of first sphere = 4(2R)2

Volume of second sphere = 4/3 R3

Given; 4(2R)2 = 4/3 R3

16R2 = 4/3 R3

12 = R So, radius of first sphere = 2R = 24 units

65. A

Volume of right circular cylinder = r2h = 22 x (21/2)2 x 38/7 = 13167 cm3

Volume of cone = 1/3r2h = (22 x (7/2) 2 x 12)/(7 x 3) = 154 cm3

Volume of hemisphere on the top = 2/3 x r3 = 2/3 x (22/7) x (7/2)3 = 89.833 cm3

Total volume of the cone = 154 + 89.833 = 243.833 cm3

Number of cones to be filled = 13167/243.833 = 54

66. B The given equation is: [1 – (2xy/(x2+y2))] [(x3-y3)/(x-y) – 3xy] = [(x2+y2 - 2xy)/(x2+y2)] [(x-y)( x2+y2 + xy)/(x-y) – 3xy]

= (x-y)2/(x2+y2) (x-y) = 1/(x2+y2)

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67. A

[1/(a+b)(b+c)] = [1/(ab + ac + b2 + bc)] = 1[b(a + b + c) + ac] = 1/(0 + ac) = 1/ac … (I) Similarly,

[1/(b+c)(c+a)] = 1/ab … (II) And, [1/(c+a)(a+b)] = 1/bc … (III) So,

[1/(a+b)(b+c)] + [1/(b+c)(c+a)] + [1/(c+a)(a+b)] = 1/ab + 1/ac + 1/bc … (IV) Multiplying by abc on both sides; abc x [1/(a+b)(b+c)] + [1/(b+c)(c+a)] + [1/(c+a)(a+b)] = a + b + c = 0 [1/(a+b)(b+c)] + [1/(b+c)(c+a)] + [1/(c+a)(a+b)] = 0/abc = 0

68. A x2+y2+2x+1 = 0 (x+1) 2 + y2 = 0 … (I) So, x = -1 and y = 0

x31 + y35 = (-1)31 + (0)35

= -1 + 0 = -1

69. B (√5+1) 2 = 6 + 2√5 (√5-1) 2 = 6 - 2√5 Now, (x2+xy+y2) = [(6 + 2√5)/(6 - 2√5) + 1 + (6 - 2√5)/(6 + 2√5)] = [(6 + 2√5)2 + (6 - 2√5) 2 + (6 + 2√5)(6 - 2√5)]/[(6 - 2√5)/(6 + 2√5)]] = 128/16 Similarly, (x2-xy+y2) = [(6 + 2√5)/(6 - 2√5) - 1 + (6 - 2√5)/(6 + 2√5)] = [(6 + 2√5)2 + (6 - 2√5) 2 - (6 + 2√5)(6 - 2√5)]/[(6 - 2√5)/(6 + 2√5)]] 96/16 (x2+xy+y2)/(x2-xy+y2) = 128/96 = 4/3

70. C (x-(1/x))2 = 3 x2 + 1/x2 - 2 = 3

x2 + 1/x2 = 5 … (1) (x2 + 1/x2)3 = 125 x6 + 1/x6 + 3x4/x2 + 3x2/x4 = 125 x6 + 1/x6 + 3x2 + 3/x2 = 125

x6 + 1/x6 + 3(x2 + 1/x2) = 125 x6 + 1/x6 = 125 – 15 = 110

71. C x4+2x3+ax2+bx+9 = (x2+nx+3)2

= x4+n2x2+ 9 + 2x2nx+ 6x2 + 6nx = x4+2nx3 + (n2 + 6)x2 + 6nx + 9 … (I) Comparing both sides of the equation; We get,

2n = 2, a = (n2 + 6) and b = 6n So, n = 1, a = 7 and b = 6

72. C As, 16 and 25 are perfect squares of 4 and 5.

So, let a = 4 and x = 5. b = c = y = z = 0 (a+b+c)/(x+y+z) = (4+0+0)/(5+0+0) = 4/5

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73. B

By putting values from options; Option 2: - ((x+a2+2c2)/(b+c) + ((x+b2+2a2 )/(c+a) + ((x+c+2b2 )/(a+b) = (b2-c2)/ (b+c) + (c2-a2)/ (c+a) + (a2-b2)/ (a+b) = (b-c)(b+c)/(b+c) + (c-a)(c+a)/(c+a) + (a-b)(a+b)/(a+b)

= b – c + c – a + a – b = 0 Hence option 2 satisfies the given equation. Hence, option 2 is the correct answer.

74. A a = b + 3 taking cube from both the sides; a3 = b3 + 9b2 + 27b + 27 117 + b3 = b3 + 9b2 + 27b + 27

9b2 + 27b – 90 = 0 b2 + 3b – 10 = 0 b = -5, 2 When, b = 2;

a = 3 + 2 = 5 a + b = 7 When, b = -5; a = 3 + (-5) = -2 a + b = -2 + (-5) = -7 So, a + b = ±7

75. A a + 1/a = -2 … (I) (a + 1/a)2 = 4 a2 + 1/a2 + 2 = 4 a2 + 1/a2 = 2 … (II) (a + 1/a)3 = -8 a3 + 1/a3 + 3a + 3/a = -8 a3 + 1/a3 + 3(a + 1/a) = -8 a3 + 1/a3 + (-6) = -8 a3 + 1/a3 = -2 … (III) taking square of equation (II); (a2 + 1/a2) 2 = 2

a4 + 1/a4 + 2 = 4 a4 + 1/a4 = 2 … (IV) From above equations, it is evident that; If, a + 1/a = -2 then,

an + 1/an = -2 (for n = odd) / 2 (for n = even) So, a1000 + a-1000 = 2

76. B

For similar triangles, area of the triangles and square of length of the corresponding sides are in proportion. So, Area of ABC/Area of DEF = BC2/EF2

9/16 = (2.1)2/(EF)2

3/4 = 2.1/EF EF = 2.1 * 4/3 = 2.8 cm

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77. C

Chord AB = Radius AC = Radius BC So, in equilateral triangle ABC;

∠BAC = 600

Radius is always perpendicular to the radius. So,

∠DAC = 900

At the point of contact (i.e. A), circumference of the circle is parallel to the tangent AD. So, required angle = ∠DAB = 90 – 60 = 300

78. B

XZ(radius) is perpendicular to AB (tangent). In triangle AXZ, (AX)2 = (AZ)2 - (XZ)2 … (I)

Similarly, in triangle BXZ, (BX)2 = (BZ)2 - (XZ)2 … (II) But, AZ = BZ … [radius of the same circle] So, AX = BX … (III) AB = AX + BX = 2AX Similarly, AC = 2AY As, 2 sides of both the triangles are in proportion and angles subtended by both the sides are equal. So, both the triangles are similar triangles. Therefore, AX/AB = AY/AC = XY/BC 1/2 = XY/BC XY = 1/2 BC

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79. B

Perimeter of the equilateral triangle = 24 cm

AB = 24/3 = 8 cm Centroid is the common point of three medians of the triangle. Medians are also perpendicular bisectors in the equilateral triangle. AD is perpendicular to BC and it equally divides BC. Therefore, BC = 4 cm

In triangle ABD; AD2 = 82 - 42

AD = 43 … (I) Centroid divides the median in the ratio 2: 1. So, AG = 2/3 x AD = 83/3 = 8/3 cm

80. D

Two circles with centre A and B and radius 6 and 11 respectively are drawn. Line BC drawn parallel to the common tangent PQ.

So, PQBC is a rectangle. CQ = 6 cm and AC = 5 cm In triangle ABC, BC2 = 132 - 52

BC = 12 cm

As, PQBC is a rectangle. PQ = BC = 12 cm

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81. B

As, ABC is isosceles triangle, the line passing from AO will be perpendicular to BC. So, in triangle ADC,

AD2 = (12√5)2 - 122 AD = 24 cm AO = OC = r OD = 24 – r

In triangle ODC, OC2 = (OD)2 + 122

r2 = (12)2 + (24 - r)2

r2 = 144 + r2 + 576 – 48r 48r = 720 R = 720/48 = 15 cm

82. A

Tangents drawn from the same point are equal in the length which are denoted by X, Y and Z in the diagram. Given that;

AB = AC X + Y = X + Z Y = Z … (I) So, BP = PC

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83. D

D and E are midpoints of AB and AC respectively. AD/AB = AE/AC = 1/2 … (I) In triangle ADE and ABC; 2 sides are in proportion and angle included by both the sides is equal. Therefore, ADE and ABC are similar triangles.

Let, AP is perpendicular to DE and BC. So, in similar triangle; AO/AP = AE/AC = DE/BC = 1/2 Area of ABC = ½ x BC x AP … (I)

Area of ADE = ½ x DE x AO = ½ x BC/2 x AP/2 = 1/8 x BC x AP … (II) Area of DECB = Area of ABC - Area of ADE = 3/8 x BC x AP … (III) Required ratio = (1/8 x BC x AP)/(3/8 x BC x AP) = 1/3

84. D

We have; AB = AC = 17 cm BC = 6 cm OR will divide BC in 2 equal parts.

BR = RC = 3 cm Also, BP = BR = 3 cm and RC = CQ = 3 cm … (tangents to a circle from a common point are equal in magnitude) So, AP = AQ = 14 cm

In triangle ARB; AR2 = AB2 – BR2

= 289 – 9 = 280 AR = 280 cm … (I) Let, radius of the circle = r cm

Now, In triangle APO; AP2 + PO2 = AO2 (Since, OP is perpendicular to AB) 196 + r2 = (280 - r)2

196 + r2 = 280 + r2 – 2r280

r = 42/280

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85. B

The diagram can be drawn as follows:

Where, AA1 || BB1 || CC1

As we know, all lines intersecting 3 parallel lines will gets divided into the same ratio. Let, AC, AC1 and A1C gets divided into the ratio x : y. So, AB/BC1 = AB1/B1C = x/y AB/AC1 = AB1/AC = x/(x+y)

In triangle ABB1 and AC1C; 2 sides are in the same proportion and angle included by both the sides is equal. Therefore, both the triangles are similar triangles. AB/AC1 = AB1/AC = BB1/C1C = x/(x+y) … (I) So, BB1/C1C = x/(x+y) … (II) Similarly, If we consider the triangles CBB1 and CA1A; We get, BB1/A1A = y/(x + y) … (III) Adding equation (II) and (III); BB1 (1/AA1 + 1/CC1) = x/(x+y) + y/(x+y) [(1/AA1) + (1/CC1)] = 1/BB1

86. D

(1+sec22°+cot68°)(1-cosec22°+tan68°) = (1 + sec22 + tan22)(1 – cosec22 + cot22) = (cos22 + 1 + sin22)/cos22 x (sin22 – 1 + cos22)/sin22 = ((Cos22 + sin22)2 – 1)/sin22.cos22 = (Cos222 + sin222 + 2sin22.cos22 -1)/sin22.cos22 = (1 + 2sin22.cos22 -1)/ sin22.cos22 = 2

87. A Xsin3𝜃 + Ycos3𝜃 = sin 𝜃cos𝜃 … (I) Xsin𝜃 - Ycos𝜃 = 0 … (II)

So, X = Y cot𝜃 Putting this values in equation (I); Xsin3𝜃 + Ycos3𝜃 = Ycos𝜃sin2𝜃 + Ycos3𝜃 = Ycos𝜃(sin2𝜃 + cos2𝜃) Y = sin𝜃 … (III) Similarly, putting Y = X tan𝜃;

We will get, X = cos𝜃 … (IV) x2+y2 = (sin𝜃)2+(cos𝜃)2 = 1

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88. B

Sec𝜃 + tan𝜃 = m Sec𝜃 = m – tan𝜃 Sec2𝜃 = m2 + tan2𝜃 – 2mtan𝜃

Sec2𝜃 – tan2𝜃 = m2 – 2mtan𝜃 1-m2 = -2mtan𝜃 (m2 -1)/2m = tan𝜃

So, AC2 = (2m)2 + (m2 -1)2

= m4 -2m2 + 1 + 4m2

= m4 + 2m2 + 1 = (m2 + 1) 2

AC = m2 + 1 Sin𝜃 = BC/AC = (m2 - 1)/(m2 + 1)

89. B (a2-b2)sin𝜃 + 2abcos 𝜃 = a2 + b2

Let, a = kcosx and b = ksinx. So, a2 + b2 = k2

k2 (cos2x – sin2x)sin𝜃 + 2k2cosx.sin𝜃 = k2

cos2x.sin𝜃 + sin2x.cos𝜃 = 1 sin (2x+𝜃) = 1 So, it is clear that; 𝜃 = /2 - 2x So, tan𝜃 = cot2x = (1 + tan2x) /2.tanx

= (a2 + b2)/2ab

90. C

Let, angle ABE = 450 and angle ABD = 600

So, angle EBC = 450 and angle DBC = 300

In triangle EBC; EC = BC = x In triangle BDC; DC = x/3

Distance covered in 10 minutes = EC – DC = x (1-1/3) = x (3-1)/ 3 Time required to travel remaining distance (t) is given by; ED/DC = 10/t [x(3-1)/ 3]/[ x/3] = 10/t

t = 10/(3-1) = 10/0.732 = 13.66 = 13 minutes 40 seconds

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91. D

2ycos𝜃 = xsin𝜃 2ycosec𝜃 – xsec𝜃 = 0 4ycosec𝜃 – 2xsec𝜃 = 0 … (I)

2xsec𝜃 – ycosec𝜃 = 3 … (II) Adding equation, I and II; 3ycosec𝜃 = 3 y = sin𝜃 … (III) Putting value of y in equation I; x = 2 cos𝜃 … (IV)

x2+4y2 = 4cos2𝜃 + 4sin2𝜃 = 4 (cos2𝜃 + sin2𝜃) = 4

92. A

In above figure; dbz = 450 dby = 600 in triangle bmz ; bm = zm = l … (I)

In triangle bay; ba = √3/2 x by by = 200/√3 ay = ½ x by

ay = 100/√3 … (II) Height of pole = yz = ba – bm = ba – zm = ba = ay = 100 - 100/√3 = (100/3)(3-√3) metre

93. B

Given, DC = h Tan⍺ = BC/AB Tan 𝛽 = (h +BC)/AB Tan 𝛽 - tan⍺ = h/AB … (I) tan⍺ = BC/AB h.tan⍺ = h.BC/AB … (II)

Divinding equation (II) by (I); [h tan⍺/(tan 𝛽 -tan⍺)] = BC

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94. A

Angle EDC = 300 and angle EDA = 600

In triangle BCD, Angle CDB = 600

Tan60 = 3 BC = BD. 3

In triangle BAD, Angle ADB = 300

Tan60 = 1/3 AB = BD.(1/3) Distance covered in 10 minutes; AC = BC – AB = BD (3 – 1/3) = 2/3. BD Let time taken to reach the shore is ‘t’.

AC/AB = 10/t (2/3)/(1/3) = 10/t t = 5 minutes

95. A

(cot𝜃 + cosec𝜃 - 1)/(cot𝜃 - cosec𝜃 + 1) = [(cotA + cosecA) – (cosec2A – cot2A)]/(cotA – cosecA + 1) = cotA + cosecA = (1 + cosA)/sinA

96. B Monthly expenditure on clothing = 10% of the income = Rs.825 Total monthly income = 100% = Rs.8250

97. A From given pie chart, it is clearly apparent that 15% of total income family saves.

98. C

Expenditure on food = 30% = Rs.2475 Expenditure on miscellaneous = 20% = Rs.1650 Required ratio = 2475/1650 = 3/2

99. A

Expenditure on clothing = 10% = Rs.825 Expenditure on rent = 25% = Rs.2062.5 Average expenditure = 2887.5/2 = Rs.1443.75

100. D

Expenditure on food = 30% = Rs.2475 Expenditure on miscellaneous = 20% = Rs.1650 Expenditure on clothing = 10% = Rs.825 Average expenses = (2475+1650+825)/3 = Rs.1650

Expenditure on rent = 25% = Rs.2062.5 Expenditure on saving = 15% = Rs.1237.5 Average expenses = (2062.5+1237.5)/2 = Rs.1650 Require ratio = 1650/1650 = 1: 1

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