Quantitative Aptitude


DIRECTIONS (Qs. 36-40): Find the missing term.
36. 0.5, 1.5, 5, 8,76,?
(a) 380
(b) 385
(c) 390
(d) 395
(e) None of these
Ans.b

0.5 × 1 + 1 = 1.5
1.5 × 2 + 2 = 5
5 × 3 + 3 = 18
18 × 4 + 4 = 76
76 × 5 + 5 = 385

37. 65, 72, 86, 114?
(a) 160
(b) 165
(c) 170
(d) 175
(e) None of these
Ans.c

65 + 7; 72 + 14; 86 + 28 ; 114+ 56= 170

38. 63, 31, 15, 7, 3 ?
(a) 0
(b) 1
(c) 2
(d) 3
(e) None of these
Ans.b

(63-1) / 2; (31-1)/2 ;(15-1)/2; (7-1)/2; (3-1)/2 =1

39. 13. 70, 71, 76, ?, 81, 86, 70, 91
(a) 70
(b) 71
(c) 80
(d) 96
(e) None of these
Ans.a

In this series,5 is added to the previous number; the number 70 is inserted as every third number.

40. 8, 43, 11, 41, ?, 39, 17
(a) 8
(b) 14
(c) 43
(d) 44
(e) None of these
Ans.b

This is a simple alternating addition and subtraction series. The first series begins with 8 and adds 3; the second begins with 43 and subtracts 2

DIRECTIONS (Qs. 41-45): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer

(a) if x > y (b)if x < y
(c) if x\geq y (d) if x\leq y

(e) if x=yor relation cannot be established between 'x' and 'y'.

41. I. 8x + y =10

II. 4x + 2y = 13

Ans.b

From both equation

x = 7/12, y= 16/3

y > x

42.I.(x + 3) (y + 2) = 12

II. 2xy + 4x + 5y = 11

Ans.e

xy + 3y + 2x + 6 = 12
2xy + 6y + 4x = 12 ...(i)
2xy + 5y + 4x = 11 ...(ii)
From equation (i) and (ii)
y = 1
From equation (i)
x = 1
x = y

43. I. (3x – 2)/y = (3x + 6)/(y + 16)

II. (x + 2)/(y + 4) = (x + 5)/(Y + 10)

Ans.b

(3x – 2)/y =(3x + 6)/(y + 16)48x – 8y = 32 ...(i)

(x + 2)/(y + 4) = (x + 5)/(y + 10)

y = 2x ...(ii)

From equation (i) & (ii)
x = 1, y = 2

y > x

44. I.x^{2}+ 20x + 4 = 50 – 25x

II.y^{2}– 10y – 24 = 0

Ans.b

From the given equation

x = 1, – ,6

& y = – 4, 6

45. I.(x^{2}– 10x + 16)/x^{2}– 12x + 24) = 2/3

II.y^{2} – y – 20 = 0

Ans.a

From 1st equation

x^{2} – 6x = 0

x = 0,6

From 2nd equation

(y + 4) (y – 5)y = – 4, 5

x > y

DIRECTIONS (Qs. 46-50) : Study the given table carefully to answer the following questions.

Field Name Shapes side (in m) base (in m) Height (in m) Radius (in m) Cost of flooring (in Rs. per sq.metre) Cost of fencing (in Rs. per m)
A Triangle 16 12 50 20
B Rectangle 10*20 30 15
C Square 15 40 18
D Parallelogram 20 12 60 25
E Circle 10 45 22

 

46. What is the cost of flooring of A?
(a)`4000
(b)`4600
(c)`4800
(d)`5000
(e)`4400
Ans.c

A is a triangle

So, area of A =1/2 × 16 × 12 = 96 sqm

So, cost of flooring of A = 96 × 50 = `4800

47. What is the difference between the cost of fencing of C and that of B?
(a)`180
(b)`120
(c)`240
(d)`360
(e)`480
Ans.a

Perimeter of B = 2 (10 + 20) = 60 m

So, cost of fencing of B = 60 × 15 = 900

Perimeter of C = 4 × 15 = 60 m

So, cost of fencing of C = 60 × 18 = 1080

So, required difference = 1080 – 900 = 180

48. What is the ratio of the cost of flooring to that of fencing of field D?
(a) 4 : 1
(b) 6 : 1
(c) 8 : 1
(d) 9 : 1
(e) 5 : 1
Ans.d

Area of D = Base × Height = 20 × 12 = 240 mtr sq

So, cost of flooring of D= 240 × 60 = 14400

Perimeter of D = 2 (20 + 12) = 64 m

So, cost of fencing of D = 64 × 25 = 1600

So, required ratio = 14400 : 1600 = 9 : 1

49. The cost of fencing of field E is approximately what percent of the cost of flooring of field C?
(a) 10.5%
(b) 19.46%
(c) 18.71%
(d) 15.36%
(e) 13.82%
Ans.d

Perimeter of E = 2pr = 2 × 22/7 × 10 = 440/7 m

Cost of fencing of E = 440/7 × 22 = 1382.85

Area of C = 15 × 15 = 225 mtr square

So, cost of flooring of C = 225 × 40 = 9000

So, required % = 1382.85 ×100 / 9000= 15.36% of flooring cost of C.

50. The cost of fencing of field C is what percent of the cost of fencing of field D?
(a) 87.54%
(b) 67.5%
(c) 72.13%
(d) 54.36%
(e) 46.5%
Ans.b

Fencing cost of C = 1080

Fencing cost of D = 1600

Required % = 1080/1600 × 100 = 67.5%

DIRECTIONS (Qs. 51-58) : Study the following table and pie chart carefully to answer the given questions.

The table shows the ratio of Hindu religion soldiers to soldiers of other religions

51. What is the number of Hindu soldiers in J at regiment?
(a) 2600
(b) 2700
(c) 3200
(d) 2800
(e) 2350
Ans.d

Number of soldiers in Jat regiment = 10000 × 35% =3500

Number of Hindu soldiers in Jat regiment = 3500 × 4/5= 2800

52. What is the difference between Hindu soldiers in Madras regiment and soldiers of other religions in Bihar regiment?
(a) 485
(b) 550
(c) 520
(d) 510
(e) 490
Ans.b

Number of Hindu soldiers in Madras regiment=10000 × 15% × 2/3= 1000

Number of soldiers of other religions in Biharregiment = 10000 × 12% × 3/8= 450

So, difference = 1000 –3/8= 450 = 550

53. The number of Hindu soldiers in Sikh regiment is what percent of the number of other soldiers in Maratha regiment?
(a) 97.12%
(b) 99.56%
(c) 102%
(d) 104.16%
(e) 25%
Ans.d

Number of Hindu soldiers in Sikh regiment = 10000× 20% × 3/8 = 750

Number of soldier so f other religions in Maratha regiment = 10000 × 18% × 2/5 = 720

54. In which regiment is the number of non-Hindu soldiers the maximum?
(a) Maratha regiment
(b) Sikh regiment
(c) Madras regiment
(d) Jat regiment
(e) Bihar regiment
Ans.a

Number of non-Hindu soldiers in Jat regiment= 3500 – 2800 = 700

Similary in Sikh regiment = 10000 × 20% × 5/8 = 125

In Madras regiment = 10000 × 15% × 1/3 = 500

In Maratha regiment = 10000 × 18% × 2/5 = 720

In Bihar regiment = 10000 × 12% × 3/8 = 450

In Maratha regiment the number of non-Hindu soldiers is the maximum.

55. What is the ratio of the number of Hindu soldiers in Bihar regiment to the number of non-Hindu soldiers in Jat regiment?
(a) 11 : 10
(b) 12 : 11
(c) 13 : 12
(d) 14 : 13
(e) 15 : 14
Ans.e

Number of Hindu soldiers in Bihar regiment = 10000

× 12% × 58= 750

Number of non-Hindu soldiers in Jat regiment = 700

So, required ratio = 750 : 700 = 15 : 14

56. If the compound interest on an amount of `29000 in two years is `9352.5, what is the rate of interest?
(a) 11
(b) 9
(c) 15
(d) 18
(e) None of these
Ans.c

P = 29000 CI = 9352.5 N = 2 years A = P + I = 38,352.50

Substituting the values in

A = P(\frac{R}{100})^{n}

Solving we get R = 15%

57. Three friends A, B and C start running around a circular stadium and complete a single round in 8, 18 and 15 seconds respectively. After how many minutes will they meet again at the starting point for the first time?
(a) 12
(b) 6
(c) 8
(d) 15
(e) 18
Ans.b

The required time will be the LCM of 8, 18 and 15 which is 360 sec or 6 minutes.

58. The perimeter of a square is equal to the radius of a circle having area 39424 sq cm, what is the area of square?
(a) 1225 sq cm
(b) 441 sq cm
(c) 784 sq cm
(d) Can't say
(e) None of these
Ans.c

r^{2}= 39424

R = 112

Perimeter of square = 4a = 112

Side of square = 112/4 = 28

Area of square = 28^{2} = 784cm^{2}

DIRECTIONS (Qs. 59-61) : Study the following information carefully to answer the questions that follow-

A committee of five members is to be formed out of 5 Males, 6 Females and 3 Children. In how many different way scan it be done if-?

59. The committee should consist of 2 Males, 2 Females and 1 Child?
(a) 450
(b) 225
(c) 55
(d) 90
(e) None of these
Ans.a

Number of ways= ^{2}{C_{5}}\times^{6}{C_{2}} \times ^{3}{C_{1}}

60. The committee should include all the 3 Childs?
(a) 90
(b) 180
(c) 21
(d) 55
(e) None of these
Ans.d

Number of ways = ^{11}{C_{2}}\times^{3}{C_{3}} = 55

61.Thirty men can complete a work in 36 days. In how many days can 18 men complete the same piece of work?
(a) 48
(b) 36
(c) 60
(d) 72
(e) None of these
Ans.c

Required number of days= (30 × 36)/18 = 60

62. Ram spends 50% of his monthly income on house hold items, 20% of his monthly income on buying clothes, 5% of his monthly income on medicines and saves remaining`11,250. What is Ram's monthly income?
(a) 38,200
(b) 34,000
(c) 41,600
(d) 45,000
(e) None of these
Ans.d

Let total income of Ram be x. Then

(100 – 50 – 20 – 5)% of x = 11250

x = 45000.

63. The number obtained by interchanging the two digits of a two digit number is lesser than the original number by 54.If the sum of the two digits of the number is 12, then what is the original number?
(a) 28
(b) 39
(c) 82
(d) Can't say
(e) None of these
Ans.e

Let the number be

xy (10x + y)– (10y + x) = 54

x – y = 6 And x + y = 12

Solving the equations we get x = 9 and y = 3

So the number is 93

64. At present Geeta is eight times her daughter's age. Eight years from now, the ratio of the ages of Geeta and her daughter will be 10 : 3 respectively.What is Geeta's presentage ?
(a) 32 years
(b) 40 years
(c) 36 years
(d) Can't say
(e) None of these
Ans.a

Let the age of Geeta's daughter be x. Then Geeta's age is 8x.

(8x + 8)/(x + 8) = 10/3

x = 4

Geeta's present age = 8x = 32 years.

65. In how many different ways can 4 boys and 3 girls be arranged in a row such that all the boys stand together and all the girls stand together ?
(a) 75
(b) 576
(c) 288
(d) 24
(e) None of these
Ans.c

Required number of ways = 4! × 3! × 2! = 288.

DIRECTIONS (Qs. 66-70) : What approximate value will come in place of question mark (?) in the following questions (You are not expected to calculate the exact value).

66. 9228.789 – 5021.832 + 1496.989 = ?
(a) 6500
(b) 6000
(c) 6300
(d) 5700
(e) 5100
Ans.d

9228.789 ~ 9230; 5021.832~5020 and 1496.989 ~1500

Now the equation will become 9230 – 5020 + 1500 = ?

? = 5710

But the nearest value is 5700.

[Note: Even rounding of the numbers to nearest hundred places gives the same

67. 1002 ÷ 49 × 99 – 1299 = ?
(a) 700
(b) 600
(c) 900
(d) 250
(e) 400
Ans.a

1002 ~1000; 49~ 50; 99~ 100 and 1299~ 1300

Now the equation will become

1000 ÷ 50 × 100 – 1300 = ?

20 × 100 – 1300 = ?

2000 – 1300 = ?

? = 700

68. 29.8% of 260 + 60.01% of 510 - 103.57 = ?
(a) 450
(b) 320
(c) 210
(d) 280
(e) 350
Ans.d

The difference between two nearest values is 70 (210 and 280). So round off the numbers to the nearest integers.

29.8% of 260 ~ 30% of 260; 60.01% of 510~ 60% of 510 and 103.57 ~104

Now the equation will become

30% of 260 + 60% of 510 – 104 = ?

30/100 × 260 + 60/100 × 510 – 104=?

78 + 306 – 104 = ?

? = 384 – 104 = 280

69. \left (21.98 \right )^{2}-\left (25.02 \right )^{2}+\left (13.03 \right )^{2}= ?
(a) 25
(b) 120
(c) 10
(d) 65
(e) 140
Ans.a

\left ( 21.98 \right )^{2}=\left ( 22 \right )^{2} \left ( 25.02 \right )^{2}=\left ( 25 \right )^{2}

and \left ( 13.03 \right )^{2}=\left ( 13 \right )^{2}

The equation will becomes

22^{2}-25^{2}+13^{2} = ?

484 – 625 + 169 = ?

653 – 625 = ?

? = 28 so the nearest value is 25

70. \sqrt{24.98}\times\sqrt{625}\times\sqrt{99} = ?
(a) 110
(b) 90
(c) 200
(d) 160
(e) 125
Ans.e

\sqrt{24.98}\times\sqrt{6.25}\times\sqrt{9.9} = ?

5 × 2.5 × 10 = 125