# Quantitative Aptitude

**DIRECTIONS (Qs. 36-40):**Find the missing term.

(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

(b) 165

(c) 170

(d) 175

(e) None of these

**Ans.c**

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

(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

(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.

(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 (d) if

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

II. 4x + 2y = 13

**Ans.b**

From both equation

x = 7/12, y= 16/3

y > x

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

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

II.– 10y – 24 = 0

**Ans.b**

From the given equation

x = 1, – ,6

& y = – 4, 6

II. – y – 20 = 0

**Ans.a**

From 1st equation

– 6x = 0x = 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 |

(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

(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

(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

(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.

(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

(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

(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

(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

(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.

(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

(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

Solving we get R = 15%

(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.

(b) 441 sq cm

(c) 784 sq cm

(d) Can't say

(e) None of these

**Ans.c**= 39424

R = 112

Perimeter of square = 4a = 112

Side of square = 112/4 = 28

Area of square = = 784

**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-?

(b) 225

(c) 55

(d) 90

(e) None of these

**Ans.a**

Number of ways=

(b) 180

(c) 21

(d) 55

(e) None of these

**Ans.d**

Number of ways =

(b) 36

(c) 60

(d) 72

(e) None of these

**Ans.c**

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

(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.

(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

(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.

(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).

(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(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

(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

(b) 120

(c) 10

(d) 65

(e) 140

**Ans.a**

and

The equation will becomes

= ?484 – 625 + 169 = ?

653 – 625 = ?

? = 28 so the nearest value is 25