﻿ SSC - CGL Quantitative Aptitude Solved Question Paper 2012 | Previous Year Papers | Zigya

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Quantitative Aptitude

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# SSCCGL Quantitative Aptitude Solved Question Paper 2012

#### Multiple Choice Questions

3.

The product of two number is 1575 and their quotient is 9/7. Then the sum of the numbers is

• 74

• 78

• 80

• 90

C.

80

Let the numbers be x and y.
∴   (x) x (y) = 1575 (given)  ...(i)
And, (given)  ...(ii)
Multiply (i) x (ii)  297 Views

7.

When (6767 + 67) is divided by 68, the remainder is

• 1

• 63

• 66

• 67

C.

66

Shortcut Method:
First see power values, i.e. 6767 divide this value by 68.
If power is even then remainder would be 1, but if power is odd then divide it in usual manner. Here 67 is less than 68 so we take 67 as remainder.
Now (Remainder  + 67) should be divided by 68
= (67 + 67) = 134 should be divided by 68, so remainder would be 66.
364 Views

2.

If 738A6A is divisible by 11, then the value of A is

• 6

• 3

• 9

• 1

C.

9

A number is exactly divisible by 11, if the difference between the sum of digits at even and odd places be either 0 or a multiple of 11.
∴  (A + A + 3) - (6 + 8 + 7) = 0
⟹   2A + 3 = 21
⟹   2A = 21 - 3 = 18 279 Views

10.

Given that the value of is

• 70.4

• 70.464

• 71.104

• 71.4

B.

70.464 ∴  Expression = 64 + 6.4 + 0.064 = 70.464
145 Views

4.

The value of is

• 3

• 6

• 9

• 8.2

C.

9   126 Views

5. is equal to
• 2

• 5

• 4

• 3

D.

3

Let On squaring both sides,   Shortcut Method:
Co-Prime factors of 6 are (3, 2) as number is positive, therefore, answer would be highest co-prime factor i.e. 3.

116 Views

8.

A can do a piece of work in 24 days, B in 52 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

• 720/29

• 20

• 18

• 30

A.

720/29

Let the work be finished in x days.
∴ Work done by A in 6 days + Work done by B in (x - 6) days  + Work done by C in x days = 1
Use formula:  240 Views

9.

The square root of: is

• 1

• 2

• 3

• 4

D.

4 ∴ Required square root 143 Views

1.

In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend, then

• • • • D. Divisor  = a,  Dividend  = b (given)
Quotient = Remainder = ∴   Dividend = Divisor X Quotient + Remainder  552 Views

6.

The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is

• 24

• 32

• 40

• 28

D.

28

Let x and x+2 be the two consecutive odd numbers.  So, the numbers are (x = 13), (x+2 = 13+2 = 15)
∴  The required sum = 13 + 15 = 28

Shortcut Method:
By mental operation,
132 + 152 = 169 + 225  = 394
∴  The required sum = 13 + 15 = 28

174 Views