Sometimes Quantitative aptitude section of SSC CHSL syllabus is referred as maths.

**SSC CHSL Maths Questions and Answers PDF:**

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**Question 1:**

The average of the first 100 positive integers is

a) 100

b) 51

c) 50.5

d) 49.5

**Question 2:**

Two natural numbers are in the ratio 3:5 and their product is 2160. The smaller of the numbers is

a) 36

b) 24

c) 18

d) 12

**Question 3:**

By selling an article, a man makes a profit of 25% of its selling price. His profit per cent is

a) 20

b) 25

c) 30

d) 33.33

Question 4:

The ninth term of the sequence 0, 3, 8, 15, 24, 35, is

a) 63

b) 70

c) 80

d) 99

**Question 5:**

If the ratio of cost price and selling price of an article be as 10:11, the percentage of profit is

a) 8

b) 10

c) 11

d) 15

Question 6:

By what least number should 675 be multiplied so as to obtain a perfect cube number ?

a) 3

b) 5

c) 24

d) 40

**Question 7:
**

The number 0. 121212…. in the P form p/q is equal to

a) 4/11

b) 2/11

c) 4/33

d) 2/33

**Question 8:
**

The square root of 0.09 is

a) 0.30

b) 0.03

c) 0.81

d) 0.081

**Question 9:
**

The H.C.F. and L.C.M. of, two numbers are 8 and 48 respectively. If one of the numbers is 24, then the other number is

a) 48

b) 36

c) 24

d) 16

Question 10:

The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35, is

a) 1120

b) 4714

c) 5200

d) 5600

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**Solutions (1 to 10)**

1) Answer (c)

summation of first n natural numbers is $$\frac{n\times(n+1)}{2}$$

now putting n = 100

summation will be 5050

hence average will be 50.5

2) Answer (a)

Given:

Ratio of two numbers = 3:5

Hence we can say first number is $$3x$$ and second number is $$5x$$

Product of two numbers is = 2160

or $$15x^{2}$$ = 2160

or $$x$$ = 12

hence numbers will be 30 and 60

3) Answer (d)

Given Profit on selling price is 25%

Suppose selling price is y

hence profit will be $$\frac{y}{4}$$ and cost price will be $$\frac{3y}{4}$$

Now profit percentage on cost price will be $$\frac{\frac{y}{4}}{\frac{3y}{4}}\times100$$

i.e. $$\frac{100}{3}$$ = 33.33

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4) Answer (c)

Given: Sequence of numbers 0,3,8,15,24,35

To know:

9th term of the sequence

As we can see difference of sequence is in arthmatic progression of 3,5,7,9,11,13 and so on. 6th term of sequence is 35

and adding common difference for it (that is 13) , hence 7th term will be 48

accordingly 9th term will be 80

5) Answer (b)

Let the cost price be 10*x* and selling price be 11*x*

Percentage Profit = $$\frac{11x – 10x}{10x}$$ x 100 = 10

6) Answer (b)

Factorising 675 we get

675 = 5 $$\times$$ 5 $$\times$$ 3 $$\times$$ 3 $$\times$$ 3

There in order to make 675 a perfect cube, it has to be multiplied by 5.

7) Answer (c)

Let

*x* = 0.121212 be equation 1

100*x* = 12.121212 be equation 2

Subtracting 1 from 2

99*x = *12

*x = *$$\frac{12}{99}$$

= $$\frac{4}{33}$$

8) Answer (a)

0.09 = $$\frac{9}{100}$$

$$\sqrt{\frac{9}{100}} = \frac{3}{10}$$ = 0.3

9) Answer (d)

Given:-

Numbers- First = 24

Second = x (suppose)

H.C.F. of numbers = 8

L.C.M. of numbers = 48

As we know:

H.C.F.* L.C.M. = Product of numbers

Hence

48*8 = 24*x

x = 16

10) Answer (b)

Given: Numbers- First = 5834

Second = x (Suppose)

And number (5834 – x) is divisible by each of 20,28,32,35

Let’s say it is y

Hence 5834 – x = y

or x = 5834 – y

Now for x to be greatest y should be least

hence y should be least common multiple of 20,28,32,35

y = 1120

now x = 5834 – 1120

x = 4714

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