**Average Questions for SSC CGL With Solutions :**

Average Questions and answers for SSC CGL with solutions download PDF of important asked questions in previous papers. Important maths averages problems

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**Question 1:Â **The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is equal to

a) 32%

b) 34%

c) 42%

d) 44%

**Question 2:** Three utensils contain equal quantity of mixtures of milk and water in the ratio 6:1, 5: 2 and 3: 1 respectively. If all the solutions are mixed together, the ratio of milk and water in the final mixture Is

a) 65 : 28

b) 65 : 19

c) 19 : 65

d) 19 : 28

**Question 3:**Â The average of 6 consecutive natural numbers is K. If the next two natural numbers are also included, how much more than K will the average of these 8 numbers be?

a) 3

b) 1

c) 2

d) 1.8

**Question 4:Â **The incomes of A and B are in the ratio 3 : 2 and their expenditures are in the ratio 5:3. If each saves Rs. 1000, then Aâ€™s income is

a) Rs. 6000

b) Rs. 4000

c) Rs. 2000

d) Rs. 5000

**Question 5:**The mean of $1^{3} , 2^{3} , 3^{3} , 4^{3} , 5^{3} , 6^{3} , 7^{3}$ is

a) 20

b) 112

c) 56

d) 28

**Question 6:Â **What is the Arithmetic mean of the first â€˜nâ€™ natural numbers ?

a)Â $\frac{n+1}{3}$

b)$\frac{n+1}{2}$

c)$\frac{n+1}{4}$

d)Â Â 2(n + 1)

**Question 7:Â **Eighteen years ago, the ratio of Aâ€™s age to Bâ€™s age was 8:13 Their present ratios are 5 :7 What is the present age of A ?

**a) 60 years
b) 70 years
c) 50 years
d) 40 years**

**Question 8:Â **729 ml of a mixture contains milk and water in the ratio 7 : 2. How much more water is to be added to get a new mixture containing milk and water in the ratio 7:3 ?

**a) 81 ml
b) 60 ml
c) 71 ml
d) 52 ml**

**Question 9:Â **Out of four numbers the average of the first three is 16 and that of the last three is 15. If the last number is 20 then the first number is

**a) 23
b) 25
c) 28
d) 21**

**Question 10:Â **A man spends 75% of his income. His income is increased by 20% and he increased his expenditure by 10%. His savings are increased by

**a) $37\frac{1}{2}$ %
b) 50%
c) 25%
d) 10%**

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**Answers & Solutions:**

**1) Answer (D)**

let the length of rectangle be L units and breadth be B units.

So area of rectangle = L x B

As both length and breadth has been increased by 20%

So new length = 1.2L

New breadth = 1.2B

New area = 1.2L x 1.2B = 1.44LB

Percentage increase in area = $\frac{1.44LB – LB}{LB}$Ã—100 = 44%

**2) Answer (B)**

Let each vessel contain 28 litres of mixture

=> Total quantity of mixture in 3 containers = 3*28 = 84 litres

=> Total quantity of milk = $(\frac{6}{7} + \frac{5}{7} + \frac{3}{4})$ * 28

= $\frac{24 + 20 + 21}{28}$ * 28 = 65 litres

Now, total quantity of water = 84-65 = 19 litres

=> Required ratio = 65 : 19

**3) Answer (B)**

Let the 6 consecutive numbers be a-3,a-2,a-1,a,a+1,a+2

average = $\frac{SumofElements}{NumberofElements}$

It is given that average of 6 consecutive numbers be k and hence

k = $\frac{a-3+a-2+a-1+a+a+1+a+2}{6}$ = $\frac{6a-3}{6}$ = a – $\frac{1}{2}$

now next two numbers (a+3, a+4) are also added

Sum of 8 numbers = a-3+a-2+a-1+a+a+1+a+2+a+3+a+4 = 8a +4

average of 8 numbers = $\frac{8a+4}{8}$ =Â a + $\frac{1}{2}$Â = k + 1

so average of 8 numbers is more than average of 6 numbers by = k+1 – K = 1

**4) Answer (A)**

As incomes of A and B are in the ratio 3 : 2. So , assume incomes of A and B be 3z and 2z respectively

As the expenditure are in the ratio 5:3. So assume expenditure of A and B be 5y and 3y respectively

Now it is given that every one saves Rs 1000

3z – 5y = 1000 ……….(1)

2z – 3y = 1000……….(2)

Solving equations 1 and 2

y = Rs 1000

z = Rs 2000

hence income of A = 3z = 3 x 2000 = Rs 6000

**5) Answer (B)**

We know that sum of cubes of first n natural numbers is = $(\frac{n(n+1)}{2})^2$

and we are provided with $1^{3} , 2^{3} , 3^{3} , 4^{3} , 5^{3} , 6^{3} , 7^{3}$

number of elements = 7

Sum of the given elements = $(\frac{7(7+1)}{2})^2$Â = 784

Mean = $\frac{SumofElements}{NumberofElements}$ = $\frac{784}{7}$ = 112

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**6) Answer (B)**

Arithmetic Mean = $\frac{Sumof Elements}{NumberofElements}$

Sum of first n natural numbers = $\frac{n(n+1)}{2}$

Arithmetic Mean of first n natural numbers = $\frac{n(n+1)}{2\times n}$ = $\frac{n+1}{2}$

**7) Answer (C)**

let the ages of A and B 18 years ago be 8x and 13x

So there present ages will be

A= 8x +18

B = 13x + 18

It is given that ,

$\frac{8x+18}{13x + 18}$ = $\frac{5}{7}$

x = 4

So present age of A = 8x + 18 = 32 +18 = 50 years

**8) Answer (A)**

As it is mentioned that in a 729 ml mixture of milk and water ,the ratio of Milk : water = 7:2

So amount of milk = $\frac{7}{9}$ x 729 ml = 567 ml

amount of water = $\frac{2}{9}$ x 729 ml = 162 ml

Let the amount of water added be y ml

So it is given that :

=> $\frac{567}{162+y}$ = $\frac{7}{3}$

=> $162 + y = 243$

=> $y$ = 243-162 = 81 ml

**9) Answer (A)**

let the four numbers be a,b,c,d

Using formula average = $\frac{SumofElements}{NumberofElements}$

Now, the average of first 3 numbers = 16

=> Sum of first 3 numbers=a+b+c = 16 Ã— 3 = 48………. (1)

Also, average of last three numbers = 15

=> Sum of last three numbers = b+c+d = 15 Ã— 3 = 45……(2)

Subtracting equation (2) from (1)

=> a – d = 3

Since, d = 20

=> a – 20 = 3

=> a = 23

**10) Answer (B)**

let the income be Rs 100

As it is given that he spends 75% of his income so his spending is = Rs 75

Saving =100-75=Rs25

Now his income is increased by 20% .So his new income is = 1.2×100 = 120 Rs

His expenditure increased by 10 % .So his new expenditure is = 1.1x 75 = 82.5 Rs

New savings= 120 – 82.5 = Rs 37.5

Increment in savings = $\frac{37.5-25}{25}$ * 100 = 50% .

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