SSC CGL Average Questions with Detailed Solutions

​$$\textbf{Given:} \\ \space \, \\ Avg_{even} = Avg_{odd} + 4 \\ \space \, \\ Sum_{even} = Sum_{odd} + 24 \\ \space \, \\ \textbf{Solution:} \\ \space \, \\ \text{Let } Avg_{odd} = x. \\ \space \, \\ Sum_{odd} = 5x. \\ \space \, \\ Avg_{even} = x + 4\\ \ \\ \implies Sum_{even} = 6(x + 4) = 6x + 24. \\ \space \, \\ \text{Difference in Sums:} \\ \space \, \\ (6x + 24) - 5x = 24 \\ \space \, \\ x + 24 = 24 \implies x = 0. \\ \space \, \\ \textbf{Final Answer} \\ \space \, \\ 0$$​​

Given:

Average of 15 numbers = 72

Average of first 6 numbers = 65

Average of next 5 numbers is 20% higher than the first 6

12th number = 8 less than the 15th

13th number = 5 greater than the 15th

14th number = 78

Formula Used:

Sum = Average × Number of terms

Required average = (Sum of required terms) / (Number of terms)

Solution:

Total sum of 15 numbers

= 15 × 72 = 1080

Sum of first 6 numbers

= 6 × 65 = 390

Average of next 5 numbers

= 65 + 20% of 65 = 65 × 1.2 = 78

=> Sum of next 5 numbers = 5 × 78 = 390

Sum of first 11 numbers

= 390 + 390 = 780

Sum of last 4 numbers

= 1080 − 780 = 300

Let 15th number = x

Then

12th = x − 8

13th = x + 5

14th = 78

Sum of last 4:

(x − 8) + (x + 5) + 78 + x = 300

=> 3x + 75 = 300

=> x = 75

Hence,

12th number = 75 − 8 = 67

13th number = 75 + 5 = 80

Required average = (67 + 80) / 2 = 73.5  

Exam Hall Method: 

Given:

Average of 5 consecutive odd numbers is 63

Formula Used:

Average = Sum of observations /Number of observations

Solution:

Let the consecutive odd numbers be$$x ,x+2,x+4,x+6,x+8$$​ respectively

Then total sum of numbers = $$x+x+2+x+4+x+6+x+8= 5x +20$$​​

Average of numbers = 63

Total of numbers = $$63 \times 5 = 315$$​​

So,

​$$5x+20 =315$$​​

​$$5x=315-20$$​​

​$$x=\frac{295}{5}$$​​

​$$x =59$$​​

Greatest number is$$x+ 8 = 59+8 = 67$$​​

Given:

First 120 odd natural numbers

Formula Used:

The $$n^\text{th}$$​ odd number = 2n - 1

Average of first nn odd numbers = $$\frac{\text{1st} + \text{nth}}{2}$$

$$= \frac{1 + (2n - 1)}{2} = n$$​​​

Solution:

For n = 120,

Average = 120

Thus, The average of first 120 odd natural numbers is 120.

Given:
Initial number of hens = 120
Added hens = 140
Increase in expenses = Rs. 4050
Average expenditure decreased by Rs. 3
Solution:
Let the original average expenditure per hen be Rs. x.
Then the total original expenditure = 120x.
After adding hens, new average expenditure per hen = x - 3.
New total expenditure = 260(x - 3).
Increase in expenditure = New total - Original total = Rs. 4050
260(x - 3) - 120x = 4050
260x - 780 - 120x = 4050
140x = 4830
x = 34.5
Thus, the original expenditure was 120 ×34.5 = Rs. 4140.

Given:

Total sum of five numbers = 655

Average of first two numbers = 77 

Third number = 123

Formula Used:

Sum of remaining two numbers = Total sum – (sum of first three numbers)

Average = $$\frac{\text{Sum}}{2}$$​​

Solution:

Sum of the first two numbers = 77 × 2 = 154

Sum of remaining 2 numbers = 655 − (154 + 123)

= 655 − 277 = 378

Average = $$\frac{378}{2}$$​ = 189

Given:
Average runs scored by 4 players = 61
Runs scored by 3 players = 68, 31, 27
Formula Used:
Total Runs = $$Average \times Number \text{ of players}$$​​
Runs by Fourth Player = $$Total Runs - \text{Sum of first 3 players' runs}$$​​
Solution:
Total Runs = $$61 \times 4$$​​
Total Runs = 244
Sum of Runs by 3 Players = 68 + 31 + 27 = 126
Runs by Fourth Player = 244 - 126
Runs = 118
Therefore, the fourth player scored 118 runs.

Given:

The average price of 80 mobile phones is Rs. 30,000.

After selling the highest and lowest price mobile phones, the average price of the remaining 78 mobile phones is Rs. 29,500.

The cost of the highest mobile phone is Rs. 80,000.

Formula Used:

Average = $$\frac{\text{Sum of all items}}{\text{Number of items}}$$​​

Solution:

Sum of prices of 80 mobile phones = 30,000 × 80 = Rs. 24,00,000

Sum of prices of 78 mobile phones = 29,500 × 78 = Rs. 23,01,000

The difference between the sum of prices before and after selling the two phones is Rs. 24,00,000 - Rs. 23,01,000 = Rs. 99,000

The sum of the highest and lowest priced mobile phones is Rs. 99,000.

Since the highest priced mobile is Rs. 80,000, the cost of the lowest priced mobile is:

= 99,000 - 80,000 = Rs. 19,000

Therefore, the cost of the lowest priced mobile is Rs. 19,000.

Given:

Average height of 15 boys = 160 cm

Average height of remaining 35 boys = 168 cm

Total number of boys = 50

Formula Used:

Average = $$\frac{(A_1 \times N_1) + (A_2 \times N_2)}{N_1 + N_2}$$

A₁ and A₂ are Average height 

N₁ and N₂ are number of boys

Solution:

Sum of first 15 boys = 160 × 15 = 2400 cm

Sum of remaining 35 boys = 168 × 35 = 5880 cm

Total sum = 2400 + 5880 = 8280 cm

Average height = $$\frac{8280}{50}$$​ = 165.6 cm

Given:

Average age of three persons = 56 years

Ratio of their ages = 2 : 5 : 7

Concept Used:

Average =$$\frac{\text{ (Sum of all observations) }}{\text{ (Number of observations)}}$$​​

Solution:

Let the ages of the three persons be 2x, 5x, and 7x, respectively.

The sum of their ages is 2x + 5x + 7x = 14x.

The average age of the three persons is given as 56 years. Therefore,

​$$\frac{\text{Sum of their ages}}{3} = 56$$​​

​$$\frac{14x }{ 3} = 56$$​​

14x = 168

x = $$\frac{168 }{ 14}$$​​

x = 12

Youngest person's age = 2x = 2 $$\times$$​ 12 = 24 years

The age of the youngest person is 24 years

Given:

In the first year, 75 employees appeared, and 84% passed.

In the second year, 50 employees appeared, and 52% passed.

Formula Used:

Average Pass Percentage = $$\frac{(n_1 \times p_1) + (n_2 \times p_2)}{n_1 + n_2}$$​

Solution:

n$$_1 = 75, p_1 = 84\% = 0.84$$​​

​$$n_2 = 50, p_2 = 52\% = 0.52$$​​

Average Pass Percentage = $$\frac{(75 \times 0.84) + (50 \times 0.52)}{75 + 50}$$​

$$= \frac{63 + 26}{125}\\ \ \\= \frac{89}{125} \times 100\\ \ \\= \frac{89}{5} \times 4\\ \ \\= \frac{356}{5}$$​

= 71$$\frac15$$​%

Given:

Average age of Ruby and Soni = 40 years

Ratio of their ages = 11:5

Formula Used:

The average age formula is:

​$$\text{Average} = \frac{\text{Sum of ages}}{\text{Number of people}}$$​

Solution:

Since the average age is given, the sum of their ages is:

Sum of ages = 40 × 2=80

Let Ruby's age = 11x and Soni's age = 5x.
Thus,

11x + 5x = 80

16x = 80

x = 5

Soni’s age = 5x = 5 × 5 = 25 years 

Given:
Total students in the class = 30
Average score of 12 students = 62
Average score of remaining students = 74
Formula Used:
Total Score = Number of Students × Average Score
​$$\text{Overall Average Score} = \frac{\text{Total Score of All Students}}{\text{Total Number of Students}}$$​​
Solution:
Total Score of 12 Students = 12 × 62 = 744
Total Score of Remaining 18 Students = 18 × 74 = 1332
Total Score of Class = 744 + 1332 = 2076
Overall Average Score = $$\frac{2076 }{ 30}$$​ = 69.2

Given:

Total strength of class = 50

Average marks of 20 students is 70

Average marks of remaining 30 is 80

Formula Used:

Average =$$\frac{Sum\ of\ observations}{Number\ of\ observations}$$​​

Solution:

Total marks of 20 students is 20 $$\times$$​70 = 1400

Total marks of 30 students is 30 $$\times$$​ 80 = 2400

Total marks of whole class = 1400+2400 = 3800

Average marks =$$\frac{3800}{50} =76$$​​