The average of 41 numbers is 16. The average of the first 17 numbers is 27 and the average of the last 25 numbers is 26. Ifthe 17th number from the beginning is excluded, then what is the average of the remaining numbers?

Correct option is A

Given:

Average of 41 numbers = 16

Average of the first 17 numbers = 27

Average of the last 25 numbers = 26

Formula Used:

Sum= Average × Number of terms

Solution:

total sum of 41 numbers:

Total sum = $$16 \times 41$$​ = 656

the sum of the first 17 numbers:

Sum of first 17 numbers = $$27 \times 17$$​ = 459

the sum of the last 25 numbers:

Sum of last 25 numbers = $$26 \times 25$$​ = 650

The 17th number is included in both the first 17 numbers and the last 25 numbers.

Let the 17th number be x

Total sum = Sum of first 17 numbers + Sum of last 25 numbers - x

656 = 459 + 650 - x

656 = 1109 - x

x = 1109 - 656 = 453

Exclude the 17th number and find the new sum:

New sum = 656 - 453 = 203

New average = $$\frac{203}{40}$$​ = 5.075

The average of the remaining numbers is 5.075.