The average of 21 numbers is 74. The average of the first 11 numbers is 72.8 and that of the last 11 numbers is 77.2. If the 10th number is five less than the 11th number and if 10th and 11th numbers are excluded, then what is the average of the remaining numbers? (correct to nearest integer)

Correct option is D

Given:

Total numbers = 21

Overall average = 74 

First 11 numbers average = 72.8 

Last 11 numbers average = 77.2 

10th number = 11th number – 5

Solution: 

Total sum = 21 × 74 = 1554

Sum of fist 11 number = 11× 72.8 = 800.8 

Sum of last 11 number = 11 × 77.2 = 849.2

Let the 11th number be x, then the 10th number is x - 5

Sum of numbers 1 – 10 = 800.8 - x

Sum of numbers 12–21 = 849.2 - x

Total sum:

(800.8 - x) + x + (849.2 - x) = 1554

1650 - x = 1554

x = 96

10th number = 96 - 5 = 91

Sum of the remaining 19 numbers = 1554 - 96 - 91 = 1367

Average =$$\frac{1367}{19} \approx 71.947$$​ → 72

The average of the remaining numbers is 72