It is given that the average of 19 numbers is 8 5 If one number with value 63 was entered incorrectl...........

Quantitative Aptitude (CHSL) Average

It is given that the average of 19 numbers is 8.5. If one number with value 63 was entered incorrectly as 25,then the corrected

average is:

This questions was previously asked in

SSC CHSL 7th June 2022 Shift-3

9.5

11.5

7.5

10.5

- The average of 19 numbers is 8.5

- So, total sum = 19 × 8.5 = 161.5

- One number was entered as 25, but it should have been 63.

- The actual sum should be: 161.5 - 25 + 63 = 199.5

- The corrected average is 199.5 ÷ 19 = 10.5

Options:

1. 9.5

2. 11.5

3. 7.5

4. 10.5

- The correct answer is option 4, 10.5

- The error was in the data entry, and correcting it increased the average

- Option 4 is highlighted as it matches the calculation

Submit Error