Question

There are four children A, B, C and D. A has 6 chocolates, B has 7 chocolates, C has 8 chocolates and D has 10 chocolates. Which of these children has the number of chocolates closest to the mean?

• A. A
• B. B
• C. C
• D. D

Answer 1

The correct option is C

C

The average number of chocolates is

$\frac{6+7+8+10}{4}$
$=\frac{31}{4}=7.75$

Now the closest integral number to 7.75 is 8. Thus, we can say that the child with 8 chocolates has the average number of chocolates, which would be C.

Note that the actual average is 7.75.