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 average number of chocolates?

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Solution

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.

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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 average number of chocolates?

The correct option is C CThe average number of chocolates is 6+7+8+104 =314=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.

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.