A boy walks to his school at a distance of 6 km with constant speed of $2.5 k m g^{- 1}$ and walks back with a constant speed of $4 k m h^{- 1}$ . His average speed for round trip expressed in $k m h^{- 1}$ , is

\frac{24}{13}

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\frac{40}{13}

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3

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\frac{1}{2}

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Solution

The correct option is B $\frac{40}{13}$
Lets consider $x = 6 k m$
A boy walks $6 k m$ with speed $2.5 k m / h$ at time $t_{1}$
$t_{1} = \frac{6 k m}{2.5 k m / h} = 2.4 h r$
A boy walk back $6 k m$ with speed $4 k m / h$ at time $t_{2}$
$t_{2} = \frac{6 k m}{4 k m / h} = 1.5 h r$
Total time $t = t_{1} + t_{2}$
$t = 3.9 h r$
Average speed, $v_{a v g} = \frac{T o t a l t i m e}{T o t a l t i m e}$
$v_{a v g} = \frac{2 \times 6}{3.9}$
$v_{a v g} = \frac{40}{13} k m / h r$
The correct option is B.

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