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A boy walks to his school at a distance of $6 \mathrm{~km}$ with constant speed of 2.5 $\mathrm{km} /$ hour and walks back with a constant speed of $4 \mathrm{~km} / \mathrm{hr}$. His average speed for round trip expressed in $\mathrm{km} /$ hour, is
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The correct answer is:
40/13
Distance average speed $=\frac{2 v_1 v_2}{v_1+v_2}=\frac{2 \times 2.5 \times 4}{2.5+4}$
$=\frac{200}{65}=\frac{40}{13} \mathrm{~km} / \mathrm{hr}$
$=\frac{200}{65}=\frac{40}{13} \mathrm{~km} / \mathrm{hr}$
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