Question

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

• A. $\frac{24}{13}$
• B. $\frac{40}{13}$
• C. 3
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$\frac{40}{13}$

Answer 2

Total times $t = \frac{6}{2.5} + \frac{6}{4} = \frac{12}{5} + \frac{3}{2} = \frac{39}{10} h$
Average speed, $v ={ \frac{6+6}{39/10} = \frac{120}{39} = \frac{40}{13} km \, h^{-1}}$