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)

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