Question

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

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

Answer 1

The correct option is D

$\frac{40}{13}$

Step 1:Given data:

Distance covered by the boy while walking towards his school = 6 km

Speed of boy while walking towards the school = 2.5 km/h

Speed of boy while coming back from the school = 4 km/h

Step 2:Finding the time:

Time taken while walking towards the school = $\frac{\mathrm{Distance}}{\mathrm{Speed}}=\frac{6}{2.5}\mathrm{hours}$

Time taken while coming back from the school = $\frac{\mathrm{Distance}}{\mathrm{Speed}}=\frac{6}{4}=\frac{3}{2}\mathrm{hours}$

Total time taken = $\frac{6}{2.5}+\frac{3}{2}=\frac{12+7.5}{5}=\frac{19.5}{5}\mathrm{hours}$

Multiply time with 2 to make the calculation easier,

$\frac{19.5}{5}\times 2=\frac{39}{10}$

Total distance travelled = 6 + 6 = 12 km.

Step 3:Finding the average speed:

$\mathrm{Average}\mathrm{speed}=\frac{\mathrm{Total}\mathrm{distance}}{\mathrm{Total}\mathrm{time}}$

$=\frac{12}{{\displaystyle \frac{39}{10}}}=\frac{12\times 10}{39}=\frac{120}{39}=\frac{40}{13}\mathrm{km}/\mathrm{h}$

Thus, the average speed for his round trip, expressed in km/h, is $\frac{40}{13}$.

Hence, option D is correct.