Question

A man walks on a straight road from his home to a market $2.5\, km$ away with a speed of $5 \,km / h$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5 \,km / h$. The average speed of the man over the interval of time $0$ to $40\, min$. is

• A. $5\, km / h$
• B. $\frac{25}{4} \, km / h$
• C. $\frac{30}{4} \, km / h$
• D. $\frac{45}{8}\, km / h$

Answer 1

Answer: (D)

A man walks from his home to market with a speed of $5\, km / h$.
Distance $=2.5\, km$ and
time $=\frac{d}{v}=\frac{2.5}{5}=\frac{1}{2} h r$
and he returns back with speed of $7.5\, km / h$ in rest of time of $10$ minutes.
Distance $=7.5 \times \frac{10}{60}=1.25\, km $
Average speed $=\frac{\text { Total distance }}{\text { Total time }} $
$=\frac{(2.5+1.25)\, km }{(40 / 60) h r}=\frac{45}{8}\, km / hr$