A man walks on a straight road form his home to a market 2.5 km away with speed of 5 `(km)/(hr)`. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 `(km)/(hr)`. The average speed of the man over the intervel of time 0 to 40 min is equal to

Options

A. `5 kmh^(-1)`
B. `(25)/(4)kmh^(-1)`
C. `(30)/(4)kmh^(-1)`
D. `(45)/(8)kmh^(-1)`

Answer

C

Explanation

Man walks from his home to market with a speed of `5 kmh^(-1)`. Distance `= 2.5 km` and time `=(d)/(upsilon)=(2.5)/(5)=(1)/(2)` and he retuens back with speed of `7.5 kmh^(-1)` in rest a time of 20 min.
Distance `=7.5xx(20)/(60)=2.5 km`
So, Average speed `=(Total distance")/("Total time")=((2.5+2.5))/((50//60)h)=(30)/(4)km//h`