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A man walks on a straight road from his ...

A man walks on a straight road from his home to a market `2.5 km` away a speed of `5 kmh^(-1)`. Finnding the market closed , he instantly turns and walks back home with a speed of `7.5 kmh^(-1)`. The average speed of the man over the interval of time 0 to 50 min is equal to

A

`5(km)/(hr)`

B

`(25)/(4)(km)/(hr)`

C

`(30)/(4)(km)/(hr)`

D

`(45)/(8)(km)/(hr)`

Text Solution

Verified by Experts

The correct Answer is:
D
A man walks from his home to market with a speed of 5 `(km)/(hr)`.
Distance`=2.5` km and time`=(d)/(v)=(2.5)/(5)=(1)/(2)`hr
and he returns back with speed of 7.5 `(km)/(h)` in rest of time of 10 minutes.
distance`=(total dista nce)/(total time)`
`=((2.5+1.25))/((40)/(60)hr)=(45)/(8)(km)/(hr)`.
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