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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 kmh^(-1)`

B

`(25)/(4)kmh^(-1)`

C

`(30)/(4)kmh^(-1)`

D

`(45)/(8)kmh^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
Man walks from his home to market with a speed of `5 kmh^(-1)`. Distance `= 2.5 km` and time `=(2.5)/(5)=(1)/(2)h=30` min and he returns back with speed `7.5 km//h` in rest of time of 10 min.
Distance `=7.5xx(10)/(60)=1.25 km`
So, Average speed `=("Total distance")/("Total time")`
`=((2.5+1.25)km)/((40//60)h)=(45)/(8)kmh^(-1)`
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