Question

A boy walks to his school at a distance of $ 6km $ with a speed of $ 2.5km{h^{ - 1}} $ and walks back with a constant speed by $ 4km{h^{ - 1}} $ . His average speed for trip expressed in $ km{h^{ - 1}} $ is
 $ \left( A \right)\dfrac{{24}}{{13}} \\ $
 $ \left( B \right)\dfrac{{40}}{{13}} \\ $
 $ \left( C \right)3 \\ $
 $ \left( D \right)4.8 \\ $

Answer 1

Hint :In order to solve this question, we are going to first find the time taken for the forward journey from the values of the distance covered and the speed of the journey. After that, the time for the back journey is taken, then, by dividing the total distance and time, average speed is calculated.
The time taken for a journey of the distance $ s $ and the speed $ v $ is given by
 $ t = \dfrac{s}{v} $ .

Complete Step By Step Answer:
Distance covered by the boy towards his school is
 $ s = 6km $
The speed of the boy is
 $ v = 2.5km{h^{ - 1}} $
Thus, the time taken for the forward journey is
 $ t = \dfrac{s}{v} $
Putting the values from as given above,
 $ t = \dfrac{6}{{2.5}} = 2.4h $
Now, for the back journey, the distance covered by the boy is the same as above
 $ s = 6km $
While the speed as given in the question
 $ v = 4km{h^{ - 1}} $
Now the time for the back journey is
 $ t' = \dfrac{6}{4} = 1.5h $
Now, the average speed for the whole trip is the ratio of the whole distance and the times for the two journeys
i.e.
 $ {t_{avg}} = \dfrac{{6 + 6}}{{2.4 + 1.5}} = \dfrac{{12}}{{3.9}} = \dfrac{{40}}{{13}}km{h^{ - 1}} $
Hence, option $ \left( B \right)\dfrac{{40}}{{13}} $ is the correct answer.

Note :
The average speed for a journey can be defined as the average of the speeds with which the boy travels for the various different distances with the different speeds. It can be calculated from the total distance covered and the total time taken for the whole journey taken by the boy.