Question

a person travels from p to q at a speed of 40 kmph and returns by increasing his speed by 50%. what is his average speed for both the trips ?

Answer 1

total distance = 2pq
speed = distance/time ➡ time = distance/speed.
time taken for pq while going is pq/40.
New speed = (40+50%×40) = (40+20) = (60).
time taken for q to p is pq/60.
Average speed = total distance/total time.
= 2pq/( pq/40 + pq/60)
= 48 kmph

Answer 2

Given:

Speed of a person= 40kmph

Increase in speed=50%

To find:

The average speed for both the trips

Solution:

We can find the solution in a few simple steps-

We know that the average speed can be calculated by dividing the total distance by the total time taken.

Average speed= Total distance traveled/ Total time taken

Let the distance from p to q be D km.

So, the total distance= D+ D= 2D

Original speed=40kmph

Increased speed=1.5×40= 60kmph

The total time taken= Total distance/Total speed

=D/40+D/60

=(60D+40D)/40×60

Total time=100D/40×60

Average speed=2D÷100D/40×60

= 2D×40×60/100D

=40×60/50

=48kmph

Therefore, the average speed for both trips is 48kmph.