A person travels from X to Y a speed of 40 kmph and returns by increasing his speed by 50%. What is his average speed for both the trips ?

examrobotsa's picture
Q: 150 (IAS/2001)
A person travels from X to Y a speed of 40 kmph and returns by increasing his speed by 50%. What is his average speed for both the trips ?

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,5,7,1,3,5,3

keywords: 

{'average speed': [0, 2, 4, 2], 'speed': [0, 1, 2, 0]}

To find the average speed for both trips, we can use the formula:

Average speed = Total distance / Total time

Let`s assume the distance between X and Y is D kilometers.

For the first trip from X to Y at a speed of 40 kmph, the time taken can be calculated as:

Time for the first trip = D / 40

For the return trip, the speed is increased by 50%, which means the new speed is 40 + (50% of 40) = 40 + 20 = 60 kmph.

The time taken for the return trip can be calculated as:

Time for the return trip = D / 60

The total time for both trips is the sum of the individual times:

Total time = Time for the first trip + Time for the return trip

Total time = D / 40 + D / 60

Now, let`s calculate the average speed:

Average speed = Total distance / Total time

Average speed = 2D / (D / 40 + D / 60)

Simplifying the expression:

Average speed = 2D / [(3D + 2D) / (120)]

Average speed = 2D / (5D / 120)

Average speed = 240D / 5D

Average speed = 48 kmph

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