The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10 km/hr more than the speed of going, what was the speed per hour in each direction?

Given:

The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey.

He returned at a speed of 10 km/hr more than the speed of going.

To do:

We have to find the speed per hour in each direction.

 Solution:

Let the speed of the person while going be $x$ km/hr.

This implies,

Speed of the person while returning$=x+10$ km/hr.

Time taken by the person to travel 150 km while going$=\frac{150}{x}$ hours

Time taken by the person to travel 150 km while returning$=\frac{150}{x+10}$ hours

According to the question,

$\frac{150}{x}-\frac{150}{x+10}=2.5$

$\frac{150(x+10)-150(x)}{(x)(x+10)}=2.5$

$\frac{150(x+10-x)}{x^2+10x}=2.5$

$150(10)=2.5(x^2+10x)$   (On cross multiplication)

$600=x^2+10x$

$x^2+10x-600=0$

Solving for $x$ by factorization method, we get,

$x^2+30x-20x-600=0$

$x(x+30)-20(x+30)=0$

$(x+30)(x-20)=0$

$x+30=0$ or $x-20=0$

$x=-30$ or $x=20$

Speed cannot be negative. Therefore, the value of $x$ is $20$ km/hr.

$x+10=20+10=30$ km/hr

The speed of the person while going is $20$ km/hr and the speed while returning is $30$ km/hr. 

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Updated on: 10-Oct-2022

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