A fast train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed of the two trains.

Given:

A fast train takes one hour less than a slow train for a journey of 200 km.

The speed of the slow train is 10 km/hr less than that of the fast train.

To do:

We have to find the speed of the two trains.

 Solution:

Let the speed of the fast train be $x$ km/hr.

This implies,

Speed of the slow train$=(x-10)$ km/hr.

Time taken by the fast train to travel 200 km $=\frac{200}{x}$ hours

Time taken by the slow train to travel 200 km $=\frac{200}{x-10}$ hours

According to the question,

$\frac{200}{x-10}-\frac{200}{x}=1$

$\frac{200(x)-200(x-10)}{(x)(x-10)}=1$

$\frac{200x-200x+2000}{x^2-10x}=1$

$2000=1(x^2-10x)$   (On cross multiplication)

$2000=x^2-10x$

$x^2-10x-2000=0$

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

$x^2-50x+40x-2000=0$

$x(x-50)+40(x-50)=0$

$(x-50)(x+40)=0$

$x-50=0$ or $x+40=0$

$x=50$ or $x=-40$

Speed cannot be negative. Therefore, the value of $x$ is $50$ km/hr.

$x-10=50-10=40$ km/hr

The speed of the fast train is $50$ km/hr and the speed of the slow train is $40$ km/hr.

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Updated on: 10-Oct-2022

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