A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Given:

A train covered a certain distance at a uniform speed. If the train could have been 10 km/hr. faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/hr; it would have taken 3 hours more than the scheduled time.

To do:

We have to find the distance covered by the train.

Solution:

Let the original speed of the train be $x$ km/hr, the distance travelled by train be $y\ km/hr$ and the time taken be $t$.

This implies,

Time taken by the train to travel $y$ km at original speed$t=\frac{y}{x}$ hours.

$\Rightarrow y=xt$....(i)

If the train could have been 10 km/hr. faster, it would have taken 2 hours less than the scheduled time.

Time taken by the train to travel $y$ km when the speed is 10 km/hr more than the original speed$=\frac{y}{x+10}$ hours

According to the question,

$\frac{y}{x+10}=t-2$

$\frac{xt}{x+10}=t-2$     (From (i))

$xt=(t-2)(x+10)$

$xt=xt+10t-2x-20$

$10t-2x-20=0$......(ii)

If the train were slower by 10 km/hr; it would have taken 3 hours more than the scheduled time.

Time taken by the train to travel $y$ km when the speed is 10 km/hr less than the original speed$=\frac{y}{x-10}$ hours

According to the question,

$\frac{y}{x-10}=t+3$

$\frac{xt}{x-10}=t+3$        (From (i))

$xt=(t+3)(x-10)$

$xt=xt-10t+3x-30$

$3x-10t-30=0$......(iii)

Adding equations (ii) and (iii), we get,

$10t-2x-20+3x-10t-30=0$

$x-50=0$

$x=50$

Substituting the value of $x=50$ in equation (iii), we get,

$3(50)-10t-30=0$

$150-10t-30=0$

$10t=120$

$t=12$

Distance covered by the train $y=xt=50(12)=600$

The distance covered by the train is $600$ km.