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. Find the distance covered by the train.
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,
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