Question

# A train covers a distance at a uniform speed . If the train could have been ${\text{10 kmph}}$ faster , it would have taken two hours less than the scheduled time. And if the train were slower by ${\text{10 kmph}}$, it would have taken ${\text{3}}$ hours more than the scheduled time. Find the distance covered by the train.

Hint – Use the formula of speed and make equations according to the question .

Let the distance covered by the train $= d\;km$
Speed $= s\;kmph$
$t{\text{ }}$be the scheduled time
Distance = Speed X Time = constant in this case
$st = (s + 10)(t - 2)\; = d{\text{ }}\;{\text{ }}\;{\text{ }}......\left( i \right) \\ st = (s - 10)(t + 3) = d\;{\text{ }}\;{\text{ }}\;{\text{ }}\;.....\left( {ii} \right) \\$
Since the distance travelled is same in all the cases
Simplifying the $\left( i \right)$,
$st = st - 2s + 10t - 20 \\ \Rightarrow 2s - 10t = - 20\;{\text{ }}\;{\text{ }}.....\left( {iii} \right) \\$
Simplifying the $\left( {ii} \right)$ we get ,
$st = st + 3s + 10t - 30\; \\ \Rightarrow 3s - 10t = 30\;{\text{ }}\;............\left( {iv} \right) \\$
Multiplying $\left( {iii} \right)$ by $3\;\& {\text{ }}\left( {iv} \right)$ by $\;2$, we get,
$6s - 30t = - 60\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}..........\left( v \right) \\ 6s - 20t = 60\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;..........{\text{ }}\left( {vi} \right) \\$
Subtracting equation $\left( {vi} \right)$ from $\left( v \right)$, we get,
$- 10t = - 120 \\ t = 12hrs \\$
Putting the value of $t{\text{ }}$ in equation $\left( {iii} \right)$,
Then we get speed as,
$s = 50\,kmph\,$
And we know Distance = Speed x Time
So,
$d = st = 50 \times 12 = 600\;km\;$
Hence the distance is $600\;km$.

Note – Whenever you struck with this type of problem always try to make equations of various variables according to question , then solve it and get the asked parameter. During solving this problem we kept in mind that the distance travelled is always same whatever the speed may be.