Question

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.

Answer 1

Let $x$ be the speed of train and $y$ be the time taken by the train.

Let $x.y$ be the distance traveled by the train

Case 1

if train $10Kmh{r}^{-1}$ faster it take $2$ hrs less, then the schedule time,

$(x+10)(y-2)=xy$

$x-5y=-10$ ------ (1)

case 2:

if the train slower by $10Kmh{r}^{-1}$, it taken $3$ hrs more.

$(x-10)(y+3)=xy$

$3x-10y=30$ ----- (2)

On solving (1) and (2).

We get,

$x=50,y=12$

Hence the distance $600Km$