Question

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

Answer 1

Step 1. Let us consider

Time is taken by train$=tsec$

Speed $=xkm/h$

Distance $=dkm$

Time$=\frac{dis\tan ce}{speed}$

(or)

$dis\tan ce=speedxtime$

$d=x.t$………………………………….$\left(i\right)$

Step 2: When the speed of train would have been $10km/h$ faster, it would have taken $2hours$ less than the scheduled time

Then speed $=(x+10)km/h$

time$=(t-2)hours$

Therefore we have,

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

$$\begin{gathered} =xt-2x+10t-20 \\ =d-2x+10t-20[\sin ced=xtfrom(i\left)\right] \end{gathered}$$

$\Rightarrow 10t-2x=20$………………………………$\left(ii\right)$

Step 3: When the train were slower by $10km/h$; it would have taken $3hours$ more than the scheduled time

Then speed $=(x-10)km/h$

time$=(t+3)hours$

Therefore we have,

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

$$\begin{gathered} =xt+3x-10t-30 \\ =d+3x-10t-30[\sin ced=xtfrom(i\left)\right] \end{gathered}$$

$\Rightarrow 3x-10t=30$………………………………$\left(iii\right)$

Step 4. Adding equation $\left(ii\right)$ and $\left(iii\right)$ we have,

$$\begin{gathered} 10t-2x+3x-10t=20+30 \\ \Rightarrow x=50km/h \end{gathered}$$

Substituting the value of$x$ in $\left(ii\right)$

$$\begin{gathered} \Rightarrow 10t-2x50=20 \\ \Rightarrow 10t-100=20 \\ \Rightarrow 10t=120 \\ \Rightarrow t=12hours \end{gathered}$$

Step 5. Now substitute the value of $t$ and in $x$ in equation $\left(i\right)$

$$\begin{gathered} d=x.t \\ \Rightarrow d=50x12 \\ \Rightarrow d=600km \end{gathered}$$

Hence the distance covered by the train is $600km$