Question

After covering a distance of $30km$ with a uniform speed there is some defect in a train engine and therefore its speed is reduced to $\frac{4}{5}$ of its original speed. Consequently the train reaches its destination late by $45$ minutes. Had it happened after covering $18$ kilometers more the train would have reached $9$ minutes earlier, find the speed of the train and the distance of journey.

• A. Speed $=40km/hr$, Distance $=160km$
• B. Speed $=20km/hr$, Distance $=100km$
• C. Speed $=30km/hr$, Distance $=120km$
• D. Speed $=35km/hr$, Distance $=140km$

Answer 1

The correct option is C Speed $=30km/hr$, Distance $=120km$

Let $d$ be the total distance (in km) covered by the train, $t$ be the total time taken by the train to reach it's destination with original speed and $x$ be the original speed (in km/hr) of the train.

$t=\frac{d}{x}$

As the train covered the distance of $30km$ by its original speed

The time taken by the train to cover first $30km=\frac{30}{x}$

After covering $30km,$ the speed of the train reduced to $\frac{4}{5}$ of its original speed.

$\therefore$ the reduced speed of train $=\frac{4}{5}x$ km/hr

As the train covered the distance of $30km$ by its original speed,

distance covered by the train with reduced speed $=(d-30)km$

$\therefore$ time taken by train to cover the distance of $(d-30)km=\frac{(d-30)}{\frac{4}{5}x}$

Given that the train reaches its destination late by $45$ minutes.

$\Rightarrow \text{45 minutes}=\frac{45}{60}hr=\frac{3}{4}hr$

$\therefore$ Time taken by the train for first $30$ km $+$ time taken by the train for rest of the distance $=(t+\frac{3}{4})$ hr

$\Rightarrow \frac{30}{x}+\frac{(d-30)}{\frac{4}{5}x}=t+\frac{3}{4}$

$\Rightarrow \frac{30}{x}+\frac{(d-30)}{\frac{4}{5}x}=\frac{d}{x}+\frac{3}{4}$

$\Rightarrow d-3x=30$ ......... $\left(i\right)$

If the speed of train reduced after covering $18$ kms more, the train would have reached $9$ minutes earlier.

$\Rightarrow \text{9 minutes}=\frac{9}{60}hr$

In that case:

Total time taken by the train to reach the destination $=t+\frac{3}{4}-\frac{9}{60}$

Distance covered by the train with original speed $=30+18=48$ km

Time taken by the train to cover those $48km=\frac{48}{x}$

distance covered by the train with reduced speed $=(d-48)$ km

$\therefore$ time taken by train to cover the distance of $(d-48)$ km $=\frac{(d-48)}{\frac{4}{5}x}$

$\therefore$ Time taken by the train for first $48$ km $+$ time taken by the train for rest of the distance $=(t+\frac{3}{4}-\frac{9}{60})$ hr

$\Rightarrow$ time taken by the train for first 48 km + time taken by the train for rest of the distance $=(t+\frac{36}{60})$ hr

$\Rightarrow \frac{48}{x}+\frac{(d-48)}{\frac{4}{5}x}=t+\frac{36}{60}$

$\Rightarrow \frac{48}{x}+\frac{(d-48)}{\frac{4}{5}x}=\frac{d}{x}+\frac{36}{60}$

$\Rightarrow 5d-12x=240$ ....... $\left(ii\right)$

Multiplying ${eq}^n\left(i\right)$ by $5,$ we have

$5d-15x=150$ ......... $\left(iii\right)$

Subtracting $\left(iii\right)$ from $\left(ii\right)$, we have

$5d-12x-(5d-15x)=240-150$

$\Rightarrow 5d-12x-5d+15x=90$

$\Rightarrow 3x=90$

$\Rightarrow x=30$

$\therefore$ speed of the train $=30$ km/hr

Substituting the value of $x$ in ${eq}^n\left(i\right)$, we have

$d-3\times 30=30$

$\Rightarrow d=120$

$\therefore$ Distance covered by the train $=120$ km