Question

After travelling for $30$ minutes a train meets with an accident, due to which it has to stop for $45$ minutes. Due to the accident its speed is also reduced to $\frac{2}{3}$ of its former value and the train reaches its destination $1$ hour $30$ minutes late. Had the accident occurred $60$ km later, the train would have reached $30$ minutes earlier. The length of journey is

• A. $90$ km
• B. $120$ km
• C. $150$ km
• D. $180$ km

Answer 1

The correct option is B $120$ km

Let $d$ be the distance from point of first accident to destination and $s$ be the speed of train.

$1^{st}case$

$\frac{d}{s}+\frac{3}{4}=\frac{d}{\frac{2s}{3}}$ $(\frac{3}{4}hr=45minutes)$

$\Rightarrow \frac{d}{s}+\frac{3}{4}=\frac{3}{2}\frac{d}{s}$ $Time=\frac{Distance}{Speed}$

$\Rightarrow \frac{3}{4}\times \frac{d}{s}(\frac{3}{2}-1)=\frac{1}{2}s\Rightarrow \frac{d}{s}=t=\frac{3}{2}hr$

Train got late by $1hr30min\left(90min\right)$

$\therefore$ Actual delay due to speed resolution $=(90-45)min=45min$

$2^{nd}case$

$30minutes$ could have been saved because of train travelling $60km$ more with speed s

Time difference if it would have travelled this 60km with $\frac{25}{3}$ speed is nothing but due to same $30minutes\left(\frac{1}{2}hr\right)$

$-\frac{60}{s}+\frac{60}{\frac{6s}{3}}=\frac{1}{2}\Rightarrow -\frac{60}{s}+\frac{90}{s}=\frac{1}{2}\Rightarrow \frac{30}{s}=\frac{1}{2}\Rightarrow s=60km/hr$

Initially train travelled for $30min$

Distance travelled before accident $=\frac{60}{2}km=30km$

After accident the train travel $dkm$

$\therefore \frac{d}{s}=\frac{3}{2}\Rightarrow d=\frac{3}{2}\times s=\frac{3}{2}\times 60km=90km$

$\therefore$ Total length of journey $=30km+90km=120km$