Question

A train met with an accident $60$ km away from Anantpur station. It completed the remaining journey at ${\frac{5}{6}}^{th}$ of the previous speed and reached Barmula station $1$ hour $12$ in late. Had accident taken place $60$ km further, it would have been only $1$ hour late.
A) What is the normal speed of train?
B) What is the distance between Anantpur and Barmula ?

• A. $60$ km/hr, $420$ km
• B. $80$ km/hr, $800$ km
• C. $8$ km/hr, $80$ km
• D. None of these

Answer 1

The correct option is A $60$ km/hr, $420$ km

Let normal speed of the train be $x$ km/hr and distance between Anantpur and Barmula be $y$ km.
Normally train takes $\frac{y}{x}$ hrs to reach Trent Bridge.
Case 1:
Time taken till accident $=\frac{60}{x}$ hours
Time taken $2$ cover remaining distance $=6\times \frac{(y-60)}{5x}$
According to question, we have
$\frac{60}{x}+6\times \frac{(y-60)}{5x}=\frac{y}{x}+1.2$ (converting minute into hour)on solving this equation, we get,
$y-6x=60$ ......(i)
Case 2:
Time taken till accident $=\frac{(60+60)}{x}$ hr $=\frac{120}{x}$ hr
Time taken $2$ cover remaining distance $=6\times \frac{(y-120)}{5x}$
Now, $\frac{120}{x}+6\times \frac{(y-120)}{5x}=\frac{y}{x}+1$
On solving this equation, we get
$y-5x=120$ ........(ii)
On solving (i) and (ii), we get
$x=60$
$\Rightarrow y=420$