Question

A train met with an accident 80 km away from old Delhi railway station. It travels at ${\frac{5}{6}}^{th}$ of the usual speed and reaches the Kanpur central 1 hour 12 min late. Had the accident taken place 120 km further. It would have been only 1 hour late. What is the distance between Old Delhi Railway station and Kanpur Central?

• A. 800
• B. 720
• C. 700
• D. none of these

Answer 1

The correct option is B

720

Since the speed is decreased by 1/6. So, the time will be increased by $\frac{1}{5}$, which is equal to 1 hour 12 minutes. It means the normal time required for this remaining part of journey is 5 $\times$ 72 min = 360 min = 6 hour
Now, accident is supposed to happen 120 km further.
Since, the speed is decreased by $\frac{1}{6}$. So, the time will be increased by $\frac{1}{5}$, which is equal to 1 hour. Hence normal time required for this remaining part of journey is 5 $\times$ 1 = 5 hour.

Thus , it is clear that when the trains runs 120 km of its normal speed it takes 1 hour less, which implies that in 1 hour the train can run 120km with its normal speed. Thus the normal speed of train is 120km/h.
Since the train requires 6 hours at its normal speed of 120 km/h for the rest part of journey. So,
AK = 120 $\times$ 6 = 720.
Thus,
The total distance = Distance travelled before accident + Distance travelled after accident
= 80 + 720 = 800.
Option (a).