Question

A passenger train running at a speed of 80km/hr leaves the railway station 6 hours after a good train leaves and overtakes it is 4 hours. What is the speed of the good train goods train?
(a) 48 km/hr
(b) 60 km/hr
(c) 32 km/hr
(d) 80 km/hr

Answer 1

Hint: Let the speed of the good train be x km/hr. Now when the good train is overtaken, it has travelled for a total time of 10 hours, so using the formula $ \text{distance=speed}\times \text{time} $ we can say that distance is equal to 10x km. The passenger train has to cover the same distance in 4 hours to overtake it. So, again using the same formula we can say distance is equal to $ 80\times 4\text{ km} $ . Equate the two distances to get the answer.

Complete step-by-step answer:
Let the speed of the good train be x km/hr.
Now when the good train is overtaken, it has travelled for a total time of 10 hours,as it started 6 hours early and took 4 hours to be overtaken. So, the distance travelled by is:
 $ \text{distance=speed}\times \text{time=}10x\text{ km} $
Also, by the time the train is overtaken, the passenger train should have travelled the same distance as the goods train. It is given that the speed of the passenger train is 80 km/hr and time taken is 4 hours.
 $ \text{distance=speed}\times \text{time=80}\times \text{4=320 km} $
Now, we will equate the two distances that we found in the above case, we get
 $ 10x=320 $
 $ x=32\text{ km/hr} $
Therefore, the speed of the good train is 32 km/hr. Hence, the answer is option (c).

Note: See if this would have been a question from physics, you would have used the concept of relative speed. Let us see how we can do so:
As they are moving in the same direction the speeds get subtracted to give the relative speed. So, the relative speed is (80-x) km/hr. The relative distance between them would be the distance at the start of the passenger train, which is equal to 6x. As relative distance is equal to relative distance multiplied by time, we can say:
 $ 6x=\left( 80-x \right)\times 4 $
Solve this, the answer would be the same as the above method, i.e., 32 km/hr.