Question

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B, 75 km away from A at the same time. On the way, however, the train lost about 12.5 min. while stopping at the stations. The speed of the car is:
A) 100 kmph
B) 110 kmph
C) 120 kmph
D) 130 kmph

Answer 1

Hint:
We have given that the speed of the train is 50% more than the speed of the car. So, firstly we have to consider the speed of the car. Once we calculate the speed of the car we can find the speed of the train also. After considering the speed of the car we calculate the time taken to cover the distance between A and B. we also find the time taken by the train to cover the distance A and B. Since, we have given that both take equal times. So equate both the times and find the speed of the car.

Complete step by step solution:
Let the speed of car = x km\h
We have given the speed of the train is 50% more than the speed of the car.
∴ speed of train = $ x + 50\% of.x = x + \dfrac{{50}}{{100}} \times x = \dfrac{{3x}}{2} $ .
We have also given the total distance between A and B is 75 km.
Now calculate the total time takes to cover the distance by car and train. Since, we know that the relation between speed, distance, and time is
 $ speed = \dfrac{{dis\tan ce}}{{time}} $
So, we use this formula to calculate the time taken by car and train.
Let t1 be the time taken by the car top complete the distance between A and B. Also, the distance is 75 km.
Therefore $ x = \dfrac{{75}}{{{t_1}}} \Rightarrow {t_1} = \dfrac{{75}}{x} $ …………….(i)
This equation (i) gives the time taken to complete the distance.
Let t2 be the time taken by the train to complete the distance. We have also given that the train stops 12.5 min. into the journey. So, we also add this time to t2.
So, $ {t_2} = \dfrac{{75}}{{\dfrac{{3x}}{2}}} + 12.5\min . $
Here the time 12.5 has units minutes but the speed of the train is taken in Km\h. so we have to convert these minutes into hours.
∴ one minute = $ \dfrac{1}{{60}} $ hours
 $ 12.5\min = \dfrac{{12.5}}{{60}} $ hours
So $ {t_2} = \dfrac{{75}}{{\dfrac{{3x}}{2}}} + \dfrac{{12.5}}{{60}} $
 $ {t_2} = \dfrac{{75 \times 2}}{{3x}} + \dfrac{{125}}{{600}} $ …………………….(ii)
Equation (ii) gives the time taken by the train to complete the distance. Now the time taken by car and the train to complete the distance is equal.
So, $ {t_2} = {t_1} $
 $ \dfrac{{75}}{x} = \dfrac{{75 \times 2}}{{3x}} + \dfrac{{25}}{{120}} $
 $ \dfrac{{75}}{x}\left( {1 - \dfrac{2}{3}} \right) = \dfrac{{25}}{{120}} $
 $ \dfrac{{75}}{x}\left( {\dfrac{{3 - 2}}{3}} \right) = \dfrac{{25}}{{120}} $
 $ \dfrac{{75}}{x} \times \dfrac{1}{3} = \dfrac{{25}}{{120}} $
 $ \dfrac{{25}}{x} = \dfrac{{25}}{{120}} $
 $ x = 120 $

So the speed of the train = 120km\h. option (C).

Note:
The speed is the time rate of change of distance. If D is the distance of an object in some time T. the speed is equal to S = D/T. Remember, when the distance is constant: The average speed is equal to 2xy/x+y, where x and y are the two speeds at which the same distance has been covered.