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 km/h
B. 110 km/h
C. 120 km/h
D. 130 km/h

Answer 1

Hint: To solve this question, we should know the concept of speed that is, $\text{Speed =}\dfrac{\text{Distance}}{\text{Time}}$ and also we should have an idea about percentage. Now, to form the equation, we will assume the speed of the car as x. Then, we will simplify to get the value of x.

Complete step-by-step answer:
We have been given the question that the train can travel 50% faster than a car. So, let us assume that the speed of the car is x. So, from the question, we can write, speed of the train as x + 50% of x.
Speed of the train = $x+\dfrac{50x}{100}$
$\begin{align}
  & \Rightarrow x+\dfrac{x}{2} \\
 & \Rightarrow \dfrac{3x}{2} \\
\end{align}$
In the question, it is given that the car and the train start from the same point A and reach point B, 75 km away from point A at the same time, then we can say that, the car and the train both travelled a distance of 75 km. Now, we know that the speed of an object is the ratio of the distance travelled by it to the time taken by it, that is, $\text{Speed =}\dfrac{\text{Distance}}{\text{Time}}$.
So in order to find the time taken, we can rearrange and write the formula as $\text{Time =}\dfrac{\text{Distance}}{\text{Speed}}$.
Therefore, we can write:
Time taken by the car to travel 75 km = $\dfrac{75}{x}\ldots \ldots \ldots (i)$.
Time taken by the train to travel 75 km (excluding the stoppage time) = $\dfrac{75}{\dfrac{3}{2}x}\Rightarrow \dfrac{50}{x}\ldots \ldots \ldots (ii)$.
Now, we know that the train and the car reach point B at the same time but the train lost 12.5 minutes, that is $\dfrac{12.5}{60}$ hours while stopping at stations. So, we can write:
Time taken by car to travel 75 km = Time taken by train (including stoppage time)
Time taken by car = $\dfrac{12.5}{60}$ + time taken by train (excluding stoppage time) ………(iii)
Now, from equations (i) and (ii), we will put the values to equation (iii), so we will get, $\dfrac{75}{x}=\dfrac{12.5}{60}+\dfrac{50}{x}$. Now, we will simplify it to get the value of x. So, we get,
$\begin{align}
  & \dfrac{75}{x}-\dfrac{50}{x}=\dfrac{12.5}{60} \\
 & \Rightarrow \dfrac{75-50}{x}=\dfrac{12.5}{60} \\
 & \Rightarrow \dfrac{25}{x}=\dfrac{12.5}{60} \\
\end{align}$
On taking x to one side and the constants to the other side, we get,
\[\begin{align}
  & \Rightarrow x=\dfrac{25\times 60}{12.5} \\
 & \Rightarrow x=2\times 60 \\
 & \Rightarrow x=120 \\
\end{align}\]
Therefore, we get the speed of the car as 120 km/h.
Hence, option (C) is the correct answer.

Note: In this question, the possible mistakes that the students can make is by not changing the unit of time, that is, by not converting 12.5 minutes to $\dfrac{12.5}{60}$ hours. This will lead to wrong answers.