Question

The distance between two stations is 482 km. Two trains start simultaneously from these stations on a parallel track to cross each other. The speed of one of them is greater than that of the other by 4km/hr. If the distance between the two trains after 3 hours is 20 km, find the speed of each train.

Answer 1

Hint: Let the speed of the slower train be x km/hr. So, the speed of the other train is (x+4) km/hr. Now use the formula $\text{distance=speed}\times \text{time}$ to find the distance travelled by each train in three hours separately and subtract the results from the distance between the stations and equate it with the distance left at the end of the three hours. Solve the equation, to get the value of x.

Complete step-by-step answer:
Let the speed of the slower train be x km/hr. As it is given that the speed of one of them is greater than that of the other by 4km/hr the speed of the other train is (x+4) km/hr.
Now let us find the distance travelled by the slower train in 3 hours.
$\text{distance=speed}\times \text{time}$
$\text{distance=3x km}$
Now, we will find the distance travelled by the other train in 3 hours.
$\text{distance=speed}\times \text{time}$
$\text{distance=3}\left( x+4 \right)$
Now as the trains are approaching each other starting from two different stations and the distance at the end of the three hours between them is 20Km, we can say that 482 km minus the distance travelled by each train is equal to 20km.
$482-3x-3\left( x-4 \right)=20$
$\Rightarrow 462-3x-3x+12=0$
$\Rightarrow 474=6x$
$\Rightarrow x=79km/hr$
Therefore, the speed of the slower train is 79km/hr and the speed of the other train is 83km/hr.

Note: See if this would have been a question from physics, you would have got two pairs of answers, as the distance between them here would actually be the displacement there and you would have to take once the displacement to be positive and one negative giving 2 answers. Remember that the validity of the formula $\text{distance=speed}\times \text{time}$ is only for uniform motions and not for accelerated motions.