Question

A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km per hour less than the speed of the faster train, then, find the speed of the two trains.

Answer 1

Hint: Recall the formula for speed in terms of distance and time. Assign variables to the speed of the two trains and find the two equations relating them with the given information. Solve the equations to find the speed of the train.

Complete step-by-step answer:

There are two trains and one of the trains is the fast train and the other is the slow train.
Let the speed of the faster train be x and the speed of the slower train be y.
We know the formula for speed v in terms of distance d and time t as follows:
\[v = \dfrac{d}{t}\]
Time taken in terms of speed and distance is then given by the formula:
$\Rightarrow$ \[t = \dfrac{d}{v}\]
Time taken by the fast train to cover 600 km is given by:
$\Rightarrow$ \[{t_x} = \dfrac{{600}}{x}..............(1)\]
Time taken by the slow train to cover 600 km is given by:
$\Rightarrow$ \[{t_y} = \dfrac{{600}}{y}..............(2)\]
The time taken by fast train is 3 hours less than the time taken by the slow train, hence, we have:
$\Rightarrow$ \[{t_y} - {t_x} = 3\]
Substituting equations (1) and (2), we have:
$\Rightarrow$ \[\dfrac{{600}}{y} - \dfrac{{600}}{x} = 3..............(3)\]
It is given that the speed of the fast train is 10 km per hour faster than the speed of the slow train, hence, we have:
$\Rightarrow$ \[x - y = 10\]
Writing x in terms of y, we get:
$\Rightarrow$ \[x = y + 10............(4)\]
Substituting equation (4) in equation (3), we get:
$\Rightarrow$ \[\dfrac{{600}}{y} - \dfrac{{600}}{{y + 10}} = 3\]
Cross multiplying, we have:
$\Rightarrow$ \[600(y + 10) - 600y = 3y(y + 10)\]
$\Rightarrow$ \[600y + 6000 - 600y = 3{y^2} + 30y\]
Simplifying, we have:
$\Rightarrow$ \[3{y^2} + 30y - 6000 = 0\]
Solving for y, we have:
$\Rightarrow$ \[{y^2} + 10y - 2000 = 0\]
$\Rightarrow$ \[{y^2} + 50y - 40y - 2000 = 0\]
$\Rightarrow$ \[y(y + 50) - 40(y + 50) = 0\]
$\Rightarrow$ \[(y - 40)(y + 50) = 0\]
$\Rightarrow$ \[y = 40;y = - 50\]
Speed of the train is a positive quantity.
$\Rightarrow$ \[y = 40.............(5)\]
Hence, the speed of slow trains is 40 km per hour.
Substituting equation (5) in equation (4), we get:
$\Rightarrow$ \[x = 40 + 10\]
$\Rightarrow$ \[x = 50\]
The speed of the fast train is 50 km per hour.
The speed of the trains are 50 km per hour and 40 km per hour respectively.

Note: We get a quadratic equation with a positive and a negative root. Speed is a positive quantity and hence, we need to choose only the positive root.