Question

Roohi travels $300$km to her home partly by train and partly by bus. She takes four hours if she
travels $60$km by train and the remaining by bus. If she travels $100$km by train and the
remaining by bus, she takes $10$minutes longer. Find the speed of the train and the bus
respectively.

Answer 1

Hint: First assume the speed of train as x and the speed of bus as y, and then use the given
conditions in the problem to form the equation of two variables. Then, solve these equations to
find the value of the speed of train and the bus.
It is given in the problem that Roohi travels $300$km to her home partly by train and partly by
bus. She takes four hours if she travels $60$km by train and the remaining by bus. If she travels
$100$km by train and the remaining by bus, she takes $10$minutes longer.
The goal of the problem is to find the speed of the train and the bus respectively.
Assume that the speed of the train is$x$and the speed of the bus is$y$
Then according to the question,
She takes 4 hours if she travels 60km by tram and the remaining by bus. So, we have
The distance travelled by train \[ = {\rm{ }}60\]km
The distance travelled by bus \[ = {\rm{ }}240\]km

We know that:
Time$ = \dfrac{{{\rm{Distance}}}}{{{\rm{Speed}}}}$
So, the time taken by train is given as:
${T_1} = \dfrac{{60}}{x}$
And the time taken by the bus is given as:
${T_2} = \dfrac{{240}}{y}$
It is given that the total time taken is 4 hours so we can write:
${T_1} + {T_2} = 4$
Substitute the values${T_1} = \dfrac{{60}}{x}$and${T_2} = \dfrac{{240}}{y}$, so we have
$\dfrac{{60}}{x} + \dfrac{{240}}{y} = 4$

It is also given that, if she travels 100 km by train and the remaining by bus, she takes 10 minutes
longer.
The distance travelled by train\[ = {\rm{ }}100\]km
The distance travelled by bus\[ = {\rm{ }}200\]km
We know that:
Time$ = \dfrac{{{\rm{Distance}}}}{{{\rm{Speed}}}}$
So, the time taken by train is given as:
${T_3} = \dfrac{{100}}{x}$
And the time taken by the bus is given as:
${T_4} = \dfrac{{200}}{y}$
It is given that the total time taken is 10 minutes longer, so convert the minutes in hour:
Time$ = \dfrac{{10}}{{60}} = \dfrac{1}{6}$hours
So, we can write:
$\begin{array}{l}{T_3} + {T_4} = 4 + \dfrac{1}{6}\\ \Rightarrow {T_3} + {T_4} =
\dfrac{{25}}{6}\end{array}$
Substitute the values${T_3} = \dfrac{{100}}{x}$and${T_4} = \dfrac{{200}}{y}$, so we have
$\dfrac{{100}}{x} + \dfrac{{200}}{y} = \dfrac{{25}}{6}$
Now, we have two equations of the form:
$\dfrac{{60}}{x} + \dfrac{{240}}{y} = 4$
$\dfrac{{100}}{x} + \dfrac{{200}}{y} = \dfrac{{25}}{6}$

Let us assume that $\dfrac{1}{x} = p$ and$\dfrac{1}{y} = q$, then the above equations have
the form:
$60p + 240q = 4$
$100p + 200q = \dfrac{{25}}{6}$
Simplify both the equations:
$60p + 240q = 4$ … (1)
$600p + 1200q = 25$ … (2)
Multiply the equation (1) by 10, and then subtract the equation (1) by (2):
\[\left( {600p + 2400q} \right) - \left( {600p + 1200q} \right) = 40 - 25\]
\[ \Rightarrow 1200q = 15\]
\[ \Rightarrow q = \dfrac{{15}}{{1200}}\]
\[ \Rightarrow q = \dfrac{1}{{80}}\]
Substitute the value of q in the equation (1):
$60p + 240\left( {\dfrac{1}{{80}}} \right) = 4$
Solve the equation for the value of p:
$60p + 3 = 4$
$ \Rightarrow 60p = 4 - 3$
$ \Rightarrow p = \dfrac{1}{{60}}$
As we were assuming that:
$\dfrac{1}{x} = p$ and$\dfrac{1}{y} = q$
Substitute the value of $p$ and$q$:
$\dfrac{1}{x} = \dfrac{1}{{60}}$ and$\dfrac{1}{y} = \dfrac{1}{{80}}$
$ \Rightarrow x = 60$km per hour and$y = 80$km per hour.
Therefore, the speed of the train is $60$km per hour and the speed of the bus is $80$ km per
hour.

Note: Speed of the train and the bus are given and the total time is taken, while forming the
equation, we have to find the tame taken by the train and the bus using the distance travelled and
speed and then use them to form an equation.