# An express train takes 1 hour less than a passenger train to travel 132 km

Question:

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of two trains.

Solution:

Let the speed of the passenger train be $x \mathrm{~km} / \mathrm{hr}$. Then,

Speed of the express train $=(x+11) \mathrm{km} / \mathrm{hr}$

Time taken by the passenger train to cover $132 \mathrm{~km}$ between Mysore to Bangalore $=\frac{132}{x} \mathrm{hr}$

Time taken by the express train to cover $132 \mathrm{~km}$ between Mysore to Bangalore $=\frac{132}{(x+11)}$ hr

Therefore,

$\frac{132}{x}-\frac{132}{(x+11)}=1$

$\frac{\{132(x+11)-132 x\}}{x(x+11)}=1$

$\frac{132 x+1452-132 x}{x^{2}+11 x}=1$

$1452=x^{2}+11 x$

$x^{2}+11 x-1452=0$

$x^{2}+11 x-1452=0$

$x^{2}-33 x+44 x-1452=0$

$x(x-33)+44(x-33)=0$

$(x-33)(x+44)=0$

So, either

$(x-33)=0$

$x=33$

Or

$(x+44)=0$

$x=-44$

Eut, the speed of the train can never be negative.

Thus, when $x=33$ then speed of express train

$=(x+11)$

$=(33+11)$

$=44$

Hence, the speed of the passenger train is $x=33 \mathrm{~km} / \mathrm{hr}$

and the speed of the express train is $x=44 \mathrm{~km} / \mathrm{hr}$ respectively.