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/h more than that of the passenger train, find the average speed of the two trains.

Given :

The express train takes 1 hour less than a passenger train to cover a distance of 132 km.

The average speed of the express train is 11 km/h more than the average speed of the passenger train.

To do :

We have to find the average speeds of trains.

Solution :

Let the time taken by passenger train be 't' hours.

This implies the time taken by express train is $t-1$ hours.

Let the average speed of the passenger train be $x \frac{km}{h}$.

This implies the average speed of the express train is $x+11 \frac{km}{h}$.

We know that,

Average speed $ = \frac{Distance travelled}{Time taken}$

$Time taken = \frac{Distance}{Speed}$

Therefore,

$t=\frac{132}{x}$

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

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

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

$132(x+11)-132(x) = x(x+11)$

$132x+1452-132x=x^2+11x$

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

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

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

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

$x = -44, x = 33$

Speed cannot be negative.

Therefore, $x = 33$

$x+11= 33+11=44$

Therefore, the average speed of the passenger train is 33 km/h.

The average speed of the express train is 44 km/h.

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Updated on: 10-Oct-2022

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