#### Question

A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.

#### Solution

Let the usual speed of the passenger train be x km/hr. Then,

Increased speed of the passenger train = (x + 5) km/hr

Time taken by the train under usual speed to cover 300 km = `300/x`hr

Time taken by the train under increased speed to cover 300 k m = `300/(x + 5)`hr

Therefore,

`300/x-300/(x+5)=2`

`(300(x+5)-300x)/(x(x+5))=2`

`(300x+1500-300x)/(x^2+5x)=2`

`1500/(x^2+5x)=2`

1500 = 2(x^{2} + 5x)

1500 = 2x^{2} + 10x

2x^{2} + 10 - 1500 = 0

2(x^{2} + 5x - 750) = 0

x^{2} + 5x - 750 = 0

x^{2} - 25x + 30x - 750 = 0

x(x - 25) + 30(x - 25) = 0

(x - 25)(x + 30) = 0

So, either

x - 25 = 0

x = 25

Or

x + 30 = 0

x -30

But, the speed of the passenger train can never be negative.

Hence, the usual speed of passenger train is x = 25 km/hr