#### Question

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for a journey. Find the original speed of the train.

#### Solution

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

Increased speed of the train = (x + 15)km/hr

Time taken by the train under usual speed to cover 90 km = `90/x`hr

Time taken by the train under increased speed to cover 90 km = `90/(x+15)`hr

Therefore,

`90/x-90/(x+15)=30/60`

`(90(x+15)-90x)/(x(x+15))=1/2`

`(90x+1350-90x)/(x^2+15x)=1/2`

`1350/(x^2+15x)=1/2`

1350(2) = x^{2} + 15x

2700 = x^{2} + 15x

x^{2} + 15x - 2700 = 0

x^{2} - 45x + 60x - 2700 = 0

x(x - 45) + 60(x - 45) = 0

(x - 45)(x + 60) = 0

So, either

x - 45 = 0

x = 45

Or

x + 60 = 0

x = -60

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

Hence, the original speed of train is x = 45 km/hr