Question

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

Answer 1

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`

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

⇒ `1350/(x^2 + 15x) = 1/2`

⇒ 2700 = x² + 15x

⇒ x² + 15x - 2700 = 0

⇒ x² - 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 a train is 45 km/hr.