Question

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.

Answer 1

Let original speed of train = x km/h

We know,

Time = `"Distance"/"Speed"`

According to the question, we have,

Time taken by train = `360/x` hour

And, Time taken by train its speed increase 5 km/h = `360/((x + 5))`

It is given that,

Time taken by train in first – time taken by train in 2^nd case = 48 min = `48/60` hour

`360/x - 360/((x + 5)) = 48/60 = 4/5`

3`60(1/x - 1/((x + 5))) = 4/5`

`360 xx 5/4 (5/(x^2 + 5x))` = 1

450 × 5 = x² + 5x

x² + 5x – 2250 = 0

x = `(-5 +- sqrt(25 + 9000))/2`

= `(-5 +- sqrt(9025))/2`

= `(-5 +- 95)/2`

= – 50, 45

But x ≠ – 50 because speed cannot be negative

So, x = 45 km/h

Hence, original speed of train = 45 km/h