#### Question

The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10 km/hr more than the speed of going, what was the speed per hour in each direction?

#### Solution

Let the ongoing speed of person be x km/hr. Then,

Returning speed of the person is = (x + 10)km/hr.

Time taken by the person in going direction to cover 150km = `150/x`hr

Time taken by the person in returning direction to cover 150km = `150/(x+10)`hr

Therefore,

`150/x-150/(x+10)=5/2`

`(150(x+10)-150x)/(x(x+10))=5/2`

`(150x+1500-150x)/(x^2+10x)=5/2`

`1500/(x^2+10)=5/2`

1500(2)=5(x^{2}+10x)

3000 = 5x^{2} + 50x

5x^{2} + 50x - 3000 = 0

5(x^{2} + 10x - 600) = 0

x^{2} + 10x - 600 = 0

x^{2} - 20x + 30x - 600 = 0

x(x - 20) + 30(x - 20) = 0

(x - 20)(x + 30) = 0

So, either

x - 20 = 0

x = 20

Or

x + 30 = 0

x = -30

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

Thus, when x = 20 then

= x + 10

= 20 + 10

= 30

Hence, ongoing speed of person is x = 20 km/hr

and returning speed of the person is x = 30 km/hr respectively.