#### Question

A passenger train takes 3 hours less for a journey of 360 km, if its speed is increased by 10 km/hr from its usual speed. What is the usual speed?

#### Solution

Let the usual speed of the train be x km/hr

then the increased speed of the train = (x + 10)km/hr

We know that

`"Time"="Distance"/"speed"`

Time taken by the train under usual speed to cover 360 km = `360/x hr`

Time taken by the train under increased speed to cover 3690km = `360/(x+10)hr`

Then according to the question

`rArr360/x-360/(x+10)=3`

`rArr(360(x+10)-360x)/(x(x+10))=3`

`rArr(360x+3600-360x)/(x(x+10))=3`

`rArr3600/(x(x+10))=3`

⇒ 3x(x + 10) = 3600

⇒ 3x^{2} + 30x = 3600

⇒ 3x^{2} + 30x - 3600 = 0

⇒ 3(x^{2} + 10 - 1200) = 0

⇒ x^{2} + 10 - 1200 = 0

⇒ x^{2} - 30x + 40x - 1200 = 0

⇒ x(x - 30) + 40(x - 30) = 0

⇒ (x + 40)(x - 30) = 0

⇒ x + 40 = 0

⇒ x = -40

Or

⇒ x - 30 = 0

⇒ x = 30

∵ The speed of the train cannot be negative Therefore. the speed of the train is 30km/hr.