#### Question

A passenger train takes one hour less for a journey of 150 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.

#### Solution

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

Increased speed of the train = (x + 5)km/hr

Time taken by the train under usual speed to cover 150km = `150/x`hr

Time taken by the train under increased speed to cover 150km = `150/(x + 5)`hr

Therefore,

`150/x-150/(x+5)=1`

`(150(x+5)-150x)/(x(x+5))=1`

`(150x+750-150)/(x^2+5x)=1`

`750/(x^2+5x)=1`

750 = x^{2} + 5x

x^{2} + 5x - 750 = 0

x^{2} - 25x + 30x - 750 = 0

x(x - 25) + 30(x - 25) = 0

(x - 25)(x + 30) = 0

So, either

x - 25 = 0

x = 25

Or

x + 30 = 0

x = -30

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

Hence, the usual speed of train is x = 25km/hr

Is there an error in this question or solution?

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A Passenger Train Takes One Hour Less for a Journey of 150 Km If Its Speed is Increased by 5 Km/Hr from Its Usual Speed. Find the Usual Speed of the Train. Concept: Solutions of Quadratic Equations by Factorization.

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