Question

A passenger train takes 3 hours less for a journey of 360km . If its speed is increased by 10 km/hr from its usual speed. Find its usual speed.

Answer 1

☞ Let speed of passenger train be x km/hr.

• Total distance covered by train = 360 km

If the speed of the train is increased by 10 km/he then it's speed is (x + 10) km/hr

We know that

Time = $\dfrac{Distance}{Speed}$

So, to cover 360 km the time required is $\dfrac{360}{x}$ hrs.

And to cover 369 km with increased speed (i.e. 10 km/hr) the time required is $\dfrac{360}{x\:+\:10}$ hrs.

• A.T.Q.

=> $\dfrac{360}{x\:+\:10}$ = $\dfrac{360}{x}$ - 3

=> $\dfrac{360}{x}$ - $\dfrac{360}{x\:+\:10}$ = 3

• Take 360 common from L.H.S and also take LCM

=> $\dfrac{360(x \: + \: 10 \: - \: x)}{(x) \: (x \: + \: 10)}$ = 3

=> $\dfrac{3600}{ {x}^{2} \: + \: 10x}$ = $\dfrac{3}{1}$

• Cross-multiply them

=> 3600 = 3x² + 30x

=> 3x² + 30x - 3600 = 0

• Take 3 common

=> 3(x² + 10x - 1200) = 0

=> x² + 10x - 1200 = 0

=> x² + 40x - 30x - 1200 = 0

=> x(x + 40) - 30(x - 40) = 0

=> (x - 30) (x + 40) = 0

=> x - 30 = 0

=> x = 30

Similarly

=> x + 40 = 0

=> x = - 40 (speed can't be in negative. So, neglected)

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Speed of train = 30 km/hr

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