Question

The distance between akola and bhusawal is 168 km. An express train takes 1 hour less than a passenger train to cover the distance. Find the average speed of each train if the average speed of the express train is more by 14 km/hr than the speed of the passenger train

Answer 1

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Answer 2

Solution:

Let the speed of passenger train is x kmph

and that of express train is y kmph

Distance = 168 km

Time taken by Express train to travel between akola to bhusawal = 168/y hour

Time taken by passenger train to travel between akola to bhusawal = 168/x hour

According to the question Express train taking 1 hour less

$\frac{168}{y} + 1 = \frac{168}{x} \\ \\ 168x + xy = 168y \\ \\ 168x - 168y + xy = 0...eq1$

ATQ
$y = x + 14...eq2$
Substitute from eq 2

$168x - 168y + xy = 0 \\ \\ 168x - 168(x + 14) + x(x + 14) = 0 \\ \\168x - 168x - 2352 + {x}^{2} + 14x = 0 \\ \\ {x}^{2} + 14x - 2352 = 0 \\ \\ {x}^{2} + 56x - 42x - 2352 = 0 \\ \\ x(x + 56) - 42(x - 56) = 0 \\ \\ (x + 56)(x - 42) = 0 \\ \\ x = - 56 \\ \\ x = 42$
negative value can be discarded.

Thus Speed of passenger train 42 kmph

Speed of express train = 42+14= 56 kmph

Hope it helps you.