Question

The distance between akola and bhusaval is 168 km express train take 1 he less than a passenger train to cover a distance find the average speed of express train is more by 14 km/ he than the speed of the passenger train

Answer 1

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

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

ATQ 
$\begin{lgathered}y = x + 14...eq2 \\\end{lgathered}$
Substitute from eq 2

$\begin{lgathered}168x - 168y + xy = 0 \\ \\ 168x - 168(x + 14) + x(x + 14) = 0 \\ \\168x - 168x - 2352 + {x}^{2} + 14x = 0 \\ \\ {x}^{2} + 14x - 2352 = 0 \\ \\ {x}^{2} + 56x - 42x - 2352 = 0 \\ \\ x(x + 56) - 42(x - 56) = 0 \\ \\ (x + 56)(x - 42) = 0 \\ \\ x = - 56 \\ \\ x = 42 \\ \\\end{lgathered}$

negative value can be discarded.

Thus Speed of passenger train 42 kmph

Speed of express train = 42+14= 56 kmph

Hope it helps you.