Question

the distance between Akola and Bhusaval is 168 km .an Express train takes 1 hour less than a passenger train to cover the distance .find the average speed of train if the average speed of the express is more by 14 km/hr than the speed of the passenger train​

Answer 1

Given, distance between the two places=168km

We know that speed=distance traveled/time taken

=>time taken=distance traveled/speed

Let speed of passenger train be 'x' km/hr

Then the time taken by passenger train to cover the distance=168/x

given, speed of express is 14km more than that of passenger train

=>speed of express=(x+14) km/hr

=>Time taken by express=168/X+14

Now, it is given that express train takes 1 hour less than the passenger train to cover the distance.

=>Time taken by passenger train-time taken by express train=1

$\frac{168}{x} - \frac{168}{x + 14} = 1$

$\frac{168(x + 14) - 168x}{x(x + 14)} = 1$

On simplifying, we get the equation

${x}^{2} + 14x - 2352 = 0$

${x}^{2} + 56x - 42x - 2352 = 0$

$x(x + 56) - 42(x + 56) = 0$

Which implies, x=42,-56

But x cannot be negative

=> x=42

Therefore, speed of passenger train x=42km/hr

and speed of express (x+14) = 56km/hr