Question

A train covered a certain distance at a uniform speed. If the train would have been
10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train
were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find
the distance covered by the train​

Answer 1

Answer:

Let the speed of the train be x km/h and the time taken to travel to given distance be t hours and the distance to travel be d km.

We know that,

$\\ \sf \frac{distance \: \: travelled}{time \: \: taken \: \: to \: \: travel \: \: that \: \: distance} \\ \\ \\ \sf \: \: x = \frac{d}{t} \: \: or, \: \: d = xt - - - (1)$

According to the question, we get

$\\ \sf \: (x + 10) = \frac{d}{(t - 2)} \\ \\ \\ \implies \sf \: (x + 10)(t - 2) = d \\ \\ \\ \implies \sf \: xt + 10t - 2x - 20 = d$

Now, by using equation 1, we get

$\\ \sf \: - 2x + 10t = 20 \qquad - - - (2) \\ \\ \\ \implies \sf \: (x - 10) = \frac{d}{(t + 3)} \\ \\ \\ \implies \sf \: (x - 10)(t + 3) = d \\ \\ \\ \implies \sf \: xt - 10t + 3x - 30 = d$

By using equation 1, we get

$\\ \sf \: 3x - 10t = 30 \qquad - - - (3)$

Adding equation (ii) and (iii), we get -

$\\ \sf \blue{x = 50}$

Using equation (2), we get

$\\ \sf \: ( - 2) \times (50) + 10t = 20 \\ \\ \\ \implies \sf \: - 100 + 10t = 20 \\ \\ \\ \implies \sf \: 10t = 120 \\ \\ \\ \implies \sf \: t = \frac{120}{10} \\ \\ \\ \sf \green{t = 12 \: \: hours}$

From equation (1), we get

$\\ \sf \: d = xt \\ \\ \\ \implies \sf \: 50 \times 12 \\ \\ \\ \implies \sf \red{ 600 \: km}$