Question

A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of the slower train is 10 kmph less than that of the faster train, find the speeds of two trains.

Answer 1

Given:

Distance = 200 km.

Let faster train speed = v km/h, slower = (v – 10) km/h.

Faster takes 1 hour less than slower.

Step-wise calculation:

1. Time by slower = `200/(v - 10)`.

Time by faster = `200/v`.

2. Equation from time difference:

`200/(v - 10) - 200/v = 1`

3. Multiply both sides by v(v – 10):

200v – 200(v – 10) = v(v – 10)

⇒ 200v – 200v + 2000 = v² – 10v 

⇒ v² – 10v – 2000 = 0

4. Solve quadratic:

Discriminant D = (–10)² – 4(1)(–2000) 

= 100 + 8000

= 8100, `sqrt(D) = 90`

`v = (10 ± 90)/2` 

⇒ `v = (10 + 90)/2 = 50` or `v = (10 - 90)/2 = -40` (discard).

5. Thus, faster speed v = 50 km/h; slower = v – 10 = 40 km/h.

Faster train: 50 km/h.

Slower train: 40 km/h.