Question

A train leaves the station 1 hour before the scheduled time. The driver decreases its speed by 50 km/hr. At the next station 300 km away, the train reached on time. Find the original speed of the train.

(a) 100 km/hr

(b) 150 km/hr

(c) 125 km/hr

(d) 200 km/hr

Answer 1

Correct option is (b) 150 km/hr

Let the normal speed of the train = x km/hr.

Then the normal time to cover 300 km = \(\frac{300}x\) 

Now it takes 1 hour more.

So, the new time taken = \(1 + \frac{300}{x}\) = \( \frac{x+300}{x}\)

∴ The new speed = \( 300 \div\frac{x+300}{x}\) km/hr

\(= \frac{300x}{x+ 300}\) km/hr.

But the new speed is (x = 50) km/hr.

⇒ \(\frac{300x}{x+ 300}\) = x - 50

⇒ x² − 50x − 15000 = 0

⇒ (x − 150) (x + 100) = 0

⇒ x = (150,−100) km/hr.

We reject the negative value as the train always runs in the same direction.

∴ x = 150

So, the normal speed of the train = 150 km/hr.