# Problems on Trains

A train crosses a platform 100 m long in 60 seconds at a speed of 45 km/hr. The time taken by the train to cross an electric pole is :

**Answer**: (b)

Speed = \(45 \times \frac{5}{18}\) m/sec = \frac{25}{2}[/latex] m/sec

Let the length of the train be x metres.

Then, \(\frac{x + 100}{\frac{25}{2}}\) = 60 or x = 650 m

∴ Time taken by the train to cross an electric pole = \(\left ( 650 \times \frac{2}{25} \right )\) sec = 52 sec

A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform ?

**Answer**: (b)

Speed = \(54 \times \frac{5}{18}\) m/sec = 15 m/sec

Length of the train = (15 × 20) m = 300 m.

Let the length of the platform be x metres.

Then, \(\frac{x + 300}{36} = 15\)

⇒ x + 300 = 540

⇒ x = 240 m.

A 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform ?

**Answer**: (b)

Speed = \(\frac{300}{18}\) m/sec = \(\frac{50}{3}\) m /sec

Let the length of the platform be x metres.

Then, \(\frac{x + 300}{39} = \frac{50}{3}\)

⇒ 3(*x* + 300) = 1950

⇒ *x* = 350 m

A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. its length is :

**Answer**: (b)

Let the length of the train be x metres and its speed be y m/sec.

Then, \(\frac{x}{y} = 15\) ⇒ \(y = \frac{x}{15}\)

∴ \(\frac{x + 100}{25} = \frac{x}{15}\)

⇒ x = 150 m

A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train ?

**Answer**: (d)

Let the length of the train be *x* metres and its speed be y m/sec.

They \(\frac{x}{y}\) = 8 ⇒ *x* = 8y

Now, \(\frac{x + 264}{20} = y\)

⇒ 8y + 264 = 20y

⇒ y = 22

∴ Speed = 22 m/sec = \(\left ( 22 \times \frac{18}{5} \right )\) km/hr = 79.2 km/hr

A train takes 18 seconds to pass completely through a station 162 m long and 15 seconds through another station 120 m long. The length of the train is :

**Answer**: (c)

Let the length of the train be *x* metres.

⇒ \(\left ( x + \frac{162}{18} \right ) = \frac{x + 120}{15}\)

⇒ 15(*x* + 162) = 18(*x* + 120)

⇒ *x* = 90.

A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?

**Answer**: (c)

Speed = \(\frac{240}{24}\) m/sec = 10 m/sec

∴ Required time = \(\left ( \frac{240 + 650}{10} \right )\) sec = 89 sec

How many seconds will a 500 metre long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr ?

**Answer**: (b)

Speed of train relative to man = (60 - 3) km/hr = 60 km/hr

= \(60 \times \frac{5}{18}\) m/sec = \(\frac{50}{3}\) m/sec

Time taken to pass the man = \(500 \times \frac{3}{50}\) sec = 30 sec

A jogger running at 9 kmph alongside a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the direction opposite to that in which the train is going ?

**Answer**: (c)

Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr.

\(\left ( 36 \times \frac{5}{18} \right )\) m/sec

= 10 m/sec.

Distance to be covered = (240 + 120) m = 360 m.

∴ Time taken = \(\frac{360}{10}\) sec = 36 sec.

A train 110 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ?

**Answer**: (b)

Speed of train relative to man = (60 + 6) km/hr = 66 km/hr

= \(\left ( 66 \times \frac{5}{18} \right )\) m/sec = \(\frac{55}{3}\) m/sec

∴ Time taken to pass the man = \(\left ( 110 \times \frac{3}{55} \right )\) sec = 6 sec.

0

## Attempted

0

## Correct

0