__Question:__A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. The length of the train is:

**Options:**

A. | 45 m | |

B. | 50 m | |

C. | 54 m | |

D. | 72 m |

**Answer: Option B**

2 kmph = \(\left(2\times\frac{5}{18}\right)m/sec= \frac{5}{9}m/sec\)

4 kmph =\(\left(4\times\frac{5}{18}\right)m/sec = \frac{10}{9}m/sec\)

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

Then, \(\left[\frac{x}{y-\frac{5}{9}}\right]=9 and \left[\frac{x}{y-\frac{10}{9}}\right] = 10\)

Therefore 9*y* - 5 = *x* and 10(9*y* - 10) = 9*x*

9*y* - *x* = 5 and 90*y* - 9*x* = 100.

On solving, we get: *x* = 50.

Therefore Length of the train is 50 m.

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## More Questions on This Topic :

__Question 1.__Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 seconds. What is the length of the fast train?

**Options:**

- 23 m
- \(23\frac{2}{9}m\)
- \(27\frac{7}{9}m\)
- 29m

**Answer: Option C**

Relative speed = (40 - 20) km/hr = \(\left(20\times\frac{5}{18}\right)m/sec=\left(\frac{50}{9}\right)m/sec\)

Therefore Length of faster train = \(\left(\frac{50}{9}\times5\right)m = \frac{250}{9}m=27\frac{7}{9}\)

__Question 2.__A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in 6 seconds. The speed of the second train is:

**Options:**

- 48 km/hr
- 54 km/hr
- 66 km/hr
- 82 km/hr

**Answer: Option D**

Let the speed of the second train be *x* km/hr.

Relative speed = \(\left(x+50\right)km/hr\)

= \(\left[(x+50)\times\frac{5}{18}\right]m/sec\)

=\(\left[\frac{250+5x}{18}\right]m/sec.\)

Distance covered = (108 + 112) = 220 m.

\(\therefore \frac{220}{\left[\frac{250+5x}{18}\right]}=6\)

250 + 5*x* = 660

*x* = 82 km/hr.

__Question 3.__Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction?

**Options:**

- 10
- 12
- 15
- 20

**Answer: Option B**

Speed of the first train = \(\left(\frac{120}{10}\right)m/sec = 12m/sec.\)

Speed of the second train =\(\left(\frac{120}{15}\right)m/sec = 8m/sec\)

Relative speed = (12 + 8) = 20 m/sec.

Therefore Required time = \(\left[\frac{(120+120)}{20}\right]sec = 12sec.\)

__Question 4.__A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?

**Options:**

- 66 km/hr
- 72 km/hr
- 78 km/hr
- 81 km/hr

**Answer: Option D**

4.5 km/hr =\(\left(4.5\times\frac{5}{18}\right)m/sec=\frac{5}{4}m/sec = 1.25m/sec, and\)

5.4 km/hr =\(\left(5.4\times\frac{5}{18}\right)m/sec=\frac{3}{2}m/sec = 1.5m/sec,\)

Let the speed of the train be *x* m/sec.

Then, (*x* - 1.25) x 8.4 = (*x* - 1.5) x 8.5

8.4*x* - 10.5 = 8.5*x* - 12.75

0.1*x* = 2.25

*x* = 22.5

Speed of the train =\(\left(22.5\times\frac{18}{5}\right)km/hr = 81km/hr.\)

__Question 5.__A train travelling at 48 kmph completely crosses another train having half its length and travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. The length of the platform is

**Options:**

- 400 m
- 450 m
- 560 m
- 600 m

**Answer: Option A**

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

Then, the length of the second train is\(\left(\frac{x}{2}\right)metres.\)

Relative speed = (48 + 42) kmph =\(\left(90\times\frac{5}{18}\right)m/sec = 25m/sec\)

\(\frac{[x+(\frac{x}{2})]}{25}=12 or \frac{3x}{2}=300 . or . x= 200\)

Therefore Length of first train = 200 m.

Let the length of platform be *y* metres.

Speed of the first train = \(\left(48\times\frac{5}{18}\right)m/sec = \frac{40}{3}m/sec\)

\(\therefore\left(200+y\right)\times\frac{3}{40}= 45\)

600 + 3*y* = 1800

*y* = 400 m.

__Question 6.__Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?

**Options:**

- 9 a.m.
- 10 a.m.
- 10.30 a.m.
- 11 a.m.

**Answer: Option B**

Suppose they meet *x* hours after 7 a.m.

Distance covered by A in *x* hours = 20*x* km.

Distance covered by B in (*x* - 1) hours = 25(*x* - 1) km.

Therefore 20*x* + 25(*x* - 1) = 110

45*x* = 135

*x* = 3.

So, they meet at 10 a.m.

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