Problems on Trains Questions and Answers

Answer: C

Trains moving in the same direction close at the difference of their speeds; in opposite directions, at the sum.

\[ \text{relative speed} = 60 + 90 = 150 \text{ km/hr} = 150 \times \frac{5}{18} = \frac{125}{3} \text{ m/s} \]

\[ \text{total length} = 1100 + 900 = 2000 \text{ m} \]

\[ t = \frac{2000 \times 3}{125} = 48 \text{ seconds} \]

The lengths are given in kilometres and must be converted to metres first.

Answer: C

The jogger runs the same way, so the train closes at the difference of the speeds.

\[ \text{relative speed} = 45 - 9 = 36 \text{ kmph} = 36 \times \frac{5}{18} = 10 \text{ m/s} \]

To pass him completely, the train must cover the 240 m gap plus its own 120 m:

\[ t = \frac{240 + 120}{10} = 36 \text{ seconds} \]

Covering only the 240 m gap brings the engine level with the jogger, not the whole train past him.

Answer: A

Trains moving in the same direction close at the difference of their speeds; in opposite directions, at the sum.

\[ \text{relative speed} = 120 + 80 = 200 \text{ kmph} = 200 \times \frac{5}{18} = \frac{500}{9} \text{ m/s} \]

\[ \text{total length} = \frac{500}{9} \times 9 = 500 \text{ m} \]

\[ \text{other train} = 500 - 270 = 230 \text{ m} \]

Answer: D

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]

\[ 72 \text{ kmph} = 20 \text{ m/s} \]

\[ \text{total distance} = 20 \times 26 = 520 \text{ m} \]

\[ \text{train} = 520 - 250 = 270 \text{ m} \]

Answer: C

Trains moving in the same direction close at the difference of their speeds; in opposite directions, at the sum.

\[ \text{relative speed} = \frac{100 + 100}{8} = 25 \text{ m/s} = 25 \times \frac{18}{5} = 90 \text{ km/hr} \]

With speeds \( v \) and \( 2v \):

\[ 3v = 90 \quad\Longrightarrow\quad v = 30, \qquad 2v = 60 \text{ km/hr} \]

Answer: D

Trains moving in the same direction close at the difference of their speeds; in opposite directions, at the sum.

\[ \text{relative speed} = 60 + 40 = 100 \text{ km/hr} = 100 \times \frac{5}{18} = \frac{250}{9} \text{ m/s} \]

\[ t = \frac{140 + 160}{250/9} = \frac{300 \times 9}{250} = 10.8 \text{ seconds} \]

Answer: B

The man walks toward the train, so the speeds add.

\[ \text{relative speed} = 60 + 6 = 66 \text{ kmph} = 66 \times \frac{5}{18} = \frac{55}{3} \text{ m/s} \]

A man is a point, so only the train's own length is covered:

\[ t = \frac{110 \times 3}{55} = 6 \text{ seconds} \]

Had he walked the same way as the train, the relative speed would be 54 kmph and the answer about 7.3 seconds.

Answer: C

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]

\[ 78 \text{ km/hr} = 78 \times \frac{5}{18} = \frac{65}{3} \text{ m/s} \]

\[ \text{total distance in 60 s} = \frac{65}{3} \times 60 = 1300 \text{ m} \]

\[ \text{tunnel} = 1300 - 800 = 500 \text{ metres} \]

Answer: B

A signal pole is a point, so that crossing gives the speed.

\[ v = \frac{300}{18} = \frac{50}{3} \text{ m/s} \]

\[ \text{total distance over the platform} = \frac{50}{3} \times 39 = 650 \text{ m} \]

\[ \text{platform} = 650 - 300 = 350 \text{ m} \]

The data are complete, so "data inadequate" is a trap; 650 m is the combined distance, not the platform.

Answer: B

The pole crossing covers the train alone, the platform crossing the train plus 100 m.

\[ L = 15v, \qquad L + 100 = 25v \]

Subtracting:

\[ 100 = 10v \quad\Longrightarrow\quad v = 10 \text{ m/s} \]

\[ L = 15 \times 10 = 150 \text{ m} \]