Problems on Trains Questions and Answers

Answer: A

Passing a platform means the train travels its own length plus the platform's, from the moment its front enters to the moment its rear leaves.

\[ 31 + 175 = 206 \text{ m} \]

No speed or time is needed, so "cannot be determined" is a trap - the distance follows from the two lengths alone.

Answer: B

A pole has no length, so the first crossing gives the speed.

\[ v = \frac{240}{24} = 10 \text{ m/s} \]

\[ t = \frac{240 + 650}{10} = \frac{890}{10} = 89 \text{ seconds} \]

Using only the platform's 650 m would give 65 seconds - the first distractor.

Answer: A

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

\[ 45 \text{ km/hr} = 12.5 \text{ m/s} \]

\[ t = \frac{360 + 140}{12.5} = \frac{500}{12.5} = 40 \text{ seconds} \]

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: B

Two trains that meet and then take \( t_{1} \) and \( t_{2} \) to finish have speeds in the ratio

\[ \frac{v_{1}}{v_{2}} = \sqrt{\frac{t_{2}}{t_{1}}} \]

\[ = \sqrt{\frac{16}{9}} = \frac{4}{3} \quad\Longrightarrow\quad 4 : 3 \]

The square root is essential; taking the times directly would give 9 : 16, which is offered as the last option and is also the wrong way round.

Answer: B

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

\[ 63 \text{ km/hr} = 63 \times \frac{5}{18} = 17.5 \text{ m/s} \]

A tree is a point, so the train covers only its own length:

\[ t = \frac{280}{17.5} = 16 \text{ seconds} \]

The conversion is where the answer is decided: at 17.5 m/s the 280 m takes exactly 16 seconds, not 15.

Answer: C

The pole crossing covers only the train:

\[ L = 8v \]

The bridge crossing covers the train plus 200 m:

\[ L + 200 = 28v \quad\Longrightarrow\quad 8v + 200 = 28v \]

\[ 20v = 200 \quad\Longrightarrow\quad v = 10 \text{ m/s} \]

\[ L = 80 \text{ m} \]

Only one option gives a pair consistent with both crossings.

Answer: B

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

\[ \text{Kerala express length} = \frac{2}{3} \times 1200 = 800 \text{ m} \]

\[ \text{relative speed} = 80 - 60 = 20 \text{ km/hr} = 20 \times \frac{5}{18} = \frac{50}{9} \text{ m/s} \]

To overtake, the Jammu express must cover both lengths:

\[ t = \frac{1200 + 800}{50/9} = \frac{2000 \times 9}{50} = 360 \text{ s} = 6 \text{ minutes} \]

Overtaking always needs the sum of the lengths, even though the speeds subtract.

Answer: B

Correct answer: 180 m

Since both move the same way, the speeds subtract.

\[60 - 6 = 54 \text{ km/hr} = 54 \times \frac{5}{18} = 15 \text{ m/s}\]

Passing a man means covering only the train’s length.

\[L = 15 \times 12 = 180 \text{ m}\]

Answer: B

Let the train's speed be \( v \) kmph. Passing a walking person covers the train's own length at the relative speed.

\[ L = (v - 2) \times \frac{5}{18} \times 9 = (v - 4) \times \frac{5}{18} \times 10 \]

\[ 9(v - 2) = 10(v - 4) \quad\Longrightarrow\quad 9v - 18 = 10v - 40 \]

\[ v = 22 \text{ kmph} \]

\[ L = 20 \times \frac{5}{18} \times 9 = 50 \text{ m} \]