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: B

Each crossing covers the train plus that platform.

\[ L + 120 = 15v, \qquad L + 180 = 18v \]

Subtracting:

\[ 60 = 3v \quad\Longrightarrow\quad v = 20 \text{ m/s} \]

\[ L = 15 \times 20 - 120 = 180 \text{ m} \]

Subtraction removes the unknown length in a single step, leaving the speed.

Answer: B

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

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

Crossing a pole covers only the train's own length:

\[ L = 12.5 \times 12 = 150 \text{ m} \]

Crossing a bridge covers the train plus the bridge:

\[ t = \frac{150 + 600}{12.5} = \frac{750}{12.5} = 60 \text{ seconds} \]

Answer: B

Correct answer: 250 m

Let the length be \(L\) and the speed \(v\). Passing the pole covers \(L\), and passing the platform covers \(L + 250\).

\[\frac{L}{v} = 15, \qquad \frac{L + 250}{v} = 30\]

Subtracting, \(\tfrac{250}{v} = 15\), so \(v = \tfrac{50}{3}\) m/s and \(L = 15 \times \tfrac{50}{3} = 250\) m.

The extra 15 seconds is exactly the time to cover the platform alone.

Answer: B

The first train has a half-hour head start.

\[ \text{lead at 9:00} = 60 \times 0.5 = 30 \text{ km} \]

The second closes at \( 75 - 60 = 15 \) km/hr:

\[ t = \frac{30}{15} = 2 \text{ h} \]

\[ \text{distance from Jaipur} = 75 \times 2 = 150 \text{ km} \]

Checking with the first train: \( 60 \times 2.5 = 150 \) km.

Answer: C

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

The crossing time is

\[ t = \frac{L_{1} + L_{2}}{v_{1} + v_{2}} \]

which needs both speeds and both lengths. One speed is already given in the question.

Statement A supplies the second speed but no lengths - the numerator stays unknown.

Statement B supplies both lengths but only one speed is known - the denominator stays unknown.

Together they give \( \tfrac{600}{180 \times 5/18} = 12 \) seconds, so both statements together are needed.

Answer: D

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

\[ \text{relative speed} = \frac{108 + 112}{6} = \frac{220}{6} = 36.67 \text{ m/s} \]

\[ 36.67 \times \frac{18}{5} = 132 \text{ km/hr} \]

\[ v_{2} = 132 - 50 = 82 \text{ km/hr} \]

The relative speed is the sum here, so the known 50 km/hr is subtracted from it to leave the second train's speed.

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

A telegraph post is a point, so each crossing gives that train's speed.

\[ v_{1} = \frac{120}{10} = 12 \text{ m/s}, \qquad v_{2} = \frac{120}{15} = 8 \text{ m/s} \]

Running opposite ways they close at the sum and must cover both lengths:

\[ t = \frac{120 + 120}{12 + 8} = \frac{240}{20} = 12 \text{ seconds} \]

Answer: A

Each crossing covers the train plus that platform.

\[ L + 150 = 15v, \qquad L + 250 = 20v \]

\[ 100 = 5v \quad\Longrightarrow\quad v = 20 \text{ m/s} \]

\[ L = 15 \times 20 - 150 = 150 \text{ m} \]