Problems on Trains Questions and Answers

Answer: B

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

\[ 54 \text{ km/hr} = 15 \text{ m/s} \]

The man is a point, so that crossing gives the train's length:

\[ L = 15 \times 20 = 300 \text{ m} \]

The platform crossing covers train plus platform:

\[ 15 \times 36 = 540 \quad\Longrightarrow\quad P = 540 - 300 = 240 \text{ m} \]

Answer: B

Crossing a man in the other train covers only the faster train's length.

\[ \text{relative speed} = 50 - 30 = 20 \text{ km/hr} = 20 \times \frac{5}{18} = \frac{50}{9} \text{ m/s} \]

\[ L = \frac{50}{9} \times 18 = 100 \text{ m} \]

The slower train's length is irrelevant, because the reference point is a single person, not the whole train.

Answer: B

Three stops of 20 minutes cost exactly one hour, so the journey with stops takes an hour longer.

\[ \frac{D}{40} - \frac{D}{60} = 1 \]

\[ D\left(\frac{3 - 2}{120}\right) = 1 \quad\Longrightarrow\quad D = 120 \text{ km} \]

Checking: 3 hours with stops against 2 hours of pure running.

Answer: C

Total distance = addition of length of the two trains = 140 + 120 = 260 metres
As the two trains are travelling in the same direction, their relative speed is:
v = | v1 - v2 | = | 40 - 60 | = 20 km/hr = 20*1000/60 = 1000/3 metres/min
t = 260/ 1000*3
t = 0.78 minutes

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.