Problems on Trains Questions and Answers

Answer: C

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

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

\[ \text{total distance} = 12.5 \times 30 = 375 \text{ m} \]

\[ \text{bridge} = 375 - 130 = 245 \text{ m} \]

That total covers the train plus the bridge, so the train's own length is subtracted at the end.

Answer: B

The distance is the same both ways, so equate the two expressions for it.

\[ 3v = 2(v + 6) \]

\[ 3v = 2v + 12 \quad\Longrightarrow\quad v = 12 \text{ km/hr} \]

\[ D = 3 \times 12 = 36 \text{ km} \]

Checking: at 18 km/hr the same 36 km takes 2 hours.

Answer: D

Passing the driver - a single point - means covering only the faster train's own length.

\[ \text{relative speed} = 50 - 20 = 30 \text{ kmph} = 30 \times \frac{5}{18} = \frac{25}{3} \text{ m/s} \]

\[ L = \frac{25}{3} \times 20 = \frac{500}{3} = 166.67 \approx 166 \text{ m} \]

Had it passed the whole slower train, both lengths would have counted.

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: C

Correct answer: 7.5 seconds

Convert the speed to metres per second by multiplying by \(\tfrac{5}{18}\).

\[72 \times \frac{5}{18} = 20 \text{ m/s}\]

Crossing a pole means covering just the train’s own length, since a pole has no length of its own.

\[t = \frac{150}{20} = 7.5 \text{ seconds}\]

Answer: D

A telegraph post has no length, so the train covers only its own 120 m.

\[ v = \frac{120}{6} = 20 \text{ m/s} \]

\[ 20 \times \frac{18}{5} = 72 \text{ km/hr} \]

The factor \( \tfrac{18}{5} \) converts m/s to km/hr.

Answer: B

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

Passing the driver of the other train means covering only the slower train's own length.

\[ \text{relative speed} = 45 + 30 = 75 \text{ km/hr} = 75 \times \frac{5}{18} = \frac{125}{6} \text{ m/s} \]

\[ t = \frac{500 \times 6}{125} = 24 \text{ seconds} \]

Passing the whole train would need 1000 m and take 48 seconds - which is why that value is on the list.

Answer: B

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

\[ 13 - 8 = 5 \text{ km/hr} \]

Because they travel the same way, only the excess of B's speed over A's brings them together. In opposite directions the answer would be their sum, 21 km/hr - which is why that value is offered.

Answer: D

The post crossing covers the train alone; the bridge crossing adds 264 m.

\[ L = 8v, \qquad L + 264 = 20v \]

\[ 12v = 264 \quad\Longrightarrow\quad v = 22 \text{ m/s} \]

\[ 22 \times \frac{18}{5} = 79.2 \text{ km/hr} \]

The train works out at 176 m long, though the question asks only for the speed.

Answer: B

A tree is a point, so the first crossing gives the speed.

\[ v = \frac{1200}{120} = 10 \text{ m/s} \]

\[ t = \frac{1200 + 700}{10} = 190 \text{ seconds} \]

Timing only the platform's 700 m would give 70 seconds; the train's own length is part of the crossing.