Problems on Trains Questions and Answers

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

The four stops cost \( 4 \times 15 = 60 \) minutes, or one hour, so the stopped journey takes exactly one hour longer than the non-stop one.

\[ \frac{D}{45} - \frac{D}{50} = 1 \]

\[ D\left(\frac{50 - 45}{2250}\right) = 1 \quad\Longrightarrow\quad \frac{5D}{2250} = 1 \]

\[ D = 450 \text{ km} \]

Checking: 10 hours with stops against 9 hours without.

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: B

A telegraph post has no length, so the first crossing covers only the train.

\[ L = 8v \]

The bridge crossing covers the train plus 264 m:

\[ L + 264 = 20v \]

Substituting the first into the second:

\[ 8v + 264 = 20v \quad\Longrightarrow\quad 12v = 264 \quad\Longrightarrow\quad v = 22 \text{ m/s} \]

\[ L = 8 \times 22 = 176 \text{ m} \]

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: D

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

They approach at \( 120 + 100 = 220 \) km/hr, but the 40 km gap in distance comes from the difference in their speeds.

\[ 120t - 100t = 40 \quad\Longrightarrow\quad 20t = 40 \quad\Longrightarrow\quad t = 2 \text{ h} \]

\[ PQ = 220 \times 2 = 440 \text{ km} \]

Checking: one train covers 240 km and the other 200 - exactly 40 km apart.

Answer: C

Let each train be \( L \) metres long. Crossing a pole gives each speed:

\[ v_{P} = \frac{L}{5}, \qquad v_{Q} = \frac{L}{6} \]

Moving the same way, they close at the difference:

\[ v_{P} - v_{Q} = \frac{L}{5} - \frac{L}{6} = \frac{L}{30} \]

To cross, the faster must cover both lengths, \( 2L \):

\[ t = \frac{2L}{L/30} = 60 \text{ seconds} \]

The length cancels, so it never needs a value.

Answer: A

Find the speed from the first crossing.

\[ v = \frac{150 + 550}{56} = \frac{700}{56} = 12.5 \text{ m/s} \]

\[ t = \frac{150 + 250}{12.5} = \frac{400}{12.5} = 32 \text{ seconds} \]

The train's own length must be added to both platform lengths; leaving it out of the second would give 20 seconds.

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: 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.