Problems on Trains Questions and Answers

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

The man walks the same way, so the speeds subtract.

\[ \text{relative speed} = 63 - 3 = 60 \text{ km/hr} = 60 \times \frac{5}{18} = \frac{50}{3} \text{ m/s} \]

\[ t = \frac{500 \times 3}{50} = 30 \text{ seconds} \]

Only the train's own length counts, since the man occupies a single point.

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: C

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

\[ \text{relative speed} = \frac{120 + 120}{12} = 20 \text{ m/s} \]

Both trains have the same speed \( v \), so

\[ 2v = 20 \quad\Longrightarrow\quad v = 10 \text{ m/s} = 10 \times \frac{18}{5} = 36 \text{ km/hr} \]

Forgetting to halve the relative speed gives 72 km/hr - the last option.

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: 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: B

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

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

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

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

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

Answer: D

Both walkers move the same way as the train, so the relative speeds are differences.

\[ L = (v - 4.5) \times \frac{5}{18} \times 8.4 = (v - 5.4) \times \frac{5}{18} \times 8.5 \]

\[ 8.4(v - 4.5) = 8.5(v - 5.4) \]

\[ 8.4v - 37.8 = 8.5v - 45.9 \quad\Longrightarrow\quad 0.1v = 8.1 \]

\[ v = 81 \text{ km/hr} \]

The train's length works out at 75.6 m, though the question asks only for the speed.

Answer: A

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

Let the first train be \( L \) long; the second is \( \tfrac{L}{2} \).

\[ \text{relative speed} = 48 + 42 = 90 \text{ kmph} = 25 \text{ m/s} \]

\[ L + \frac{L}{2} = 25 \times 12 = 300 \quad\Longrightarrow\quad 1.5L = 300 \]

\[ L = 200 \text{ m} \]

Crossing the platform, the train travels alone at 48 kmph \( = \tfrac{40}{3} \) m/s:

\[ 200 + P = \frac{40}{3} \times 45 = 600 \quad\Longrightarrow\quad P = 400 \text{ m} \]

Answer: B

Suppose they meet x hours after 7 a.m.

Distance covered by A in x hours = 20x km.

Distance covered by B in (x - 1) hours = 25(x - 1) km.

 20x + 25(x - 1) = 110

" src="https://www.indiabix.com/_files/images/aptitude/1-sym-imp.gif" /> 45x = 135

" src="https://www.indiabix.com/_files/images/aptitude/1-sym-imp.gif" /> x = 3.

So, they meet at 10 a.m.