Problems on Trains Questions and Answers

Answer: B

Correct answer: 50 m

The relative speed is the difference, since they travel the same way.

\[46 - 36 = 10 \text{ km/hr} = 10 \times \frac{5}{18} = \frac{25}{9} \text{ m/s}\]

\[\text{Distance covered} = \frac{25}{9} \times 36 = 100 \text{ m}\]

That 100 m is the combined length of both trains, so each is 50 m long.

Answer: C

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

\[ 78 \text{ km/hr} = 78 \times \frac{5}{18} = \frac{65}{3} \text{ m/s} \]

\[ \text{total distance in 60 s} = \frac{65}{3} \times 60 = 1300 \text{ m} \]

\[ \text{tunnel} = 1300 - 800 = 500 \text{ metres} \]

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: B

The pole crossing covers the train alone, the platform crossing the train plus 100 m.

\[ L = 15v, \qquad L + 100 = 25v \]

Subtracting:

\[ 100 = 10v \quad\Longrightarrow\quad v = 10 \text{ m/s} \]

\[ L = 15 \times 10 = 150 \text{ m} \]

Answer: A

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

\[ 144 \text{ km/hr} = 144 \times \frac{5}{18} = 40 \text{ m/s} \]

An electric pole has no length, so only the train's own 100 m is covered:

\[ t = \frac{100}{40} = 2.5 \text{ seconds} \]

Answer: D

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

\[ \text{relative speed} = 12 + 23 = 35 \text{ m/s} \]

\[ t = \frac{126 + 119}{35} = \frac{245}{35} = 7 \text{ seconds} \]

The speeds are already in m/s, so no conversion is needed.

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

The extra 60 km comes from the difference in speeds.

\[ 25t - 20t = 60 \quad\Longrightarrow\quad 5t = 60 \quad\Longrightarrow\quad t = 12 \text{ h} \]

They approach at \( 20 + 25 = 45 \) km/hr:

\[ D = 45 \times 12 = 540 \text{ km} \]

Checking: 240 km and 300 km, differing by exactly 60.

Answer: B

The first train has a half-hour head start.

\[ \text{lead at 8:30} = 90 \times 0.5 = 45 \text{ km} \]

The second closes at \( 120 - 90 = 30 \) km/hr:

\[ t = \frac{45}{30} = 1.5 \text{ h} \]

\[ \text{distance from Kolkata} = 120 \times 1.5 = 180 \text{ km} \]

Checking with the first train: \( 90 \times 2 = 180 \) km in the two hours since 8 pm.

Answer: B

Length of the two trains = 600m + 400m 

Speed of the first train = X

Speed of the second train= 48 Kmph 

1000/X - 48 = 180

1000/x - 48 * 5/18 = 180

50 = 9X - 120

X = 68 Kmph