Problems on Trains Questions and Answers

Answer: C

The first train has a one-hour head start.

\[ \text{covered by 11 pm} = 40 \times 1 = 40 \text{ km} \]

\[ \text{gap remaining} = 220 - 40 = 180 \text{ km} \]

From 11 pm they approach at \( 40 + 50 = 90 \) km/hr:

\[ t = \frac{180}{90} = 2 \text{ hours} \]

So they meet at 1 am. Ignoring the head start would give a meeting at about 12:27 am.

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: D

Let us assume for this question that the length of train is x metres and it is assumed to be running at the speed of y m/sec.

A pole is assumed as a point object.

=>Time taken by the train to pass the pole= x/y

=>15=x/y

=>y=x/15

Now, the train passes the platform which is 100 m in length in 25 seconds.

=>x+100/y=25

=>x+100/25=y

Equating speed generated from both cases:

x+100/25=x/15

Therefore x = 150 metres.

Answer: C

The jogger runs the same way, so the train closes at the difference of the speeds.

\[ \text{relative speed} = 45 - 9 = 36 \text{ kmph} = 36 \times \frac{5}{18} = 10 \text{ m/s} \]

To pass him completely, the train must cover the 240 m gap plus its own 120 m:

\[ t = \frac{240 + 120}{10} = 36 \text{ seconds} \]

Covering only the 240 m gap brings the engine level with the jogger, not the whole train past him.

Answer: C

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

\[ \text{relative speed} = \frac{100 + 100}{8} = 25 \text{ m/s} = 25 \times \frac{18}{5} = 90 \text{ km/hr} \]

With speeds \( v \) and \( 2v \):

\[ 3v = 90 \quad\Longrightarrow\quad v = 30 \text{ km/hr} \]

\[ \text{faster train} = 60 \text{ km/hr} \]

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: C

Speed = [54 * 5/18] m/sec = 15 m/sec.

Length of the train = (15 * 20) m = 300 m. 

Let the length of the platform be x meters. 

Then, x + 300 / 36 = 15

x + 300 = 540

x = 240 m. 

Answer: D

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

\[ 72 \text{ kmph} = 20 \text{ m/s} \]

\[ \text{total distance} = 20 \times 26 = 520 \text{ m} \]

\[ \text{train} = 520 - 250 = 270 \text{ m} \]

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: 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.