Problems on Trains Questions and Answers

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 

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 \( 2v \):

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

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