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.

\[ v_{1} + v_{2} = \frac{276}{12} = 23 \text{ km/hr} \]

The slower train is 14 km/hr behind the faster, so

\[ v_{\text{slow}} = \frac{23 - 14}{2} = 4.5 \text{ km/hr} \]

The faster is then 18.5 km/hr, and \( 4.5 + 18.5 = 23 \) as required.

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

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

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

\[ 13 - 8 = 5 \text{ km/hr} \]

Because they travel the same way, only the excess of B's speed over A's brings them together. In opposite directions the answer would be their sum, 21 km/hr - which is why that value is offered.

Answer: A

In one hour of clock time the train covers 36 km instead of 40, losing 4 km of travel.

\[ t = \frac{4}{40} \text{ h} = \frac{1}{10} \text{ h} = 6 \text{ minutes} \]

The lost distance is timed at the running speed of 40 km/hr, since that is the speed it would have been covered at.

Answer: A

Passing a platform means the train travels its own length plus the platform's, from the moment its front enters to the moment its rear leaves.

\[ 31 + 175 = 206 \text{ m} \]

No speed or time is needed, so "cannot be determined" is a trap - the distance follows from the two lengths alone.

Answer: C

Let the train's speed be \( v \) km/hr. Passing each boy covers the train's own length at the relative speed.

\[ L = (v - 8) \times \frac{5}{18} \times 36 = (v - 16) \times \frac{5}{18} \times 40 \]

The factor \( \tfrac{5}{18} \) is common:

\[ 36(v - 8) = 40(v - 16) \quad\Longrightarrow\quad 36v - 288 = 40v - 640 \]

\[ 4v = 352 \quad\Longrightarrow\quad v = 88 \text{ km/hr} \]

\[ L = (88 - 8) \times \frac{5}{18} \times 36 = 80 \times 10 = 800 \text{ m} \]

Answer: B

Two trains that meet and then take \( t_{A} \) and \( t_{B} \) to finish have speeds in the ratio

\[ \frac{v_{A}}{v_{B}} = \sqrt{\frac{t_{B}}{t_{A}}} \]

Here \( t_{A} = 24\tfrac{1}{2} \) h and \( t_{B} = 18 \) h:

\[ \frac{v_{A}}{v_{B}} = \sqrt{\frac{18}{24.5}} = \sqrt{\frac{36}{49}} = \frac{6}{7} \]

\[ v_{B} = 120 \times \frac{7}{6} = 140 \text{ km/hr} \]

The square root is essential: taking the times' ratio directly would give about 163 km/hr.

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

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.