Problems on Trains Questions and Answers

Answer: C

A signal pole is a point, so that crossing gives the speed.

\[ v = \frac{300}{18} = \frac{50}{3} \text{ m/s} \]

\[ \text{total distance over the platform} = \frac{50}{3} \times 39 = 650 \text{ m} \]

\[ \text{platform} = 650 - 300 = 350 \text{ m} \]

The data are complete, so "data inadequate" is a trap; 650 m is the combined distance, not the platform.

Answer: A

Find the speed from the first crossing.

\[ v = \frac{150 + 550}{56} = \frac{700}{56} = 12.5 \text{ m/s} \]

\[ t = \frac{150 + 250}{12.5} = \frac{400}{12.5} = 32 \text{ seconds} \]

The train's own length must be added to both platform lengths; leaving it out of the second would give 20 seconds.

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: C

Let the speed of the slower train be x m/sec.

Then, speed of the train = 2x m/sec.

Relative speed =  ( x + 2x) = 3x m/sec.

(100 + 100)/8 = 3x => x = 25/3.

So, speed of the faster train = 50/3 = 50/3 * 18/5 = 60 km/hr.

Answer: C

Find the speed from the long run, then use it on the post.

\[ v = \frac{12 \text{ km}}{10 \text{ min}} = \frac{12000}{600} = 20 \text{ m/s} \]

A telegraph post is a point, so the 6 seconds covers only the train:

\[ L = 20 \times 6 = 120 \text{ m} \]

Converting 12 km in 10 minutes to 72 km/hr and then to m/s gives the same 20.

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: A

Overtaking covers both trains' lengths at the difference of the speeds.

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

\[ 2L = \frac{25}{9} \times 36 = 100 \text{ m} \]

\[ L = 50 \text{ m} \]

The 100 m is the combined length; each train is half of it.

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

The four stops cost \( 4 \times 15 = 60 \) minutes, or one hour, so the stopped journey takes exactly one hour longer than the non-stop one.

\[ \frac{D}{45} - \frac{D}{50} = 1 \]

\[ D\left(\frac{50 - 45}{2250}\right) = 1 \quad\Longrightarrow\quad \frac{5D}{2250} = 1 \]

\[ D = 450 \text{ km} \]

Checking: 10 hours with stops against 9 hours without.

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.