Problems on Trains Questions and Answers

Answer: B

The faster train's speed is 22 m/sec.

To pass each other completely, the trains must cover their combined length, \(140 + 100 = 240\) m. In the same direction they close at the difference of their speeds, and in opposite directions at the sum:

\[ \begin{aligned} u - v &= \frac{240}{60} = 4\ \text{m/s} \\ u + v &= \frac{240}{6} = 40\ \text{m/s} \end{aligned} \]

Adding, \(2u = 44\), so \(u = 22\) m/s, and the slower train runs at 18 m/s.

20 m/s is the average of the two speeds, and 42 and 28 m/s do not satisfy both equations.

Answer: C

The train's speed is 44 km/h. The train covers its own length of 450 m in 45 s relative to the man, so the relative speed is 450 / 45 = 10 m/s, which is 36 km/h.

They move in the same direction, so the man's speed is added to the relative speed to get the train's speed; 42, 43 and 45 do not satisfy this.

Check: a train at 44 km/h and a man running at 8 km/h in the same direction are 36 km/h apart, which is 10 m/s, and the train's 450 m take 45 s to pass him. If the man were running against the train the speeds would be added instead, and the train would then be 28 km/h, which is not an option.

Answer: D

Train B crosses the bridge in 45 seconds. Train A passes a pole in 27 seconds, so its speed is 360/27 = 40/3 m/s = 48 km/h. Train B runs the opposite way at 60 km/h, so the relative speed is 108 km/h, which is 30 m/s. They cross in 19 seconds, so the sum of their lengths is 30 x 19 = 570 m, and B is 570 - 360 = 210 m long.

\[ \text{Speed of B} = 60 \times \frac{5}{18} \]

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

\[ \text{Time} = \frac{210 + 540}{50/3} = 45 \text{ s} \]

The options 36, 40 and 42 seconds do not equal 750 m divided by the speed of 50/3 m/s, so the fourth option is correct.

Answer: B

The other train is 185 m long. The first train covers its own length of 240 m past the pole in 18 s, so its speed is \(\frac{240}{18} = \frac{40}{3}\) m/s, which is 48 km/h. The trains move in opposite directions, so the relative speed is \(48 + 42 = 90\) km/h, which is 25 m/s. In 17 s they cover \[ 25\times 17 = 425\text{ m} \] which is the sum of the two lengths. The other train is \(425 - 240 = 185\) m. Using the difference of speeds (as for the same direction) would give a different length, so 170, 195 and 200 do not arise.

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.