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: D

The train is 175 m long and the platform 200 m.

54 km/h is 15 m/s and 9 km/h is 2.5 m/s. Passing the man, the train covers its own length at the relative speed:

\[ \begin{aligned} \text{train} &= (15 - 2.5) \times 14 = 175\ \text{m} \\ \text{train + platform} &= 15 \times 25 = 375\ \text{m} \\ \text{platform} &= 375 - 175 = 200\ \text{m} \end{aligned} \]

210 and 140 come from ignoring the man's speed (15 × 14 = 210). The other pairs add up to 375 or less but give a train length that does not match the 14 seconds at the relative speed.

Answer: D

The train is 160 m long and the platform is 140 m: (160, 140m).

Convert the speeds to m/s: 54 km/h is 15 m/s and 6 km/h is 5/3 m/s. Passing a man walking the same way uses the relative speed.

\[ \begin{aligned} \text{relative} &= 15 - \tfrac{5}{3} = \tfrac{40}{3}\ \text{m/s} \\ \text{train} &= \tfrac{40}{3} \times 12 = 160\ \text{m} \end{aligned} \]

\[ \begin{aligned} \text{train + platform} &= 15 \times 20 = 300 \\ \text{platform} &= 300 - 160 = 140\ \text{m} \end{aligned} \]

All four options agree on 160 m for the train, so the question turns on the platform. 120, 130 and 150 m would need the train to cover 280, 290 or 310 m in 20 seconds, which is not 15 m/s.

Answer: C

The trains cross each other in 9 seconds.

Moving in opposite directions, their relative speed is the sum of their speeds, and to cross they must cover the sum of their lengths:

\[ \begin{aligned} v &= 72 + 36 = 108 \text{ km/h} \\ &= 108 \times \tfrac{5}{18} = 30 \text{ m/s} \\ t &= \frac{150 + 120}{30} = 9 \text{ s} \end{aligned} \]

If the trains moved in the same direction, the relative speed would be 36 km/h (10 m/s) and the time 27 seconds, which is not offered. 5, 7 and 11 seconds do not fit the given lengths and speeds.

Answer: B

Both Statement I and Statement II are false.

Statement I: \(30 \text{ km/h} = 30 \times \tfrac{5}{18} = \tfrac{25}{3}\) m/s, so passing a man takes \(200 \div \tfrac{25}{3} = 24\) seconds, not 12.

Statement II: the speeds are \(180/20 = 9\) m/s and \(180/30 = 6\) m/s. In opposite directions they close at 15 m/s and must cover both lengths:

\[ \frac{180 + 180}{9 + 6} = \frac{360}{15} = 24 \text{ s} \]

So the time is 24 seconds, not 20. Both statements are false, which rules out every option calling either one true.

Answer: C

The speed of the train is 72 kmph.

To cross a platform, the train must cover its own length plus the platform's:

\[ \begin{aligned} v &= \frac{240 + 300}{27} = 20\ \text{m/s} \\ &= 20 \times \frac{18}{5} = 72\ \text{km/h} \end{aligned} \]

Using only the platform length gives \(300/27 \times 18/5 = 40\) km/h, and only the train's length gives 32 km/h, neither listed. 66, 60 and 64 do not follow from the total distance.

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.