Problems on Trains Questions and Answers

Answer: D

Each train is 50 m long.

Moving the same way, the faster train gains at the difference of the speeds, and to pass completely it must cover both lengths, \(2L\):

\[ \begin{aligned} v &= 46 - 36 = 10\ \text{km/h} \\ &= 10 \times \tfrac{5}{18} = \tfrac{25}{9}\ \text{m/s} \\ 2L &= \tfrac{25}{9} \times 36 = 100\ \text{m} \\ L &= 50\ \text{m} \end{aligned} \]

100 m is the combined length, so 80 m and 72 m do not fit. 36 m comes from treating the relative speed as 1 m/s. Adding the speeds would be right only for trains moving in opposite directions.

Answer: C

The ratio of their speeds is 3:2.

Let the speeds be \(x\) and \(y\). Passing a standing man, each train covers its own length, so the lengths are \(27x\) and \(17y\). Crossing each other in opposite directions, they cover the sum of their lengths at the sum of their speeds:

\[ \begin{aligned} \frac{27x + 17y}{x + y} &= 23 \\ 27x + 17y &= 23x + 23y \\ 4x = 6y &\Rightarrow x : y = 3 : 2 \end{aligned} \]

2:3 is the same ratio reversed; the working shows the train that takes 27 seconds to pass the man is the faster one, so it must also be much the longer. 1:3 and 3:4 do not satisfy the equation.

Answer: B

The train passes the platform in 24 sec.

To pass the platform completely, the train covers its own length plus the platform's: \(140 + 260 = 400\) m. Convert the speed to metres per second:

\[ \begin{aligned} 60\ \text{km/h} &= 60 \times \frac{5}{18} = \frac{50}{3}\ \text{m/s} \\ \text{time} &= 400 \div \frac{50}{3} = 24\ \text{s} \end{aligned} \]

Using only the platform's 260 m would give 15.6 s, and only the train's 140 m gives 8.4 s; neither is offered, and 36, 44 and 58 s do not follow from the 400 m distance.

Answer: A

The platform is 360 meter long.

Convert the speed: \(72 \times \frac{5}{18} = 20\) m/s. Passing a man, the train covers its own length; passing a platform, its length plus the platform's.

\[ \begin{aligned} \text{train} &= 20 \times 14 = 280\ \text{m} \\ \text{train + platform} &= 20 \times 32 = 640\ \text{m} \\ \text{platform} &= 640 - 280 = 360\ \text{m} \end{aligned} \]

280 m is the train's own length, the usual trap.

Answer: A

The train is 250 m long.

Crossing a pole, the train covers its own length \(L\) in 15 s. Crossing the bridge, it covers \(L + 500\) in 45 s at the same speed.

\[ \begin{aligned} \frac{L}{15} &= \frac{L + 500}{45} \\ 3L &= L + 500 \\ L &= 250\ \text{m} \end{aligned} \]

Check: speed is \(250 / 15 = 16.67\) m/s, and \(750 / 16.67 = 45\) s for the bridge. The bridge takes three times as long, so it adds twice the train's length.

Answer: D

The platform is 175m long.

To cross a platform, the train must cover its own length plus the platform's. Convert the speed to metres per second first:

\[ \begin{aligned} 90\ \text{km/h} &= 90 \times \frac{5}{18} = 25\ \text{m/s} \\ \text{distance} &= 25 \times 15 = 375\ \text{m} \\ \text{platform} &= 375 - 200 = 175\ \text{m} \end{aligned} \]

375 m is the total distance, which forgets to subtract the train's length; 225 m and 150 m do not follow from the data.

Answer: B

The train is 300 m long.

Passing a pole, the train covers its own length \(L\) in 30 s. Passing the platform, it covers \(L + 200\) in 50 s, at the same speed.

\[ \begin{aligned} \frac{L}{30} &= \frac{L + 200}{50} \\ 50L &= 30L + 6000 \\ L &= 300\ \text{m} \end{aligned} \]

Shortcut: the extra 20 seconds are spent covering the 200 m platform, so the speed is 10 m/s, and in 30 s the train covers 300 m. 320, 280 and 250 m do not give the same speed in both cases.

Answer: A

The faster train is 100 m long.

The trains move in the same direction, so the faster train passes the man at the difference of their speeds. In crossing him, it covers its own length.

\[ \begin{aligned} v &= 50 - 30 = 20 \text{ km/h} \\ &= 20 \times \frac{5}{18} = \frac{50}{9} \text{ m/s} \\ L &= \frac{50}{9} \times 18 = 100 \text{ m} \end{aligned} \]

Using the sum of the speeds (80 km/h), as for trains moving in opposite directions, would give 400 m, which is not offered. 120 m, 90 m and 150 m do not fit a relative speed of 20 km/h over 18 seconds.

Answer: A

The speed is 45 km/h.

To cross a pole, the train covers its own length.

\[ \begin{aligned} v &= \frac{150}{12} = 12.5\ \text{m/s} \\ &= 12.5 \times \frac{18}{5} = 45\ \text{km/h} \end{aligned} \]

Multiplying by \(\frac{5}{18}\) instead of \(\frac{18}{5}\) gives about 3.5, and 50 or 60 come from rounding 12.5 m/s up before converting.

Answer: B

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

\[ \text{Kerala express length} = \frac{2}{3} \times 1200 = 800 \text{ m} \]

\[ \text{relative speed} = 80 - 60 = 20 \text{ km/hr} = 20 \times \frac{5}{18} = \frac{50}{9} \text{ m/s} \]

To overtake, the Jammu express must cover both lengths:

\[ t = \frac{1200 + 800}{50/9} = \frac{2000 \times 9}{50} = 360 \text{ s} = 6 \text{ minutes} \]

Overtaking always needs the sum of the lengths, even though the speeds subtract.