Problems on Trains Questions and Answers

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: 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.