Problems on Trains Questions and Answers

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.

Answer: A

In one hour of clock time the train covers 36 km instead of 40, losing 4 km of travel.

\[ t = \frac{4}{40} \text{ h} = \frac{1}{10} \text{ h} = 6 \text{ minutes} \]

The lost distance is timed at the running speed of 40 km/hr, since that is the speed it would have been covered at.

Answer: B

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

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]

\[ 60 \text{ km/hr} = 60 \times \frac{5}{18} = \frac{50}{3} \text{ m/s} \]

A pole has no length, so the train covers only itself:

\[ L = \frac{50}{3} \times 9 = 150 \text{ metres} \]

Answer: D

Correct answer: 360 m

\[\text{Speed} = 72 \times \frac{5}{18} = 20 \text{ m/s}\]

\[\text{Total distance covered} = 20 \times 30 = 600 \text{ m}\]

That total is the train plus the bridge, so the bridge is \(600 - 240 = 360\) m.

Answer: B

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]

\[ 72 \text{ km/hr} = 20 \text{ m/s} \]

\[ t = \frac{110 + 132}{20} = \frac{242}{20} = 12.1 \text{ seconds} \]

Both the train and the bridge must be cleared, so their lengths add.

Answer: B

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]

\[ 60 \text{ km/hr} = \frac{50}{3} \text{ m/s} \]

\[ L = \frac{50}{3} \times 9 = 150 \text{ m} \]

150 m is not among the listed values, so the answer is none of these. The data are complete, which rules out "cannot be determined".

Answer: B

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]

\[ 71 \text{ km/hr} = 71 \times \frac{5}{18} = 19.72 \text{ m/s} \]

\[ t = \frac{90 + 132}{19.72} = \frac{222}{19.72} = 11.26 \text{ seconds} \]

Crossing a bridge covers the train plus the bridge; the train's own 90 m must not be dropped.

Answer: B

Suppose they meet x hours after 7 a.m.

Distance covered by A in x hours = 20x km.

Distance covered by B in (x - 1) hours = 25(x - 1) km.

 20x + 25(x - 1) = 110

" src="https://www.indiabix.com/_files/images/aptitude/1-sym-imp.gif" /> 45x = 135

" src="https://www.indiabix.com/_files/images/aptitude/1-sym-imp.gif" /> x = 3.

So, they meet at 10 a.m.