Problems on Trains Questions and Answers

Answer: D

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

They approach at \( 120 + 100 = 220 \) km/hr, but the 40 km gap in distance comes from the difference in their speeds.

\[ 120t - 100t = 40 \quad\Longrightarrow\quad 20t = 40 \quad\Longrightarrow\quad t = 2 \text{ h} \]

\[ PQ = 220 \times 2 = 440 \text{ km} \]

Checking: one train covers 240 km and the other 200 - exactly 40 km apart.

Answer: C

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

\[ \text{relative speed} = \frac{100 + 100}{8} = 25 \text{ m/s} = 25 \times \frac{18}{5} = 90 \text{ km/hr} \]

With speeds \( v \) and \( 2v \):

\[ 3v = 90 \quad\Longrightarrow\quad v = 30, \qquad 2v = 60 \text{ km/hr} \]

Answer: C

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

\[ 78 \text{ km/hr} = 78 \times \frac{5}{18} = \frac{65}{3} \text{ m/s} \]

\[ \text{total distance in 60 s} = \frac{65}{3} \times 60 = 1300 \text{ m} \]

\[ \text{tunnel} = 1300 - 800 = 500 \text{ metres} \]

Answer: C

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

\[ \text{relative speed} = 30 + 36 = 66 \text{ km/hr} = 66 \times \frac{5}{18} = 18.33 \text{ m/s} \]

\[ t = \frac{156.62 + 100}{18.33} = \frac{256.62}{18.33} \approx 14 \text{ seconds} \]

Both lengths are covered, since the trains must clear each other completely.

Answer: A

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

\[ \text{relative speed} = 120 + 80 = 200 \text{ kmph} = 200 \times \frac{5}{18} = \frac{500}{9} \text{ m/s} \]

\[ \text{total length} = \frac{500}{9} \times 9 = 500 \text{ m} \]

\[ \text{other train} = 500 - 270 = 230 \text{ m} \]

Answer: D

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

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

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

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

Answer: 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: B

The distance is the same both ways, so equate the two expressions for it.

\[ 3v = 2(v + 6) \]

\[ 3v = 2v + 12 \quad\Longrightarrow\quad v = 12 \text{ km/hr} \]

\[ D = 3 \times 12 = 36 \text{ km} \]

Checking: at 18 km/hr the same 36 km takes 2 hours.

Answer: D

Passing the driver - a single point - means covering only the faster train's own length.

\[ \text{relative speed} = 50 - 20 = 30 \text{ kmph} = 30 \times \frac{5}{18} = \frac{25}{3} \text{ m/s} \]

\[ L = \frac{25}{3} \times 20 = \frac{500}{3} = 166.67 \approx 166 \text{ m} \]

Had it passed the whole slower train, both lengths would have counted.

Answer: B

A pole has no length, so the first crossing gives the speed.

\[ v = \frac{240}{24} = 10 \text{ m/s} \]

\[ t = \frac{240 + 650}{10} = \frac{890}{10} = 89 \text{ seconds} \]

Using only the platform's 650 m would give 65 seconds - the first distractor.