Time & Distance Questions and Answers

Answer: D

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

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

\[ \text{relative speed} = 50 + 70 = 120 \text{ kmph} = \frac{100}{3} \text{ m/s} \]

\[ t = \frac{150 + 100}{100/3} = \frac{250 \times 3}{100} = 7.5 = 7\tfrac{1}{2} \text{ sec} \]

Answer: C

Crossing the platform covers the train plus the platform.

\[ v = \frac{100 + 125}{15} = 15 \text{ m/s} \]

\[ 15 \times \frac{18}{5} = 54 \text{ kmph} \]

Answer: C

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

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

\[ \text{total distance} = 20 \times 25 = 500 \text{ m} \]

\[ \text{platform} = 500 - 150 = 350 \text{ m} \]

Answer: B

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

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

\[ d = 20 \times 30 = 600 \text{ m} \]

Answer: D

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

\[ 36 \text{ kmph} = 10 \text{ m/s} \]

An electric pole has no length, so the train covers only its own 100 m:

\[ t = \frac{100}{10} = 10 \text{ sec} \]

Answer: B

Equal distances each way call for the harmonic mean, not the arithmetic one.

\[ \bar{v} = \frac{2 \times 20 \times 30}{20 + 30} = \frac{1200}{50} = 24 \text{ kmph} \]

The arithmetic mean of 25 kmph is too high, because more time is spent on the slower outward leg.

Answer: A

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

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

\[ \text{relative speed} = 72 - 36 = 36 \text{ kmph} = 10 \text{ m/s} \]

\[ t = \frac{200 + 300}{10} = 50 \text{ sec} \]

Both lengths count, because the whole of one train must pass the whole of the other.

Answer: C

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

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

\[ \text{relative speed} = 54 + 72 = 126 \text{ kmph} = 35 \text{ m/s} \]

To meet, only the 100 m gap between them has to close:

\[ t = \frac{100}{35} = \frac{20}{7} \text{ sec} \]

Their lengths play no part until they begin to pass each other.

Answer: B

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

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

\[ \text{relative speed} = 54 + 72 = 126 \text{ kmph} = 35 \text{ m/s} \]

To cross completely, the gap and both lengths must be covered:

\[ 100 + 100 + 200 = 400 \text{ m} \]

\[ t = \frac{400}{35} = \frac{80}{7} \text{ sec} \]

Meeting takes only \( \tfrac{20}{7} \) seconds; crossing takes four times as long.

Answer: A

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

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

Crossing a man in the other train covers only the faster train's length.

\[ \text{relative speed} = 72 - 36 = 36 \text{ kmph} = 10 \text{ m/s} \]

\[ L = 10 \times 27 = 270 \text{ m} \]

The slower train's own length is irrelevant, since the reference point is a single person.