Time & Distance Questions and Answers

Answer: A

In one minute the train travels

\[ 120 \text{ kmph} = \frac{120000}{60} = 2000 \text{ m/min} \]

To pass the field, the train must cover the field's length plus its own length:

\[ \text{side} = 2000 - 800 = 1200 \text{ m} = 1.2 \text{ km} \]

\[ \text{area} = (1.2)^{2} = 1.44 \text{ sq. km} \]

Forgetting to subtract the train's own length would give a 2 km side and an area of 4 sq. km - exactly the second option.

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.

Answer: B

Distance travelled by Ramu = 45 * 4 = 180 km

Somu travelled the same distance in 6 hours.

His speed = 180/6 = 30 km/hr

Hence in the conditional case, Ramu's speed = 45 - 9 = 36 km/hr and Somu's speed = 30 + 10 = 40km/hr.

Therefore travel time of Ramu and Somu would be 5 hours and 4.5 hours respectively. Hence difference in the time taken = 0.5 hours = 30 minutes. 

Answer: C

In one hour of clock time the train covers 36 km instead of 45, losing 9 km of travel.

\[ t = \frac{9}{45} \text{ h} = \frac{1}{5} \text{ h} = 12 \text{ minutes} \]

The lost distance is timed at the running speed of 45 kmph, since that is the speed it would have been covered at.

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: C

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

\[ 54 \text{ kmph} = 15 \text{ m/s} \]

\[ t = \frac{540 + 180}{15} = \frac{720}{15} = 48 \text{ sec} \]

Answer: A

Crossing the bridge covers the train plus the bridge.

\[ v = \frac{100 + 300}{45} = \frac{400}{45} = 8.89 \text{ m/s} \]

\[ 8.89 \times \frac{18}{5} = 32 \text{ kmph} \]

Answer: D

\[ v_{2} = \frac{400}{4} = 100 \text{ km/hr} \]

The speeds are in the ratio \( 7 : 8 \), so

\[ v_{1} = 100 \times \frac{7}{8} = 87.5 \text{ km/hr} \]

The first train is the slower of the two, so its speed must fall below 100 - which rules out one option without any calculation.

Answer: D

Each half of the distance is \( \tfrac{d}{2} \).

\[ \frac{d/2}{30} + \frac{d/2}{25} = 11 \]

\[ \frac{d}{60} + \frac{d}{50} = \frac{5d + 6d}{300} = \frac{11d}{300} \]

\[ \frac{11d}{300} = 11 \quad\Longrightarrow\quad d = 300 \text{ km} \]

Equivalently, the average speed is the harmonic mean \( \tfrac{2 \times 30 \times 25}{55} = 27.27 \) kmph, and \( 27.27 \times 11 = 300 \).

Answer: D

Speed and time are inversely proportional over a fixed distance, so walking at \( \tfrac{7}{6} \) of his rate takes \( \tfrac{6}{7} \) of the time.

\[ t - \frac{6}{7}t = \frac{1}{7}t = 4 \text{ min} \]

\[ t = 28 \text{ minutes} \]

Checking: the faster walk takes 24 minutes, exactly 4 less.