Time & Distance Questions and Answers

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: B

Let the distance to be travelled to catch the bus be x km

x/30 - x/40 = 30/60 => (4x - 3x)/120 = 1/2 => x = 60 km

By traavelling 30 kmph time taken = 60/30 = 2 hours

By taking 2 hours, he is late by 40 min. So, he has to cover 60 km in at most speed = 60/(4/3) = 45 kmph.

Answer: C

The slower car takes one hour longer over the same distance.

\[ \frac{d}{60} - \frac{d}{64} = 1 \]

\[ d\left(\frac{64 - 60}{3840}\right) = 1 \quad\Longrightarrow\quad \frac{4d}{3840} = 1 \]

\[ d = 960 \text{ km} \]

The unit matters: the answer is in kilometres, since the speeds were given in kmph.

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.

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

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