Time & Distance Questions and Answers

Answer: C

Each man takes four hours, so their speeds are equal - call each \( \tfrac{d}{4} \).

By 8 a.m. the first has already covered half the distance:

\[ 2 \times \frac{d}{4} = \frac{d}{2} \]

The gap is then \( \tfrac{d}{2} \), closing at their combined speed \( \tfrac{d}{2} \) per hour:

\[ t = \frac{d/2}{d/2} = 1 \text{ hour} \]

So they meet at 9 a.m. The distance cancels, so it never needs a value.

Answer: D

Average speed is total distance over total time, which for two equal legs is the harmonic mean, not the arithmetic one:

\[ \bar{v} = \frac{2v_{1}v_{2}}{v_{1} + v_{2}} \]

\[ \bar{v} = \frac{2 \times 30 \times 60}{30 + 60} = \frac{3600}{90} = 40 \text{ kmph} \]

The arithmetic mean of 45 kmph is the trap; the return trip at double the speed takes half the time, so the slow outward leg dominates.

Answer: B

Vijay reaches B first:

\[ t = \frac{72}{19} \text{ h} \]

By then Ajay has covered

\[ 17 \times \frac{72}{19} = \frac{1224}{19} = 64.42 \text{ km} \]

so they are \( 72 - 64.42 = 7.58 \) km apart and now approaching each other at \( 17 + 19 = 36 \) km/hr:

\[ t' = \frac{7.58}{36} = 0.2105 \text{ h} \]

\[ BC = 19 \times 0.2105 = 4 \text{ km} \]

After the turn the two are closing, so the speeds add; before it they were separating, so only Ajay's progress mattered.

Answer: D

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

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

\[ t = \frac{110 + 170}{50/3} = \frac{280 \times 3}{50} = 16.8 \text{ sec} \]

Answer: D

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

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

A man is a point, so only the train's own length is covered:

\[ t = \frac{170}{20} = 8.5 = 8\tfrac{1}{2} \text{ sec} \]

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: 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

A has a one-hour head start:

\[ 20 \times 1 = 20 \text{ km covered by 8 a.m.} \]

\[ \text{gap remaining} = 110 - 20 = 90 \text{ km} \]

They approach at \( 20 + 25 = 45 \) kmph:

\[ t = \frac{90}{45} = 2 \text{ hours} \]

So they meet at 10 a.m.

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: A

Both journeys cover the same distance and both are late, so the difference in lateness is the difference in travel time.

\[ \frac{D}{30} - \frac{D}{45} = \frac{20 - 8}{60} = 0.2 \text{ h} \]

\[ D\left(\frac{3 - 2}{90}\right) = 0.2 \quad\Longrightarrow\quad \frac{D}{90} = 0.2 \]

\[ D = 18 \text{ km} \]

The scheduled arrival time never has to be found - only the 12-minute gap between the two lateness figures matters.