Time & Distance Questions and Answers

Answer: A

Account for the distance and time already used, then divide what is left.

\[ \text{first stage} = 60 \times 3 = 180 \text{ km in } 3 \text{ h} \]

\[ \text{second stage} = 200 \text{ km at } 50 \text{ km/hr} = 4 \text{ h} \]

\[ \text{covered} = 380 \text{ km in } 7 \text{ h} \]

\[ \text{remaining} = 120 \text{ km in } 3 \text{ h} \quad\Longrightarrow\quad 40 \text{ km/hr} \]

Both the distance and the time budgets have to be tracked together; either one alone would give the wrong final speed.

Answer: D

Distance travelled by each of them is to be made equal.

=> Distance travelled walking = Distance covered cycling.

Let after 4 hours of walking, A walk for x hours more before B catches up with him.

=> Distance = Speed * Time

(4+x)4 = 10x

x = 8/3

Therefore, It takes B 8/3 hours to catch up with A. Distance: 8/3 x 10 = 80/3 km = 26.67

Answer: A

A signal post is a point, so the first crossing gives the speed.

\[ v = \frac{600}{40} = 15 \text{ m/s} \]

\[ t = \frac{600 + 3000}{15} = \frac{3600}{15} = 240 \text{ s} = 4 \text{ minutes} \]

The data are complete, so "data inadequate" is a trap; the 3 km must be converted to 3000 m.

Answer: C

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

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

\[ \text{relative speed} = 40 - 22 = 18 \text{ kmph} = 5 \text{ m/s} \]

\[ \text{combined length} = 5 \times 60 = 300 \text{ m} \]

\[ L_{2} = 300 - 125 = 175 \text{ m} \]

Passing completely means covering both lengths, which is why the first train's 125 m is subtracted at the end.

Answer: A

Ten minutes is one sixth of an hour.

\[ \text{distance} = 90 \times \frac{10}{60} = 15 \text{ km} \]

The answer is a distance, not a speed, so it is measured in kilometres - a point the units of the options are there to test.

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:

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