Time & Distance Questions and Answers

Answer: D

Speed and time are inversely proportional over a fixed distance, so multiplying the speed by \( \tfrac{9}{5} \) multiplies the time by \( \tfrac{5}{9} \).

\[ \text{time saved} = t - \frac{5}{9}t = \frac{4}{9}t \]

\[ \frac{4}{9}t = 30 \quad\Longrightarrow\quad t = \frac{270}{4} = 67.5 \text{ min} \]

Checking: at the higher speed the trip takes 37.5 minutes, exactly 30 less.

Answer: B

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

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

\[ \bar{v} = \frac{2 \times 18 \times 45}{18 + 45} = \frac{1620}{63} = 25.71 \text{ km/hr} \]

The arithmetic mean of 31.5 km/hr overstates it, because the slower cycle leg takes far longer than the bus leg and so weighs more heavily in the total time.

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.

Answer: C

One trip is 30 minutes late and the other 10 minutes early, so the travel times differ by 40 minutes.

\[ \frac{D}{50} - \frac{D}{60} = \frac{40}{60} = \frac{2}{3} \text{ h} \]

\[ D\left(\frac{6 - 5}{300}\right) = \frac{2}{3} \quad\Longrightarrow\quad \frac{D}{300} = \frac{2}{3} \]

\[ D = 200 \text{ km} \]

Late and early must be added to get the gap: 30 + 10 = 40 minutes, not their difference.

Answer: D

The gap closes at the difference of the two speeds.

\[ \text{relative speed} = 7 - 5 = 2 \text{ m/s} \]

\[ t = \frac{80}{2} = 40 \text{ s} \]

\[ \text{thief's distance} = 5 \times 40 = 200 \text{ m} \]

Ramesh covers 280 m in the same time - exactly 80 m more, which is the head start he had to make up. The question asks for the thief's distance, not Ramesh's.

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: A

Over a fixed distance, speed and time are inversely proportional.

\[ v_{2} = v_{1} \times \frac{t_{1}}{t_{2}} = 24 \times \frac{6}{8} = 18 \text{ km/hr} \]

More time means less speed, so the answer must be below 24 km/hr - which rules out two options immediately. The distance itself (2.4 km) is never needed.

Answer: A

Correct answer: 48 km/hr

Average speed is total distance divided by total time, not the mean of the two speeds.

\[\text{Time} = \frac{240}{60} + \frac{240}{40} = 4 + 6 = 10 \text{ hours}\]

\[\text{Average speed} = \frac{480}{10} = 48 \text{ km/hr}\]

It falls below 50 because more of the journey’s time is spent at the slower speed.

Answer: C

Convert to metres per second with the factor \( \tfrac{5}{18} \).

\[ 90 \text{ kmph} = 90 \times \frac{5}{18} = 25 \text{ m/s} \]

\[ \text{distance} = 25 \times 15 = 375 \text{ m} \]

The data are complete, so "cannot be determined" is a trap. Working in kilometres and hours would need \( 15 \) s \( = \tfrac{1}{240} \) h, giving 0.375 km - the same answer.

Answer: C

Track distance and time together.

\[ \text{first stage} = 60 \times 1.5 = 90 \text{ km in } 1.5 \text{ h} \]

\[ \text{second stage} = \frac{100}{50} = 2 \text{ h for } 100 \text{ km} \]

\[ \text{covered} = 190 \text{ km in } 3.5 \text{ h} \]

\[ \text{remaining} = 110 \text{ km in } 2.5 \text{ h} \quad\Longrightarrow\quad \frac{110}{2.5} = 44 \text{ kmph} \]

The 90 minutes must become 1.5 hours before it can be combined with the rest.