Time & Distance Questions and Answers

Answer: C

Work rates add when people work together.

\[ \text{Ajay's rate} = \frac{1}{8}, \qquad \text{Vijay's rate} = \frac{1}{24} \text{ of the job per day} \]

\[ \text{combined} = \frac{1}{8} + \frac{1}{24} = \frac{3 + 1}{24} = \frac{1}{6} \]

\[ t = 6 \text{ days} \]

Averaging the two times to 16 days is wrong: rates add, times do not. The joint time must always be less than the faster worker's 8 days.

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: C

The two speeds produce arrival times two hours apart, so the travel times differ by 2 hours.

\[ \frac{D}{10} - \frac{D}{15} = 2 \quad\Longrightarrow\quad \frac{D}{30} = 2 \quad\Longrightarrow\quad D = 60 \text{ km} \]

At 10 kmph the trip takes 6 hours and ends at 2 p.m., so he starts at 8 a.m. To arrive at 1 p.m. he has 5 hours:

\[ v = \frac{60}{5} = 12 \text{ kmph} \]

Averaging the two given speeds to 12.5 kmph happens to come close, but it is not the reasoning - the times, not the speeds, are what average out here.

Answer: A

Two travellers who start together and meet, then finish in times \( t_{A} \) and \( t_{B} \), have speeds in the ratio

\[ \frac{v_{A}}{v_{B}} = \sqrt{\frac{t_{B}}{t_{A}}} \]

\[ \frac{v_{A}}{v_{B}} = \sqrt{\frac{289}{361}} = \frac{17}{19} \]

The square root is the part most often missed: taking the times' ratio directly would give 289 : 361, which is offered as a distractor. It is a square root because each traveller's remaining distance is the distance the other has already covered.

Answer: A

The two walks differ by 4 minutes in arrival: 2 late against 2 early.

\[ \frac{d}{3} - \frac{d}{4} = \frac{4}{60} = \frac{1}{15} \text{ h} \]

\[ \frac{d}{12} = \frac{1}{15} \quad\Longrightarrow\quad d = \frac{12}{15} = \frac{4}{5} \text{ km} \]

Late and early add to give the gap of 4 minutes; taking their difference would give zero.

Answer: A

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

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

\[ \bar{v} = \frac{2 \times 4 \times 6}{4 + 6} = \frac{48}{10} = 4.8 \text{ km/hr} \]

The arithmetic mean of 5 km/hr is too high, because more time is spent on the slow leg - the average always falls below it.

Answer: B

A man standing is a point, so the train covers only its own length.

\[ v = \frac{180}{6} = 30 \text{ m/s} \]

\[ 30 \times \frac{18}{5} = 108 \text{ kmph} \]

Answer: B

Convert the time to hours first.

\[ 1 \text{ h } 40 \text{ min } 48 \text{ s} = 1 + \frac{40}{60} + \frac{48}{3600} = 1.68 \text{ h} \]

\[ \text{reduced speed} = \frac{42}{1.68} = 25 \text{ km/hr} \]

That 25 is \( \tfrac{5}{7} \) of the actual speed:

\[ \frac{5}{7}v = 25 \quad\Longrightarrow\quad v = 35 \text{ km/hr} \]

Multiplying 25 by \( \tfrac{5}{7} \) instead of dividing gives \( 17\tfrac{6}{7} \) - the first distractor.

Answer: D

Walking at \( \tfrac{4}{5} \) of the usual speed takes \( \tfrac{5}{4} \) of the usual time, since speed and time are inversely proportional.

\[ \frac{5}{4}t - t = \frac{1}{4}t = 5 \text{ min} \]

\[ t = 20 \text{ minutes} \]

Checking: the slower walk takes 25 minutes, exactly 5 more.