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

In one hour of clock time the bus covers only 45 km instead of 54, so it loses 9 km of travel.

\[ \text{time to cover 9 km at 54 kmph} = \frac{9}{54} \text{ h} = \frac{1}{6} \text{ h} = 10 \text{ minutes} \]

The stoppage time is worked out at the running speed, not the average one - those 9 km would have taken 10 minutes at 54 kmph, and that is the time spent standing still.

Answer: B

Add the times for the two portions.

\[ t = \frac{d/4}{26} + \frac{3d/4}{24} = \frac{d}{104} + \frac{d}{32} \]

Over the common denominator 416:

\[ t = \frac{4d + 13d}{416} = \frac{17d}{416} \]

\[ \bar{v} = \frac{d}{17d/416} = \frac{416}{17} = 24\tfrac{8}{17} \text{ kmph} \]

The average must lie between 24 and 26, and nearer 24 since three quarters of the distance is covered at that speed - a check that rules out anything below 24.

Answer: C

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

\[ \frac{13}{36} \times \frac{18}{5} = \frac{13 \times 18}{36 \times 5} = \frac{13}{10} = 1.3 \text{ kmph} \]

The 18 cancels half of the 36, which is what makes the arithmetic so clean.

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.