Time & Distance Questions and Answers

Answer: A

Take the total distance as \( D \) and add the three times.

\[ t = \frac{D/4}{20} + \frac{D/4}{10} + \frac{D/2}{80} = \frac{D}{80} + \frac{D}{40} + \frac{D}{160} \]

\[ = \frac{2D + 4D + D}{160} = \frac{7D}{160} \]

\[ \bar{v} = \frac{D}{7D/160} = \frac{160}{7} = 22.85 \text{ km/hr} \]

Averaging the three speeds would give \( \tfrac{110}{3} = 36.7 \) km/hr, far too high - the slow 10 km/hr leg dominates the total time even though it covers only a quarter of the distance.

Answer: B

Convert the time to seconds and multiply.

\[ 3 \text{ h } 45 \text{ min} = 3.75 \times 3600 = 13{,}500 \text{ s} \]

\[ d = 12 \times 13{,}500 = 162{,}000 \text{ m} = 162 \text{ km} \]

Equivalently, 12 m/s is \( 12 \times \tfrac{18}{5} = 43.2 \) kmph, and \( 43.2 \times 3.75 = 162 \).

Answer: C

Crossing the platform covers the train plus the platform.

\[ v = \frac{100 + 125}{15} = 15 \text{ m/s} \]

\[ 15 \times \frac{18}{5} = 54 \text{ kmph} \]

Answer: D

Speed of man = 4kmph

Time for which he cycled = 45 minutes.

Distance covered = speed*time

=> 4*45/60

=> 3km

Total time to cover 3 km = 45 minutes + 15 minutes = 60 minutes = 1 hour

Thus 3 km is covered in 1 hour

Required distance to be covered = 12 km

Beginning from origin point, he covers 9 km in 3 hours taking 15 minutes rest after every 45 minutes. To complete 12 kms he goes another 3 km in 45 minutes.

Thus, total time he takes is 3 hours 45 minutes.

Answer: A

Let the usual speed be \( v \). The slowed trip takes half an hour longer.

\[ \frac{600}{v - 200} - \frac{600}{v} = 0.5 \]

\[ \frac{600v - 600(v - 200)}{v(v-200)} = \frac{120000}{v^{2} - 200v} = 0.5 \]

\[ v^{2} - 200v - 240000 = 0 \]

\[ v = \frac{200 + \sqrt{40000 + 960000}}{2} = \frac{200 + 1000}{2} = 600 \text{ km/hr} \]

\[ \text{duration} = \frac{600}{600} = 1 \text{ hour} \]

The slowed flight would take 1½ hours at 400 km/hr - exactly the 30 minutes more the question describes.

Answer: B

Here the two times are equal - one hour each - not the distances, so the plain arithmetic mean is correct.

\[ \bar{v} = \frac{90 + 60}{1 + 1} = 75 \text{ kmph} \]

Had the two distances been equal instead, the harmonic mean \( \tfrac{2 \times 90 \times 60}{150} = 72 \) kmph would apply - which is exactly the first distractor.

Answer: A

Both journeys take the same time; only the speed and the distance differ.

\[ \frac{x}{10} = \frac{x + 20}{14} \]

\[ 14x = 10x + 200 \quad\Longrightarrow\quad 4x = 200 \]

\[ x = 50 \text{ km} \]

Checking: 50 km at 10 km/hr and 70 km at 14 km/hr both take 5 hours. The 70 on the list is the longer distance, not the actual one.

Answer: D

In equal times, distances are in proportion to speeds.

\[ \frac{\text{Amar}}{\text{car}} = \frac{18}{48} = \frac{3}{8} \]

\[ 1.6 \text{ km} = 1600 \text{ m} \]

\[ \text{Amar} = 1600 \times \frac{3}{8} = 600 \text{ m} \]

Answer: B

Each half of the journey is \( \tfrac{D}{2} \) km.

\[ \frac{D/2}{21} + \frac{D/2}{24} = 10 \]

\[ \frac{D}{42} + \frac{D}{48} = \frac{8D + 7D}{336} = \frac{15D}{336} \]

\[ \frac{15D}{336} = 10 \quad\Longrightarrow\quad D = 224 \text{ km} \]

Equivalently, the average speed is the harmonic mean \( \tfrac{2 \times 21 \times 24}{45} = 22.4 \) kmph, and \( 22.4 \times 10 = 224 \).

Answer: D

Let \( D \) be the distance from the overtaking point to Bangalore.

The cab covers it between 11 am and 1:30 pm, so \( v_{\text{cab}} = \tfrac{D}{2.5} \). After a one-hour break it leaves at 2:30 pm and meets the auto at 3 pm, half an hour later:

\[ \text{cab's return distance} = \frac{D}{2.5} \times 0.5 = \frac{D}{5} \]

So they meet \( D - \tfrac{D}{5} = \tfrac{4D}{5} \) from the overtaking point. The auto took from 11 am to 3 pm - four hours - to get there:

\[ v_{\text{auto}} = \frac{4D/5}{4} = \frac{D}{5} \text{ per hour} \]

The auto still has \( \tfrac{D}{5} \) to go, which takes exactly one more hour:

\[ \text{arrival} = 4 \text{ p.m.} \]

The distance \( D \) cancels throughout, so it never needs a value.