Time & Distance Questions and Answers

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.

Answer: D

Track the two axes separately. Starting north, a left turn faces west, another left faces south, then a right turn faces west again.

• 20 m north
• 40 m west
• 20 m south
• 20 m west

\[ \text{north--south} = 20 - 20 = 0 \]

\[ \text{east--west} = 40 + 20 = 60 \text{ m west} \]

The vertical movements cancel exactly, so the displacement is a straight 60 metres - no Pythagoras needed.

Answer: A

\[ t_{\text{motorcycle}} = \frac{192}{32} = 6 \text{ hours} \]

The car starts \( 2\tfrac{1}{2} \) hours later and arrives half an hour earlier:

\[ t_{\text{car}} = 6 - 2.5 - 0.5 = 3 \text{ hours} \]

\[ v_{\text{car}} = \frac{192}{3} = 64 \text{ kmph} \]

\[ 32 : 64 = 1 : 2 \]

Answer: C

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

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

The two legs are equal in distance, so

\[ \bar{v} = \frac{2 \times 64 \times 80}{64 + 80} = \frac{10240}{144} = 71.11 \text{ km/hr} \]

The arithmetic mean of 72 is close but slightly high. The 35.55 option is exactly half the right answer - the result of dividing by the total distance instead of one leg.

Answer: C

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

\[ 36 \text{ kmph} = 10 \text{ m/s} \]

\[ t = \frac{100 + 150}{10} = 25 \text{ seconds} \]

Answer: B

Convert both quantities to the units the answer needs.

\[ 600 \text{ m} = 0.6 \text{ km}, \qquad 5 \text{ min} = \frac{1}{12} \text{ h} \]

\[ v = \frac{0.6}{1/12} = 7.2 \text{ km/hr} \]

Equivalently, 2 m/s multiplied by \( \tfrac{18}{5} \) gives the same 7.2 km/hr.