Time & Distance Questions and Answers

Answer: B

Let the total distance be \( D \). The two legs take

\[ \frac{2D/3}{4} + \frac{D/3}{5} = 1 \text{ h } 24 \text{ min} = 1.4 \text{ h} \]

\[ \frac{D}{6} + \frac{D}{15} = \frac{5D + 2D}{30} = \frac{7D}{30} \]

\[ \frac{7D}{30} = 1.4 \quad\Longrightarrow\quad D = 6 \text{ km} \]

Checking: 4 km at 4 kmph is 1 hour and 2 km at 5 kmph is 24 minutes.

Answer: C

Each two-minute cycle gains a net 3 metres, but the final climb is not followed by a slip - so do not simply divide 26 by 3.

After 14 minutes (seven full cycles):

\[ 7 \times 3 = 21 \text{ m} \]

In the 15th minute the squirrel climbs another 5 m:

\[ 21 + 5 = 26 \text{ m} \]

So it takes 15 minutes. Dividing \( \tfrac{26}{3} \times 2 \approx 17.3 \) minutes over-counts, because it charges a slip for the last climb too.

Answer: C

Overtaking means the car's own length is covered at the relative speed. Let the car do \( c \) km/hr and be \( L \) metres long.

\[ L = (c - 6) \times 9 \times \frac{5}{18} = (c - 12) \times 10 \times \frac{5}{18} \]

The factor \( \tfrac{5}{18} \) is common, so

\[ 9(c - 6) = 10(c - 12) \quad\Longrightarrow\quad 9c - 54 = 10c - 120 \]

\[ c = 66 \text{ km/hr} \]

The car's length never has to be found - it cancels because the same car passes both cyclists.

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

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

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

\[ \bar{v} = \frac{2 \times 30 \times 60}{30 + 60} = \frac{3600}{90} = 40 \text{ kmph} \]

The arithmetic mean of 45 kmph is the trap; the return trip at double the speed takes half the time, so the slow outward leg dominates.

Answer: A

In one minute the train travels

\[ 120 \text{ kmph} = \frac{120000}{60} = 2000 \text{ m/min} \]

To pass the field, the train must cover the field's length plus its own length:

\[ \text{side} = 2000 - 800 = 1200 \text{ m} = 1.2 \text{ km} \]

\[ \text{area} = (1.2)^{2} = 1.44 \text{ sq. km} \]

Forgetting to subtract the train's own length would give a 2 km side and an area of 4 sq. km - exactly the second option.

Answer: C

A pole has no length, so the train covers only its own length in passing it.

\[ 60 \text{ km/hr} = 60 \times \frac{5}{18} = \frac{50}{3} \text{ m/s} \]

\[ t = \frac{125}{50/3} = \frac{375}{50} = 7.5 \text{ s} \]

Passing a platform would require adding the platform's length; a pole adds nothing.

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: B

Let t be his usual time to reach his office and v be his usual speed.
v = d/t ……….(d is the distance Ajay travels while going to his office)
vt = d
At v1 = 4v/5 ; t1 = t + 6
4v/5 = d/(t + 6)
4v/5* (t + 6) = d
4v/5* (t + 6) = vt
On solving we get,
t = 24 minutes

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.