Time & Distance Questions and Answers

Answer: A

Let Abhay's speed be \( a \) and Sameer's be \( s \).

\[ \frac{30}{a} - \frac{30}{s} = 2 \qquad \text{(i)} \]

\[ \frac{30}{2a} = \frac{30}{s} - 1 \qquad \text{(ii)} \]

From (i), \( \tfrac{30}{s} = \tfrac{30}{a} - 2 \). Substituting into (ii):

\[ \frac{15}{a} = \frac{30}{a} - 2 - 1 \quad\Longrightarrow\quad \frac{15}{a} = 3 \]

\[ a = 5 \text{ kmph} \]

Checking: Abhay takes 6 hours, Sameer 4; at 10 kmph Abhay would take 3 hours - one less than Sameer, as required.

Answer: B

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

\[ 90 \text{ kmph} = 25 \text{ m/s} \]

A pole has no length, so the train covers only itself:

\[ L = 25 \times 5 = 125 \text{ m} \]

Answer: A

Trains in the same direction close at the difference of their speeds; in opposite directions, at the sum.

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

\[ \text{relative speed} = 72 - 36 = 36 \text{ kmph} = 10 \text{ m/s} \]

\[ t = \frac{200 + 300}{10} = 50 \text{ sec} \]

Both lengths count, because the whole of one train must pass the whole of the other.

Answer: D

Each half is \( 5\tfrac{1}{4} = 5.25 \) km.

\[ t = \frac{5.25}{3} + \frac{5.25}{7} = 1.75 + 0.75 = 2.5 \text{ hours} \]

That is \( 2\tfrac{1}{2} \) hours, which appears nowhere on the list, so the answer is none of these. The \( 2\tfrac{1}{4} \) offered is close enough to catch a rounded calculation.

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: 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.