Time & Distance Questions and Answers

Answer: C

The first train has a \( 1\tfrac{1}{2} \) hour head start:

\[ 70 \times 1.5 = 105 \text{ km} \]

The gap closes at \( 85 - 70 = 15 \) kmph:

\[ t = \frac{105}{15} = 7 \text{ hours} \]

\[ 3{:}30 \text{ p.m.} + 7 \text{ h} = 10{:}30 \text{ p.m.} \]

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: 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: D

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

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

\[ \text{relative speed} = 80 + 65 = 145 \text{ kmph} = \frac{725}{18} = 40.28 \text{ m/s} \]

\[ t = \frac{121 + 165}{40.28} = \frac{286}{40.28} \approx 7.1 \text{ sec} \]

"Clear of each other" means both lengths must pass, which is why 286 m and not 165 m is the distance.

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.