Time & Distance Questions and Answers

Answer: C

Let the car do \( c \) kmph; the train then does \( 1.5c \). Both leave together and arrive together, but the train idles 12.5 minutes.

\[ \frac{75}{c} = \frac{75}{1.5c} + \frac{12.5}{60} \]

\[ \frac{75}{c} - \frac{50}{c} = 0.20833 \quad\Longrightarrow\quad \frac{25}{c} = 0.20833 \]

\[ c = 120 \text{ kmph} \]

Checking: the car takes 37.5 minutes and the train 25 minutes of running plus 12.5 minutes stopped - both 37.5 in total.

Answer: B

Let the distances between city A to B and B to C be 2x km and 3x km respectively. 

Total time taken to cover from A to C

 = (2x)/40 + (3x)/60 = (6x + 6x)/120 = 12x/120 = x/10 Average speed = (2x + 3x)/(x/10) = 50 kmph.

Answer: D

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

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

An electric pole has no length, so the train covers only its own 100 m:

\[ t = \frac{100}{10} = 10 \text{ sec} \]

Answer: B

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

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

\[ \bar{v} = \frac{2 \times 18 \times 45}{18 + 45} = \frac{1620}{63} = 25.71 \text{ km/hr} \]

The arithmetic mean of 31.5 km/hr overstates it, because the slower cycle leg takes far longer than the bus leg and so weighs more heavily in the total time.

Answer: A

\[ v = \frac{\text{distance}}{\text{time}} = \frac{624}{6} = 104 \text{ kmph} \]

The mph option is a units trap: the distance is given in kilometres, so the speed must be in kilometres per hour.

Answer: D

The gap closes at the difference of the two speeds.

\[ \text{relative speed} = 7 - 5 = 2 \text{ m/s} \]

\[ t = \frac{80}{2} = 40 \text{ s} \]

\[ \text{thief's distance} = 5 \times 40 = 200 \text{ m} \]

Ramesh covers 280 m in the same time - exactly 80 m more, which is the head start he had to make up. The question asks for the thief's distance, not Ramesh's.

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.