Time & Distance Questions and Answers

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.

Answer: C

Start from R, the only cyclist whose time is given.

\[ v_{R} = \frac{4500}{45} = 100 \text{ m/min} \]

Since \( v_{R} = \tfrac{1}{3}v_{Q} \) and \( v_{P} = 2v_{Q} \):

\[ v_{Q} = 300 \text{ m/min}, \qquad v_{P} = 600 \text{ m/min} \]

\[ t_{P} = \frac{4500}{600} = 7.5 \text{ min} \]

In that time Q has covered \( 300 \times 7.5 = 2250 \) m, so

\[ 4500 - 2250 = 2250 \text{ m from the finish} \]

P is twice as fast as Q, so Q is exactly halfway when P finishes - a check that needs no arithmetic at all.

Answer: C

One trip is 30 minutes late and the other 10 minutes early, so the travel times differ by 40 minutes.

\[ \frac{D}{50} - \frac{D}{60} = \frac{40}{60} = \frac{2}{3} \text{ h} \]

\[ D\left(\frac{6 - 5}{300}\right) = \frac{2}{3} \quad\Longrightarrow\quad \frac{D}{300} = \frac{2}{3} \]

\[ D = 200 \text{ km} \]

Late and early must be added to get the gap: 30 + 10 = 40 minutes, not their difference.

Answer: B

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

The whole train is the engine plus 24 carriages, each 60 m:

\[ 25 \times 60 = 1500 \text{ m} \]

\[ 60 \text{ kmph} = \frac{50}{3} \text{ m/s} \]

\[ t = \frac{1500 + 1500}{50/3} = \frac{3000 \times 3}{50} = 180 \text{ s} = 3 \text{ minutes} \]

The engine counts as a twenty-fifth 60 m unit, and the bridge's 1.5 km must be converted to 1500 m.

Answer: D

A telegraph post has no length, so the train covers only its own 60 m.

\[ 36 \text{ kmph} = 36 \times \frac{5}{18} = 10 \text{ m/s} \]

\[ t = \frac{60}{10} = 6 \text{ seconds} \]

Passing a platform or bridge would require adding its length as well; a post adds nothing.

Answer: C

Add the times, not the speeds.

\[ t = \frac{x}{40} + \frac{2x}{20} = \frac{x}{40} + \frac{4x}{40} = \frac{5x}{40} = \frac{x}{8} \]

\[ \bar{v} = \frac{3x}{x/8} = 24 \text{ kmph} \]

Averaging the two speeds gives 30 kmph, too high - two thirds of the journey is at the slower rate.

Answer: C

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

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

\[ \text{relative speed} = 80 + 70 = 150 \text{ kmph} = \frac{125}{3} \text{ m/s} \]

\[ t = \frac{250 + 250}{125/3} = \frac{500 \times 3}{125} = 12 \text{ sec} \]

Both lengths are covered, since the trains must clear each other completely.