Time & Distance Questions and Answers

Answer: D

The distance is 50 km.

Let the distance be \(d\) km. The two journeys differ by 40 minutes, \(\frac{2}{3}\) hour, because the train is 10 minutes late in one case and 50 in the other:

\[ \begin{aligned} \frac{d}{30} - \frac{d}{50} &= \frac{2}{3} \\ \frac{5d - 3d}{150} &= \frac{2}{3} \\ 2d &= 100,\ d = 50 \end{aligned} \]

Check: at 50 km/h it takes 60 minutes, at 30 km/h 100 minutes, a difference of 40 minutes, so the right time is 50 minutes in both cases. 40, 48 and 60 km give time differences of 32, 38.4 and 48 minutes.

Answer: B

54 km/hr is 15 m|s, that is, 15 metres per second.

To convert km/hr to m/s, multiply by \(\frac{1000}{3600} = \frac{5}{18}\):

\[ 54 \times \frac{5}{18} = 15\ \text{m/s} \]

A quick check: 18 km/hr is exactly 5 m/s, and 54 is three times 18, so the speed is three times 5. The option 18 is the conversion factor's denominator, not the answer, and 10 and 12 do not come from the correct factor.

Answer: A

The average speed is 40 kmph.

Take the total distance as 100 km: 30 km at 20 km/h and 70 km at 70 km/h.

\[ \begin{aligned} t &= \frac{30}{20} + \frac{70}{70} = 1.5 + 1 = 2.5\ \text{h} \\ v_{avg} &= \frac{100}{2.5} = 40\ \text{km/h} \end{aligned} \]

Averaging the two speeds gives 45, and weighting them by distance gives 55; average speed must be total distance divided by total time.

Answer: A

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

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

\[ \bar{v} = \frac{2 \times 4 \times 6}{4 + 6} = \frac{48}{10} = 4.8 \text{ km/hr} \]

The arithmetic mean of 5 km/hr is too high, because more time is spent on the slow leg - the average always falls below it.

Answer: A

Account for the distance and time already used, then divide what is left.

\[ \text{first stage} = 60 \times 3 = 180 \text{ km in } 3 \text{ h} \]

\[ \text{second stage} = 200 \text{ km at } 50 \text{ km/hr} = 4 \text{ h} \]

\[ \text{covered} = 380 \text{ km in } 7 \text{ h} \]

\[ \text{remaining} = 120 \text{ km in } 3 \text{ h} \quad\Longrightarrow\quad 40 \text{ km/hr} \]

Both the distance and the time budgets have to be tracked together; either one alone would give the wrong final speed.

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: B

Convert the time to hours first.

\[ 1 \text{ h } 40 \text{ min } 48 \text{ s} = 1 + \frac{40}{60} + \frac{48}{3600} = 1.68 \text{ h} \]

\[ \text{reduced speed} = \frac{42}{1.68} = 25 \text{ km/hr} \]

That 25 is \( \tfrac{5}{7} \) of the actual speed:

\[ \frac{5}{7}v = 25 \quad\Longrightarrow\quad v = 35 \text{ km/hr} \]

Multiplying 25 by \( \tfrac{5}{7} \) instead of dividing gives \( 17\tfrac{6}{7} \) - the first distractor.

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: A

Two travellers who start together and meet, then finish in times \( t_{A} \) and \( t_{B} \), have speeds in the ratio

\[ \frac{v_{A}}{v_{B}} = \sqrt{\frac{t_{B}}{t_{A}}} \]

\[ \frac{v_{A}}{v_{B}} = \sqrt{\frac{289}{361}} = \frac{17}{19} \]

The square root is the part most often missed: taking the times' ratio directly would give 289 : 361, which is offered as a distractor. It is a square root because each traveller's remaining distance is the distance the other has already covered.

Answer: C

Add the distances and the times separately.

\[ \text{distance} = 96 + 124 + 105 = 325 \text{ km} \]

\[ t = \frac{96}{16} + \frac{124}{31} + \frac{105}{7} = 6 + 4 + 15 = 25 \text{ h} \]

\[ \bar{v} = \frac{325}{25} = 13 \text{ km/hr} \]

Averaging the three speeds gives 18 km/hr, well above the truth - the slow 7 km/hr leg accounts for 15 of the 25 hours.