Time & Distance Questions and Answers

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: 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: C

The two speeds produce arrival times two hours apart, so the travel times differ by 2 hours.

\[ \frac{D}{10} - \frac{D}{15} = 2 \quad\Longrightarrow\quad \frac{D}{30} = 2 \quad\Longrightarrow\quad D = 60 \text{ km} \]

At 10 kmph the trip takes 6 hours and ends at 2 p.m., so he starts at 8 a.m. To arrive at 1 p.m. he has 5 hours:

\[ v = \frac{60}{5} = 12 \text{ kmph} \]

Averaging the two given speeds to 12.5 kmph happens to come close, but it is not the reasoning - the times, not the speeds, are what average out here.

Answer: C

Write \( x = \tfrac{1}{v_{\text{train}}} \) and \( y = \tfrac{1}{v_{\text{car}}} \) to make both conditions linear.

\[ 120x + 480y = 8, \qquad 200x + 400y = 8\tfrac{1}{3} = \frac{25}{3} \]

Dividing the first by 8 and the second by \( \tfrac{25}{3} \):

\[ 15x + 60y = 1, \qquad 24x + 48y = 1 \]

Equating and clearing:

\[ 24(1 - 60y) = 15(1 - 48y) \quad\Longrightarrow\quad 9 = 720y \quad\Longrightarrow\quad y = \frac{1}{80} \]

\[ x = \frac{1 - 60/80}{15} = \frac{1}{60} \]

So the train does 60 kmph and the car 80 kmph:

\[ 60 : 80 = 3 : 4 \]

The reciprocal substitution is what turns two awkward equations into a pair of linear ones.

Answer: C

Let \( a \) be the hours on foot and \( b \) the hours on the bicycle.

\[ a + b = 9, \qquad 4a + 9b = 61 \]

Substituting \( b = 9 - a \):

\[ 4a + 81 - 9a = 61 \quad\Longrightarrow\quad -5a = -20 \quad\Longrightarrow\quad a = 4 \text{ h} \]

\[ \text{distance on foot} = 4 \times 4 = 16 \text{ km} \]

Setting up the unknowns as times rather than distances is what makes the second equation linear; the question then asks for a distance, so one more multiplication is needed at the end.

Answer: A

\[ 1 \text{ km/hr} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s} \]

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

The factor \( \tfrac{5}{18} \) converts kmph to m/s, and its reciprocal \( \tfrac{18}{5} \) goes the other way.

Answer: C

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

\[ 25 \times \frac{18}{5} = 90 \text{ kmph} \]

Going from m/s to kmph the number always grows, since a metre per second is a little over 3.6 kmph - so any option below 25 can be discarded at once.

Answer: A

Ten minutes is one sixth of an hour.

\[ \text{distance} = 90 \times \frac{10}{60} = 15 \text{ km} \]

The answer is a distance, not a speed, so it is measured in kilometres - a point the units of the options are there to test.

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

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.