Time & Distance Questions and Answers

Answer: A

Let the time taken to cover from A to B in car and cycle be x hours and y hours respectively.

x + y = 7 --- (1) ; 2x = 4 --- (2)

solving both the equations, we get y = 5

So, time taken to cover both ways by cycle = 2y hours = 10 hours.

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: C

Crossing a telegraph post means covering the train's own length.

\[ v = \frac{240}{16} = 15 \text{ m/s} \]

\[ 15 \times \frac{18}{5} = 54 \text{ kmph} \]

The conversion factor \( \tfrac{18}{5} \) turns m/s into kmph; using \( \tfrac{5}{18} \) by mistake would give 4.17 kmph, far too slow for a train.

Answer: D

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

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

\[ \text{relative speed} = 42 + 30 = 72 \text{ kmph} = 20 \text{ m/s} \]

\[ t = \frac{120 + 280}{20} = \frac{400}{20} = 20 \text{ sec} \]

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: B

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

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

\[ \text{relative speed} = 54 + 72 = 126 \text{ kmph} = 35 \text{ m/s} \]

To cross completely, the gap and both lengths must be covered:

\[ 100 + 100 + 200 = 400 \text{ m} \]

\[ t = \frac{400}{35} = \frac{80}{7} \text{ sec} \]

Meeting takes only \( \tfrac{20}{7} \) seconds; crossing takes four times as long.

Answer: B

Let the distance to be travelled to catch the bus be x km

x/30 - x/40 = 30/60 => (4x - 3x)/120 = 1/2 => x = 60 km

By traavelling 30 kmph time taken = 60/30 = 2 hours

By taking 2 hours, he is late by 40 min. So, he has to cover 60 km in at most speed = 60/(4/3) = 45 kmph.

Answer: C

The slower car takes one hour longer over the same distance.

\[ \frac{d}{60} - \frac{d}{64} = 1 \]

\[ d\left(\frac{64 - 60}{3840}\right) = 1 \quad\Longrightarrow\quad \frac{4d}{3840} = 1 \]

\[ d = 960 \text{ km} \]

The unit matters: the answer is in kilometres, since the speeds were given in kmph.

Answer: A

In one minute the train travels

\[ 120 \text{ kmph} = \frac{120000}{60} = 2000 \text{ m/min} \]

To pass the field, the train must cover the field's length plus its own length:

\[ \text{side} = 2000 - 800 = 1200 \text{ m} = 1.2 \text{ km} \]

\[ \text{area} = (1.2)^{2} = 1.44 \text{ sq. km} \]

Forgetting to subtract the train's own length would give a 2 km side and an area of 4 sq. km - exactly the second option.

Answer: A

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

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

Crossing a man in the other train covers only the faster train's length.

\[ \text{relative speed} = 72 - 36 = 36 \text{ kmph} = 10 \text{ m/s} \]

\[ L = 10 \times 27 = 270 \text{ m} \]

The slower train's own length is irrelevant, since the reference point is a single person.