Time & Distance Questions and Answers

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.

Answer: C

Walking the same way, the gap grows at the difference of their speeds.

\[ 5\tfrac{3}{4} - 4\tfrac{1}{2} = 1.25 \text{ kmph} \]

\[ 1.25 \times 3.5 = 4.375 = 4\tfrac{3}{8} \text{ km} \]

Had they walked in opposite directions the gap would grow at the sum, 10.25 kmph, giving \( 35\tfrac{7}{8} \) km - which is the last option.

Answer: B

Let t be his usual time to reach his office and v be his usual speed.
v = d/t ……….(d is the distance Ajay travels while going to his office)
vt = d
At v1 = 4v/5 ; t1 = t + 6
4v/5 = d/(t + 6)
4v/5* (t + 6) = d
4v/5* (t + 6) = vt
On solving we get,
t = 24 minutes

Answer: C

Each two-minute cycle gains a net 3 metres, but the final climb is not followed by a slip - so do not simply divide 26 by 3.

After 14 minutes (seven full cycles):

\[ 7 \times 3 = 21 \text{ m} \]

In the 15th minute the squirrel climbs another 5 m:

\[ 21 + 5 = 26 \text{ m} \]

So it takes 15 minutes. Dividing \( \tfrac{26}{3} \times 2 \approx 17.3 \) minutes over-counts, because it charges a slip for the last climb too.

Answer: A

The thief's half-hour head start puts him

\[ 40 \times 0.5 = 20 \text{ km} \]

ahead when the owner sets off. The gap then closes at the difference of the speeds:

\[ 50 - 40 = 10 \text{ kmph} \]

\[ t = \frac{20}{10} = 2 \text{ hours} \]

That is two hours from the owner's start - two and a half from the theft itself.