Time & Distance Questions and Answers

Answer: C

The track is 95 m long.

Let the track length be \(L\). In the given time she covers \(L - 20\) at her first speed, and 40% more distance, \(L + 10\), at 1.4 times that speed.

\[ \begin{aligned} 1.4(L - 20) &= L + 10 \\ 1.4L - 28 &= L + 10 \\ 0.4L &= 38 \\ L &= 95 \end{aligned} \]

Check: 75 m at the first speed becomes \(75 \times 1.4 = 105\) m, which is 10 m past 95 m. The common slip is to equate the 30 m gap to 40% of the track, \(0.4L = 30\), which gives 75 m and matches none of the options.

Answer: C

P is late by 25 minutes.

The time needed to cover 100 km at 60 km/h is

\[ \begin{aligned} t &= \frac{100}{60}\ \text{h} = 1\frac{2}{3}\ \text{h} \\ &= 1\ \text{h}\ 40\ \text{min} \end{aligned} \]

Starting at 9:45 AM, P arrives at 11:25 AM, which is 25 minutes after the 11:00 AM start. 15 minutes is what you get by taking 100 km at 60 km/h as one and a half hours, and 5 or 35 minutes come from errors in converting the fraction of an hour into minutes.

Answer: A

The ratio of times is 3 : 2.

For a fixed distance, time is inversely proportional to speed, \(t = d/v\):

\[ \begin{aligned} t_K : t_A &= \frac{d}{40} : \frac{d}{60} \\ &= 60 : 40 = 3 : 2 \end{aligned} \]

Karan is slower, so he takes longer, and the ratio of their times is the reverse of the ratio of their speeds (40 : 60 = 2 : 3).

2 : 3 is the speed ratio itself, the usual slip. 1 : 2 matches neither. The data are sufficient, because the distance cancels out, so 'Insufficient data' is wrong.

Answer: D

The journey takes 10 hours.

• 120 km at 20 km/hr takes 120/20 = 6 hours.
• 90 km at 30 km/hr takes 90/30 = 3 hours.
• The refreshment stop adds 60 minutes = 1 hour.

Total = 6 + 3 + 1 = 10 hours.

9 hours leaves out the stop. 7 or 8 hours would come from using a single average speed for the whole 210 km, but the two stretches are driven at different speeds and must be timed separately.

Answer: B

The tiger must run 800 m.

Convert leaps to metres. The tiger covers \(5 \times 8 = 40\) m per minute and the deer \(4 \times 5 = 20\) m per minute. The gap is 50 tiger leaps:

\[ \begin{aligned} \text{gap} &= 50 \times 8 = 400\ \text{m} \\ \text{time} &= \frac{400}{40 - 20} = 20\ \text{min} \\ \text{distance} &= 40 \times 20 = 800\ \text{m} \end{aligned} \]

400 m is the gap, not the distance run. 1000 m would come from a 25-minute chase, and 600 m and 700 m do not match the closing speed of 20 m per minute.

Answer: D

Riding both ways takes 4 hours 45 minutes.

Let one walk take \(w\) and one ride take \(r\). In minutes:

\[ \begin{aligned} w + r &= 375 \\ 2w &= 465, \quad w = 232.5 \\ r &= 375 - 232.5 = 142.5 \\ 2r &= 285\ \text{min} = 4\ \text{h}\ 45\ \text{min} \end{aligned} \]

A shortcut: riding both ways saves, compared with walking both ways, twice the difference saved on one leg. Walking both ways takes 1 h 30 min more than walk-plus-ride, so riding both ways takes 1 h 30 min less: 6 h 15 min − 1 h 30 min = 4 h 45 min. The other options do not satisfy both conditions.

Answer: B

The average speed is 48 km/h.

The distance is the same each way, so the time spent at the slower speed is longer and the average is not the simple mean. With distance \(d\) each way:

\[ \begin{aligned} \text{time} &= \frac{d}{40} + \frac{d}{60} = \frac{d}{24} \\ \text{average} &= \frac{2d}{d/24} = 48\ \text{km/h} \end{aligned} \]

This is the harmonic mean, \(\frac{2 \times 40 \times 60}{40 + 60} = 48\). The trap is 50 km/h, the simple average of the two speeds; 45 and 52 come from no valid method.

Answer: D

The distance is 1296 km.

Let the speed be \(v\) and the distance \(d\). The two conditions give

\[ \begin{aligned} \frac{d}{v} - \frac{d}{v + 216} &= 1 \Rightarrow 216d = v(v + 216) \\ \frac{d}{v - 108} - \frac{d}{v} &= 1 \Rightarrow 108d = v(v - 108) \end{aligned} \]

Dividing, \(2 = \frac{v + 216}{v - 108}\), so \(v = 432\) km/h. Then \(d = \frac{432 \times 648}{216} = 1296\) km, a 3-hour flight. Check: at 648 km/h it takes 2 hours, at 324 km/h 4 hours.

Answer: A

Running the same way, they close at \( 60 - 40 = 20 \) kmph \( = \tfrac{50}{9} \) m/s, and cover both lengths:

\[ 2L = \frac{50}{9} \times 50 = \frac{2500}{9} \text{ m} \]

Running opposite ways they close at \( 100 \) kmph \( = \tfrac{250}{9} \) m/s over the same total length:

\[ t = \frac{2500/9}{250/9} = 10 \text{ sec} \]

The combined length cancels, so it never has to be found - the answer is simply 50 seconds scaled by the ratio of the two relative speeds.

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.