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: 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.

Answer: B

The 60 km difference comes from the difference in speeds.

\[ 21t - 16t = 60 \quad\Longrightarrow\quad 5t = 60 \quad\Longrightarrow\quad t = 12 \text{ h} \]

They approach at \( 16 + 21 = 37 \) kmph:

\[ d = 37 \times 12 = 444 \text{ km} \]

Checking: 192 km and 252 km, differing by exactly 60.

Answer: B

The hourly distances form an arithmetic progression: 30, 32, 34, … with \( a = 30 \), \( d = 2 \) and \( n = 10 \).

\[ S_{n} = \frac{n}{2}\left[2a + (n-1)d\right] \]

\[ = \frac{10}{2}\left[60 + 9 \times 2\right] = 5 \times 78 = 390 \text{ km} \]

Multiplying the first hour's 30 km by 10 would give 300 km, ignoring the acceleration entirely.

Answer: B

Let the speed of man in still water be x km/hr

Now, speed with current=15 km/hr

Speed of current: 2.5 km/hr

Speed of man with current: Speed of man in still water+Speed of current

=>15=x+2.5

=>x=12.5

Speed of man against the current: Speed of man in still water-Speed of current

=>12.5-2.5

=>10 km/hr

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

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

\[ 54 \text{ kmph} = 15 \text{ m/s} \]

\[ t = \frac{165 + 660}{15} = \frac{825}{15} = 55 \text{ sec} \]