Time & Distance Questions and Answers

Answer: D

The distance is 50 km.

Let the distance be \(d\) km. The two journeys differ by 40 minutes, \(\frac{2}{3}\) hour, because the train is 10 minutes late in one case and 50 in the other:

\[ \begin{aligned} \frac{d}{30} - \frac{d}{50} &= \frac{2}{3} \\ \frac{5d - 3d}{150} &= \frac{2}{3} \\ 2d &= 100,\ d = 50 \end{aligned} \]

Check: at 50 km/h it takes 60 minutes, at 30 km/h 100 minutes, a difference of 40 minutes, so the right time is 50 minutes in both cases. 40, 48 and 60 km give time differences of 32, 38.4 and 48 minutes.

Answer: B

54 km/hr is 15 m|s, that is, 15 metres per second.

To convert km/hr to m/s, multiply by \(\frac{1000}{3600} = \frac{5}{18}\):

\[ 54 \times \frac{5}{18} = 15\ \text{m/s} \]

A quick check: 18 km/hr is exactly 5 m/s, and 54 is three times 18, so the speed is three times 5. The option 18 is the conversion factor's denominator, not the answer, and 10 and 12 do not come from the correct factor.

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

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

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

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

Answer: C

Let the distance and original speed be d km and k kmph respectively.

d/0.8k - d/k = 20/60 => 5d/4k - d/k = 1/3

=> (5d - 4d)/4k = 1/3 => d = 4/3 k

Time taken to cover the distance at original speed

= d/k = 4/3 hours = 1 hour 20 minutes.