Time & Distance Questions and Answers

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.

Answer: C

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

\[ 72 \text{ kmph} = 20 \text{ m/s} \]

\[ \text{total distance} = 20 \times 25 = 500 \text{ m} \]

\[ \text{platform} = 500 - 150 = 350 \text{ m} \]

Answer: A

Correct answer: 48 km/hr

Average speed is total distance divided by total time, not the mean of the two speeds.

\[\text{Time} = \frac{240}{60} + \frac{240}{40} = 4 + 6 = 10 \text{ hours}\]

\[\text{Average speed} = \frac{480}{10} = 48 \text{ km/hr}\]

It falls below 50 because more of the journey’s time is spent at the slower speed.

Answer: C

Add the distances and the times separately.

\[ \text{distance} = 96 + 124 + 105 = 325 \text{ km} \]

\[ t = \frac{96}{16} + \frac{124}{31} + \frac{105}{7} = 6 + 4 + 15 = 25 \text{ h} \]

\[ \bar{v} = \frac{325}{25} = 13 \text{ km/hr} \]

Averaging the three speeds gives 18 km/hr, well above the truth - the slow 7 km/hr leg accounts for 15 of the 25 hours.

Answer: C

Track distance and time together.

\[ \text{first stage} = 60 \times 1.5 = 90 \text{ km in } 1.5 \text{ h} \]

\[ \text{second stage} = \frac{100}{50} = 2 \text{ h for } 100 \text{ km} \]

\[ \text{covered} = 190 \text{ km in } 3.5 \text{ h} \]

\[ \text{remaining} = 110 \text{ km in } 2.5 \text{ h} \quad\Longrightarrow\quad \frac{110}{2.5} = 44 \text{ kmph} \]

The 90 minutes must become 1.5 hours before it can be combined with the rest.