Time & Distance Questions and Answers

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: D

Track the two axes separately. Starting north, a left turn faces west, another left faces south, then a right turn faces west again.

• 20 m north
• 40 m west
• 20 m south
• 20 m west

\[ \text{north--south} = 20 - 20 = 0 \]

\[ \text{east--west} = 40 + 20 = 60 \text{ m west} \]

The vertical movements cancel exactly, so the displacement is a straight 60 metres - no Pythagoras needed.

Answer: D

Distance travelled by each of them is to be made equal.

=> Distance travelled walking = Distance covered cycling.

Let after 4 hours of walking, A walk for x hours more before B catches up with him.

=> Distance = Speed * Time

(4+x)4 = 10x

x = 8/3

Therefore, It takes B 8/3 hours to catch up with A. Distance: 8/3 x 10 = 80/3 km = 26.67

Answer: B

Convert both quantities to the units the answer needs.

\[ 600 \text{ m} = 0.6 \text{ km}, \qquad 5 \text{ min} = \frac{1}{12} \text{ h} \]

\[ v = \frac{0.6}{1/12} = 7.2 \text{ km/hr} \]

Equivalently, 2 m/s multiplied by \( \tfrac{18}{5} \) gives the same 7.2 km/hr.

Answer: A

Both journeys take the same time; only the speed and the distance differ.

\[ \frac{x}{10} = \frac{x + 20}{14} \]

\[ 14x = 10x + 200 \quad\Longrightarrow\quad 4x = 200 \]

\[ x = 50 \text{ km} \]

Checking: 50 km at 10 km/hr and 70 km at 14 km/hr both take 5 hours. The 70 on the list is the longer distance, not the actual one.

Answer: C

Let the car do \( c \) kmph; the train then does \( 1.5c \). Both leave together and arrive together, but the train idles 12.5 minutes.

\[ \frac{75}{c} = \frac{75}{1.5c} + \frac{12.5}{60} \]

\[ \frac{75}{c} - \frac{50}{c} = 0.20833 \quad\Longrightarrow\quad \frac{25}{c} = 0.20833 \]

\[ c = 120 \text{ kmph} \]

Checking: the car takes 37.5 minutes and the train 25 minutes of running plus 12.5 minutes stopped - both 37.5 in total.

Answer: B

In one hour of clock time the bus covers only 45 km instead of 54, so it loses 9 km of travel.

\[ \text{time to cover 9 km at 54 kmph} = \frac{9}{54} \text{ h} = \frac{1}{6} \text{ h} = 10 \text{ minutes} \]

The stoppage time is worked out at the running speed, not the average one - those 9 km would have taken 10 minutes at 54 kmph, and that is the time spent standing still.

Answer: A

Let the usual speed be \( v \). The slowed trip takes half an hour longer.

\[ \frac{600}{v - 200} - \frac{600}{v} = 0.5 \]

\[ \frac{600v - 600(v - 200)}{v(v-200)} = \frac{120000}{v^{2} - 200v} = 0.5 \]

\[ v^{2} - 200v - 240000 = 0 \]

\[ v = \frac{200 + \sqrt{40000 + 960000}}{2} = \frac{200 + 1000}{2} = 600 \text{ km/hr} \]

\[ \text{duration} = \frac{600}{600} = 1 \text{ hour} \]

The slowed flight would take 1½ hours at 400 km/hr - exactly the 30 minutes more the question describes.

Answer: B

Each half of the journey is \( \tfrac{D}{2} \) km.

\[ \frac{D/2}{21} + \frac{D/2}{24} = 10 \]

\[ \frac{D}{42} + \frac{D}{48} = \frac{8D + 7D}{336} = \frac{15D}{336} \]

\[ \frac{15D}{336} = 10 \quad\Longrightarrow\quad D = 224 \text{ km} \]

Equivalently, the average speed is the harmonic mean \( \tfrac{2 \times 21 \times 24}{45} = 22.4 \) kmph, and \( 22.4 \times 10 = 224 \).

Answer: D

\[ v_{2} = \frac{400}{4} = 100 \text{ km/hr} \]

The speeds are in the ratio \( 7 : 8 \), so

\[ v_{1} = 100 \times \frac{7}{8} = 87.5 \text{ km/hr} \]

The first train is the slower of the two, so its speed must fall below 100 - which rules out one option without any calculation.