Time & Distance Questions and Answers

Q: Amar takes as much time in running 18 meters as a car takes in covering 48 meters. What will be the distance covered by Amar during the time the car covers 1.6 km?

In equal times, distances are in proportion to speeds.\[ \frac{\text{Amar}}{\text{car}} = \frac{18}{48} = \frac{3}{8} \]\[ 1.6 \text{ km} = 1600 \text{ m} \]\[ \text{Amar} = 1600 \times \frac{3}{8} = 600 \text{ m} \]

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141.

A man misses a bus by 40 minutes if he travels at 30 kmph. If he travels at 40 kmph, then also he misses the bus by 10 minutes. What is the minimum speed required to catch the bus on time?

A.48 kmph
B.45 kmph
C.43 kmph
D.44 kmph

Answer: B

Let the distance to be travelled to catch the bus be x km

x/30 - x/40 = 30/60 => (4x - 3x)/120 = 1/2 => x = 60 km

By traavelling 30 kmph time taken = 60/30 = 2 hours

By taking 2 hours, he is late by 40 min. So, he has to cover 60 km in at most speed = 60/(4/3) = 45 kmph.

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142.

Ramu rides his bike at an average speed of 45 km/hr and reaches his desitination in four hours. Somu covers the same distance in six hours. If Ramu covered his journey at an average speed which was 9 km/hr less and Somu covered his journey at an average speed which was 10 km/hr more, then the difference in their times taken to reach the destination would be (in minutes).

A.36
B.30
C.40
D.45

Answer: B

Distance travelled by Ramu = 45 * 4 = 180 km

Somu travelled the same distance in 6 hours.

His speed = 180/6 = 30 km/hr

Hence in the conditional case, Ramu's speed = 45 - 9 = 36 km/hr and Somu's speed = 30 + 10 = 40km/hr.

Therefore travel time of Ramu and Somu would be 5 hours and 4.5 hours respectively. Hence difference in the time taken = 0.5 hours = 30 minutes.

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143.

Amar takes as much time in running 18 meters as a car takes in covering 48 meters. What will be the distance covered by Amar during the time the car covers 1.6 km?

A.800 m
B.480 m
C.520 m
D.600 m

Answer: D

In equal times, distances are in proportion to speeds.

\[ \frac{\text{Amar}}{\text{car}} = \frac{18}{48} = \frac{3}{8} \]

\[ 1.6 \text{ km} = 1600 \text{ m} \]

\[ \text{Amar} = 1600 \times \frac{3}{8} = 600 \text{ m} \]

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144.

A car started running at a speed of 30 km/hr and the speed of the car was increased by 2 km/hr at the end of every hour. Find the total distance covered by the car in the first 10 hours of the journey.

A.380 km
B.390 km
C.400 km
D.410 km

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.

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