Speed, Time and Distance
- A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they next meet at the starting point again ?
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Required time = LCM of 252,
308 and 198 seconds.
Now, 252 = 2 × 2 × 3 × 3 × 7
308 = 2 × 2 × 7 × 11
198 = 2 × 3 × 3 × 11
∴ LCM = 2 × 2 × 3 × 3 × 7 × 11
= 36 × 77 seconds= 36 × 77 minutes 60 = 231 = 46 minutes 12 seconds 5 Correct Option: A
Required time = LCM of 252,
308 and 198 seconds.
Now, 252 = 2 × 2 × 3 × 3 × 7
308 = 2 × 2 × 7 × 11
198 = 2 × 3 × 3 × 11
∴ LCM = 2 × 2 × 3 × 3 × 7 × 11
= 36 × 77 seconds= 36 × 77 minutes 60 = 231 = 46 minutes 12 seconds 5
- A man walks a certain distance and rides back in 4 hours 30 minutes. He could ride both
ways in 3 hours. The time required by the man to walk both ways is
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Suppose, time taken while walking be x hours
And, time taken on riding be y hours
∴ According to questionx + y = 4 1 hours ...(i) 2
Then, 2y = 3 hoursy = 1 1 hours 2
From equation (i)x = 4 1 − 1 1 = 3 hours 2 2
Time required to walk both ways = 6 hoursCorrect Option: D
Suppose, time taken while walking be x hours
And, time taken on riding be y hours
∴ According to questionx + y = 4 1 hours ...(i) 2
Then, 2y = 3 hoursy = 1 1 hours 2
From equation (i)x = 4 1 − 1 1 = 3 hours 2 2
Time required to walk both ways = 6 hours
- A person, who can walk down a
and up the hill at the rate of 3 km/hour, ascends and comeshill at the rate of 4 1 km/hour 2
down to his starting point in 5 hours. How far did he ascend ?
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Let the required distance be x km.
∴ x + x = 5 9/2 3 ⇒ x 
2 + 1 
= 5 9 3 ⇒ x 2 + 3 = 5 9 ⇒ x = 5 × 9 = 9 km. 5 Correct Option: D
Let the required distance be x km.
∴ x + x = 5 9/2 3 ⇒ x 2 + 1 = 5 9 3 ⇒ x 2 + 3 = 5 9 ⇒ x = 5 × 9 = 9 km. 5
- A walks at a uniform rate of 4 km an hour; and 4 hours after his start, B bicycles after him at the uniform rate of 10 km an hour. How far from the starting point will B catch A ?
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Distance covered by A in 4 hours = 4 × 4 = 16 km
Relative speed of B with respect to A = 10 – 4 = 6 km/hr
∴ Time taken to catch A= 16 = 8 hours 6 3 ∴ Required distance = 8 × 10 = 80 3 3
= 26.67 km. ≈ 26.7 kmCorrect Option: D
Distance covered by A in 4 hours = 4 × 4 = 16 km
Relative speed of B with respect to A = 10 – 4 = 6 km/hr
∴ Time taken to catch A= 16 = 8 hours 6 3 ∴ Required distance = 8 × 10 = 80 3 3
= 26.67 km. ≈ 26.7 km
- A car completes a journey in 10 hours. If it covers half of the journey at 40 kmph and the remaining half at 60 kmph, the distance covered by car is
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Suppose distance be x km.
x + x = 10 2 × 40 2 × 60 ⇒ x + x = 10 80 120 ⇒ 3x + 2x = 10 240 ⇒ 5x = 10 240
x = 480 kmCorrect Option: B
Suppose distance be x km.
x + x = 10 2 × 40 2 × 60 ⇒ x + x = 10 80 120 ⇒ 3x + 2x = 10 240 ⇒ 5x = 10 240
x = 480 km
