Question:
In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same starting point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?
In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same starting point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?
Show Hint
When two runners on a loop are at the same point for the first time, the faster one has gained exactly one lap on the slower one. Use that to find the speeds.
Updated On: Nov 7, 2026
- 20 min
- 15 min
- 10 min
- 5 min
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
Two runners start together from the same point on a circular track of length \(1000\) m. The fast runner is exactly twice as quick as the slow one. They are at the same point again, for the first time, at the end of \(5\) minutes. We need the time the fast runner takes to run \(4000\) m.
Step 2: What meeting at the same point means:
Both run in the same direction on the loop. The fast runner gains distance on the slow one. They are at the same point on the track whenever the gain is a whole number of laps.
The first time this happens after the start is when the gain is exactly one lap, which is \(1000\) m.
Step 3: Use relative speed:
Let the slow runner's speed be \(v\) m/min. Then the fast runner's speed is \(2v\) m/min, so the fast runner gains \(2v - v = v\) metres every minute.
The gain of \(1000\) m takes \(5\) minutes:
\[ v \times 5 = 1000 \Rightarrow v = 200 \text{ m/min} \]
So the slow runner goes at \(200\) m/min and the fast runner at \(400\) m/min.
Step 4: Time for the whole race:
The fast runner needs to cover \(4000\) m at \(400\) m/min:
\[ \text{time} = \dfrac{4000}{400} = 10 \text{ minutes} \]
Step 5: Check the options:
\(20\) min would mean a speed of only \(200\) m/min, which is the slow runner's speed. \(15\) min would mean about \(267\) m/min, which fits neither runner. \(5\) min would mean \(800\) m/min, four times the slow speed, which breaks the "twice" condition. Only \(10\) min fits.
Final Answer:
The fastest runner finishes the race in \(10\) minutes, which is option 3. \[ \boxed{10 \text{ min (Option 3)}} \]
Two runners start together from the same point on a circular track of length \(1000\) m. The fast runner is exactly twice as quick as the slow one. They are at the same point again, for the first time, at the end of \(5\) minutes. We need the time the fast runner takes to run \(4000\) m.
Step 2: What meeting at the same point means:
Both run in the same direction on the loop. The fast runner gains distance on the slow one. They are at the same point on the track whenever the gain is a whole number of laps.
The first time this happens after the start is when the gain is exactly one lap, which is \(1000\) m.
Step 3: Use relative speed:
Let the slow runner's speed be \(v\) m/min. Then the fast runner's speed is \(2v\) m/min, so the fast runner gains \(2v - v = v\) metres every minute.
The gain of \(1000\) m takes \(5\) minutes:
\[ v \times 5 = 1000 \Rightarrow v = 200 \text{ m/min} \]
So the slow runner goes at \(200\) m/min and the fast runner at \(400\) m/min.
Step 4: Time for the whole race:
The fast runner needs to cover \(4000\) m at \(400\) m/min:
\[ \text{time} = \dfrac{4000}{400} = 10 \text{ minutes} \]
Step 5: Check the options:
\(20\) min would mean a speed of only \(200\) m/min, which is the slow runner's speed. \(15\) min would mean about \(267\) m/min, which fits neither runner. \(5\) min would mean \(800\) m/min, four times the slow speed, which breaks the "twice" condition. Only \(10\) min fits.
Final Answer:
The fastest runner finishes the race in \(10\) minutes, which is option 3. \[ \boxed{10 \text{ min (Option 3)}} \]
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