Two runners are running clockwise around a circular track. If they start at the same time and at the same position, after 4.5 minutes, the faster runner will have covered 1 more lap than the slower runner, with each having covered an integer number of laps in that time. The slower runner doesn't get passed until after he finishes a lap. The time it takes each of them to finish one lap is an integer number of seconds.
How many more seconds does it take the slower runner to finish one lap compared with the faster runner?
Answer (2)
Define variables for the times and number of laps for both runners.
Set up the equation based on the given information: t f 270 = t s 270 + 1 .
Solve for integer solutions of t f and t s .
The difference in time is t s − t f = 3 seconds.
Explanation
Define variables Let t f be the time in seconds for the faster runner to complete one lap, and t s be the time in seconds for the slower runner to complete one lap. Let n f be the number of laps the faster runner completes in 4.5 minutes (270 seconds), and n s be the number of laps the slower runner completes in 270 seconds. We are given that n f = n s + 1 , and we want to find t s − t f .
Set up the equation We know that n f = t f 270 and n s = t s 270 . Substituting these into the equation n f = n s + 1 , we get t f 270 = t s 270 + 1
Rearrange the equation Multiplying both sides by t f t s , we have 270 t s = 270 t f + t f t s Rearranging the terms, we get t f t s − 270 t s + 270 t f = 0 Adding 27 0 2 to both sides to factor, we have t f t s − 270 t s + 270 t f − 27 0 2 = − 27 0 2 This doesn't seem to help. Let's go back to the original equation: t f 270 = t s 270 + 1 Multiplying by t f t s , we get 270 t s = 270 t f + t f t s , so t f t s = 270 ( t s − t f ) .
Thus, t s − t f = 270 t f t s . Since t s and t f are integers, t f t s must be divisible by 270.
Solve for integer solutions We are looking for integer solutions for t f and t s . From t f 270 = t s 270 + 1 , we can write t f 270 − t s 270 = 1 , which means 270 t s − 270 t f = t f t s , or 270 ( t s − t f ) = t f t s . We also know that t_f"> t s > t f .
Let t s − t f = x . Then t s = t f + x , and 270 x = t f ( t f + x ) . So t f 2 + x t f − 270 x = 0 . Using the quadratic formula, we have t f = 2 − x ± x 2 + 4 ( 270 x ) = 2 − x ± x 2 + 1080 x Since t f must be positive, we take the positive root. Also, x 2 + 1080 x must be a perfect square.
Find the difference in time We need to find an integer x such that x 2 + 1080 x is a perfect square. Let x 2 + 1080 x = y 2 for some integer y . Completing the square, we have ( x + 540 ) 2 − 54 0 2 = y 2 , so ( x + 540 ) 2 − y 2 = 54 0 2 , which factors as ( x + 540 − y ) ( x + 540 + y ) = 54 0 2 = 2 4 ⋅ 3 6 ⋅ 5 2 .
We can test values for t s starting from 2. If t s = 30 , then n s = 30 270 = 9 . Then n f = 10 , so t f = 10 270 = 27 . Then t s − t f = 30 − 27 = 3 .
So the difference in time is 3 seconds.
Final Answer The difference in time it takes the slower runner to finish one lap compared with the faster runner is 3 seconds.
Examples
Understanding relative speeds and lap times is crucial in various real-world scenarios, such as optimizing race strategies, coordinating logistics in delivery services, or even managing traffic flow. For instance, if you're planning a relay race, knowing the individual lap times and the differences between runners can help you predict the overall race time and optimize team composition. Similarly, in logistics, understanding the speed differences between delivery vehicles can help in planning routes and estimating delivery times more accurately. This problem demonstrates how mathematical relationships can be applied to improve efficiency and decision-making in everyday situations.
The slower runner takes 3 more seconds than the faster runner to complete one lap. By using equations based on lap times and the given conditions, we found this difference through calculations involving integer solutions. This demonstrates the relationship between lap time and number of laps completed in a fixed time frame.
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