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Difference between revisions of "1973 AHSME Problems/Problem 29"

(Solution to Problem 29 — track problem)
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==Problem==
 
==Problem==
 
   
 
   
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Two boys start moving from the same point A on a circular track but in opposite directions. Their speeds are 5 ft. per second and 9 ft. per second. If they start at the same time and finish when they first me at the point A again, then the number of times they meet, excluding the start and finish, is  
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Two boys start moving from the same point A on a circular track but in opposite directions. Their speeds are 5 ft. per second and 9 ft. per second. If they start at the same time and finish when they first meet at the point A again, then the number of times they meet, excluding the start and finish, is  
  
 
<math> \textbf{(A)}\ 13 \qquad
 
<math> \textbf{(A)}\ 13 \qquad
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==See Also==
 
==See Also==
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{{AHSME 35p box|year=1973|num-b=28|num-a=30}}
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{{AHSME 30p box|year=1973|num-b=28|num-a=30}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 13:04, 20 February 2020

Problem

Two boys start moving from the same point A on a circular track but in opposite directions. Their speeds are 5 ft. per second and 9 ft. per second. If they start at the same time and finish when they first meet at the point A again, then the number of times they meet, excluding the start and finish, is

$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 44 \qquad \textbf{(D)}\ \text{infinity} \qquad \textbf{(E)}\ \text{none of these}$

Solution

Let $d$ be the length of the track in feet and $x$ be the number of laps that one of the boys did, so time one of the boys traveled before the two finish is $\tfrac{dx}{5}$. Since the time elapsed for both boys is equal, one boy ran $5(\tfrac{dx}{5})$ feet while the other boy ran $9(\tfrac{dx}{5})$ feet. Because both finished at the starting point, both ran an integral number of laps, so $5(\tfrac{dx}{5})$ and $9(\tfrac{dx}{5})$ are multiples of $d$. Because both stopped when both met at the start for the first time, $x = 5$.

Note that between the time a runner finishes a lap and a runner (can be same) finishes a lap, both runners must meet each other. When $0 < x \le 5$ and either $5(\tfrac{dx}{5})$ or $9(\tfrac{dx}{5})$ is a multiple of $d$, one of the runners completed a lap. This is achieved when $x = \tfrac59, 1, \tfrac{10}{9}, \tfrac{15}{9}, 2, \tfrac{20}{9}, \tfrac{25}{9}, 3, \tfrac{30}{9}, \tfrac{35}{9}, 4, \tfrac{40}{9}, 5$, so the two meet each other (excluding start and finish) a total of $\boxed{\textbf{(A)}\ 13}$ times.

See Also

1973 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions
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