Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "==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...")
 
(→‎See Also)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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 me at the point A again, then the number of times they meet, excluding the start and finish, is  
+
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
Line 9: Line 9:
 
\textbf{(E)}\ \text{none of these}</math>
 
\textbf{(E)}\ \text{none of these}</math>
  
−
==Solution (WIP)==
+
==Solution==
  
−
Let <math>d</math> be the length of the track in feet and <math>x</math> be the number of laps that one of the boys did, so time one of the boys traveled before the two finish is <math>\frac{dx}{5}</math>.  Since the time elapsed for both boys is equal, one boy ran <math>5(\frac{dx}{5})</math> feet while the other boy ran <math>9(\frac{dx}{5})</math> feet.
+
Let <math>d</math> be the length of the track in feet and <math>x</math> be the number of laps that one of the boys did, so time one of the boys traveled before the two finish is <math>\tfrac{dx}{5}</math>.  Since the time elapsed for both boys is equal, one boy ran <math>5(\tfrac{dx}{5})</math> feet while the other boy ran <math>9(\tfrac{dx}{5})</math> feet.  Because both finished at the starting point, both ran an integral number of laps, so <math>5(\tfrac{dx}{5})</math> and <math>9(\tfrac{dx}{5})</math> are multiples of <math>d</math>.  Because both stopped when both met at the start for the first time, <math>x = 5</math>.
  
 +
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 <math>0 < x \le 5</math> and either <math>5(\tfrac{dx}{5})</math> or <math>9(\tfrac{dx}{5})</math> is a multiple of <math>d</math>, one of the runners completed a lap.  This is achieved when <math>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</math>, so the two meet each other (excluding start and finish) a total of <math>\boxed{\textbf{(A)}\ 13}</math> times.
  
 
==See Also==
 
==See Also==
−
{{AHSME 35p box|year=1973|num-b=28|num-a=30}}
+
{{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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1973_AHSME_Problems/Problem_29&oldid=118211"