Difference between revisions of "1973 AHSME Problems/Problem 29"
Rockmanex3 (talk | contribs) (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...") |
Rockmanex3 (talk | contribs) (Solution to Problem 29 — track problem) |
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\textbf{(E)}\ \text{none of these}</math> | \textbf{(E)}\ \text{none of these}</math> | ||
| − | ==Solution | + | ==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>\ | + | 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== | ||
Revision as of 20:16, 7 July 2018
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
Solution
Let 
be the length of the track in feet and 
be the number of laps that one of the boys did, so time one of the boys traveled before the two finish is 
. Since the time elapsed for both boys is equal, one boy ran 
feet while the other boy ran 
feet. Because both finished at the starting point, both ran an integral number of laps, so and are multiples of . Because both stopped when both met at the start for the first time, 
.
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 
and either or is a multiple of , one of the runners completed a lap. This is achieved when 
, so the two meet each other (excluding start and finish) a total of 
times.
See Also
| 1973 AHSC (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 | ||
