Difference between revisions of "1970 AHSME Problems/Problem 32"
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== Solution == | == Solution == | ||
<math>\fbox{C}</math> | <math>\fbox{C}</math> | ||
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| + | Let <math>x</math> be half the circumference of the track. They first meet after <math>B</math> has run <math>100</math> yards, meaning that in the time <math>B</math> has run <math>100</math> yards, <math>A</math> has run <math>x-100</math> yards. The second time they meet is when <math>A</math> is 60 yards before he completes the lap. This means that in the time that <math>A</math> has run <math>2x-60</math> yards, <math>B</math> has run <math>x+60</math> yards. | ||
| + | Because they run at uniform speeds, we can write the equation | ||
| + | <cmath> \frac{100}{x-100}=\frac{x+60}{2x-60} .</cmath> | ||
| + | Cross multiplying, | ||
| + | <cmath> 200x-6000=x^2-40x-6000</cmath> | ||
| + | Adding <math>6000</math> to both sides and simplifying, we have | ||
| + | <cmath>200x=x^2-40x </cmath> | ||
| + | <cmath> 240x=x^2 </cmath> | ||
| + | <cmath>x=240.</cmath> | ||
| + | Because <math>x</math> is only half of the circumference of the track, the answer we are looking for is <math>2 \cdot 240 = 480, \text{or } \fbox{C} </math>. | ||
== See also == | == See also == | ||
Revision as of 18:13, 31 March 2020
Problem
and 
travel around a circular track at uniform speeds in opposite directions, starting from diametrically opposite points. If they start at the same time, meet first after has travelled 
yards, and meet a second time 
yards before completes one lap, then the circumference of the track in yards is
Solution
Let 
be half the circumference of the track. They first meet after has run yards, meaning that in the time has run yards, has run 
yards. The second time they meet is when is 60 yards before he completes the lap. This means that in the time that has run 
yards, has run 
yards.
Because they run at uniform speeds, we can write the equation
![\[\frac{100}{x-100}=\frac{x+60}{2x-60} .\]](http://latex.artofproblemsolving.com/8/0/4/8040013fe37a853fa8018cae068271f415cb57ac.png)
Cross multiplying,
![\[200x-6000=x^2-40x-6000\]](http://latex.artofproblemsolving.com/0/9/5/0958c7089aacc65a64158e9079ea77b3438c056a.png)
Adding 
to both sides and simplifying, we have
![\[200x=x^2-40x\]](http://latex.artofproblemsolving.com/1/c/d/1cd171d6b1539c9baaeefaaeaee90ff5246aca21.png)
![\[240x=x^2\]](http://latex.artofproblemsolving.com/e/2/5/e252e202d977708482ceb9f923c55aed3e2b4843.png)
![\[x=240.\]](http://latex.artofproblemsolving.com/4/5/5/45573b162a77e38330a8d0dc300f2f4f7edc62b4.png)
Because is only half of the circumference of the track, the answer we are looking for is 
.
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 31 |
Followed by Problem 33 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
