Difference between revisions of "1968 AHSME Problems/Problem 25"
m (→See also) |
(→Solution 2) |
||
| Line 14: | Line 14: | ||
== Solution 2 == | == Solution 2 == | ||
| − | Let <math>k</math> denotes the distance Ace needs to run after the <math>y</math> yard. Since the distance | + | Let <math>k</math> denotes the distance Ace needs to run after the <math>y</math> yard. Since the distance they run with same amount of time is proportional to their speed, we have |
<cmath>\frac{1}{x}=\frac{k}{y+k}</cmath> | <cmath>\frac{1}{x}=\frac{k}{y+k}</cmath> | ||
<cmath>k=\frac{y}{x-1}</cmath> | <cmath>k=\frac{y}{x-1}</cmath> | ||
Revision as of 09:24, 23 February 2024
Contents
Problem
Ace runs with constant speed and Flash runs 
times as fast, 
. Flash gives Ace a head start of 
yards, and, at a given signal, they start off in the same direction. Then the number of yards Flash must run to catch Ace is:
Solution
Solution 2
Let 
denotes the distance Ace needs to run after the yard. Since the distance they run with same amount of time is proportional to their speed, we have
![\[\frac{1}{x}=\frac{k}{y+k}\]](http://latex.artofproblemsolving.com/6/7/6/676fb9755840e3d663be39211115de1e4fba112d.png)
![\[k=\frac{y}{x-1}\]](http://latex.artofproblemsolving.com/4/6/5/46559ef09e4144af8a696534779640a7eecb03eb.png)
Thus the total distance ran by Flash is
![\[y+k=y+\frac{y}{x-1}=\frac{xy}{x-1}\boxed{C}\]](http://latex.artofproblemsolving.com/9/8/b/98be3a53ebcf5a85a21c0107c18d09b694651022.png)
~ Nafer
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 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. 
