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 10:24, 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
Thus the total distance ran by Flash is
~ 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.