Difference between revisions of "1968 AHSME Problems/Problem 25"
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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 | 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 |
Latest revision as of 21:22, 17 July 2024
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
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.