Difference between revisions of "2011 AIME II Problems/Problem 1"
(→Problem) |
|||
(6 intermediate revisions by 5 users not shown) | |||
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
− | Let <math>x</math> be the [[fraction]] consumed, then <math>(1-x)</math> is the fraction wasted. We have <math>1 | + | Let <math>x</math> be the [[fraction]] consumed, then <math>(1-x)</math> is the fraction wasted. We have <math>\frac{1}{2} - 2x =\frac{2}{9} (1-x)</math>, or <math>9 - 36x = 4 - 4x</math>, or <math>32x = 5</math> or <math>x = 5/32</math>. Therefore, <math>m + n = 5 + 32 = \boxed{037}</math>. |
==See also== | ==See also== | ||
{{AIME box|year=2011|n=II|before=First Problem|num-a=2}} | {{AIME box|year=2011|n=II|before=First Problem|num-a=2}} | ||
− | [[Category:Intermediate | + | [[Category:Intermediate Algebra Problems]] |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:09, 28 January 2024
Problem
Gary purchased a large beverage, but only drank of it, where
and
are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only
as much beverage. Find
.
Solution
Let be the fraction consumed, then
is the fraction wasted. We have
, or
, or
or
. Therefore,
.
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.