2011 AIME II Problem 1

Attempt Problem 1 of the 2011 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AIME II solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

Gary purchased a large beverage, but drank only mn\frac{m}{n} of this beverage, where mm and nn are relatively prime positive integers. If Gary had purchased only half as much and drunk twice as much, he would have wasted only 29\frac{2}{9} as much beverage. Find m+n.m + n.

Answer: 37
Concepts:linear equationratio and proportion
Difficulty rating: 1710
Small Hint:

Let xx be the amount purchased and yy the amount drunk; the wasted amount is x−yx - y

Big Hint:

The second scenario wastes x2−2y,\frac{x}{2} - 2y, so set x2−2y=29(x−y)\frac{x}{2} - 2y = \frac{2}{9}(x - y) and solve for yx\frac{y}{x}

Solution:

Say Gary purchased an amount xx and drank an amount y,y, wasting x−y.x - y. In the second scenario he would have purchased x2\frac{x}{2} and drunk 2y,2y, wasting x2−2y.\frac{x}{2} - 2y. The condition is x2−2y=29(x−y).\frac{x}{2} - 2y = \frac{2}{9}(x - y).

Multiplying by 1818 gives 9x−36y=4x−4y,9x - 36y = 4x - 4y, so 5x=32y5x = 32y and yx=532.\frac{y}{x} = \frac{5}{32}. Since gcd⁡(5,32)=1,\gcd(5, 32) = 1, the answer is 5+32=37.5 + 32 = 37.

Full Exam

Problem 1 in Other Years