2011 AIME II Problem 1

Question

Gary purchased a large beverage, but drank only

\frac{m}{n}

of this beverage, where

m

and

n

are relatively prime positive integers. If Gary had purchased only half as much and drunk twice as much, he would have wasted only

\frac{2}{9}

as much beverage. Find

m + n.

Answer

Let

x

be the amount purchased and

y

the amount drunk; the wasted amount is

x - y

The second scenario wastes

\frac{x}{2} - 2y,

so set

\frac{x}{2} - 2y = \frac{2}{9}(x - y)

and solve for

\frac{y}{x}

Say Gary purchased an amount

x

and drank an amount

y,

wasting

x - y.

In the second scenario he would have purchased

\frac{x}{2}

and drunk

2y,

wasting

\frac{x}{2} - 2y.

The condition is

\frac{x}{2} - 2y = \frac{2}{9}(x - y).

Multiplying by

18

gives

9x - 36y = 4x - 4y,

so

5x = 32y

and

\frac{y}{x} = \frac{5}{32}.

Since

\gcd(5, 32) = 1,

the answer is

5 + 32 = 37.

Problem 1 in other years