# 2011 AMC 12B Problems/Problem 7

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## Problem

Let $x$ and $y$ be two-digit positive integers with mean $60$. What is the maximum value of the ratio $\frac{x}{y}$?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac{33}{7} \qquad \textbf{(C)}\ \frac{39}{7} \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ \frac{99}{10}$

## Solution

If $x$ and $y$ have a mean of $60$, then $\frac{x+y}{2}=60$ and $x+y=120$. To maximize $\frac{x}{y}$, we need to maximize $x$ and minimize $y$. Since they are both two-digit positive integers, the maximum of $x$ is $99$ which gives $y=21$. $y$ cannot be decreased because doing so would increase $x$, so this gives the maximum value of $\frac{x}{y}$, which is $\frac{99}{21}=\boxed{\frac{33}{7}\ \textbf{(B)}}$

 2011 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 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 All AMC 12 Problems and Solutions