Difference between revisions of "2011 AMC 12B Problems/Problem 7"

Latest revision as of 10:02, 4 July 2013

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 by Problem 6	Followed by Problem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 7: / Line 7: @@
 ==See also==
 {{AMC12 box|year=2011|num-b=6|num-a=8|ab=B}}
+{{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "2011 AMC 12B Problems/Problem 7"

Latest revision as of 10:02, 4 July 2013

Problem

Solution

See also