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

(Created page with '==Problem== Let <math>x</math> and <math>y</math> be two-digit positive integers with mean <math>60</math>. What is the maximum value of the ratio <math>\frac{x}{y}</math>? <mat…')
 
(→‎See also)
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If <math>x</math> and <math>y</math> have a mean of <math>60</math>, then <math>\frac{x+y}{2}=60</math> and <math>x+y=120</math>. To maximize <math>\frac{x}{y}</math>, we need to maximize <math>x</math> and minimize <math>y</math>. Since they are both two-digit positive integers, the maximum of <math>x</math> is <math>99</math> which gives <math>y=21</math>. <math>y</math> cannot be decreased because doing so would increase <math>x</math>, so this gives the maximum value of <math>\frac{x}{y}</math>, which is <math>\frac{99}{21}=\boxed{\frac{33}{7}\ \textbf{(B)}}</math>
 
If <math>x</math> and <math>y</math> have a mean of <math>60</math>, then <math>\frac{x+y}{2}=60</math> and <math>x+y=120</math>. To maximize <math>\frac{x}{y}</math>, we need to maximize <math>x</math> and minimize <math>y</math>. Since they are both two-digit positive integers, the maximum of <math>x</math> is <math>99</math> which gives <math>y=21</math>. <math>y</math> cannot be decreased because doing so would increase <math>x</math>, so this gives the maximum value of <math>\frac{x}{y}</math>, which is <math>\frac{99}{21}=\boxed{\frac{33}{7}\ \textbf{(B)}}</math>
 
==See also==
 
==See also==
 +
{{AMC12 box|year=2011|num-b=6|num-a=8|ab=B}}

Revision as of 18:35, 6 March 2011

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 $\frac{99}{21}=\boxed{\frac{33}{7}\ \textbf{(B)}}$

See also

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
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