Difference between revisions of "2004 AMC 10A Problems/Problem 15"

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==Problem==
 
==Problem==
Given that <math>-4\leq x\leq-2</math> and <math>2\leq y\leq4</math>, what is the largest possible value of (x+y)/x?
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Given that <math>-4\leq x\leq-2</math> and <math>2\leq y\leq4</math>, what is the largest possible value of <math>\frac{x+y}{x}</math>?
  
 
<math> \mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1  </math>
 
<math> \mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1  </math>
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Rewrite <math>\frac{(x+y)}x</math> as <math>\frac{x}x+\frac{y}x=1+\frac{y}x</math>.
 
Rewrite <math>\frac{(x+y)}x</math> as <math>\frac{x}x+\frac{y}x=1+\frac{y}x</math>.
  
We also know that <math>\frac{y}x<0</math> because <math>x</math> and <math>y</math> are of opposite parity.
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We also know that <math>\frac{y}x<0</math> because <math>x</math> and <math>y</math> are of opposite sign.
  
Therefore, <math>1+\frac{y}x</math> is maximized when <math>\frac{y}x</math> is minimized, which occurs when <math>|x|</math> is the largest and <math>|y|</math> is the smallest.
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Therefore, <math>1+\frac{y}x</math> is maximized when <math>|\frac{y}x|</math> is minimized, which occurs when <math>|x|</math> is the largest and <math>|y|</math> is the smallest.
  
This occurs at (-4,2), so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow \mathrm{(D)}</math>.
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This occurs at <math>(-4,2)</math>, so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{\mathrm{(D)}\ \frac{1}{2}}</math>.
  
==See Also==
 
  
*[[2004 AMC 10A Problems]]
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== Solution 2==
  
*[[2004 AMC 10A Problems/Problem 14|Previous Problem]]
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If the answer choice is valid, then it must satisfy <math>\frac{(x+y)}x</math>. We use answer choices from greatest to least since the question asks for the greatest value.
  
*[[2004 AMC 10A Problems/Problem 16|Next Problem]]
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Answer choice <math>\text{(E)}</math>. We see that if <math>\frac{(x+y)}x = 1</math> then
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<math>x+y=x</math> and <math>y=0</math>. However, <math>0</math> is not in the domain of <math>y</math>, so <math>\text{(E)}</math> is incorrect.
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Answer choice <math>\text{(D)}</math>, however, we can find a value that satisfies <math>\frac{x+y}{x}=\frac{1}{2}</math> which simplifies to <math>x+2y=0</math>, such as <math>(-4,2)</math>.
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Therefore, <math>\boxed{\text{(D)}}</math> is the greatest.
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==See also==
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{{AMC10 box|year=2004|ab=A|num-b=14|num-a=16}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 12:46, 14 December 2019

Problem

Given that $-4\leq x\leq-2$ and $2\leq y\leq4$, what is the largest possible value of $\frac{x+y}{x}$?

$\mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1$

Solution

Rewrite $\frac{(x+y)}x$ as $\frac{x}x+\frac{y}x=1+\frac{y}x$.

We also know that $\frac{y}x<0$ because $x$ and $y$ are of opposite sign.

Therefore, $1+\frac{y}x$ is maximized when $|\frac{y}x|$ is minimized, which occurs when $|x|$ is the largest and $|y|$ is the smallest.

This occurs at $(-4,2)$, so $\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{\mathrm{(D)}\ \frac{1}{2}}$.


Solution 2

If the answer choice is valid, then it must satisfy $\frac{(x+y)}x$. We use answer choices from greatest to least since the question asks for the greatest value.

Answer choice $\text{(E)}$. We see that if $\frac{(x+y)}x = 1$ then

$x+y=x$ and $y=0$. However, $0$ is not in the domain of $y$, so $\text{(E)}$ is incorrect.

Answer choice $\text{(D)}$, however, we can find a value that satisfies $\frac{x+y}{x}=\frac{1}{2}$ which simplifies to $x+2y=0$, such as $(-4,2)$.

Therefore, $\boxed{\text{(D)}}$ is the greatest.

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 10 Problems and Solutions

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