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

Revision as of 21:39, 22 January 2015

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

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

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 $\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{\mathrm{(D)}\ \frac{1}{2}}$ .

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

@@ Line 9: / Line 9: @@
 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.
+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 <math>(-4,2)</math>, so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{\mathrm{(D)}\ \frac{1}{2}}</math>.