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

Revision as of 00:50, 21 July 2014

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
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 10 Problems and Solutions

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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>.
+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==