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

Revision as of 12:12, 5 November 2006

Given that $-4\leq x\leq-2$ and $2\leq y\leq4$ , what is the largest possible value of (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 parity.

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

@@ Line 2: / Line 2: @@
 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?
-<math> \mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 {2}\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>
 ==Solution==
@@ Line 20: / Line 20: @@
 *[[2004 AMC 10A Problems/Problem 16|Next Problem]]
+[[Category:Introductory Algebra Problems]]