2004 AMC 10A Problems/Problem 15: Difference between revisions

Revision as of 19:11, 28 February 2007

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

@@ Line 7: / Line 7: @@
 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.
+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.