2004 AMC 10A Problems/Problem 15

Revision as of 18:32, 4 November 2006 by B-flat (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Given that $-4\leq x\leq-2$ and $2\leq y\leq4$, what is the largest possible value of (x+y)/x?

$\mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 {2}\qquad \mathrm{(E) \ } 1$

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

See Also

Invalid username
Login to AoPS