Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2004 AMC 10A Problems/Problem 15

Revision as of 12:12, 5 November 2006 by Xantos C. Guin (talk | contribs) (added category; fixed typo in answer choice)

Problem

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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_15&oldid=11029"