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

Revision as of 07:03, 22 December 2020

Problem

Given that $-4\leq x\leq-2$ and $2\leq y\leq4$ , what is the largest possible value of $\frac{x+y}{x}$ ?

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

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 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 $\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{\mathrm{(D)}\ \frac{1}{2}}$ .

Solution 2

If the answer choice is valid, then it must satisfy $\frac{(x+y)}x$ . We use answer choices from greatest to least since the question asks for the greatest value.

Answer choice $\text{(E)}$ . We see that if $\frac{(x+y)}x = 1$ then

$x+y=x$ and $y=0$ . However, $0$ is not in the domain of $y$ , so $\text{(E)}$ is incorrect.

Answer choice $\text{(D)}$ , however, we can find a value that satisfies $\frac{x+y}{x}=\frac{1}{2}$ which simplifies to $x+2y=0$ , such as $(-4,2)$ .

Therefore, $\boxed{\text{(D)}}$ is the greatest.

Solution 3

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 25: / Line 25: @@
 Therefore, <math>\boxed{\text{(D)}}</math> is the greatest.
+== Solution 3==
 ==See also==

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