2004 AMC 10A Problems/Problem 15

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

As $-4\leq x\leq-2$ , we know that the denominator of our given fraction is negative. So to achieve the greatest value possible, our numerator, or $x+y$ must also be as small as possible.

So we pick our smallest value for $y$ , which is $2$ .

Now if we if set our value of $x$ to its lowest, our expression becomes $\frac{(-2+2)}{-2} = \frac{0}{2}$ As we decrease our $x$ value, we see that our numerator decrease from $0$ and our denominator decrease from $-2$ at the same rate. So decreasing our $x$ value decreases the overwhelming gap between our denominator and numerator, which gives us an overall bigger number.

So we also pick the smallest value for $x$ , which is $-4$ . We know have $\frac{(-4+2)}{-4} = \frac{-2}{-4} = \frac{1}{2}$ .

Therefore, $\boxed{\text{(D)}}$ is our greatest possible value.

-JinhoK

Video Solution

https://youtu.be/LBgCCFdCvYc

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