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

Revision as of 20:00, 28 February 2007

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

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

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

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

@@ Line 1: / Line 1: @@
 ==Problem==
-Given that <math>-4\leq x\leq-2</math> and <math>2\leq y\leq4</math>, what is the largest possible value of (x+y)/x?
+Given that <math>-4\leq x\leq-2</math> and <math>2\leq y\leq4</math>, what is the largest possible value of <math>\frac{x+y}{x}</math>?
 <math> \mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1  </math>
@@ Line 13: / Line 13: @@
 This occurs at (-4,2), so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow \mathrm{(D)}</math>.
-==See Also==
+==See also==
+{{AMC10 box|year=2004|ab=A|num-b=14|num-a=16}}
-*[[2004 AMC 10A Problems]]
-*[[2004 AMC 10A Problems/Problem 14|Previous Problem]]
-*[[2004 AMC 10A Problems/Problem 16|Next Problem]]
 [[Category:Introductory Algebra Problems]]