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

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

(→‎Solution)
(→‎Solution)
Line 12: Line 12:
  
 
This occurs at <math>(-4,2)</math>, so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{\mathrm{(D)}\ \frac{1}{2}}</math>.
 
This occurs at <math>(-4,2)</math>, so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{\mathrm{(D)}\ \frac{1}{2}}</math>.
 +
 +
 +
== Solution 2==
 +
 +
If the answer choice is valid, then it must satisfy <math>\frac{(x+y)}x</math>. We use answer choices from greatest to least since the question asks for the greatest value.
 +
 +
Answer choice <math>\text{(E)}</math>. We see that if <math>\frac{(x+y)}x = 1</math> then
 +
 +
<math>x+y=x</math> and <math>y=0</math>. However, <math>0</math> is not in the domain of <math>y</math>, so <math>\text{(E)}</math> is incorrect.
 +
 +
Answer choice <math>\text{(D)}</math>, however, we can find a value that satisfies <math>\frac{x+y}{x}=\frac{1}{2}</math> which simplifies to <math>x+2y=0</math>, such as <math>(-4,2)</math>.
 +
 +
Therefore, <math>\boxed{\text{(D)}</math> is the greatest.
  
 
==See also==
 
==See also==

Revision as of 11:44, 14 December 2019

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, $\boxed{\text{(D)}$ (Error compiling LaTeX. Unknown error_msg) is the greatest.

See also

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_15&oldid=112846"