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

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==Problem==
 
==Problem==
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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?
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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>
 
<math> \mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1  </math>
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Rewrite <math>\frac{(x+y)}x</math> as <math>\frac{x}x+\frac{y}x=1+\frac{y}x</math>.
 
Rewrite <math>\frac{(x+y)}x</math> as <math>\frac{x}x+\frac{y}x=1+\frac{y}x</math>.
  
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We also know that <math>\frac{y}x<0</math> because <math>x</math> and <math>y</math> are of opposite parity.
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We also know that <math>\frac{y}x<0</math> because <math>x</math> and <math>y</math> are of opposite sign.
  
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Therefore, <math>1+\frac{y}x</math> is maximized when <math>\frac{y}x</math> is minimized, which occurs when <math>|x|</math> is the largest and <math>|y|</math> is the smallest.
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Therefore, <math>1+\frac{y}x</math> is maximized when <math>|\frac{y}x|</math> is minimized, which occurs when <math>|x|</math> is the largest and <math>|y|</math> is the smallest.
  
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This occurs at (-4,2), so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow \mathrm{(D)}</math>.
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This occurs at <math>(-4,2)</math>, so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{\mathrm{(D)}\ \frac{1}{2}}</math>.
  
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==See Also==
 
  
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*[[2004 AMC 10A Problems]]
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== Solution 2==
  
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*[[2004 AMC 10A Problems/Problem 14|Previous Problem]]
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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.
  
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*[[2004 AMC 10A Problems/Problem 16|Next Problem]]
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Answer choice <math>\text{(E)}</math>. We see that if <math>\frac{(x+y)}x = 1</math> then
 +
 
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<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>.
 +
 
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Therefore, <math>\boxed{\text{(D)}}</math> is the greatest.
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 +
 
 +
 
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== Solution 3==
 +
 
 +
 
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As <math>-4\leq x\leq-2</math>, we know that the denominator of our given fraction is negative. So to achieve the greatest value possible, our numerator, or <math>x+y</math> must also be as small as possible.
 +
 
 +
So we pick our smallest value for <math>y</math>, which is <math>2</math>.
 +
 
 +
Now if we if set our value of <math>x</math> to its lowest, our expression becomes <math>\frac{(-2+2)}{-2} = \frac{0}{2}</math> As we decrease our <math>x</math> value, we see that our numerator decrease from <math>0</math> and our denominator decrease from <math>-2</math> at the same rate. So decreasing our <math>x</math> 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 <math>x</math>, which is <math>-4</math>. We know have <math>\frac{(-4+2)}{-4} = \frac{-2}{-4} = \frac{1}{2}</math>.
 +
 
 +
Therefore, <math>\boxed{\text{(D)}}</math> is our greatest possible value.
 +
 
 +
-JinhoK
 +
 
 +
==Video Solution==
 +
https://youtu.be/LBgCCFdCvYc
 +
 
 +
Education, the Study of Everything
 +
 
 +
 
 +
 
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==See also==
 +
{{AMC10 box|year=2004|ab=A|num-b=14|num-a=16}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 01:16, 5 July 2021

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

Education, the Study of Everything


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

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