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

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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.
 
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>.  
+
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.
 
Therefore, <math>\boxed{\text{(D)}}</math> is our greatest possible value.

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 (ProblemsAnswer KeyResources)
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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