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Difference between revisions of "2014 AMC 10B Problems/Problem 9"

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==Problem==
 
==Problem==
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For real numbers <math> w </math> and <math> z </math>, <cmath> \cfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014. </cmath> What is <math> \frac{w+z}{w-z} </math>? \\
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For real numbers <math> w </math> and <math> z </math>, <cmath> \cfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014. </cmath> What is <math> \frac{w+z}{w-z} </math>?  
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<math> \textbf{(A) }-2014\qquad\textbf{(B) }\frac{-1}{2014}\qquad\textbf{(C) }\frac{1}{2014}\qquad\textbf{(D) }1\qquad\textbf{(E) }2014 </math>
 
<math> \textbf{(A) }-2014\qquad\textbf{(B) }\frac{-1}{2014}\qquad\textbf{(C) }\frac{1}{2014}\qquad\textbf{(D) }1\qquad\textbf{(E) }2014 </math>
  
 
==Solution==
 
==Solution==
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Multiply the numerator and denominator of the LHS by <math>wz</math> to get <math>\frac{z+w}{z-w}=2014</math>. Then since <math>z+w=w+z</math> and <math>w-z=-(z-w)</math>, <math>\frac{w+z}{w-z}=-\frac{z+w}{z-w}=-2014</math>, or choice <math>\boxed{A}</math>.
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==Solution 2==
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Muliply both sides by <math>\left(\frac{1}{w}-\frac{1}{z}\right)</math> to get <math>\frac{1}{w}+\frac{1}{z}=2014\left(\frac{1}{w}-\frac{1}{z}\right)</math>. Then, add <math>2014\cdot\frac{1}{z}</math> to both sides and subtract <math>\frac{1}{w}</math> from both sides to get <math>2015\cdot\frac{1}{z}=2013\cdot\frac{1}{w}</math>. Then, we can plug in the most simple values for z and w (<math>2015</math> and <math>2013</math>, respectively), and find <math>\frac{2013+2015}{2013-2015}=\frac{2(2014)}{-2}=-2014</math>, or answer choice <math>\boxed{A}</math>.
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==Solution 3==
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Let <math>a = \frac{1}{w}</math> and <math>b = \frac{1}{z}</math>. To find values for a and b, we can try <math>a+b = 2014</math> and <math>a-b=1</math>. However, that leaves us with a fractional solution, so scaling it by 2, we get <math>a+b = 4028</math> and <math>a-b=2</math>. Solving by adding the equations together, we get <math>b = 2015</math> and <math>a = 2013</math>. Now, substituting back in, we get <math>w = \frac{1}{2015}</math> and <math>z = \frac{1}{2013}</math>. Now, putting this into the desired equation with <math>n = 2015 \cdot 2013</math> (since it will cancel out), we get <math>\frac{\frac{2013+2015}{n}}{\frac{2013-2015}{n}}</math>. Dividing, we get <math>\frac{4028}{-2} = \boxed{\textbf{(A)}\ -2014}</math>.
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~idk12345678
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==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/Y37KozgBEXg
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 +
~Education, the Study of Everything
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==Video Solution==
 +
https://youtu.be/6Uh77bue0bE
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 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2014|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:47, 15 April 2024

Problem

For real numbers $w$ and $z$, \[\cfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014.\] What is $\frac{w+z}{w-z}$?

$\textbf{(A) }-2014\qquad\textbf{(B) }\frac{-1}{2014}\qquad\textbf{(C) }\frac{1}{2014}\qquad\textbf{(D) }1\qquad\textbf{(E) }2014$

Solution

Multiply the numerator and denominator of the LHS by $wz$ to get $\frac{z+w}{z-w}=2014$. Then since $z+w=w+z$ and $w-z=-(z-w)$, $\frac{w+z}{w-z}=-\frac{z+w}{z-w}=-2014$, or choice $\boxed{A}$.

Solution 2

Muliply both sides by $\left(\frac{1}{w}-\frac{1}{z}\right)$ to get $\frac{1}{w}+\frac{1}{z}=2014\left(\frac{1}{w}-\frac{1}{z}\right)$. Then, add $2014\cdot\frac{1}{z}$ to both sides and subtract $\frac{1}{w}$ from both sides to get $2015\cdot\frac{1}{z}=2013\cdot\frac{1}{w}$. Then, we can plug in the most simple values for z and w ($2015$ and $2013$, respectively), and find $\frac{2013+2015}{2013-2015}=\frac{2(2014)}{-2}=-2014$, or answer choice $\boxed{A}$.

Solution 3

Let $a = \frac{1}{w}$ and $b = \frac{1}{z}$. To find values for a and b, we can try $a+b = 2014$ and $a-b=1$. However, that leaves us with a fractional solution, so scaling it by 2, we get $a+b = 4028$ and $a-b=2$. Solving by adding the equations together, we get $b = 2015$ and $a = 2013$. Now, substituting back in, we get $w = \frac{1}{2015}$ and $z = \frac{1}{2013}$. Now, putting this into the desired equation with $n = 2015 \cdot 2013$ (since it will cancel out), we get $\frac{\frac{2013+2015}{n}}{\frac{2013-2015}{n}}$. Dividing, we get $\frac{4028}{-2} = \boxed{\textbf{(A)}\ -2014}$.

~idk12345678

Video Solution (CREATIVE THINKING)

https://youtu.be/Y37KozgBEXg

~Education, the Study of Everything


Video Solution

https://youtu.be/6Uh77bue0bE

~savannahsolver

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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