2014 AMC 10B Problems/Problem 9

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 $\frac{4028}{-2} = \boxed{\textbf{(A)}\ -2014}$ .

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Video Solution (CREATIVE THINKING)

https://youtu.be/Y37KozgBEXg

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Video Solution

https://youtu.be/6Uh77bue0bE

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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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