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

Latest revision as of 00:25, 3 November 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 (left hand side) 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

Basic algebra at the end of the day, so simplify the numerator and the denominator. The numerator simplifies out to $\frac{w+z}{wz}$ and the denominator simplifies out to $\frac{z-w}{wz}$ .

This results in $\cfrac{\frac{w+z}{zw}}{\frac{z-w}{zw}} = 2014$ .

Division results in the elimination of $zw$ , so we get $\frac{w+z}{z-w} = 2014$ .

$z-w$ is just $-(w-z)$ so the equation above is $-(\frac{w+z}{w-z} = 2014$ .

Solving this results in $\frac{w+z}{w-z} = \boxed{\textbf{(A)}\ -2014}$ .

~AkCANdo

Solution 3

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 4

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

Solution 5

Set $ w = 2 $ and $ z = 1 $.

Substitute the new values into the first equation

$1/2 + 1 = 3/2$ ,

$1/2 - 1 = -1/2$ ,

$(3/2) / (-1/2) = -3$

Substitute in the second equation with new values of $ w $ and $ z $:

(2 + 1) / (2 - 1) = 3.

Answers of each equation (where X is the quotient): $x$ and $-x$

Therefore, the answers to the equations are the negatives of each other. Thus the answer is (A)

~WalkEmDownTrey

Video Solution (CREATIVE THINKING)

https://youtu.be/Y37KozgBEXg

~Education, the Study of Everything

Video Solution

https://youtu.be/6Uh77bue0bE

~savannahsolver

@@ Line 11: / Line 11: @@
 Basic algebra at the end of the day, so simplify the numerator and the denominator. The numerator simplifies out to
-<math>\frac{w+z}{wz}</math> and the denominator simplifies out to <math>/frac{z-w}{wz}</math>.
+<math>\frac{w+z}{wz}</math> and the denominator simplifies out to <math>\frac{z-w}{wz}</math>.
 This results in <math>\cfrac{\frac{w+z}{zw}}{\frac{z-w}{zw}} = 2014</math>.

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