Difference between revisions of "2017 AMC 12A Problems/Problem 2"

Revision as of 05:07, 15 January 2018

The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$

Let $x, y$ be our two numbers. Then $x+y = 4xy$ . Thus,

$\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = 4$ .

$\boxed{ \textbf{C}}$ .

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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

2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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 12 Problems and Solutions

@@ Line 9: / Line 9: @@
 Let <math>x, y</math> be our two numbers. Then <math>x+y = 4xy</math>. Thus,
-<math>\frac{x+y}{xy} = \frac{1}{x} + \frac{1}{y} = 4</math>.
+<math> \frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = 4</math>.
 <math>\boxed{ \textbf{C}}</math>.
 ==See Also==