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

Revision as of 15:35, 17 October 2021

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,

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

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

We can let $x=y.$ Then, we have that $2x=4x^2$ making $x=\tfrac{1}{2}.$ The answer is $\dfrac{2}{x}=4=\boxed{ \textbf{C}}.$ ~ Solasky (talk)

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 12: / Line 12: @@
 <math>\boxed{ \textbf{C}}</math>.
+==Solution 2==
+We can let <math>x=y.</math> Then, we have that <math>2x=4x^2</math> making <math>x=\tfrac{1}{2}.</math> The answer is <math>\dfrac{2}{x}=4=\boxed{ \textbf{C}}.</math> ~ [[User:Solasky|Solasky]] ([[User talk:Solasky|talk]])
 ==See Also==