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

Latest revision as of 14:42, 9 June 2023

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)

~Education, the Study of Everything

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>.
+==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]])
+==Video Solution (HOW TO THINK CREATIVELY!!!)==
+https://youtu.be/fOakoTBF_cE
+~Education, the Study of Everything
 ==See Also==
-{{AMC10 box|year=2017|ab=A|num-b=1|num-a=3}}
+{{AMC10 box|year=2017|ab=A|num-b=4|num-a=6}}
 {{AMC12 box|year=2017|ab=A|num-b=1|num-a=3}}
 {{MAA Notice}}