Difference between revisions of "2017 AMC 10A Problems/Problem 5"

Revision as of 19:32, 8 February 2017

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 the two real numbers be $x,y$ . We are given that

$x+y=4xy,$

and dividing both sides by $xy$ ,

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

$\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.$

Note: we can easily verify that this is the correct answer; for example, 1/2 and 1/2 work, and the sum of their reciprocals is 4.

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

@@ Line 15: / Line 15: @@
 <cmath>\frac{x}{xy}+\frac{y}{xy}=4</cmath>
-<cmath>\frac{1}{y}+\frac{1}{x}=\boxed{4 (\textbf{C})}.</cmath>
+<cmath>\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.</cmath>
+Note: we can easily verify that this is the correct answer; for example, 1/2 and 1/2 work, and the sum of their reciprocals is 4.
 ==See Also==
 {{AMC10 box|year=2017|ab=A|num-b=4|num-a=6}}
 {{MAA Notice}}