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

Revision as of 14:46, 18 June 2018

Problem

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

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

Solution

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.

Solution 2

Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction. See for yourself. And by looking into fractions, we immediately see that $\frac{1}{3}$ and $1$ would fit the rule. $\boxed{\textbf{(C)} 4}.$

Solution 3

Notice that from the information given above, $x+y=4xy$

Because the sum of the reciprocals of two numbers is just the sum of the two numbers over the product of the two numbers or $\frac{x+y}{xy}$

We can solve this by substituting $x+y\implies 4xy$ .

Our answer is simply $\frac{4xy}{xy}\implies4$ .

Therefore, the answer is $\boxed{\textbf{(C) } 4}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 33: / Line 33: @@
 {{AMC10 box|year=2017|ab=A|num-b=4|num-a=6}}
 {{MAA Notice}}
+[[Category:Introductory Algebra Problems]]

Page

Toolbox

Search