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

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m (Solution)
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==Solution==
 
==Solution==
  
Let the two real numbers be <math>x,y</math>. We are given that
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Let the two real numbers be <math>x,y</math>. We are given that <math>x+y=4xy,</math> and dividing both sides by <math>xy</math>, <math>\frac{x}{xy}+\frac{y}{xy}=4.</math>
 
 
<cmath>x+y=4xy,</cmath>
 
 
 
and dividing both sides by <math>xy</math>,
 
 
 
<cmath>\frac{x}{xy}+\frac{y}{xy}=4</cmath>
 
  
 
<cmath>\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.</cmath>
 
<cmath>\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.</cmath>

Revision as of 15:15, 9 February 2017

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.

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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