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

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Therefore, the answer is <math>\boxed{\textbf{(C) } 4}</math>.
 
Therefore, the answer is <math>\boxed{\textbf{(C) } 4}</math>.
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==Solution 4==
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Let the two numbers be <math>a</math> and <math>b</math>, respectively. We wish to find <math>\frac{1}{a} + \frac{1}{b}</math>. Note that <math>\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}</math>. We are given that <math>a+b = 4ab</math>.
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Subsituting, we have <math>\frac{a+b}{ab} = \frac{4ab}{ab} = \boxed{\textbf{(C) } 4}</math>
  
 
==See Also==
 
==See Also==

Revision as of 11:35, 3 August 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 $\boxed{\textbf{(C) } 4}$.


Solution 4

Let the two numbers be $a$ and $b$, respectively. We wish to find $\frac{1}{a} + \frac{1}{b}$. Note that $\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$. We are given that $a+b = 4ab$.

Subsituting, we have $\frac{a+b}{ab} = \frac{4ab}{ab} = \boxed{\textbf{(C) } 4}$

See Also

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

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