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Difference between revisions of "2017 AMC 10A Problems/Problem 5"
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==Problem==
 
==Problem==
  
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The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
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The sum of two nonzero real numbers is <math>4</math> times their product. What is the sum of the reciprocals of the two numbers?
  
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12</math>
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12</math>
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<cmath>\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.</cmath>
 
<cmath>\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.</cmath>
  
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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.
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Note: we can easily verify that this is the correct answer; for example, <math>\left(\frac{1}{2}, \frac{1}{2}\right)</math> works, and the sum of their reciprocals is <math>4</math>.
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==Solution 2==
 
==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.
 
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 <math>\frac{1}{3}</math> and <math>1</math> would fit the rule. <math>\boxed{\textbf{(C)} 4}.</math>
 
See for yourself. And by looking into fractions, we immediately see that <math>\frac{1}{3}</math> and <math>1</math> would fit the rule. <math>\boxed{\textbf{(C)} 4}.</math>
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 +
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==Solution 3==
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Notice that from the information given above, <math>x+y=4xy</math>
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 +
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 <math>\frac{x+y}{xy}</math>
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We can solve this by substituting <math>x+y\implies 4xy</math>.
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Our answer is simply <math>\frac{4xy}{xy}\implies4</math>.
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Therefore, the answer is <math>\boxed{\textbf{(C) } 4}</math>.
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 +
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==Video Solution 1==
 +
https://youtu.be/str7kmcRMY8
 +
 +
(TheBeautyofMath)
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 +
==Video Solution 2==
 +
https://youtu.be/TooKNMK3slY
 +
 +
~savannahsolver
 +
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=4|num-a=6}}
 
{{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}}
 
{{MAA Notice}}
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 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 09:43, 14 September 2024

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, $\left(\frac{1}{2}, \frac{1}{2}\right)$ works, 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 $\boxed{\textbf{(C) } 4}$.


Video Solution 1

https://youtu.be/str7kmcRMY8

(TheBeautyofMath)

Video Solution 2

https://youtu.be/TooKNMK3slY

~savannahsolver

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

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

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