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

 
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==Problem==
 
==Problem==
  
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>
  
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>.
 
  
 
==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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==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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==Solution 4==
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*NOTE: SAME AS PREVIOUS SOLUTION*
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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>
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==Video Solution==
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https://youtu.be/str7kmcRMY8
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https://youtu.be/TooKNMK3slY
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~savannahsolver
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==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 13:54, 13 January 2021

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 $\boxed{\textbf{(C) } 4}$.


Solution 4

  • NOTE: SAME AS PREVIOUS SOLUTION*

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

Video Solution

https://youtu.be/str7kmcRMY8

https://youtu.be/TooKNMK3slY

~savannahsolver

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

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