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Difference between revisions of "2012 AMC 10A Problems/Problem 6"

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== Problem 6 ==
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== Problem ==
  
 
The product of two positive numbers is 9. The reciprocal of one of these numbers is 4 times the reciprocal of the other number. What is the sum of the two numbers?
 
The product of two positive numbers is 9. The reciprocal of one of these numbers is 4 times the reciprocal of the other number. What is the sum of the two numbers?
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==Solution==
 
==Solution==
  
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Let the two number equal <math>x</math> and <math>y</math>. From the information given in the problem, two equations can be written:
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Let the two numbers equal <math>x</math> and <math>y</math>. From the information given in the problem, two equations can be written:
  
 
<math>xy=9</math>
 
<math>xy=9</math>
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The sum would be <math>\frac{3}{2}+6</math> = <math>\boxed{\textbf{(D)}\ \frac{15}{2}}</math>
 
The sum would be <math>\frac{3}{2}+6</math> = <math>\boxed{\textbf{(D)}\ \frac{15}{2}}</math>
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== See Also ==
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{{AMC10 box|year=2012|ab=A|num-b=5|num-a=7}}

Revision as of 22:47, 8 February 2012

Problem

The product of two positive numbers is 9. The reciprocal of one of these numbers is 4 times the reciprocal of the other number. What is the sum of the two numbers?

$\textbf{(A)}\ \frac{10}{3}\qquad\textbf{(B)}\ \frac{20}{3}\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ \frac{15}{2}\qquad\textbf{(E)}\ 8$

Solution

Let the two numbers equal $x$ and $y$. From the information given in the problem, two equations can be written:

$xy=9$

$\frac{1}{x}=4(\frac{1}{y})$

Therefore, $4x=y$

Replacing $y$ with $4x$ in the equation,

$4x^2=9$

So $x=\frac{3}{2}$ and $y$ would then be $4 \times$ $\frac{3}{2}=6$

The sum would be $\frac{3}{2}+6$ = $\boxed{\textbf{(D)}\ \frac{15}{2}}$

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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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