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

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<math>4x^2=9</math>
 
<math>4x^2=9</math>
  
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So <math>x=\frac{3}{2}</math> and <math>y</math> would then be <math>\frac{9}{\frac{3}{2}}=6</math>
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So <math>x=\frac{3}{2}</math> and <math>y</math> would then be <math>4 \times</math> <math>\frac{3}{2}=6</math>
  
 
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>

Revision as of 21:43, 8 February 2012

Problem 6

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 number 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 $\frac{3}{2}+6$ = $\boxed{\textbf{(D)}\ \frac{15}{2}}$

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