2012 AMC 10A Problems/Problem 6: Difference between revisions

Revision as of 12:36, 7 November 2025

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:

$xy=9$

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

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

Solution

Set up your first equation which is $xy=9$ and your second being $\dfrac{1}{x} = \dfrac4y$ . Then, in the first equation, rearrange it to become $x=\dfrac9y$ . Now plug this in to your second equation, and you should get $\dfrac{y}{9}=\dfrac{4}{y}$ . Cross multiply to get $y^2=36$ , and simplify to get $y=6$ . Notice how the problem mentioned that they were positive integers, so we don't consider $-6$ . We plug $6$ back into our first equation to get $6x=9$ which makes $x$ come out to be $1.5$ . We add $x+y=6+1.5=7.5$ . Notice answer choice $D$ is the only fraction that simplifies to $1.5$ , so answer option $\boxed{D}$ must be the answer.

Video Solution (CREATIVE THINKING)

https://youtu.be/jyA5_tjDOjc

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2012 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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 ==Solution==
-Set up your first equation which is <imath>xy=9</imath> and your second being <imath>\dfrac{1}{x} = \dfrac4y</imath>. Then, in the first equation, rearrange it to become <imath>x=\dfrac9y</imath>. Now plug this in to your second equation, and you should get <imath>\dfracy9=\dfrac4y</imath>. Cross multiply to get <imath>y^2=36</imath>, and simplify to get <imath>y=6</imath>. Notice how the problem mentioned that they were positive integers, so we don't consider <imath>-6</imath>. We plug <imath>6</imath> back into our first equation to get <imath>6x=9</imath> which makes <imath>x</imath> come out to be <imath>1.5</imath>. We add <imath>x+y=6+1.5=7.5</imath>. Notice answer choice <imath>D</imath> is the only fraction that simplifies to <imath>1.5</imath>, so answer option <imath>\boxed{D}</imath> must be the answer.
+Set up your first equation which is <imath>xy=9</imath> and your second being <imath>\dfrac{1}{x} = \dfrac4y</imath>. Then, in the first equation, rearrange it to become <imath>x=\dfrac9y</imath>. Now plug this in to your second equation, and you should get <imath>\dfrac{y}{9}=\dfrac{4}{y}</imath>. Cross multiply to get <imath>y^2=36</imath>, and simplify to get <imath>y=6</imath>. Notice how the problem mentioned that they were positive integers, so we don't consider <imath>-6</imath>. We plug <imath>6</imath> back into our first equation to get <imath>6x=9</imath> which makes <imath>x</imath> come out to be <imath>1.5</imath>. We add <imath>x+y=6+1.5=7.5</imath>. Notice answer choice <imath>D</imath> is the only fraction that simplifies to <imath>1.5</imath>, so answer option <imath>\boxed{D}</imath> must be the answer.
 ==Video Solution (CREATIVE THINKING)==

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2012 AMC 10A Problems/Problem 6: Difference between revisions

Revision as of 12:36, 7 November 2025

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Problem

Solution

Solution

Video Solution (CREATIVE THINKING)

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