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

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==Solution==
==Solution==


Let the two numbers equal <math>x</math> and <math>y</math>. From the information given in the problem, two equations can be written:
Let the two numbers equal <imath>x</imath> and <imath>y</imath>. From the information given in the problem, two equations can be written:


<math>xy=9</math>
<imath>xy=9</imath>


<math>\frac{1}{x}=4 \left( \frac{1}{y} \right)</math>
<imath>\frac{1}{x}=4 \left( \frac{1}{y} \right)</imath>


Therefore, <math>4x=y</math>
Therefore, <imath>4x=y</imath>


Replacing <math>y</math> with <math>4x</math> in the equation,
Replacing <imath>y</imath> with <imath>4x</imath> in the equation,


<math>4x^2=9</math>
<imath>4x^2=9</imath>


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


The sum would be <math>\frac{3}{2}+6</math> = <math>\boxed{\textbf{(D)}\ \frac{15}{2}}</math>
The sum would be <imath>\frac{3}{2}+6</imath> = <imath>\boxed{\textbf{(D)}\ \frac{15}{2}}</imath>
 
==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 \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}$ must be the answer.


==Video Solution (CREATIVE THINKING)==
==Video Solution (CREATIVE THINKING)==

Revision as of 12:35, 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 \dfracy9=\dfrac4y$. Cross multiply to get$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

~Education, the Study of Everything

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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