Difference between revisions of "2012 AMC 10A Problems/Problem 6"
Mattchu386 (talk | contribs) (→Problem 6) |
|||
| Line 4: | Line 4: | ||
<math> \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 </math> | <math> \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 </math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | Let the two number 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>\frac{1}{x}=4(\frac{1}{y})</math> | ||
| + | |||
| + | Therefore, <math>4x=y</math> | ||
| + | |||
| + | Replacing <math>y</math> with <math>4x</math> in the equation, | ||
| + | |||
| + | <math>4x^2=9</math> | ||
| + | |||
| + | So <math>x=\frac{3}{2}</math> and <math>y</math> would then be <math>\frac{9}{\frac{3}{2}}=6</math> | ||
| + | |||
| + | The sum would be <math>\frac{3}{2}+6</math> = <math>\boxed{\textbf{(D)}\ \frac{15}{2}}</math> | ||
Revision as of 20:38, 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?
Solution
Let the two number equal 
and 
. From the information given in the problem, two equations can be written:
Therefore, 
Replacing with 
in the equation,
So 
and would then be 
The sum would be 
= 
