Difference between revisions of "2012 AMC 10A Problems/Problem 6"
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| Line 11: | Line 11: | ||
<math>xy=9</math> | <math>xy=9</math> | ||
| − | <math>\frac{1}{x}=4(\frac{1}{y})</math> | + | <math>\frac{1}{x}=4 \left( \frac{1}{y} \right)</math> |
Therefore, <math>4x=y</math> | Therefore, <math>4x=y</math> | ||
| Line 22: | Line 22: | ||
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> | ||
| + | |||
== See Also == | == See Also == | ||
{{AMC10 box|year=2012|ab=A|num-b=5|num-a=7}} | {{AMC10 box|year=2012|ab=A|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Revision as of 12:04, 4 July 2013
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?
Solution
Let the two numbers 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 
= 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
