Difference between revisions of "2012 AMC 10A Problems/Problem 6"
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− | == Problem | + | == 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? | 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? | ||
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==Solution== | ==Solution== | ||
− | Let the two | + | Let the two numbers 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>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 19: | Line 19: | ||
<math>4x^2=9</math> | <math>4x^2=9</math> | ||
− | So <math>x=\frac{3}{2}</math> and <math>y</math> would then be <math>\ | + | 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> | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/jyA5_tjDOjc | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2012|ab=A|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:55, 1 July 2023
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 =
Video Solution (CREATIVE THINKING)
~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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