Difference between revisions of "2017 AMC 10A Problems/Problem 5"
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Note: we can easily verify that this is the correct answer; for example, 1/2 and 1/2 work, and the sum of their reciprocals is 4. | Note: we can easily verify that this is the correct answer; for example, 1/2 and 1/2 work, and the sum of their reciprocals is 4. | ||
| + | ==Solution 2== | ||
| + | Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction. | ||
| + | See for yourself. And by looking into fractions, we immediately see that <math>\frac{1}{3}</math> and <math>1</math> would fit the rule. <math>\boxed{\textbf{(C)} 4}.</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2017|ab=A|num-b=4|num-a=6}} | ||
{{AMC12 box|year=2017|ab=A|num-b=1|num-a=3}} | {{AMC12 box|year=2017|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:07, 14 February 2017
Contents
Problem
The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
Solution
Let the two real numbers be 
. We are given that 
and dividing both sides by 
, 
![\[\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.\]](http://latex.artofproblemsolving.com/e/a/1/ea1dadbf53587b5968e46e59c6f61f87125eeed4.png)
Note: we can easily verify that this is the correct answer; for example, 1/2 and 1/2 work, and the sum of their reciprocals is 4.
Solution 2
Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction.
See for yourself. And by looking into fractions, we immediately see that 
and 
would fit the rule. 
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
