Difference between revisions of "2017 AMC 10A Problems/Problem 5"
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==Problem== | ==Problem== | ||
| − | The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers? | + | The sum of two nonzero real numbers is <math>4</math> times their product. What is the sum of the reciprocals of the two numbers? |
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12</math> | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12</math> | ||
| Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
| − | Let the two real numbers be <math>x,y</math>. We are given that | + | Let the two real numbers be <math>x,y</math>. We are given that <math>x+y=4xy,</math> and dividing both sides by <math>xy</math>, <math>\frac{x}{xy}+\frac{y}{xy}=4.</math> |
| − | <cmath>x | + | <cmath>\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.</cmath> |
| − | and | + | Note: we can easily verify that this is the correct answer; for example, <math>\left(\frac{1}{2}, \frac{1}{2}\right)</math> works, and the sum of their reciprocals is <math>4</math>. |
| − | < | + | ==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> | ||
| − | < | + | |
| + | ==Solution 3== | ||
| + | Notice that from the information given above, <math>x+y=4xy</math> | ||
| + | |||
| + | Because the sum of the reciprocals of two numbers is just the sum of the two numbers over the product of the two numbers or <math>\frac{x+y}{xy}</math> | ||
| + | |||
| + | We can solve this by substituting <math>x+y\implies 4xy</math>. | ||
| + | |||
| + | Our answer is simply <math>\frac{4xy}{xy}\implies4</math>. | ||
| + | |||
| + | Therefore, the answer is <math>\boxed{\textbf{(C) } 4}</math>. | ||
| + | |||
| + | |||
| + | ==Video Solution 1== | ||
| + | https://youtu.be/str7kmcRMY8 | ||
| + | |||
| + | (TheBeautyofMath) | ||
| + | |||
| + | ==Video Solution 2== | ||
| + | https://youtu.be/TooKNMK3slY | ||
| + | |||
| + | ~savannahsolver | ||
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 09:43, 14 September 2024
Contents
Problem
The sum of two nonzero real numbers is 
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, 
works, and the sum of their reciprocals is .
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. 
Solution 3
Notice that from the information given above, 
Because the sum of the reciprocals of two numbers is just the sum of the two numbers over the product of the two numbers or 
We can solve this by substituting 
.
Our answer is simply 
.
Therefore, the answer is 
.
Video Solution 1
(TheBeautyofMath)
Video Solution 2
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
