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> | ||
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<cmath>\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.</cmath> | <cmath>\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.</cmath> | ||
− | Note: we can easily verify that this is the correct answer; for example, 1 | + | 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== | ==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. | 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 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>. | ||
+ | |||
+ | |||
+ | ==Solution 4== | ||
+ | *NOTE: SAME AS PREVIOUS SOLUTION* | ||
+ | Let the two numbers be <math>a</math> and <math>b</math>, respectively. We wish to find <math>\frac{1}{a} + \frac{1}{b}</math>. Note that <math>\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}</math>. We are given that <math>a+b = 4ab</math>. | ||
+ | |||
+ | Subsituting, we have <math>\frac{a+b}{ab} = \frac{4ab}{ab} = \boxed{\textbf{(C) } 4}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/str7kmcRMY8 | ||
+ | |||
+ | 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 13:54, 13 January 2021
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 ,
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 .
Solution 4
- NOTE: SAME AS PREVIOUS SOLUTION*
Let the two numbers be and , respectively. We wish to find . Note that . We are given that .
Subsituting, we have
Video Solution
~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 |
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