2017 AMC 10A Problems/Problem 5

Problem

The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$

Solution

Let the two real numbers be $x,y$. We are given that $x+y=4xy,$ and dividing both sides by $xy$, $\frac{x}{xy}+\frac{y}{xy}=4.$

\[\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.\]

Note: we can easily verify that this is the correct answer; for example, $\left(\frac{1}{2}, \frac{1}{2}\right)$ works, 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 $\frac{1}{3}$ and $1$ would fit the rule. $\boxed{\textbf{(C)} 4}.$


Solution 3

Notice that from the information given above, $x+y=4xy$

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 $\frac{x+y}{xy}$

We can solve this by substituting $x+y\implies 4xy$.

Our answer is simply $\frac{4xy}{xy}\implies4$.

Therefore, the answer is $\boxed{\textbf{(C) } 4}$.


Solution 4

Let the two numbers be $a$ and $b$, respectively. We wish to find $\frac{1}{a} + \frac{1}{b}$. Note that $\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$. We are given that $a+b = 4ab$.

Subsituting, we have $\frac{a+b}{ab} = \frac{4ab}{ab} = \boxed{\textbf{(C) } 4}$

Video Solution

https://youtu.be/str7kmcRMY8

https://youtu.be/TooKNMK3slY

~savannahsolver

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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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