Difference between revisions of "2017 AMC 10A Problems/Problem 5"

Revision as of 13:17, 23 June 2020

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

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

Page

Toolbox

Search