2012 AMC 10A Problem 6

1. Call the two numbers $x$ and $y$. We know $x \cdot y = 9$.
2. The reciprocal of $x$ is $4$ times the reciprocal of $y$: $\dfrac{1}{x} = \dfrac{4}{y}$.
3. Before going on, see what this says about the numbers themselves. Multiply both sides by $xy$ to clear the fractions: $y = 4x$. So one number is $4$ times the other. (Test: if $x = 1$ and $y = 4$, then $\frac11 = 1$ and $\frac44 = 1$. It fits.)
4. Put $y = 4x$ into the product: $x \cdot 4x = 9$, so $4x^2 = 9$ and $x^2 = \frac94$.
5. The numbers are positive, so $x = \frac32$. Then $y = 4 \cdot \frac32 = 6$.
6. Check both facts: $\frac32 \cdot 6 = 9$, and $\frac{1}{3/2} = \frac23$ while $4 \cdot \frac16 = \frac23$. Both hold.
7. The sum is $\frac32 + 6 = \frac{15}{2}$.

The answer is $\boxed{\textbf{(D)}\ \frac{15}{2}}$.

A fact about reciprocals is really a fact about the numbers. If $\frac1x$ is $4$ times bigger than $\frac1y$, then $x$ is $4$ times smaller than $y$. Once one number is written as a multiple of the other, the product gives an equation with only one unknown. Checking the final pair against both facts catches a flipped relation.