2006 AMC 10B Problem 13

1. Joe. He drinks $2$ ounces of plain coffee first, then pours in $2$ ounces of cream. No cream ever leaves his cup, so he ends with exactly $2$ ounces of cream.
2. JoAnn. She adds cream first. Her cup now holds $12 + 2 = 14$ ounces, and $2$ of those ounces are cream. She stirs it, so the cream is spread evenly.
3. She drinks $2$ of the $14$ ounces. That is $\frac{2}{14} = \frac{1}{7}$ of the cup, so she drinks $\frac17$ of the cream too, and keeps $\frac67$ of it:
$$\frac{6}{7} \cdot 2 = \frac{12}{7} \text{ ounces of cream.}$$
4. Compare Joe's cream to JoAnn's:
$$2 \div \frac{12}{7} = 2 \cdot \frac{7}{12} = \frac{14}{12} = \frac{7}{6}.$$

The answer is $\boxed{\textbf{(E)}\ \frac{7}{6}}$.

When you pour out part of a well-stirred mix, you pour out the same fraction of every ingredient in it. So track what fraction is left, not the ounces. The only difference between the two cups is the order of "drink" and "add," and that order decides whether any cream is lost. The $16$-ounce cup size does not matter.

Use the answer choices. JoAnn drank some cream and Joe did not, so Joe has more. The ratio of Joe's to JoAnn's must be bigger than $1$, which leaves (D) and (E). JoAnn keeps $\frac67$ of what Joe has, so Joe has $\frac76$ times as much as JoAnn: (E).