2006 AMC 10B Problem 13

Joe: he drinks pure coffee first, then adds $2$ ounces of cream. No cream leaves his cup, so he has exactly $2$ ounces of cream.

JoAnn: after adding cream she has $12 + 2 = 14$ ounces of uniform mixture, of which $2$ ounces is cream. Drinking $2$ of the $14$ ounces removes $\frac{2}{14} = \frac{1}{7}$ of the mixture, and hence $\frac{1}{7}$ of the cream. She keeps $\frac{6}{7}$ of it:
$$2 \cdot \frac{6}{7} = \frac{12}{7} \text{ ounces of cream}.$$

Ratio of Joe's cream to JoAnn's: $\dfrac{2}{12/7} = \dfrac{14}{12} = \dfrac{7}{6}$.

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

Removing part of a well-stirred mixture removes the same fraction of every ingredient, so track the fraction that remains rather than the volumes. The order of operations (drink then add, versus add then drink) is the entire difference between the two cups; the $16$-ounce cup size is irrelevant.

Sanity check with the choices: Joe has more cream than JoAnn (she drank some of hers), so the ratio exceeds $1$, leaving only (D) and (E). JoAnn keeps $\frac{6}{7}$ of $2$ ounces, so the ratio is $\frac{7}{6}$.

• Assuming JoAnn also ends with 2 ounces of cream (ratio 1), forgetting that drinking a stirred mixture removes some cream.
• Using $\frac{2}{12}$ or $\frac{2}{16}$ instead of $\frac{2}{14}$ for the fraction of cream in JoAnn's cup, or inverting the final ratio to get $\frac{6}{7}$.