2005 AMC 10B Problem 3

It is easier to track the paint that is kept each day than the paint that is used.

1. Day one. One third of the gallon is used, so two thirds is kept. Paint left: $\frac{2}{3}$ gallon.
2. Day two. One third of what is left is used. That means two thirds of what is left is kept:
$$\frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9} \text{ gallon}.$$
3. So $\frac{4}{9}$ of the original gallon is ready for day three.

Check by subtracting instead: on day two, the amount used is $\frac{1}{3}$ of $\frac{2}{3}$, which is $\frac{2}{9}$. Then $\frac{2}{3} - \frac{2}{9} = \frac{6}{9} - \frac{2}{9} = \frac{4}{9}$. Same answer.

The answer is $\boxed{\textbf{(D)}\ \frac{4}{9}}$.

When each step uses a fraction of what is left, the amounts multiply rather than subtract. Each day the paint shrinks to two thirds of its size. Watch for the words "of the remaining" next time: they tell you to multiply.

Pretend the gallon is $9$ cups, since $9$ splits nicely into thirds twice. Day one uses $3$ cups, leaving $6$. Day two uses one third of $6$, which is $2$, leaving $4$. So $4$ of the $9$ cups remain, which is $\frac{4}{9}$.

Using one third of a quantity leaves two thirds of it.

After day one, $\frac{2}{3}$ of the gallon remains.

On day two, one third of that amount is used, so two thirds of it is kept:
$$\frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}.$$

So $\frac{4}{9}$ of the original gallon is available on the third day.

The answer is $\boxed{\textbf{(D)}\ \frac{4}{9}}$.