AMC12 2009 B

Let $x$ be the weight of the bucket and let $y$ be the weight of the water in a full bucket. Then we are given that $x + \frac 23y = a$ and $x + \frac 12y = b$. Hence $\frac 16y = a-b$, so $y = 6a-6b$. Thus $x = b - \frac 12 (6a-6b) = -3a + 4b$. Finally $x + y = \boxed {3a-2b}$. The answer is $\mathrm{(E)}$. Imagine that we take three buckets of the first type, to get rid of the fraction. We will have three buckets and two buckets' worth of water. On the other hand, if we take two buckets of the second type, we will have two buckets and enough water to fill one bucket. The difference between these is exactly one bucket full of water, hence the answer is $3a-2b$. We are looking for an expression of the form $xa + yb$. We must have $x+y=1$, as the desired result contains exactly one bucket. Also, we must have $\frac 23 x + \frac 12 y = 1$, as the desired result contains exactly one bucket of water. At this moment, it is easiest to check that only the options (A), (B), and (E) satisfy $x+y=1$, and out of these only (E) satisfies the second equation. Alternatively, we can directly solve the system, getting $x=3$ and $y=-2$. Since $a$ is one bucket plus two-thirds of the total amount of water, and $b$ is one bucket plus one-half of the total amount of water, $a-b$ would equal $\cancel{\text{bucket}}+\frac23\cdot\text{water}-(\cancel{\text{bucket}}+\frac12\cdot\text{water})=\frac16\cdot\text{water}$. Therefore, $a-b$ is one-sixth of the total mass of the water. Starting from $a$, we add two one-sixths of the total amount of the water to become full. Therefore, the full bucket of water is $a+2(a-b)=a+2a-2b=3a-2b\Rightarrow\textbf{(E)}$