AMC12 2013 A

Answer (E): Let $a$, $b$, and $c$ be the number of rocks in piles $A$, $B$, and $C$, respectively. Then $$\frac{40a+50b}{a+b}=43\ \text{and}\ 7b=3a.$$ Because $7$ and $3$ are relatively prime, there is a positive integer $k$ such that $a=7k$ and $b=3k$. Let $\mu_C$ equal the mean weight in pounds of the rocks in $C$ and $\mu_{BC}$ equal the mean weight in pounds of the rocks in $B$ and $C$. Then $$\frac{40\cdot 7k+\mu_C\cdot c}{7k+c}=44,\ \text{so}\ \mu_C=\frac{28k+44c}{c},$$ and $$\mu_{BC}=\frac{50\cdot 3k+(28k+44c)}{3k+c}=\frac{178k+44c}{3k+c}.$$ Clearing the denominator and rearranging yields $(\mu_{BC}-44)c=(178-3\mu_{BC})k$. Because the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds, and the mean weight of the rocks in $B$ is greater than the mean weight of the rocks in $A$, it follows that the mean weight of the rocks in $B$ and $C$ must be greater than $44$ pounds. Thus $(\mu_{BC}-44)c>0$ and therefore $178-3\mu_{BC}$ must be greater than zero. This implies that $\mu_{BC}<\frac{178}{3}=59\frac{1}{3}$. If $k=15c$ and $\mu_C=464$, then $\mu_{BC}=59$. Thus the greatest possible integer value for the weight in pounds of the combined piles $B$ and $C$ is $59$.