Question

$A,$ $B,$ and $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles $B$ and $C?$

$55$

$56$

$57$

$58$

$59$

Answer: E

Concepts:mean optimization

Difficulty rating: 1980

Solution:

Let $a, b, c$ be the numbers of rocks in the piles. From $\dfrac{40a + 50b}{a + b} = 43,$ we get $7b = 3a,$ so $a = 7k$ and $b = 3k.$

Let $\mu_{BC}$ be the mean of $B$ and $C.$ Using the $A, C$ mean $44$ to express $\mu_C = \dfrac{28k + 44c}{c},$ we find $\mu_{BC} = \dfrac{178k + 44c}{3k + c},$ so $(\mu_{BC} - 44)c = (178 - 3\mu_{BC})k.$

Since $B$ is heavier than $A,$ the mean of $B$ and $C$ exceeds $44,$ forcing $178 - 3\mu_{BC} \gt 0,$ i.e. $\mu_{BC} \lt \tfrac{178}{3} = 59\tfrac13.$ The value $59$ is attainable by taking $k=15c;$ then pile $C$ has mean $464$ and the displayed formula gives $\mu_{BC}=59.$ Thus the greatest integer mean is $59.$

Thus, the correct answer is E.

2013 AMC 12A Problem 16

Answer: E

Difficulty rating: 1980

Solution:

Problem 16 in Other Years