2013 AMC 12A Problems/Problem 16

Problem

$A$ , $B$ , $C$ 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 $C$ ?

$\textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59$

Solution 1

Let pile $A$ have $A$ rocks, and so on.

The total weight of $A$ and $C$ can be expressed as $44(A + C)$ .

To get the total weight of $B$ and $C$ , we add the weight of $B$ and subtract the weight of $A$ : $44(A + C) + 50B - 40A = 4A + 44C + 50B$

Therefore, the mean of $B$ and $C$ is $\frac{4A + 44C + 50B}{B + C}$ , which is simplified to $44 + \frac{4A + 6B}{B + C}$ .

We now need to eliminate $A$ in the numerator. Since we know that $40A + 50B = 43(A + B)$ , we have $A = \frac{7}{3}B$

Substituting,

$44 + \frac{4(\frac{7}{3}B) + 6B}{B + C}=44 + \frac{46}{3}*\frac{B}{B + C}$

In order to maximize $\frac{B}{B + C}$ , we can minimize the denominator by letting $C = 1$ (C must be a positive integer). Since $\frac{46}{3}$ must cancel to give an integer, and the only fraction that satisfies both conditions is $\frac{45}{46}$

Plugging in, we get

$44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59$ , which is choice E

Solution 2

Suppose there are $A,B,C$ rocks in the three piles, and that the mean of pile C is $x$ , and that the mean of the combination of $B$ and $C$ is $y$ . We are going to maximize $y$ , subject to the following conditions:

$40A+50B=43(A+B)$ $40A+xC=44(A+C)$ $50B+xC=y(B+C)$

which can be rearranged as:

$7B=3A$ $(x-44)C=4A$ $(x-y)C=(y-50)B$

Let us test $y=59$ is possible. If so, it is already the answer. If not, there will be some contradiction. So the third equation becomes

$(x-59)C=9B.$

So $15C = (x-44)C - (x-59)C = 4A - 9B$ , $45C=4(3A)-27B=28B-27B$ , $105C=28A-9(7B)=A$ , therefore,

$A=105C, B=45C, x=4(105)+44=464$ , which gives us a consistent solution. Therefore $y=59$ is the answer.

(Note: To further illustrate the idea, let us look at $y=60$ and see what happens. We then get $7\cdot 16C = 4A-30A<0$ , which is a contradiction!)

Solution 3

Obtain the 3 equations as in solution 2.

$7B=3A$ $(x-44)C=4A$ $(x-y)C=(y-50)B$

Our goal is to try to isolate $y$ into an inequality. The first equation gives $A=\frac{7}{3}B$ , which we plug into the second equation to get

$(x-44)C=\frac{28}{3}B$

To eliminate $x$ , subtract equation 3 from equation 2:

$(x-44)C-(x-y)C=\frac{28}{3}B-(y-50)B$ $(y-44)C=(\frac{178}{3}-y)B$

In order for the coefficients to be positive, $44<y<\frac{178}{3}$

Thus, the greatest integer value is $y=59$ , choice $(E)$ .

Solution 4

Let the number of rocks in $A$ be $a$ , $B$ be $b$ , $C$ be $c$ . The total weight of $A$ be $40a$ , $B$ be $50b$ , $C$ be $kc$ .

$\frac{40a + 50b}{a+b} = 43$ , $\frac{40a + kc}{a+c} = 44$ , $\frac{50b + kc}{b+c} = ?$

$40a + 50b = 43 a + 43 b$ , $3a = 7b$

$40a + kc = 44a + 44 c$ , $kc = 4a + 44c = \frac{28}{3}b + 44c$

$\frac{50b + kc}{b+c} = \frac{50b + \frac{28}{3}b + 44c}{b+c} = \frac{\frac{178}{3}b + 44c}{b + c}$

$\frac{50b + kc}{b+c} = \frac{50 \cdot \frac{3}{7}a + kc}{\frac{3}{7}a+c} = \frac{150a + 7kc}{3a + 7c} = \frac{150a + 28a + 308c}{3a+7c} = \frac{178a + 308c}{3a+7c} = \frac{44(3a+7c)+46}{3a+7c} = 44 + \frac{46a}{3a+7c} = 44 + \frac{46}{3 + \frac{7}{a}} < 44 + \frac{46}{3} = \boxed{\textbf{(E) } 59}$

~isabelchen

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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