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 $kc$ .

We can write the information given as, $\frac{40a + 50b}{a+b} = 43$ , $\frac{40a + kc}{a+c} = 44$ , $\frac{50b + kc}{b+c} = ?$

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

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

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

$= 44 + \frac{46a}{3a+7c} = 44 + \frac{46}{3 + \frac{7}{a}} < 44 + \frac{46}{3} \approx 59.3$

$\frac{50b + kc}{b+c} \le \boxed{\textbf{(E) } 59}$

~isabelchen

Solution 5

Let the total number of rocks in pile $A$ be $A_n$ , and the total number of rocks in pile $B$ be $B_n$ . Then, by restriction 3 (the average of $A$ and $B$ ), we can establish the equation: $\frac{40A_n+50B_n}{A_n+B_n}=43$ . Cross-multiplying, we get: $40A_n+50B_n=43A_n+43B_n \implies 3A_n=7B_n$ . Let's say we have $7k$ rocks in $A$ and $3k$ rocks in $B$ . Hence, we have $280k$ and $150k$ as the total weight of piles $A$ and $B$ , respectively. Let the total weight of $C$ be $m$ , and the total number of rocks in $C$ be $n$ . Using the last restriction regarding the average of piles $A$ and $C$ , we have: $\frac{280k+m}{7k+n}=44 \implies 280k + m=280k + 28k + 44n \implies m=28k+44n$ . To find the average of piles $B$ and $C$ , we can establish the expression: $\frac{150k+28k+44n}{3k+n}=\frac{178k+44n}{3k+n}=\frac{132k+44n+46k}{3k+n}=\frac{44(3k+n)+46k}{3k+n}=44+\frac{46k}{3k+n}.$ When we let the final expression equal $59$ , we get: $44+\frac{46k}{3k+n}=59\implies\frac{46k}{3k+n}=15$ Cross-multiplying, we get: $46k=45k+14n\implies k=14n$ $k$ is still positive here, so 59 works. As this is the greatest option, we can circle $\textbf{(E)}$ immediately. To show why $59$ is the greatest, consider the following: When we let the final expression equal $60$ , we get: $44+\frac{46k}{3k+n}=60\implies\frac{46k}{3k+n}=16$ Cross-multiplying, we get: $46k=48k+14n\implies k=-7n$ Since $k$ is positive, the final expression could not equal 60. It further implies that the final expression could not equal any other integer greater than 60. Therefore, we have our final answer $\boxed{59}$ .

Solution 6

Denote the number of rocks in each of the three piles as $a$ , $b$ , and $c$ , respectively, and denote the weight of each rock in pile $C$ as $x$ . We are given: \begin{align} \frac{40a + 50b}{a+b} &= 43\\ \frac{40a + xc}{b+c} &= 44\\ \end{align} From these equations, we get $b = \frac{3}{7}a$ and $c = \frac{4}{x-44}a$ .

Now, we can compute the average weight in combined piles B and C: \begin{align} \frac{50b + xc}{b+c} &= \frac{\frac{150}{7}a + \frac{4x}{x-44}a}{\frac{3}{7}a + \frac{4}{x-44}a}\\ &= \frac{178x-6600}{3x-104}\\ &= 59 + \frac{1}{3} - \frac{C}{3x-104}\\ \end{align} Where $C$ is just some constant that I'm too lazy to compute. As $x$ approaches infinity, we have that $\frac{C}{3x-104}$ becomes negligibly small, and at some point, $59$ is attained. ~akliu

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