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

Solution 1

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

The mean weight of $A$ and $C$ together is $44$ , so the total weight of $A$ and $C$ is $44(A + C)$

To get the total weight of $B$ and $C$ , we need to add the total weight of $B$ and subtract the total weight of $A$

$44A + 44C + 50B - 40A = 4A + 44C + 50B$

And then dividing by the number of rocks $B$ and $C$ together, to get the mean of $B$ and $C$ ,

$\frac{4A + 44C + 50B}{B + C}$

Simplifying,

$\frac{4A + 44C + 44B + 6B}{B + C}$

$44 + \frac{4A + 6B}{B + C}$

Now, to get rid of the $A$ in the numerator, we use two definitions of the total weight of $A$ and $B$

$40A + 50B = 43A + 43B$

$3A = 7B$

$A = \frac{7}{3}B$

Substituting back in,

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

$44 + \frac{46}{3}*\frac{B}{B + C}$

Note that $\frac{B}{B + C} < 1$ , and the maximal value of this factor occurs when $C = 1$

Also note that $\frac{46}{3}$ must cancel to give an integer value, and the only fraction that satisfies both these conditions is $\frac{45}{46}$

Plugging in, we get

$44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59$

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.

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

Combining the 1st and 2nd equations, we see that

$(x-44)C=4A=\frac{28}{3}B$

Subtracting equation 3 from equation 2, we have

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

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

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

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