Difference between revisions of "2013 AMC 12A Problems/Problem 16"

Revision as of 20:24, 27 October 2021

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 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 $(E)$ .

@@ Line 22: / Line 22: @@
 <math>44 + \frac{4(\frac{7}{3}B) + 6B}{B + C}=44 + \frac{46}{3}*\frac{B}{B + C}</math>
-<math>\frac{B}{B + C} < 1</math>, so the maximum value occurs when <math>C = 1</math>. Since <math>\frac{46}{3}</math> must cancel to give an integer, and the only fraction that satisfies both conditions is <math>\frac{45}{46}</math>
+In order to maximize <math>\frac{B}{B + C}</math>, we can minimize the denominator by letting <math>C = 1</math> (C must be a positive integer). Since <math>\frac{46}{3}</math> must cancel to give an integer, and the only fraction that satisfies both conditions is <math>\frac{45}{46}</math>
 Plugging in, we get

2013 AMC 12A (Problems • Answer Key • Resources)
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