2013 AMC 12A Problems/Problem 16

Revision as of 16:00, 3 July 2013 by Kj2002 (talk | contribs) (See also)

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 $(E)$.

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS