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Difference between revisions of "2013 AMC 12A Problems/Problem 16"
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Cross-multiplying, we get: <cmath>46k=48k+14n\implies k=-7n</cmath>
 
Cross-multiplying, we get: <cmath>46k=48k+14n\implies k=-7n</cmath>
 
Since <math>k</math> 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 <math>\boxed{59}</math>.
 
Since <math>k</math> 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 <math>\boxed{59}</math>.
 +
 +
==Solution 6==
 +
Denote the number of rocks in each of the three piles as <math>a</math>, <math>b</math>, and <math>c</math>, respectively, and denote the weight of each rock in pile <math>C</math> as <math>x</math>. We are given:
 +
\begin{align}
 +
\frac{40a + 50b}{a+b} &= 43\\
 +
\frac{40a + xc}{b+c} &= 44\\
 +
\end{align}
 +
From these equations, we get <math>b = \frac{3}{7}a</math> and <math>c = \frac{4}{x-44}a</math>.
 +
 +
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 <math>C</math> is just some constant that I'm too lazy to compute. As <math>x</math> approaches infinity, we have that <math>C</math> becomes negligibly small, and at some point, <math>59</math> is attained. ~akliu
  
 
== See also ==
 
== See also ==

Revision as of 21:36, 13 October 2024

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 $C$ becomes negligibly small, and at some point, $59$ is attained. ~akliu

See also

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

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