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

m (Solution 2)
(Solution 1)
(15 intermediate revisions by 6 users not shown)
Line 1: Line 1:
Let pile <math>A</math> have <math>A</math> rocks, and so on.
+
== Problem==
  
The mean weight of <math>A</math> and <math>C</math> together is <math>44</math>, so the total weight of <math>A</math> and <math>C</math> is <math>44(A + C)</math>
+
<math>A</math>, <math>B</math>, <math>C</math> are three piles of rocks. The mean weight of the rocks in <math>A</math> is <math>40</math> pounds, the mean weight of the rocks in <math>B</math> is <math>50</math> pounds, the mean weight of the rocks in the combined piles <math>A</math> and <math>B</math> is <math>43</math> pounds, and the mean weight of the rocks in the combined piles <math>A</math> and <math>C</math> is <math>44</math> pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles <math>B</math> and <math>C</math>?
  
To get the total weight of <math>B</math> and <math>C</math>, we need to add the total weight of <math>B</math> and subtract the total weight of <math>A</math>
+
<math> \textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59</math>
  
<math>44A + 44C + 50B - 40A = 4A + 44C + 50B</math>
+
==Solution==
 
+
===Solution 1===
And then dividing by the number of rocks <math>B</math> and <math>C</math> together, to get the mean of <math>B</math> and <math>C</math>,
+
Let pile <math>A</math> have <math>A</math> rocks, and so on.
 
 
<math>\frac{4A + 44C + 50B}{B + C}</math>
 
 
 
Simplifying,
 
 
 
<math>\frac{4A + 44C + 44B + 6B}{B + C}</math>
 
 
 
 
 
<math>44 + \frac{4A + 6B}{B + C}</math>
 
 
 
Now, to get rid of the <math>A</math> in the numerator, we use two definitions of the total weight of <math>A</math> and <math>B</math>
 
 
 
<math>40A + 50B = 43A + 43B</math>
 
  
<math>3A = 7B</math>
+
The total weight of <math>A</math> and <math>C</math> can be expressed as <math>44(A + C)</math>.
  
<math>A = \frac{7}{3}B</math>
+
To get the total weight of <math>B</math> and <math>C</math>, we add the weight of <math>B</math> and subtract the weight of <math>A</math>: <math>44(A + C) + 50B - 40A = 4A + 44C + 50B</math>
  
Substituting back in,
+
Therefore, the mean of <math>B</math> and <math>C</math> is <math>\frac{4A + 44C + 50B}{B + C}</math>, which is simplified to <math>44 + \frac{4A + 6B}{B + C}</math>.
  
<math>44 + \frac{4(\frac{7}{3}B) + 6B}{B + C}</math>
+
We now need to eliminate <math>A</math> in the numerator. 
 +
Since we know that <math>40A + 50B = 43(A + B)</math>, we have <math>A = \frac{7}{3}B</math>
  
 +
Substituting,
  
<math>44 + \frac{46}{3}*\frac{B}{B + C}</math>
+
<math>44 + \frac{4(\frac{7}{3}B) + 6B}{B + C}=44 + \frac{46}{3}*\frac{B}{B + C}</math>
  
Note that <math>\frac{B}{B + C} < 1</math>, and the maximal value of this factor occurs when <math>C = 1</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>
 
 
Also note that <math>\frac{46}{3}</math> must cancel to give an integer value, and the only fraction that satisfies both these conditions is <math>\frac{45}{46}</math>
 
  
 
Plugging in, we get
 
Plugging in, we get
  
<math>44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59</math>
+
<math>44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59</math>, which is choice E
  
==Solution 2==
+
===Solution 2===
 
Suppose there are <math>A,B,C</math> rocks in the three piles, and that the mean of pile C is <math>x</math>, and that the mean of the combination of <math>B</math> and <math>C</math> is <math>y</math>. We are going to maximize <math>y</math>, subject to the following conditions:
 
Suppose there are <math>A,B,C</math> rocks in the three piles, and that the mean of pile C is <math>x</math>, and that the mean of the combination of <math>B</math> and <math>C</math> is <math>y</math>. We are going to maximize <math>y</math>, subject to the following conditions:
  
Line 52: Line 39:
 
<cmath>7B=3A</cmath>
 
<cmath>7B=3A</cmath>
 
<cmath>(x-44)C=4A</cmath>
 
<cmath>(x-44)C=4A</cmath>
<cmath>(x-y)C=(y-50)B.</cmath>
+
<cmath>(x-y)C=(y-50)B</cmath>
  
 
Let us test <math>y=59</math> is possible. If so, it is already the answer. If not, there will be some contradiction. So the third equation becomes  
 
Let us test <math>y=59</math> is possible. If so, it is already the answer. If not, there will be some contradiction. So the third equation becomes  
Line 63: Line 50:
  
 
(Note: To further illustrate the idea, let us look at <math>y=60</math> and see what happens. We then get <math>7\cdot 16C = 4A-30A<0</math>, which is a contradiction!)
 
(Note: To further illustrate the idea, let us look at <math>y=60</math> and see what happens. We then get <math>7\cdot 16C = 4A-30A<0</math>, which is a contradiction!)
 +
 +
===Solution 3===
 +
Obtain the 3 equations as in '''solution 2'''.
 +
 +
<cmath>7B=3A</cmath>
 +
<cmath>(x-44)C=4A</cmath>
 +
<cmath>(x-y)C=(y-50)B</cmath>
 +
 +
Our goal is to try to isolate <math>y</math> into an inequality.
 +
The first equation gives <math>A=\frac{7}{3}B</math>, which we plug into the second equation to get
 +
 +
<cmath>(x-44)C=\frac{28}{3}B</cmath>
 +
 +
To eliminate <math>x</math>, subtract equation 3 from equation 2:
 +
 +
<cmath>(x-44)C-(x-y)C=\frac{28}{3}B-(y-50)B</cmath>
 +
<cmath>(y-44)C=(\frac{178}{3}-y)B</cmath>
 +
 +
In order for the coefficients to be positive, <cmath>44<y<\frac{178}{3}</cmath>
 +
 +
Thus, the greatest integer value is <math>y=59</math>, choice <math>(E)</math>.
 +
 +
== See also ==
 +
{{AMC12 box|year=2013|ab=A|num-b=15|num-a=17}}
 +
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 01:14, 11 October 2020

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,

$44 + \frac{4(\frac{7}{3}B) + 6B}{B + C}=44 + \frac{46}{3}*\frac{B}{B + C}$

$\frac{B}{B + C} < 1$, so the maximum value occurs when $C = 1$. 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 $(E)$.

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