Difference between revisions of "2013 AMC 12A Problems/Problem 16"
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<math> \textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59</math> | <math> \textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59</math> | ||
− | + | ==Solution 1== | |
− | |||
Let pile <math>A</math> have <math>A</math> rocks, and so on. | Let pile <math>A</math> have <math>A</math> rocks, and so on. | ||
− | The | + | The total weight of <math>A</math> and <math>C</math> can be expressed as <math>44(A + C)</math>. |
− | To get the total weight of <math>B</math> and <math>C</math>, we | + | 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> |
− | <math> | + | 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>. |
− | + | 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{4(\frac{7}{3}B) + 6B}{B + C}=44 + \frac{46}{3}*\frac{B}{B + C}</math> | |
− | <math>\frac | + | 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> |
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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== | |
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 72: | 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. | + | Obtain the 3 equations as in '''solution 2'''. |
<cmath>7B=3A</cmath> | <cmath>7B=3A</cmath> | ||
Line 79: | Line 57: | ||
<cmath>(x-y)C=(y-50)B</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 | + | <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>(x-44)C-(x-y)C=\frac{28}{3}B-(y-50)B</cmath> | ||
− | <cmath>(y-44)C | + | <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> | 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>. | Thus, the greatest integer value is <math>y=59</math>, choice <math>(E)</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Let the number of rocks in <math>A</math> be <math>a</math>, <math>B</math> be <math>b</math>, <math>C</math> be <math>c</math>. The total weight of <math>A</math> be <math>40a</math>, <math>B</math> be <math>50b</math>, <math>C</math> be <math>kc</math>. | ||
+ | |||
+ | We can write the information given as, <math>\frac{40a + 50b}{a+b} = 43</math>, <math>\frac{40a + kc}{a+c} = 44</math>, <math>\frac{50b + kc}{b+c} = ?</math> | ||
+ | |||
+ | <math>40a + 50b = 43 a + 43 b</math>, <math>3a = 7b</math> | ||
+ | |||
+ | <math>40a + kc = 44a + 44 c</math>, <math>kc = 4a + 44c = \frac{28}{3}b + 44c</math> | ||
+ | |||
+ | <math>\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}</math> | ||
+ | |||
+ | <math> = 44 + \frac{46a}{3a+7c} = 44 + \frac{46}{3 + \frac{7}{a}} < 44 + \frac{46}{3} \approx 59.3</math> | ||
+ | |||
+ | <math>\frac{50b + kc}{b+c} \le \boxed{\textbf{(E) } 59}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 5== | ||
+ | Let the total number of rocks in pile <math>A</math> be <math>A_n</math>, and the total number of rocks in pile <math>B</math> be <math>B_n</math>. Then, by restriction 3 (the average of <math>A</math> and <math>B</math>), we can establish the equation: <cmath>\frac{40A_n+50B_n}{A_n+B_n}=43</cmath>. | ||
+ | Cross-multiplying, we get: <cmath>40A_n+50B_n=43A_n+43B_n \implies 3A_n=7B_n</cmath>. | ||
+ | Let's say we have <math>7k</math> rocks in <math>A</math> and <math>3k</math> rocks in <math>B</math>. Hence, we have <math>280k</math> and <math>150k</math> as the total weight of piles <math>A</math> and <math>B</math>, respectively. Let the total weight of <math>C</math> be <math>m</math>, and the total number of rocks in <math>C</math> be <math>n</math>. | ||
+ | Using the last restriction regarding the average of piles <math>A</math> and <math>C</math>, we have: <cmath>\frac{280k+m}{7k+n}=44 \implies 280k + m=280k + 28k + 44n \implies m=28k+44n</cmath>. | ||
+ | To find the average of piles <math>B</math> and <math>C</math>, we can establish the expression: <cmath>\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}.</cmath> | ||
+ | When we let the final expression equal <math>59</math>, we get: <cmath>44+\frac{46k}{3k+n}=59\implies\frac{46k}{3k+n}=15</cmath> | ||
+ | Cross-multiplying, we get: <cmath>46k=45k+14n\implies k=14n</cmath> | ||
+ | <math>k</math> is still positive here, so 59 works. As this is the greatest option, we can circle <math>\textbf{(E)}</math> immediately. | ||
+ | To show why <math>59</math> is the greatest, consider the following: | ||
+ | When we let the final expression equal <math>60</math>, we get: <cmath>44+\frac{46k}{3k+n}=60\implies\frac{46k}{3k+n}=16</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>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=A|num-b=15|num-a=17}} | {{AMC12 box|year=2013|ab=A|num-b=15|num-a=17}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 01:43, 29 October 2023
Problem
,
,
are three piles of rocks. The mean weight of the rocks in
is
pounds, the mean weight of the rocks in
is
pounds, the mean weight of the rocks in the combined piles
and
is
pounds, and the mean weight of the rocks in the combined piles
and
is
pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles
and
?
Solution 1
Let pile have
rocks, and so on.
The total weight of and
can be expressed as
.
To get the total weight of and
, we add the weight of
and subtract the weight of
:
Therefore, the mean of and
is
, which is simplified to
.
We now need to eliminate in the numerator.
Since we know that
, we have
Substituting,
In order to maximize , we can minimize the denominator by letting
(C must be a positive integer). Since
must cancel to give an integer, and the only fraction that satisfies both conditions is
Plugging in, we get
, which is choice E
Solution 2
Suppose there are rocks in the three piles, and that the mean of pile C is
, and that the mean of the combination of
and
is
. We are going to maximize
, subject to the following conditions:
which can be rearranged as:
Let us test is possible. If so, it is already the answer. If not, there will be some contradiction. So the third equation becomes
So ,
,
, therefore,
, which gives us a consistent solution. Therefore
is the answer.
(Note: To further illustrate the idea, let us look at and see what happens. We then get
, which is a contradiction!)
Solution 3
Obtain the 3 equations as in solution 2.
Our goal is to try to isolate into an inequality.
The first equation gives
, which we plug into the second equation to get
To eliminate , subtract equation 3 from equation 2:
In order for the coefficients to be positive,
Thus, the greatest integer value is , choice
.
Solution 4
Let the number of rocks in be
,
be
,
be
. The total weight of
be
,
be
,
be
.
We can write the information given as, ,
,
,
,
Solution 5
Let the total number of rocks in pile be
, and the total number of rocks in pile
be
. Then, by restriction 3 (the average of
and
), we can establish the equation:
.
Cross-multiplying, we get:
.
Let's say we have
rocks in
and
rocks in
. Hence, we have
and
as the total weight of piles
and
, respectively. Let the total weight of
be
, and the total number of rocks in
be
.
Using the last restriction regarding the average of piles
and
, we have:
.
To find the average of piles
and
, we can establish the expression:
When we let the final expression equal
, we get:
Cross-multiplying, we get:
is still positive here, so 59 works. As this is the greatest option, we can circle
immediately.
To show why
is the greatest, consider the following:
When we let the final expression equal
, we get:
Cross-multiplying, we get:
Since
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
.
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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