Difference between revisions of "2013 AMC 12A Problems/Problem 16"
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− | ==Solution 1== | + | == Problem== |
+ | |||
+ | <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>? | ||
+ | |||
+ | <math> \textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59</math> | ||
+ | |||
+ | ==Solution== | ||
+ | ===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. | ||
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<math>44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59</math> | <math>44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59</math> | ||
− | ==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: | ||
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(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== | + | ===Solution 3=== |
Obtain the 3 equations as in solution 2. | Obtain the 3 equations as in solution 2. | ||
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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>. | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2013|ab=A|num-b=15|num-a=17}} |
Revision as of 18:43, 22 February 2013
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
Solution 1
Let pile have
rocks, and so on.
The mean weight of and
together is
, so the total weight of
and
is
To get the total weight of and
, we need to add the total weight of
and subtract the total weight of
And then dividing by the number of rocks and
together, to get the mean of
and
,
Simplifying,
Now, to get rid of the in the numerator, we use two definitions of the total weight of
and
Substituting back in,
Note that , and the maximal value of this factor occurs when
Also note that must cancel to give an integer value, and the only fraction that satisfies both these conditions is
Plugging in, we get
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.
Combining the 1st and 2nd equations, we see that
Subtracting equation 3 from equation 2, we have
In order for the coefficients to be positive,
Thus, the greatest integer value is , choice
.
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 |