2013 AMC 12A Problems/Problem 16
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 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,
, so the maximum value occurs when . 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 .
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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All AMC 12 Problems and Solutions |
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