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 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 |
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All AMC 12 Problems and Solutions |
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