Difference between revisions of "2012 AMC 10A Problems/Problem 13"
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=== Solution 1 === | === Solution 1 === | ||
− | The minimum and maximum can be achieved with the orders <math>5, 4, 3, 2, 1</math> and <math>1, 2, 3, 4, 5</math>. | + | The minimum and maximum can be achieved with the orders <math>5, 4, 3, 2, 1</math> and <math>1, 2, 3, 4, 5</math> respectively. We can see this because the iterative average is like a weighted average that gives more weight to later numbers. |
<math>5,4,3,2,1 \Rightarrow \frac92,3,2,1 \Rightarrow \frac{15}{4},2,1 \Rightarrow \frac{23}{8},1 \Rightarrow \frac{31}{16}</math> | <math>5,4,3,2,1 \Rightarrow \frac92,3,2,1 \Rightarrow \frac{15}{4},2,1 \Rightarrow \frac{23}{8},1 \Rightarrow \frac{31}{16}</math> |
Revision as of 12:49, 26 January 2020
- The following problem is from both the 2012 AMC 12A #8 and 2012 AMC 10A #13, so both problems redirect to this page.
Problem
An iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?
Solutions
Solution 1
The minimum and maximum can be achieved with the orders and respectively. We can see this because the iterative average is like a weighted average that gives more weight to later numbers.
The difference between the two is .
Solution 2
The iterative average of any 5 integers is defined as:
Plugging in for , we see that in order to maximize the fraction,
,
and in order to minimize the fraction,
.
After plugging in these values and finding the positive difference of the two fractions, we arrive with , which is our answer of
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 10 Problems and Solutions |
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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.