Difference between revisions of "2012 AMC 10A Problems/Problem 8"
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Because <math>b</math> is the middle number, the middle number is <math>\boxed{\textbf{(D)}\ 7}</math> | Because <math>b</math> is the middle number, the middle number is <math>\boxed{\textbf{(D)}\ 7}</math> | ||
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+ | ==Solution 3== | ||
+ | Just guess and check. You can do it in your head in about 45 seconds and easily check your answer. | ||
== See Also == | == See Also == |
Revision as of 00:00, 2 December 2019
- The following problem is from both the 2012 AMC 12A #6 and 2012 AMC 10A #8, so both problems redirect to this page.
Problem
The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
Solution
Let the three numbers be equal to , , and . We can now write three equations:
Adding these equations together, we get that
and
Substituting the original equations into this one, we find
Therefore, our numbers are 12, 7, and 5. The middle number is
Solution 2 (Faster)
Let the three numbers be , and and . We get the three equations:
We add the first and last equations and then subtract the second one.
Because is the middle number, the middle number is
Solution 3
Just guess and check. You can do it in your head in about 45 seconds and easily check your answer.
See Also
2012 AMC 10A (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 10 Problems and Solutions |
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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.