2012 AMC 10A Problems/Problem 8
- The following problem is from both the 2012 AMC 12A #6 and 2012 AMC 10A #8, so both problems redirect to this page.
Contents
Problem
The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
Solution 1
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:
To isolate , We add the first and last equations and then subtract the second one.
Because is the middle number, the middle number is
Video Solution (CREATIVE THINKING)
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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.