Difference between revisions of "2012 AMC 10A Problems/Problem 8"
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<math>b+c=19</math> | <math>b+c=19</math> | ||
| − | We add the first and last equations and then subtract the second one. | + | To isolate <math>b</math>, We add the first and last equations and then subtract the second one. |
<math>(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7</math> | <math>(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7</math> | ||
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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> | ||
| − | ==Solution | + | ==Video Solution (CREATIVE THINKING)== |
| − | + | https://youtu.be/sU5NAg6bhxI | |
| + | |||
| + | ~Education, the Study of Everything | ||
== See Also == | == See Also == | ||
Latest revision as of 12:55, 1 July 2023
- 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)
~Education, the Study of Everything
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. 
