Difference between revisions of "2012 AMC 10A Problems/Problem 8"
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<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math> | <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | Let the three numbers be equal to <math>a</math>, <math>b</math>, and <math>c</math>. We can now write three equations: | ||
+ | |||
+ | <math>a+b=12</math> | ||
+ | |||
+ | <math>b+c=17</math> | ||
+ | |||
+ | <math>a+c=19</math> | ||
+ | |||
+ | Adding these equations together, we get that | ||
+ | |||
+ | <math>2(a+b+c)=48</math> and | ||
+ | |||
+ | <math>a+b+c=24</math> | ||
+ | |||
+ | Substituting the original equations into this one, we find | ||
+ | |||
+ | <math>c+12=24</math> | ||
+ | |||
+ | <math>a+17=24</math> | ||
+ | |||
+ | <math>b+19=24</math> | ||
+ | |||
+ | Therefore, our numbers are 12, 7, and 5. The middle number is <math>\boxed{\textbf{(D)}\ 7}</math> |
Revision as of 20:20, 8 February 2012
Problem 8
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