Difference between revisions of "2012 AMC 10A Problems/Problem 8"

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(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>
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==Solution==
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Let the three numbers be equal to <math>a</math>, <math>b</math>, and <math>c</math>. We can now write three equations:
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<math>a+b=12</math>
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<math>b+c=17</math>
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<math>a+c=19</math>
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Adding these equations together, we get that
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<math>2(a+b+c)=48</math> and
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<math>a+b+c=24</math>
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Substituting the original equations into this one, we find
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<math>c+12=24</math>
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<math>a+17=24</math>
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<math>b+19=24</math>
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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?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution

Let the three numbers be equal to $a$, $b$, and $c$. We can now write three equations:

$a+b=12$

$b+c=17$

$a+c=19$

Adding these equations together, we get that

$2(a+b+c)=48$ and

$a+b+c=24$

Substituting the original equations into this one, we find

$c+12=24$

$a+17=24$

$b+19=24$

Therefore, our numbers are 12, 7, and 5. The middle number is $\boxed{\textbf{(D)}\ 7}$