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

Revision as of 21:20, 8 February 2012

The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?

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

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 $\boxed{\textbf{(D)}\ 7}$

@@ Line 3: / Line 3: @@
 <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>