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

Revision as of 13:32, 12 February 2012

The following problem is from both the 2012 AMC 12A #6 and 2012 AMC 10A #8, so both problems redirect to this page.

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

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

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+{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #6]] and [[2012 AMC 10A Problems|2012 AMC 10A #8]]}}
 == Problem ==
 The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
@@ Line 32: / Line 34: @@
 {{AMC10 box|year=2012|ab=A|num-b=7|num-a=9}}
+{{AMC12 box|year=2012|ab=A|num-b=5|num-a=7}}