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

Revision as of 00:58, 16 February 2021

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

Let the three numbers be $a$ , $b$ and $c$ and $a<b<c$ . We get the three equations:

$a+b=12$

$a+c=17$

$b+c=19$

We add the first and last equations and then subtract the second one.

$(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7$

Because $b$ is the middle number, the middle number is $\boxed{\textbf{(D)}\ 7}$

Just guess and check. You can do it in your head in about 45 seconds and easily check your answer.

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

@@ Line 6: / Line 6: @@
 <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
-==Solution==
+==Solution 1==
 Let the three numbers be equal to <math>a</math>, <math>b</math>, and <math>c</math>. We can now write three equations: