# 2012 AMC 10A Problems/Problem 8

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

## Problem

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

## Solution 2 (Faster)

Let the three numbers be $a$, $b$ and $c$ and $a. 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 $\boxed{\textbf{(D)}\ 7}$

## Solution 3

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 