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

(Solution 2 (Faster))
(Solution 3)
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Because <math>b</math> is the middle number, the middle number is <math>\boxed{\textbf{(D)}\ 7}</math>
 
Because <math>b</math> is the middle number, the middle number is <math>\boxed{\textbf{(D)}\ 7}</math>
  
==Solution 3==
+
==Solution 3 (Speed)==
 
Just guess and check. You can do it in your head in about 45 seconds and easily check your answer.
 
Just guess and check. You can do it in your head in about 45 seconds and easily check your answer.
  

Revision as of 16:49, 2 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.

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 $a<b<c$. 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 $\boxed{\textbf{(D)}\ 7}$

Solution 3 (Speed)

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

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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

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