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

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Therefore, our numbers are 12, 7, and 5. The middle number is <math>\boxed{\textbf{(D)}\ 7}</math>
 
Therefore, our numbers are 12, 7, and 5. The middle number is <math>\boxed{\textbf{(D)}\ 7}</math>
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==Solution 2 (Faster)==
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Let the three numbers be <math>a</math>, <math>b</math> and <math>c</math> and <math>a<b<c</math>. We get the three equations:
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<math>a+b=12</math>
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<math>a+c=17</math>
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<math>b+c=19</math>
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We add the first and last equations and then subtract the second one.
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<math>(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7</math>
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Because <math>b</math> is the middle number, the middle number is <math>\boxed{\textbf{(D)}\ 7}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 10:15, 10 July 2018

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

See Also

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

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

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