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

Revision as of 00:00, 2 December 2019

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

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

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

@@ Line 1: / Line 1: @@
-== Problem 8 ==
+{{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?
 <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
+==Solution==
+Let the three numbers be equal to <math>a</math>, <math>b</math>, and <math>c</math>. We can now write three equations:
+<math>a+b=12</math>
+<math>b+c=17</math>
+<math>a+c=19</math>
+Adding these equations together, we get that
+<math>2(a+b+c)=48</math> and
+<math>a+b+c=24</math>
+Substituting the original equations into this one, we find
+<math>c+12=24</math>
+<math>a+17=24</math>
+<math>b+19=24</math>
+Therefore, our numbers are 12, 7, and 5. The middle number is <math>\boxed{\textbf{(D)}\ 7}</math>
+==Solution 2 (Faster)==
+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:
+<math>a+b=12</math>
+<math>a+c=17</math>
+<math>b+c=19</math>
+We add the first and last equations and then subtract the second one.
+<math>(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7</math>
+Because <math>b</math> is the middle number, the middle number is <math>\boxed{\textbf{(D)}\ 7}</math>
+==Solution 3==
+Just guess and check. You can do it in your head in about 45 seconds and easily check your answer.
+== See Also ==
+{{AMC10 box|year=2012|ab=A|num-b=7|num-a=9}}
+{{AMC12 box|year=2012|ab=A|num-b=5|num-a=7}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

Page

Toolbox

Search