Difference between revisions of "2011 AMC 12A Problems/Problem 8"

Revision as of 21:10, 31 May 2022

Problem

In the eight term sequence $A$ , $B$ , $C$ , $D$ , $E$ , $F$ , $G$ , $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$ ?

$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$

Solution 1

Let $A=x$ . Then from $A+B+C=30$ , we find that $B=25-x$ . From $B+C+D=30$ , we then get that $D=x$ . Continuing this pattern, we find $E=25-x$ , $F=5$ , $G=x$ , and finally $H=25-x$ . So $A+H=x+25-x=25 \rightarrow \boxed{\textbf{C}}$

Solution 2

Given that the sum of 3 consecutive terms is 30, we have $(A+B+C)+(C+D+E)+(F+G+H)=90$ and $(B+C+D)+(E+F+G)=60$

It follows that $A+B+C+D+E+F+G+H=85$ because $C=5$ .

Subtracting, we have that $A+H=25\rightarrow \boxed{\textbf{C}}$ .

Solution 3 (the tedious one)

From the given information, we can deduce the following equations:

$A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G =30$ , and $F+G+H=30$ .

We can then cleverly manipulate the equations above by adding and subtracting them to be left with the answer.

$(A+B)-(B+D)=25-25 \implies (A-D)=0$

$(A-D)+(D+E)=0+25 \implies (A+E)=25$

$(A+E)-(E+F+G)=25-30 \implies (A-F-G)=-5$ (Notice how we don't use $D+E+F=30$ )

$(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25$

Therefore, we have $A+H=25 \rightarrow \boxed{\textbf{C}}$

~JinhoK

Solution 4 (the cheap one)

Since all of the answer choices are constants, it shouldn't matter what we pick $A$ and $B$ to be, so let $A = 20$ and $B = 5$ . Then $D = 30 - B -C = 20$ , $E = 30 - D - C = 5$ , $F = 30 - D - E =5$ , and so on until we get $H = 5$ . Thus $A + H = \boxed{\mathbf{(C)}25}$

Note Something useful to shorten a lot of the solutions above is to notice $5 + D + E = D + E + F$ so F = 5

Video Solution

https://www.youtube.com/watch?v=6tlqpAcmbz4 ~Shreyas S

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AMC 12 Problems and Solutions

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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

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

@@ Line 46: / Line 46: @@
 Since all of the answer choices are constants, it shouldn't matter what we pick <math>A</math> and <math>B</math> to be, so let <math>A = 20</math> and <math>B = 5</math>. Then <math>D = 30 - B -C = 20</math>, <math>E = 30 - D - C = 5</math>, <math>F = 30 - D - E =5</math>, and so on until we get <math>H = 5</math>. Thus <math>A + H = \boxed{\mathbf{(C)}25}</math>
-==Useful Note==
+Note
 Something useful to shorten a lot of the solutions above is to notice <cmath>5 + D + E = D + E + F</cmath> so F = 5

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