2011 AMC 12A Problems/Problem 8: Difference between revisions

Revision as of 19:46, 16 September 2013

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

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 $A+H=25\rightarrow \boxed{\textbf{C}}$ .

2011 AMC 12A (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 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 12: / Line 12: @@
 ===Solution 1===
 Let <math>A=x</math>. Then from <math>A+B+C=30</math>, we find that <math>B=25-x</math>. From <math>B+C+D=30</math>, we then get that <math>D=x</math>. Continuing this pattern, we find <math>E=25-x</math>, <math>F=5</math>, <math>G=x</math>, and finally <math>H=25-x</math>. So <math>A+H=x+25-x=25 \rightarrow \boxed{\textbf{C}}</math>
 ===Solution 2===
-A faster technique is to assume that the problem can be solved, and thus <math>A+H</math> is an invariant.  Since <math>A + B + 5 = 30</math>, assign any value to <math>A</math>.  <math>10</math> is a simple value to plug in, which gives a value of <math>15</math> for B.  The 8-term sequence is thus <math>10, 15, 5, 10, 15, 5, 10, 15</math>.   The sum of the first and the last terms is <math>25\rightarrow \boxed{\textbf{C}}</math>
-Note that this alternate solution is not a proof.  If the sum of <math>A+G</math> had been asked for, this technique would have given <math>20</math> as an answer, when the true answer would have been "cannot be determined".
-===Solution 3===
 Given that the sum of 3 consecutive terms is 30, we have
 <math>(A+B+C)+(C+D+E)+(F+G+H)=90</math> and <math>(B+C+D)+(E+F+G)=60</math>

Page

Toolbox

Search