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Difference between revisions of "2011 AMC 10A Problems/Problem 17"

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Equating the two values we get for the sum, we get the answer <math>A+H+60=85</math> <math>\Longrightarrow</math> <math>A+H=\boxed{25 \ \mathbf{(C)}}</math>.
 
Equating the two values we get for the sum, we get the answer <math>A+H+60=85</math> <math>\Longrightarrow</math> <math>A+H=\boxed{25 \ \mathbf{(C)}}</math>.
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===Solution 3===
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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>
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===Solution 4===
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Given that the sum of 3 consecutive terms is 30, we have
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<math>(A+B+C)+(C+D+E)+(F+G+H)=90</math> and <math>(B+C+D)+(E+F+G)=60</math>
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It follows that <math>A+B+C+D+E+F+G+H=85</math> because <math>C=5</math>.
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Subtracting, we have that <math>A+H=25\rightarrow \boxed{\textbf{C}}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 01:59, 11 January 2018

Problem 17

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 $A+H$?

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

Solution

We consider the sum $A+B+C+D+E+F+G+H$ and use the fact that $C=5$, and hence $A+B=25$.

\begin{align*} &A+B+C+D+E+F+G+H=A+(B+C+D)+(E+F+G)+H=A+30+30+H=A+H+60\\ &A+B+C+D+E+F+G+H=(A+B)+(C+D+E)+(F+G+H)=25+30+30=85 \end{align*}

Equating the two values we get for the sum, we get the answer $A+H+60=85$ $\Longrightarrow$ $A+H=\boxed{25 \ \mathbf{(C)}}$.

Solution 3

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 4

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 2

We see that $A+B+C=30$, and by substituting the given $C=5$, we find that $A+B=25$. Similarly, $B+D=25$ and $D+E=25$.

\begin{align*} &(A+B)-(B+D)=A-D=0\\ &A=D\\ &(B+D)-(D+E)=B-E=0\\ &B=E\\ &A, B, 5, A, B, 5, G, H \end {align*}

Similarly, $G=A$ and $H=B$, giving us $A, B, 5, A, B, 5, A, B$. Since $H=B$, $A+H=A+B=\boxed{25 \ \mathbf{(C)}}$.

See Also

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

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