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Difference between revisions of "2011 AMC 12A Problems/Problem 8"
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\textbf{(E)}\ 43 </math>
 
\textbf{(E)}\ 43 </math>
  
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== Solution ==
+
==Solution 1==
−
===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>
 
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===
+
==Solution 2==
 
Given that the sum of 3 consecutive terms is 30, we have
 
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>
 
<math>(A+B+C)+(C+D+E)+(F+G+H)=90</math> and <math>(B+C+D)+(E+F+G)=60</math>
Line 23: Line 22:
  
  
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===Solution 3 (the tedious one)===
+
==Solution 3 (the tedious one)==
 
From the given information, we can deduct the following equations:
 
From the given information, we can deduct the following equations:
  

Revision as of 13:53, 5 May 2021

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 deduct 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

Video Solution

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

See also

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
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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