Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution)
Line 21: Line 21:
  
 
Subtracting, we have that <math>A+H=25\rightarrow \boxed{\textbf{C}}</math>.
 
Subtracting, we have that <math>A+H=25\rightarrow \boxed{\textbf{C}}</math>.
 +
 +
 +
===Solution 3 (the tedious one)===
 +
From the given information, we can deduct the following equations:
 +
 +
<math>A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G
 +
=30</math>, and <math>F+G+H=30</math>.
 +
 +
 +
We can then add and subtract the equations above to be left with the answer.
 +
 +
<math>(A+B)-(B+D)=25-25 \implies (A-D)=0</math>
 +
 +
<math>(A-D)+(D+E)=0+25 \implies (A+E)=25</math>
 +
 +
<math>(A+E)-(E+F+G)=25-30 \implies (A-F-G)=-5</math>
 +
 +
<math>(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25</math>
 +
 +
Therefore, we have that <math>A+H=25 \rightarrow \boxed{\textbf{C}}</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=7|num-a=9|ab=A}}
 
{{AMC12 box|year=2011|num-b=7|num-a=9|ab=A}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:01, 11 July 2020

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


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 add and subtract the equations above to be left with the answer.

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

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

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

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

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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_8&oldid=128123"