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Difference between revisions of "2014 AMC 10A Problems/Problem 10"

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Thus, the answer is <math> \boxed{\textbf{(B)}\ a+4} </math>
 
Thus, the answer is <math> \boxed{\textbf{(B)}\ a+4} </math>
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==Video Solutions==
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===Video Solution One===
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https://youtu.be/wBdD6Ge8FuE
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~savannahsolver
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===Video Solution Two===
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https://youtu.be/rJytKoJzNBY
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2014|ab=A|num-b=8|num-a=10}}
 
{{AMC12 box|year=2014|ab=A|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category: Introductory Algebra Problems]]

Revision as of 00:17, 26 November 2020

The following problem is from both the 2014 AMC 12A #9 and 2014 AMC 10A #10, so both problems redirect to this page.

Problem

Five positive consecutive integers starting with $a$ have average $b$. What is the average of $5$ consecutive integers that start with $b$?

$\textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}\ a+6\qquad\textbf{(E)}\ a+7$

Solution 1

Let $a=1$. Our list is $\{1,2,3,4,5\}$ with an average of $15\div 5=3$. Our next set starting with $3$ is $\{3,4,5,6,7\}$. Our average is $25\div 5=5$.

Therefore, we notice that $5=1+4$ which means that the answer is $\boxed{\textbf{(B)}\ a+4}$.

Solution 2

We are given that \[b=\frac{a+a+1+a+2+a+3+a+4}{5}\] \[\implies b =a+2\]

We are asked to find the average of the 5 consecutive integers starting from $b$ in terms of $a$. By substitution, this is \[\frac{a+2+a+3+a+4+a+5+a+6}5=a+4\]

Thus, the answer is $\boxed{\textbf{(B)}\ a+4}$

Video Solutions

Video Solution One

https://youtu.be/wBdD6Ge8FuE

~savannahsolver

Video Solution Two

https://youtu.be/rJytKoJzNBY

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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

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