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

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Five positive consecutive integers starting with <math>a</math> have average <math>b</math>. What is the average of <math>5</math> consecutive integers that start with <math>b</math>?
 
Five positive consecutive integers starting with <math>a</math> have average <math>b</math>. What is the average of <math>5</math> consecutive integers that start with <math>b</math>?
  
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<math> \textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}}\ a+6\qquad\textbf{(E)}\ a+7</math>
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<math> \textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}\ a+6\qquad\textbf{(E)}\ a+7</math>
  
 
==Solution 1==
 
==Solution 1==
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==Solution 2==
 
==Solution 2==
  
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We are given that <cmath> \begin{aligned}\frac{a+a+1+a+2+a+3+a+4}5 & =b\rightarrow \\  
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We are given that <cmath>b=\frac{a+a+1+a+2+a+3+a+4}{5}</cmath>
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b & =a+2\end{aligned}</cmath>
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<cmath>\implies b =a+2</cmath>
  
 
We are asked to find the average of the 5 consecutive integers starting from <math>b</math> in terms of <math>a</math>. By substitution, this is <cmath>\frac{a+2+a+3+a+4+a+5+a+6}5=a+4</cmath>
 
We are asked to find the average of the 5 consecutive integers starting from <math>b</math> in terms of <math>a</math>. By substitution, this is <cmath>\frac{a+2+a+3+a+4+a+5+a+6}5=a+4</cmath>
  
 
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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==Solution 3 (fast)==
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We know from experience (or logical reasoning) that the average of <math>5</math> consecutive numbers is the <math>3^\text{rd}</math> one or the <math>1^\text{st} + 2</math>. With the logic, we find that <math>b=a+2</math>. Therefore, <math>b+2=(a+2)+2=\boxed{a+4}</math>.
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~MathFun1000
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==Solution 4==
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The list of numbers is <math>\left\{a,\ a+1,\ b,\ a+3,\ a+4\right\}</math> so <math>b=a+2</math>. The new list is <math>\left\{a+2,\ a+3,\ a+4,\ a+5,\ a+6\right\}</math> and the average is <math>a+4 \Longrightarrow \boxed{\textbf{(B) } a+4}</math>.
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~JH. L
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/GonWHjrROzI
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~Education, the Study of Everything
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==Video Solutions==
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===Video Solution 1===
 +
https://youtu.be/wBdD6Ge8FuE
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 +
~savannahsolver
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===Video Solution 2===
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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]]

Latest revision as of 03:36, 1 November 2024

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 $\boxed{\textbf{(B)}\ a+4}$

Solution 3 (fast)

We know from experience (or logical reasoning) that the average of $5$ consecutive numbers is the $3^\text{rd}$ one or the $1^\text{st} + 2$. With the logic, we find that $b=a+2$. Therefore, $b+2=(a+2)+2=\boxed{a+4}$.

~MathFun1000

Solution 4

The list of numbers is $\left\{a,\ a+1,\ b,\ a+3,\ a+4\right\}$ so $b=a+2$. The new list is $\left\{a+2,\ a+3,\ a+4,\ a+5,\ a+6\right\}$ and the average is $a+4 \Longrightarrow \boxed{\textbf{(B) } a+4}$.

~JH. L

Video Solution (CREATIVE THINKING)

https://youtu.be/GonWHjrROzI

~Education, the Study of Everything



Video Solutions

Video Solution 1

https://youtu.be/wBdD6Ge8FuE

~savannahsolver

Video Solution 2

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

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

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