2014 AMC 10A Problems/Problem 10

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

We know from experience 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 $b=a+2$ . $b+2=(a+2)+2=\boxed{a+4}$ .

~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

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.

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