# 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 Solutions

~savannahsolver

### Video Solution 2

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