# Difference between revisions of "2002 AMC 10B Problems/Problem 25"

## Problem

When $15$ is appended to a list of integers, the mean is increased by $2$. When $1$ is appended to the enlarged list, the mean of the enlarged list is decreased by $1$. How many integers were in the original list? $\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$

## Solution 1

Let $x$ be the sum of the integers and $y$ be the number of elements in the list. Then we get the equations $\dfrac{x+15}{y+1}=\dfrac{x}{y}+2$ and $\dfrac{x+15+1}{y+1+1}=\dfrac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1$. With a lot of algebra, the solution is found to be $y= \boxed{\textbf{(A)}\ 4}$.

## Solution 2

We let $n$ be the original number of elements in the set and we let $m$ be the original average of the terms of the original list. Then we have $mn$ is the sum of all the elements of the list. So we have two equations: $$mn+15=(m+2)(n+1)=mn+m+2n+2$$ and $$mn+16=(m+1)(n+2)=mn+2m+n+2.$$Simplifying both equations and we get, $$13=m+2n$$ $$14=2m+n$$ Solving for $m$ and $n$, we get $m=5$ and $n=\boxed{\textbf{(A)}4}$.

## Solution 3

Warning: This solution will rarely ever work in any other case. However, seeing that you can so easily plug and chug in problem 25 it is funny to see this.

Plug and chug random numbers with the answer choices, starting with the choice of $4$ numbers. You see that if you have 4 5s and you add 15 to the set, the resulting mean will be 7; we can verify this with math $$\frac{5+5+5+5+15}{5}=7$$ adding in 1 to the set you result in the mean to be 6. $$\frac{5+5+5+5+15+1}{6}=6$$ Thus we conclude that 4 is the correct choice or $\boxed{\textbf{(A)}}$

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