2002 AMC 10B Problems/Problem 25

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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

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 $m$ be the number of elements in the set and we let $n$ be the 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, \begin{align*} 13&=m+2n\\ 14&=2m+n \end{align*} Solving for $m$ and $n$, we get $m=5$ and $n=\boxed{\textbf{(A)}4}$.

See also

2002 AMC 10B (ProblemsAnswer KeyResources)
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