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

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 original number of elements in the set and we let $n$ 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 $n=\boxed{\textbf{(A)}4}$ .

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2002 AMC 10B Problems/Problem 25

Contents

Problem

Solution

Solution 2

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