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

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== Solution ==
 
== Solution ==
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Let <math>x</math> be the sum of the list of integers and <math>y</math> be the number of elements in the list. Then we get the equations <math>\frac{x+15}{y+1}=\frac{x}{y}+2</math> and <math>\frac{x+15+1}{y+1+1}=\frac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1</math>. With a little work, the solution is found to be <math>y= \boxed{\textbf{(A)}\ 4} </math>.
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Let <math>x</math> be the sum of the integers and <math>y</math> be the number of elements in the list. Then we get the equations <math>\frac{x+15}{y+1}=\frac{x}{y}+2</math> and <math>\frac{x+15+1}{y+1+1}=\frac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1</math>. With a little work, the solution is found to be <math>y= \boxed{\textbf{(A)}\ 4} </math>.

Revision as of 02:18, 29 January 2011

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 $\frac{x+15}{y+1}=\frac{x}{y}+2$ and $\frac{x+15+1}{y+1+1}=\frac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1$. With a little work, the solution is found to be $y= \boxed{\textbf{(A)}\ 4}$.

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