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

Revision as of 10:24, 4 July 2013

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 $y= \boxed{\textbf{(A)}\ 4}$ .

2002 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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

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 == See also ==
 {{AMC10 box|year=2002|ab=B|num-b=24|after=Last problem}}
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Difference between revisions of "2002 AMC 10B Problems/Problem 25"

Revision as of 10:24, 4 July 2013

Problem

Solution

See also