# Difference between revisions of "2004 AMC 12A Problems/Problem 11"

The following problem is from both the 2004 AMC 12A #11 and 2004 AMC 10A #14, so both problems redirect to this page.

## Problem

The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?

$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$

## Solutions

### Solution 1

Let the total value, in cents, of the coins Paula has originally be $v$, and the number of coins she has be $n$. Then $\frac{v}{n}=20\Longrightarrow v=20n$ and $\frac{v+25}{n+1}=21$. Substituting yields: $20n+25=21(n+1),$ so $n=4$, $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she had $3$ quarters and $1$ nickel in her purse. Thus, she has $\boxed{\mathrm{(A)}\ 0}$ dimes.

### Solution 2

If the new coin was worth $20$ cents, adding it would not change the mean. The additional $5$ cents raise the mean by $1$, thus the new number of coins must be $5$. Therefore there were $4$ coins worth a total of $4\times20=80$ cents. As in the previous solution, we conclude that the only way to get $80$ cents using $4$ coins is $25+25+25+5$. Thus, having three quarters, one nickel, and no dimes $\boxed{\mathrm{(A)}\ 0}.$

## Video Solution

Education, the Study of Everything