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?

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

Solution

Solution 1

Let the total value (in cents) of the coins Paula has originally be $x$ , and the number of coins she has be $n$ . Then $\frac{x}{n}=20\Longrightarrowx=20n$ (Error compiling LaTeX. Unknown error_msg) and $\frac{x+25}{n+1}=21$ . Substituting yields $20n+25=21(n+1)\Longrightarrow n=4$ , $x = 80$ . It is easy to see that Paula has $3$ quarters and $1$ nickel before the coin addition, so she has $\boxed{\mathrm{(A)}\ 0}$ dimes.

Solution 2

If the new coin was worth 20 cents, adding it would not change the mean at all. The additional 5 cents raise the mean by 1, thus the new number of coins must be 5. Therefore initially there were 4 coins worth a total of $4\cdot 20=80$ cents. As in the previous solution, we conclude that the only way to get 80 cents using 4 coins is 25+25+25+5.

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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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2004 AMC 12A Problems/Problem 11

Contents

Problem

Solution

Solution 1

Solution 2

See also