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

https://youtu.be/9Q463wjzCzQ

Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=1649

~ pi_is_3.14

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.

Thank you for reading these solutions by the users of AoPS!

Page

Toolbox

Search