Difference between revisions of "2004 AMC 12A Problems/Problem 11"
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== Solution 1 == | == Solution 1 == | ||
| − | Let the total value (in cents) of the coins Paula has originally be <math>x</math>, and the number of coins she has be <math>n</math>. Then <math>\frac{x}{n}=20\Longrightarrowx=20n</math> and <math>\frac{x+25}{n+1}=21</math>. Substituting yields <math>20n+25=21(n+1)\Longrightarrow n=4</math>, <math>x = 80</math>. It is easy to see that Paula has <math>3</math> quarters and <math>1</math> nickel | + | Let the total value (in cents) of the coins Paula has originally be <math>x</math>, and the number of coins she has be <math>n</math>. Then <math>\frac{x}{n}=20\Longrightarrowx=20n</math> and <math>\frac{x+25}{n+1}=21</math>. Substituting yields <math>20n+25=21(n+1)\Longrightarrow n=4</math>, <math>x = 80</math>. It is easy to see that Paula has <math>3</math> quarters and <math>1</math> nickel, so she has <math>\boxed{\mathrm{(A)}\ 0}</math> dimes. |
===Solution 2=== | ===Solution 2=== | ||
Revision as of 23:47, 20 July 2014
- 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 
cents. If she had one more quarter, the average value would be 
cents. How many dimes does she have in her purse?
Solution
Solution 1
Let the total value (in cents) of the coins Paula has originally be 
, and the number of coins she has be 
. Then $\frac{x}{n}=20\Longrightarrowx=20n$ (Error compiling LaTeX. Unknown error_msg) and 
. Substituting yields 
, 
. It is easy to see that Paula has 
quarters and 
nickel, so she has 
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 
cents. As in the previous solution, we conclude that the only way to get 80 cents using 4 coins is 25+25+25+5.
See also
| 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. 
