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>v</math>, and the number of coins she has be <math>n</math>. Then <math>\frac{v}{n}=20\Longrightarrow v=20n</math> and <math>\frac{v+25}{n+1}=21</math>. Substituting yields: <math>20n+25=21(n+1),</math> so <math>n=4</math>, <math>v = 80</math> | + | Let the total value, in cents, of the coins Paula has originally be <math>v</math>, and the number of coins she has be <math>n</math>. Then <math>\frac{v}{n}=20\Longrightarrow v=20n</math> and <math>\frac{v+25}{n+1}=21</math>. Substituting yields: <math>20n+25=21(n+1),</math> so <math>n=4</math>, <math>v = 80.</math> Then, we see that the only way Paula can satisfy this rule is if she had <math>3</math> quarters and <math>1</math> nickel in her purse. Thus, she has <math>\boxed{\mathrm{(A)}\ 0}</math> dimes. |
== Solution 2 == | == Solution 2 == | ||
Revision as of 14:25, 12 February 2017
- The following problem is from both the 2004 AMC 12A #11 and 2004 AMC 10A #14, so both problems redirect to this page.
Contents
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 1
Let the total value, in cents, of the coins Paula has originally be 
, and the number of coins she has be 
. Then 
and 
. Substituting yields: 
so 
, 
Then, we see that the only way Paula can satisfy this rule is if she had 
quarters and 
nickel in her purse. Thus, she has 
dimes.
Solution 2
If the new coin was worth cents, adding it would not change the mean. The additional 
cents raise the mean by , thus the new number of coins must be . Therefore there were 
coins worth a total of 
cents. As in the previous solution, we conclude that the only way to get 
cents using coins is 
.
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. 
