Difference between revisions of "2004 AMC 10A Problems/Problem 14"
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Therefore, the total value of the coins was <math>20\times4=80</math> cents, which can only be made by using 3 quarters and 1 nickel, so there aren't any dimes <math>\Rightarrow\mathrm{(A)}</math>. | Therefore, the total value of the coins was <math>20\times4=80</math> cents, which can only be made by using 3 quarters and 1 nickel, so there aren't any dimes <math>\Rightarrow\mathrm{(A)}</math>. | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 02:40, 11 September 2007
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 would be 21 cents. How many dimes does she have in her purse?
Solution
Let 
be the number of coins in Paula's purse. Thus, the total value of the coins in her purse is 
. When 1 more quarter is added, there are 
coins, with an average of 21, or 
total cents. This can also be expressed as 
, so we set them equal and solve for .
Therefore, the total value of the coins was 
cents, which can only be made by using 3 quarters and 1 nickel, so there aren't any dimes 
.
See also
| 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 | ||
