Difference between revisions of "2004 AMC 12A Problems/Problem 11"
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== Solution == | == Solution == | ||
| − | 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 xn = 20 \Longrightarrow x = 20n</math> and <math>\frac {x+25}{n+1} = 21</math>. Substituting yields <math>20n + 25 = 21(n+1) \Longrightarrow n = 4, x = 80</math>. It is easy to see now that Paula has 3 quarters, 1 nickel, so she has <math>0\ \mathrm{(A)}</math> dimes. | + | 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 xn = 20 \Longrightarrow x = 20n</math> and <math>\frac {x+25}{n+1} = 21</math>. [[substitution|Substituting]] yields <math>20n + 25 = 21(n+1) \Longrightarrow n = 4, x = 80</math>. It is easy to see now that Paula has 3 quarters, 1 nickel, so she has <math>0\ \mathrm{(A)}</math> dimes. |
== See also == | == See also == | ||
| − | {{AMC12 box|year=2004|ab=A|num-b= | + | {{AMC12 box|year=2004|ab=A|num-b=10|num-a=12}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 18:16, 3 December 2007
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
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 
. It is easy to see now that Paula has 3 quarters, 1 nickel, so she has 
dimes.
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 | |
