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Difference between revisions of "2004 AMC 12A Problems/Problem 11"

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{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #11]] and [[2004 AMC 10A Problems/Problem 14|2004 AMC 10A #14]]}}
 
== Problem ==
 
== Problem ==
 
The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is <math>20</math> cents. If she had one more quarter, the average value would be <math>21</math> cents. How many dimes does she have in her purse?
 
The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is <math>20</math> cents. If she had one more quarter, the average value would be <math>21</math> cents. How many dimes does she have in her purse?
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2004|ab=A|num-b=10|num-a=12}}
 
{{AMC12 box|year=2004|ab=A|num-b=10|num-a=12}}
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{{AMC10 box|year=2004|ab=A|num-b=13|num-a=15}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 20:15, 3 December 2007

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?

$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$

Solution

Let the total value (in cents) of the coins Paula has originally be $x$, and the number of coins she has be $n$. Then $\frac xn = 20 \Longrightarrow x = 20n$ and $\frac {x+25}{n+1} = 21$. Substituting yields $20n + 25 = 21(n+1) \Longrightarrow n = 4, x = 80$. It is easy to see now that Paula has 3 quarters, 1 nickel, so she has $0\ \mathrm{(A)}$ 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
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
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