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) \text{ ,implying } 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.
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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 ==
  
If the new coin was worth <math>20</math> cents, adding it would not change the mean. The additional <math>5</math> cents raise the mean by <math>1</math>, thus the new number of coins must be <math>5</math>. Therefore there were <math>4</math> coins worth a total of <math>4\times20=80</math> cents. As in the previous solution, we conclude that the only way to get <math>80</math> cents using <math>4</math> coins is <math>25+25+25+5</math>. <math>\boxed{\mathrm{(A)}\ 0}</math>
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If the new coin was worth <math>20</math> cents, adding it would not change the mean. The additional <math>5</math> cents raise the mean by <math>1</math>, thus the new number of coins must be <math>5</math>. Therefore there were <math>4</math> coins worth a total of <math>4\times20=80</math> cents. As in the previous solution, we conclude that the only way to get <math>80</math> cents using <math>4</math> coins is <math>25+25+25+5</math>. Thus, having three quarters, one nickel, and no dimes <math>\boxed{\mathrm{(A)}\ 0}</math>
  
 
== See also ==
 
== See also ==

Revision as of 15:28, 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.

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 1

Let the total value, in cents, of the coins Paula has originally be $v$, and the number of coins she has be $n$. Then $\frac{v}{n}=20\Longrightarrow v=20n$ and $\frac{v+25}{n+1}=21$. Substituting yields: $20n+25=21(n+1),$ so $n=4$, $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she had $3$ quarters and $1$ nickel in her purse. Thus, she has $\boxed{\mathrm{(A)}\ 0}$ dimes.

Solution 2

If the new coin was worth $20$ cents, adding it would not change the mean. The additional $5$ cents raise the mean by $1$, thus the new number of coins must be $5$. Therefore there were $4$ coins worth a total of $4\times20=80$ cents. As in the previous solution, we conclude that the only way to get $80$ cents using $4$ coins is $25+25+25+5$. Thus, having three quarters, one nickel, and no dimes $\boxed{\mathrm{(A)}\ 0}$

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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