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]]}} | {{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #11]] and [[2004 AMC 10A Problems/Problem 14|2004 AMC 10A #14]]}} | ||
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== Problem == | == Problem == | ||
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<math>\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4</math> | <math>\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4</math> | ||
− | == Solution 1 == | + | ==Solutions== |
+ | === 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> 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. | 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>. Thus, having three quarters, one nickel, and no dimes <math>\boxed{\mathrm{(A)}\ 0}.</math> | 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> | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/9Q463wjzCzQ | ||
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+ | Education, the Study of Everything | ||
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== See also == | == See also == |
Revision as of 23:48, 16 January 2021
- 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 cents. If she had one more quarter, the average value would be cents. How many dimes does she have in her purse?
Solutions
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 . Thus, having three quarters, one nickel, and no dimes
Video Solution
Education, the Study of Everything
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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