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| − | ==Problem==
| + | #redirect [[2004 AMC 10A Problems/Problem 11]] |
| − | 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?
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| − | <math> \mathrm{(A) \ } 0 \qquad \mathrm{(B) \ } 1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } 3 \qquad \mathrm{(E) \ } 4 </math>
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| − | ==Solution==
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| − | Let <math>n</math> be the number of coins in Paula's purse. Thus, the total value of the coins in her purse is <math>20n</math>. When 1 more quarter is added, there are <math>n+1</math> coins, with an average of 21, or <math>21(n+1)</math> total cents. This can also be expressed as <math>20n+25</math>, so we set them equal and solve for <math>n</math>.
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| − | <math>\displaystyle21(n+1)=20n+25</math>
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| − | <math>\displaystyle n=4</math>
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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>.
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| − | == See also ==
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| − | {{AMC10 box|year=2004|ab=A|num-b=13|num-a=15}}
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| − | [[Category:Introductory Algebra Problems]] | |
