Difference between revisions of "2004 AMC 10B Problems/Problem 15"

Revision as of 23:51, 23 July 2014

Problem

Patty has $20$ coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have $70$ cents more. How much are her coins worth?

$\textbf{(A)}\ \textdollar 1.15\qquad\textbf{(B)}\ \textdollar 1.20\qquad\textbf{(C)}\ \textdollar 1.25\qquad\textbf{(D)}\ \textdollar 1.30\qquad\textbf{(E)}\ \textdollar 1.35$

Solution

Solution 1

She has $n$ nickels and $d=20-n$ dimes. Their total cost is $5n+10d=5n+10(20-n)=200-5n$ cents. If the dimes were nickels and vice versa, she would have $10n+5d=10n+5(20-n)=100+5n$ cents. This value should be $70$ cents more than the previous one. We get $200-5n+70=100+5n$ , which solves to $n=17$ . Her coins are worth $200-5n=\boxed{\mathrm{(A)\ }\textdollar1.15}$ .

Solution 2

Changing a nickel into a dime increases the sum by $5$ cents, and changing a dime into a nickel decreases it by the same amount. As the sum increased by $70$ cents, there are $70/5=14$ more nickels than dimes. As the total count is $20$ , this means that there are $17$ nickels and $3$ dimes.

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 7: / Line 7: @@
 ==Solution==
-=== Solution 1 ===
+== Solution 1 ==
-She has <math>n</math> nickels and <math>d=20-n</math> dimes. Their total cost is <math>5n+10d=5n+10(20-n)=200-5n</math> cents. If the dimes were nickels and vice versa, she would have <math>10n+5d=10n+5(20-n)=100+5n</math> cents. This value should be <math>70</math> cents more than the previous one. We get <math>200-5n+70=100+5n</math>, which solves to <math>n=17</math>. Her coins are worth <math>200-5n =\textdollar 1.15</math>.
+She has <math>n</math> nickels and <math>d=20-n</math> dimes. Their total cost is <math>5n+10d=5n+10(20-n)=200-5n</math> cents. If the dimes were nickels and vice versa, she would have <math>10n+5d=10n+5(20-n)=100+5n</math> cents. This value should be <math>70</math> cents more than the previous one. We get <math>200-5n+70=100+5n</math>, which solves to <math>n=17</math>. Her coins are worth <math>200-5n=\boxed{\mathrm{(A)\ }\textdollar1.15}</math>.
 === Solution 2 ===

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