2004 AMC 10B Problems/Problem 15

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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 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, which is equal to $\boxed{\mathrm{(A)\ }\textdollar1.15}$.

See also

2004 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 10 Problems and Solutions

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