2004 AMC 10B Problems/Problem 15

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?

$\mathrm{(A) \ } $$1.15 \qquad \mathrm{(B) \ } $ $1.20 \qquad \mathrm{(C) \ } $$1.25 \qquad \mathrm{(D) \ } $ $1.30 \qquad \mathrm{(E) \ } $$1.35$==Solution==

=== Solution 1 ===

She has$ (Error compiling LaTeX. Unknown error_msg)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 = $ $1.15$.

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.

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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