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 $n$ 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.

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2004 AMC 10B Problems/Problem 15

Contents

Problem

Solution

Solution 1

Solution 2

See also