SOLUTION: Dan has 21 coins, all nickels, dimes, and quarters worth $3.35. If he has 4 more quarters than dimes, how many of each coin does he have.

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Question 1057812: Dan has 21 coins, all nickels, dimes, and quarters worth $3.35. If he has 4 more quarters than dimes, how many of each coin does he have.
Found 2 solutions by josgarithmetic, solve_for_x:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
Each coin count, n, d, q;

system%28n%2Bd%2Bq=21%2C0.05n%2B0.1d%2B0.25q=3.35%2Cq=d%2B4%29

Simplify the money accounting as n%2B2d%2B5q=67.

Answer by solve_for_x(190) About Me  (Show Source):
You can put this solution on YOUR website!
Let D represent the number of dimes, and let N represent the number of nickels.

Since Dan has 4 more quarters than dimes, the number of quarters can be represented as D + 4.

The total number of coins is:

(Number of quarters) + (Number of dimes) + (Number of nickels) = 21

(D + 4) + D + N = 21

2D + N + 4 = 21

2D + N = 17

The total value of the coins can be written as:

($0.25)(Number of quarters) + ($0.10)(Number of dimes) + ($0.05)(Number of nickels) = $3.35

(0.25)(D + 4) + (0.10)(D) + (0.05)(N) = 3.35

0.25D + 1 + 0.10D + 0.05N = 3.35

0.35D + 0.05N = 2.35

This gives a system of two equations:

2D + N = 17
0.35D + 0.05N = 2.35

This system can be solved simultaneously to get values for D and N. With the value of D, you can get the number of quarters by adding 4 to D.