A box contains $6.70 in nickels, dimes, and quarters. There are 39 coins in all, and the sum of the numbers of nickels and dimes is 3 less than the number of quarters. How many coins of each kind are there?
Marked as Best Answer
n = nickels
d = dimes
q = quarters
1) n + d + q = 39
n+d = q - 3
Translate to: 2) n + d - q = -3
.05n + .1d + .25q = 6.7 amount of each coin times the number of coins must add to 6.7
Multiply both sides by 20...
3) 1n + 2d + 5q = 134
Combine 1) & 2)
n + d + q = 39
n + d - q = -3
Add together
2n + 2d = 36
Divide by 2:
4) n + d = 18
Combine 1) and 3)
n + d + q = 39 Multiply by -5 => -5n -5d -5q = -195
n + 2d + 5q = 134 n + 2d + 5q = 134
Add together-> 5) -4n - 3d = -61
Combine 4) and 5)
n + d = 18 Multiply by 4 => 4n + 4d = 72
-4n - 3d = 61 -4n - 3d = -61
Add: d = 11
11 dimes...
Plug that into 4)
n + 11 = 18
n = 18-11 = 7 nickels.
Plug that into 1.
n + d + q = 39
7 + 11 + q = 39
18 + q = 39
q = 39-18 = 21 quarters
21 quarters, 7 nickels, and 11 dimes.
21(.25) + 7(.05) + 11(.1) = 5.25 + .35 + 1.1 = 5.6 + 1.1 = 6.7 Check.
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