Question 933517
Find the number of nickels, dimes, and quarters in a collection of 80 such coins if the nickels and the quarters are worth $4.50 and the value of the quarters and the dimes is $5.50.
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Quantity:: n + d + q = 80 coins
Value:: 5n + 25q = 450 cents
Value:: 10d + 25q = 550 cents
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Using the bottom 2 equations,
Subtract and solve for 10d-5n::
10d-5n = 100
Modify::
2d - n = 20
n = 2d-20
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Substitute for "n" in equation #1 to get::
2d-20 + d + q = 80
3d + q = 100
q = -3d+100
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Substitute for "n" and for "q" in 5n+25q=450 to solve for "d"
5(2d-20) + 25(-3d+100) = 450
10d - 100 -75d + 2500 = 450
-65d = 450-2400
-65d = -1950
d = 30 (# of dimes)
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Solve for "q" using q = -3d+100
q = -3*30+100 = 10 (# of quarters)
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Solve for "n" using n = 2d-20
n = 2*30-20 = 40 (# of quarters)
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Checking with the 1st equation:
30 + 10 + 40 = 80 coins
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Cheers,
Stan H.
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