Question 493471: I cannot figure this one out. Can you please help me?
there was 48 nickles, dimes, and quarters whose value was $8.25. If there were twice as many quarters as dimes, how many coins of each kind were there?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! there was 48 nickles, dimes, and quarters whose value was $8.25. If there were twice as many quarters as dimes, how many coins of each kind were there?
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Equations:
Quantity: n + d + q = 48 coins
Quantity:: q = 2d
Value::: 5n + 10d + 25q = 825 cents
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Substitute for "q":
n + d + 2d = 48
5n + 10d + 25(2d) = 825
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Simplify:
n + 3d = 48
5n+60d = 825
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Multiply thru the 1st equation by 5
5n + 15d = 240
5n + 60d = 825
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Subtract and solve for "d":
45d = 585
d = 13 (# of dimes)
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q = 2d = 26 (# of quarters)
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n + 3d = 48
n + 39 = 48
n = 9 (# of nickels)
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Cheers,
Stan H.
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