SOLUTION: A clerk was asked to change a $10 bill. She returned 9 more dimes than nickels and twenty-one more quarters than dimes. How many coins of each did she return?

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Question 175615: A clerk was asked to change a $10 bill. She returned 9 more dimes than nickels and twenty-one more quarters than dimes. How many coins of each did she return?
Found 2 solutions by EMStelley, stanbon:
Answer by EMStelley(208) About Me  (Show Source):
You can put this solution on YOUR website!
Call the number of nickels n. Then she returned n+9 dimes and (n+9)+21 quarters. The dollar amounts of each can be represented by 0.05n, 0.10(n+9) and 0.25((n+9)+21). So you need to solve
0.05n + 0.10(n+9) + 0.25((n+9)+21) = 10.00

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A clerk was asked to change a $10 bill. She returned 9 more dimes than nickels and twenty-one more quarters than dimes. How many coins of each did she return?
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Quantity Eq.: d = n+9
Quantity Eq.: q = d+21
Value Equation: 5n + 10d + 25q = 1000
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Rearrange:
d = n+9
q = d+21 = n+9 + 21 = n+30
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Substitute to get:
5n + 10(n+9) + 25(n+30) = 1000
40n + 840 = 1000
40n = 160
n = 4 (# of nickels)
d = 4+9 = 13 (# of dimes)
q = 4 + 30 = 34 (# of quarters)
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Cheers,
Stan H.