SOLUTION: a parking lot meter contains 47 coins in dimes and quarters for a total of 5.45 . how many of each coin does the meter contain.

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Question 933770: a parking lot meter contains 47 coins in dimes and quarters for a total of 5.45 . how many of each coin does the meter contain.
Found 2 solutions by Fombitz, Alan3354:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1.D%2BQ=47
2.10D%2B25Q=545
From eq. 1,
10D%2B10Q=470
10D=470-10Q
Substitute,
470-10Q%2B25Q=545
15Q=75
Q=5
Then,
D%2B5=47
D=42
Forty two dimes and five quarters

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
a parking lot meter contains 47 coins in dimes and quarters for a total of 5.45 . how many of each coin does the meter contain.
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Another method:
47 dimes would be $4.70
Each dime replaced by a quarter add 15 cents.
545 - 470 = 75 cents
75/15 = 5 quarters
42 dimes