SOLUTION: A boy has 93 cent in pennies, nickels, dimes, and quarters. He has two more quarters than dimes. If he has as many nickels as dimes and as many pennies as quarters, how many of eac
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-> SOLUTION: A boy has 93 cent in pennies, nickels, dimes, and quarters. He has two more quarters than dimes. If he has as many nickels as dimes and as many pennies as quarters, how many of eac
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Question 175800: A boy has 93 cent in pennies, nickels, dimes, and quarters. He has two more quarters than dimes. If he has as many nickels as dimes and as many pennies as quarters, how many of each coin does he have? Answer by EMStelley(208) (Show Source):
You can put this solution on YOUR website! Call the number of dimes d. Than the number of quarters can be represented by d+2. The number of nickels is d and the number of pennies is d+2. So then we can represent the money amount of each by
dimes - 0.10d
quarters - 0.25(d+2)
nickels = 0.05d
pennies = 0.01(d+2)
So
So the boy had 1 dime, 3 quarters, 1 nickel and 3 pennies.
