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Question 158308: If Eric had nineteen fewer nickels and twenty-one fewer dimes, he would have the same number of nickels, dimes, and quarters. Eric has a total of $7.45. How many of each coin does he have?
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! If Eric had nineteen fewer nickels and twenty-one fewer dimes, he would have the same number of nickels, dimes, and quarters. Eric has a total of $7.45. How many of each coin does he have?
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Let q = number of quarters
then from: "Eric had nineteen fewer nickels and twenty-one fewer dimes, he would have the same number of nickels, dimes, and quarters" we get:
q+19 = number of nickels
q+21 = number of dimes
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Since, he has a total of $7.45 we get our formula:
.05(q+19) + .10(q+21) + .25q = 7.45
.05q + 0.95 + .10q + 2.1 + .25q = 7.45
.4q + 3.05 = 7.45
.4q = 4.4
q = 4.4/.4
q = 11 (this is the number of quarters)
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nickels:
q+19 = 11+19 = 30
dimes:
q+21 = 11+21 = 32
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