Question 1122003: Robin has a collection of nickels, dimes, and quarters worth $37.50. She has 30 more dimes than nickels and twice as many quarters as dimes. How many coins of each kind does she have?
Found 3 solutions by ikleyn, greenestamps, josgarithmetic: Answer by ikleyn(52781) (Show Source): Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!
Of course, being able to set up and solve a problem like this using formal algebra is a useful skill. But solving the problem using logical analysis can be good brain exercise.
The number of quarters is twice the number of dimes; and the number of nickels is 30 less than the number of dimes. The total value is $37.50.
(1) Add 30 more nickels, so that the number of nickels is the same as the number of dimes. The total value is now $37.50 + $1.50 = $39.
(2) Now the numbers of dimes and nickels are the same, and the number of quarters is twice as much. So make group of 2 quarters, 1 dime, and 1 nickel; each of those group has a value of $0.65.
(3) The number of those groups needed to make the new total of $39 is 39/0.65 = 60.
(4) So with the added 30 nickels, the collection has 120 quarters, 60 dimes, and 60 nickels.
(5) Now take away the 30 nickels you added in the beginning to get the final answer: 120 quarters, 60 dimes, and 30 nickels.
CHECK:
120(.25)+60(.10)+30(.05) = 30.00+6.00+1.50 = 37.50
Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! -----------------------------------------------------------
... nickels, dimes, and quarters worth $37.50. She has 30 more dimes than nickels and twice as many quarters as dimes.
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COINS COUNTS CENTS
NICKELS x 5x
DIMES x+30 10(x+30) 30 more dimes than nickels
QUARTERS 2(x+30) 25(2(x+30)) twice as many quarters as dimes
Total Cents 3750 worth $37.50
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