SOLUTION: A person exchanged 390 pennies for quarters, dimes and nickels. The number of dimes in the exchange was twice the number of quarters and the number of nickels was twice the number
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Question 961320: A person exchanged 390 pennies for quarters, dimes and nickels. The number of dimes in the exchange was twice the number of quarters and the number of nickels was twice the number of dimes. How many of each coin type were in the exchange?
I set up the problem:
390 pennies=2D(.25) + 2N(.10) + Q(.25)
390 = .50D + .20N + .25Q
And now I'm stuck. Please help. Found 2 solutions by josgarithmetic, MathTherapy:Answer by josgarithmetic(39630) (Show Source):
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A person exchanged 390 pennies for quarters, dimes and nickels. The number of dimes in the exchange was twice the number of quarters and the number of nickels was twice the number of dimes. How many of each coin type were in the exchange?
I set up the problem:
390 pennies=2D(.25) + 2N(.10) + Q(.25)
390 = .50D + .20N + .25Q
And now I'm stuck. Please help.
390 pennies = $3.90
JUST one (1) variable is NEEDED. I've chosen that variable to be Q
Let the number of quarters be Q
Then number of dimes was: 2Q
Also, number of nickels was: 2(2Q), or 4Q
We then get: .25Q + .1(2Q) + .05(4Q) = 3.9
.25Q + .2Q + .2Q = 3.9
.65Q = 3.9
Q, or number of quarters = , or
Number of dimes: 2(6), or
Number of nickels: 4(6), or
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