Question 674803: Dan has nickels, dimes and quarters. He has a total of 33.60. He has four times as many quarters as dimes and one-half as many dimes as nickels. How many of each coin does he have?
I put x nickels, y dimes, and z quarters. 5x + 10y + 25z = 3360. Next, I put z = y + 4 to represent "four times as many quarters as dimes." Next, I put either x = 1/2y or x = 0.5y to represent "one-half as many dimes as nickels.
Next, I saw how 1/2y or 0.5y is really 1/2 times 10 or 0.5 times ten which equals five (5).
Then I substituted for 5x + 10y + 25z = 3360 by plugging in the numbers as follows: 5(5) + 10y + 25(y+4) = 3360. When I solved it, it wasn't correct.
I'm supposed to get 112 quarters, 28 dimes, and 56 nickels.
Am I missing a step where I should write x + y + z? - And then substitute one at at time? This is different in that I may not be writing something correctly because this problem contains no amount but only the value as the total and that it contains 1/2 which I may be figuring wrong. Thanks in advance.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Dan has nickels, dimes and quarters.
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Write an equation for each statement, get n and q in terms of d:
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He has a total of 33.60.
.05n + .10d + .25q = 33.60
5n + 10d + 25q = 3360, get rid of the decimals
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He has four times as many quarters as dimes
q = 4d
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and one-half as many dimes as nickels.
n = 2d
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Replace n with 2d, replace q with 4d
5(2d) + 10d + 25(4d) = 3360
10d + 10d + 100d = 3360
120d = 3360
d = 3360/120
d = 28 dimes
:
See if you can find n and q, and check it in the 1st equation
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