Question 142972: Michael has $2.25 in nickels, dimes, and quarters. If he has 3 fewer dimes than quarters and as many nickels as the sum of the dimes and quarters, how many of each kind of coin does he have?
Found 2 solutions by stanbon, BrittanyM:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Michael has $2.25 in nickels, dimes, and quarters. If he has 3 fewer dimes than quarters and as many nickels as the sum of the dimes and quarters, how many of each kind of coin does he have?
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EQUATIONS:
5n + 10d + 25q = 225
0 + d - q = 3
n - d - q = 0
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Solve by any means to get:
n = 11 (# of nickels)
d = 7 (# of dimes)
q = 4 (# of quarters)
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Cheers,
Stan H.
Answer by BrittanyM(80) (Show Source):
You can put this solution on YOUR website! We first need to translate the words in this problem into equations that we can work with.
The total sum of the values of dimes, nickels, and quarters is 2.25.
Knowing this, we have
.05N +.10D +.25Q = 2.25
And we were told that the sum of dimes and quarters is equal to the number of nickels, and that there were three less dimes than quarters.
N = D + Q
Q - 3 = D
In order to solve for actual values, we will have to use the method of substitution.
From the information given above, we can find both nickels and dimes in terms of quarters:
N = 2Q -3
D = Q -3
So:
.05(2Q-3) + .10(Q - 3) + .25Q = 2.25
We end up with Q = 6.
We can plug this into the other equations to find our remaining unknown values.
N = 2Q - 3
N = 2(6) -3
N = 9
D = Q - 3
D = (6) - 3
D = 3
To check our answers, we can plug these values back into our original equation.
.05(9) = .10(3) = .25(6) = 2.25
Since the above statement is true, our answers are correct.
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