SOLUTION: Anna's grandma gave her 95 coins consisting of pennies, nickels, dimes, and quarters. She also knows there are only 5 pennies and the total value of the coins is $12.05. Also, th

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: Anna's grandma gave her 95 coins consisting of pennies, nickels, dimes, and quarters. She also knows there are only 5 pennies and the total value of the coins is $12.05. Also, th      Log On

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Question 949216: Anna's grandma gave her 95 coins consisting of pennies, nickels, dimes, and quarters. She also knows there are only 5 pennies and the total value of the coins is $12.05. Also, there are 5 more quarters than dimes. How many of each coin does Anna have?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
p,n,d,q how many of each coin;


system%28p=5%2C0.05%2B0.05n%2B0.1d%2B0.25q=12.05%2Cp%2Bn%2Bd%2Bq=95%2Cq-d=5%29

system%28p=5%2C0.05n%2B0.1d%2B0.25q=12.00%2Cp%2Bn%2Bd%2Bq=95%2Cq-d=5%29

system%28p=5%2C0.05n%2B0.1d%2B0.25q=12.00%2Cn%2Bd%2Bq=90%2Cq-d=5%29

p has already been taken care of and known, not affecting the other part of the system:

system%280.05n%2B0.1d%2B0.25q=12.00%2Cn%2Bd%2Bq=90%2Cq-d=5%29
Work with this system.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

Anna's grandma gave her 95 coins consisting of pennies, nickels, dimes, and quarters. She also knows there are only 5 pennies and the total value of the coins is $12.05. Also, there are 5 more quarters than dimes. How many of each coin does Anna have?
We already know that there are 5 pennies, valued at %22%24%220.05. Therefore,
the other coins total 90 (95 - 5), and have a value of %22%24%2212
Let the number of dimes, be D, and the number of nickels, N
Then number of quarters = D + 5
We then get:
N + D + D + 5 = 90_____N + 2D = 85 ------ eq (i)
Also, .05N + .1D + .25(D + 5) = 12________.05N + .1D + .25D + 1.25 = 12
.05N + .35D = 10.75 ------- eq (ii)
- .05N - .1D = - 4.25 -------- Multiplying eq (i) by - .05 ------- eq (iii)
.25D = 6.5 -------- Adding eqs (iii) & (ii)
D, or number of dimes = 6.5%2F.25, or highlight_green%2826%29
N + 2(26) = 85 ------- Substituting 26 for D in eq (i)
N + 52 = 85
N, or number of nickels = 85+-+52, or highlight_green%2833%29
Number of quarters: 26+%2B+5, or highlight_green%2831%29
You can do the check!!
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