SOLUTION: Joe has a collection of nickels and dimes that is worth $6.05. If the number of dimes is doubled and the number of nickels was decreased by 10, the value of the coins would be $9.8
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-> SOLUTION: Joe has a collection of nickels and dimes that is worth $6.05. If the number of dimes is doubled and the number of nickels was decreased by 10, the value of the coins would be $9.8
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Question 179941: Joe has a collection of nickels and dimes that is worth $6.05. If the number of dimes is doubled and the number of nickels was decreased by 10, the value of the coins would be $9.85. How many dimes do we have? Found 2 solutions by Mathtut, MathTherapy:Answer by Mathtut(3670) (Show Source):
You can put this solution on YOUR website! Joe has a collection of nickels and dimes that is worth $6.05. If the number of dimes is doubled and the number of nickels was decreased by 10, the value of the coins would be $9.85. How many dimes do we have?
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let the number of nickels and dimes be n and d respectively
:
.05n+.1d=6.05.........eq 1
.05(2n)+.1(d-10)=9.85.....eq 2
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rewrite eq 1 to .05n=6.05-.1d--->n=121-2d and plug the value of n which is 121-d into eq 2
:
.1(121-2d)+.1(d-10)=9.85
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12.1-.2d+.1d-1=9.85
:
-.1d=1.25
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d=12.5
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something is wrong with this problem. One of the amounts in the equation is off. there is no way to get a whole number for dimes in this scenario. Email me the correct numbers. thanks Mathtut
You can put this solution on YOUR website! Let number of dimes be d, and the number of nickels n
Then, d(.10) + n(.05) = 6.05, and
2d(.10) + (n – 10).05 = 9.85, or .2d + .05n - .5 = 9.85
.1d + .05n = 6.05
.2d + .05n = 10.35
Subtract eq (ii) from eq (i) - .1d = - 4.3
d = 43
Therefore, we have 43 dimes.
To find how many nickels there are, we just substitute 43 for d in eq (i).
This gives us: 4.3 + .05n = 6.05
.05n = 1.75
n = 35
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