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?
:
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
:
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
:
12.1-.2d+.1d-1=9.85
:
-.1d=1.25
:
d=12.5
:
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
Answer by MathTherapy(10552) (Show Source): 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
43 dimes = $4.30
35 nickels = $1.75
$4.30 + $1.75 = $6.05
Doubling dimes = 2(43) = 86 = $8.60
Decreasing nickels by 10 = (35 – 10) = 25 = $1.25
$8.60 + $1.25 = $9.85
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