SOLUTION: Could someone help me with this one? I have difficulty with word problems. You are greatly appreciated. Joe has a collection of nickels and dimes that is worth $6.05. If the num

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Question 203509: Could someone help me with this one? I have difficulty with word problems. You are greatly appreciated.
Joe has a collection of nickels and dimes that is worth $6.05. If the number of dimes was doubled and the number of nickels was decreased by 10, the value of the coins would be $9.85. How many dimes does he have?

Answer by ankor@dixie-net.com(22740) (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 was doubled and the number of nickels was decreased by 10,
the value of the coins would be $9.85. How many dimes does he have?
:
Let d = no. of dimes
Let n = no. of nickels
:
Write an equation for each statement:
:
"a collection of nickels and dimes that is worth $6.05."
.05n + .10d = 6.05
:
"If the number of dimes was doubled and the number of nickels was decreased by 10, the value of the coins would be $9.85."
.05(n-10) + .10(2d) = 9.85
:
.05n - .50 + .20d = 9.85
:
.05n + .20d = 9.85 + .50
:
.05n + .20d = 9.35
:
Subtract the 1st equation from the above equation:
.05n + .20d = 9.35
.05n + .10d = 6.05
---------------------Subtraction eliminate n, find d
0n + .10d = 3.30
d =
d = 33 dimes
:
:
Check solution
Find n:
.05n + .10d = 6.05
.05n + .10(33) = 6.05
.05n + 3.30 = 6.05
.05n = 6.05 - 3.30
.05n = 2.75
n =
n = 55 nickels
:
.05(55) + .20(33) = 9.35
2.75 + 6.60 = 9.35