SOLUTION: Joe had nickles and dimes worth $6.30. If the dimes were doubled and the number of nickles increased by 15, the value of the coins would be $10.75. How many dimes does he have?

Question 292458: Joe had nickles and dimes worth $6.30. If the dimes were doubled and the number of nickles increased by 15, the value of the coins would be $10.75. How many dimes does he have?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Let n = original no. of nickels
Let d = original no. of dimes
:
Write an equation for each statement:
:
Joe had nickles and dimes worth $6.30.
.05n + .10d = 6.30
:
If the dimes were doubled and the number of nickles increased by 15,
the value of the coins would be $10.75.
.05(n+15) + 2(.10d) = 10.75
.05n + .75 + .20d = 10.75
.05n + .20d = 10.75 -.75
.05n + .20d = 10.00
:
We can use elimination here, subtract the 1st equation from the above equation
.05n + .20d = 10.00
.05n + .10d = 6.30
----------------------subtraction eliminates n, find d
.10d = 3.70
d =
d = 37 dimes
:
:
Check solution.
Find the no. of nickels
.05n + .10(37) = 6.30
.05n = 6.30 - 3.70
.05n = 2.60
n =
n = 52 nickels
then
.05(52) = .10(37)
2.60 + 3.70 = 6.30; confirms our solution

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