SOLUTION: Joe has a collection of nickels and dimes that is worth $6.00. If the number of dimes were doubled and the number of nickels were increased by 6, the value of the coins would be $9
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: Joe has a collection of nickels and dimes that is worth $6.00. If the number of dimes were doubled and the number of nickels were increased by 6, the value of the coins would be $9.90. How many dimes does he have?
This question is from textbook
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 $6.00. If the number of dimes were doubled and the number of nickels were increased by 6, the value of the coins would be $9.90. How many dimes does he have?
:
Using the 1st statement we can write an equation:
.05n + .10d = 6.00
Rearrange so we can use for substitution
.05n = 6 - .10d
:
Multiply equation by 20 to make the coefficient of n = 1:
1n = (120 - 2d)
:
Using the 2nd statement, write another equation:
"If the number of dimes were doubled and the number of nickels were increased by 6, the value of the coins would be $9.90."
:
.05(n+6) + 2(.10d) = 9.90
.05n + .3 + .20d = 9.90
.05n + .2d = 9.9 - .3
.05n + .2d = 9.6
:
How many dimes does he have?
Substitute (120-2d) for n in the above equation:
.05(120-2d) + .2d = 9.6
6 - .10d + .2d = 9.6
-.1d + .2d = 9.6 - 6
+.1d = 3.6
d = 3.6/.1
d = 36 dimes in the collection
:
:
To confirm our solution we need to find out how many nickels:
n = 120 - 2(36)
n = 120 - 72
n = 48 dimes
:
Check solution using the 1st statement:
"Joe has a collection of nickels and dimes that is worth $6.00."
.05(48) + .10(36) =
2.40 + 3.60 = $6
:
You can check the solutions using the statement:
"If the number of dimes were doubled and the number of nickels were increased by 6, the value of the coins would be $9.90."
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