SOLUTION: Hello, I have a question I need help answering. Your time is appreciated. In Jose's coin jar there is a collection of nickels, dimes, and quarters which totals $4.05. There are t

Question 466563: Hello,
I have a question I need help answering. Your time is appreciated.
In Jose's coin jar there is a collection of nickels, dimes, and quarters which totals $4.05. There are twice as many quarters as nickels and 8 more dimes than nickels. How many coins of each kind are there?
Found 2 solutions by lwsshak3, ankor@dixie-net.com:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
In Jose's coin jar there is a collection of nickels, dimes, and quarters which totals $4.05. There are twice as many quarters as nickels and 8 more dimes than nickels. How many coins of each kind are there?
...
let x=number of nickels
2x=number of quarters
x+8= number of dimes
.05x+.25(2x)+.1(x+8)=4.05
.05x+.5x+.1x+.8=4.05
.65x=3.25
x=3.25/.65=5
ans:
number of nickels=5
number of quarters=10
number of dimes=13
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
coin jar there is a collection of nickels, dimes, and quarters which totals $4.05.
There are twice as many quarters as nickels and 8 more dimes than nickels.
How many coins of each kind are there?
:
Let n = no.of nickel
Let d = no. of dimes
let q = no. of quarters
:
Write an equation for each statement:
:
"coin jar there is a collection of nickels, dimes, and quarters which totals $4.05."
.05n + .10d + .25q = 4.05
:
"There are twice as many quarters as nickels"
q = 2n
:
"and 8 more dimes than nickels."
d = (n+8)
:
In the 1st equation, replace q with 2n, replace d with (n+8)
.05n + .10(n+8) + .25(2n) = 4.05
.05n + .10n + .8 + .50n = 4.05
.05n + .10n + .50n = 4.05 - 80
.65n = 3.25
n =
n = 5 nickels
:
I'll let you find d and q, check solutions in the total$ equation.

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