SOLUTION: Hi I have 88 coins to the value of $11.40,I have some 5 cents,25 cents and three time more 10cents then 25 cents.How many of each tipe of coins do I have? How do I put it in a e

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Question 182114: Hi
I have 88 coins to the value of $11.40,I have some 5 cents,25 cents and three time more 10cents then 25 cents.How many of each tipe of coins do I have?
How do I put it in a equation to solve it?
Thanks
Vasea
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
I have 88 coins to the value of $11.40,I have some 5 cents, 25 cents and three time more 10 cents than 25 cents. How many of each type of coins do I have?
:
Let n = no. of nickels (5 cents)
Let d = no. of dimes(10 cents)
Let q = no. of quarters{25 cents)
:
Write an equation for each statement
:
"I have 88 coins"
n + d + q = 88
:
" to the value of $11.40,"
.05n + .10d + .25q = 11.40
:
"three times more 10 cents then 25 cents."
d = 3q
:
Using the 1st equation, replace d with 3q
n + 3q + q = 88
n + 4q = 88
:
Using the 2nd equation, do the same
.05n + .10(3q) + .25q = 11.40
.05n + .30q+ .25q = 11.40
.05n + .55q = 11.40
:
Multiply the above equation by 20, this makes .05 = 1; subtract (n + 4q = 88)
n + 11q = 228
n + 4q = 88
-----------------subtraction eliminates n, find d
7q = 140
q =
q = 20 quarters
:
Use n + 4q = 88 to find n
n + 4(20) = 88
n + 80 = 88
n = 88 - 80
n = 8 nickels
:
we know d = 3q
d = 3(20)
d = 60 dimes
:
Check solution in the $ equation
.05(8) + .10(60) + .25(20) =
.40 + 6.00 + 5.00 = 11.40