SOLUTION: kyle works at a donut shop, where a 10oz. cup of coffee costs 95c, a 14 oz. cup of coffee costs $1.15, and a 20oz. cup costs $1.50. during one busy period Kyle served 24 cups of c

Question 80313: kyle works at a donut shop, where a 10oz. cup of coffee costs 95c, a 14 oz. cup of coffee costs $1.15, and a 20oz. cup costs $1.50. during one busy period Kyle served 24 cups of coffee, using 384 ounces of coffee, while collecting a total of $30.60. How many cups of each size did kyle fill
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
kyle works at a donut shop, where a 10oz. cup of coffee costs 95c, a 14 oz. cup of coffee costs $1.15, and a 20oz. cup costs $1.50. during one busy period Kyle served 24 cups of coffee, using 384 ounces of coffee, while collecting a total of $30.60. How many cups of each size did kyle fill.
:
Let x = no. of 10 oz cups sold
Let y = no. of 14 oz cups sold
Let z = no. of 20 oz cups sold
:
Equation 1: total number of cups sold:
x + y + z = 24
:
Equation 2: amt of coffee consumed:
10x + 14y + 20z = 384
:
Equation 3: total revenue from cups sold
.95x + 1.15y + 1.50z = 30.60
:
Mult the 1st equation by 20 and subtract the 2nd equation from it:
20x + 20y + 20z = 480
10x + 14y + 20z = 384
------------------------ subtracting eliminates z
10x + 6y = 96; (eq 4)
:
Mult the 1st equation by 1.5 and subtract the 3rd equation from it:
1.5x + 1.5y + 1.5z = 36.00
.95x + 1.15y+ 1.5z = 30.60
---------------------------subtracting eliminates z again
.55x + .35y = 5.40; (eq 5)
:
Multiply eq 4 by .055 and subtract from eq 5:
.55x + .35y = 5.40
.55x + .33y = 5.28
--------------------eliminates x
0x + .02y = .12
y = .12/.02
y = 6 ea 14 oz cups sold
:
Substitute 6 for y for in eq 4
10x + 6(6) = 96
10x = 96 - 36
x = 60/10
x = 6 ea 10 oz cups
:
That would leave 12 ea 20 oz cups (24 - 6 - 6 = 12)
:
Check our solutions in eq 2:
10(6) + 14(6) + 20(12) =
60 + 84 + 240 = 384 oz
:
:
A lot steps, hope it made some sense:

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