Kyle works at a donut factory, where a 10-oz cup of coffee costs 95cents, a 14-oz cup costs $1.15, and a 20-oz cup costs $1.50. during one busy period, kyle served 30 cups of coffee, using 464 ounces of coffee, while collecting a total of $37.20. how many cups of each size did kyle fill?

4, 16 and 10

Step-by-step explanation:

Here we would like to work with a system of 3 equations, we need ti find them. Lets call s the 10oz coffees, m the 14oz and l the 20oz.

As Kyle sold $37.2 selling 3 different types of coffee at their respective prices we can say:

revenue = s*0.95 + m*1.15 + l*1.5 = 37.2 [equation 1]

And as he sold 30 cups:

s + m + l = 30 [equation 2]

And as the total volume of coffee is 464 oz:

s*10 + m*14 + l*20 = 464 [equation 3]

So, we have a system of 3 equations with three unknowns s, m and l.

Take the eq 2 and subtract m and l in both sides:

s = 30 - m - l

Now, take it and replace in the other 2:

(30 - m - l)*10 + 14m + 20l = 464 [eq 3]

0.95(30 - m - l) + 1.15m + 1.5l = 37.2 [eq 1]

Working with equation 3:

300 -10m -10l + 14m + 20l = 464

300 + 4m + 10l = 464

4m + 10l = 164 [equation 3']

Now with eq 1:

0.95(30 - m - l) + 1.15m + 1.5l = 37.2

28.5 - 0.95m - 0.95l + 1.15m + 1.5 l = 37.2

28.5 + 0.2m + 0.55l = 37.2

0.2m + 0.55l = 8.7 [equaiton 1']

Now we passed to a 2-equation system with eq 1' and eq 3'.

Take eq 3' and subtract 4m in both sides:

10l = 164 - 4m

Now divide by 10 in both sides

l = [164 - 4m]/10

Now, with this value for l replace in eq 1':

0.2m + 0.55 ( [164 - 4m]/10) = 8.7

0.2m + 0.055 (164 - 4m) = 8.7

0.2m + 9.02 - 0.22m = 8.7

- 0.02m = -0.32

m = -0.32/-0.02

m = 16 > 16 14oz coffees

Now replace m=16 in eq 1' or 3' to find l:

4m +10l = 164

64 + 10l = 164

10l = 100

l = 10 > 10 20oz coffees

So, finally we replace l=10 and m=16 in equaiton 2:

s + m + l = 30

s + 16 +10 = 30

s = 30 - 26

s = 4

So, Kyle filled 4 10oz coffes, 16 14oz coffees and 10 20oz coffees.