Answer:
Cups of 120oz sold = 17
Cups of 160oz sold = 25
Cups of 200oz sold = 13
Step-by-step explanation:
We know that:
a 120z cup of coffee costs $1.65
a 160z cup costs $1.85
a 200oz cup cost $1.95
Let's define the variables:
A = # of 120z cups that she sold
B = # of 160oz cups that she sold
C = # of 200oz cups that she sold.
We know that:
"Reba served 55 cups of coffee"
A + B + C = 55.
"emptying six 1440z brewers while"
A*120oz + B*160oz + C*200oz = 6*1440oz
"collecting a total of $99.65"
A*$1.65 + B*$1.85 + C*$1.95 = $99.65
Then we have a system of equations:
A + B + C = 55
A*120oz + B*160oz + C*200oz = 6*1440oz
A*$1.65 + B*$1.85 + C*$1.95 = $99.65
To solve this, the first step is to isolate one of the variables in one of the equations, let's isolate A in the first one:
A = 55 - B - C
Now we can replace that into the other two equations:
(55 - B - C)*120oz + B*160oz + C*200oz = 6*1440oz
(55 - B - C)*$1.65 + B*$1.85 + C*$1.95 = $99.65
Now we can rewrite those equations as:
B*40oz + C*80oz = 8640oz - 55*120oz = 2040oz
B*$0.20 + C*$0.30 = $99.65 - 55*$1.65 = $8.90
Now we have the system:
B*40oz + C*80oz = 2040oz
B*$0.20 + C*$0.30 = $8.90
Now we need to isolate another variable in one of the equations, let's isolate B in the first one:
B = 2040oz/40oz - C*80oz/40oz = 51 - 2*C
Now we can replace this in the other equation and get:
(51 - 2*C)*$0.20 + C*$0.30 = $8.90
Bow we can solve this for C.
$10.20 - C*$0.40 + C*$0.30 = $8.90
$10.20 - $8.90 = C*$0.10
$1.30 = C*$0.10
($1.30/$0.10) = C = 13
Now we can replace this in the equation:
B = 51 - 2*C = 51 - 2*13 = 25
And for A, we had:
A = 55 - B - C = 55 - 25 - 13 = 17
Then:
Cups of 120oz sold = 17
Cups of 160oz sold = 25
Cups of 200oz sold = 13
