Answer:
The number of regular, large, and extra-large drinks are 12, 15, and 10 respectively.
Step-by-step explanation:
Given that the cost for regular coffee drinks (300 ml)=$2.25
The cost for large coffee drinks (500 ml)=$3.25
The cost for extra large coffee drinks (800 ml)=$5.75
Let p,q, and r be the number of regular, large, and extra-large coffee sold.
As the diner sold a total of 37 coffees, so
p+q+r=37
r=37-p-q...(i)
The volume of p regular coffee = 300p ml
The volume of q large coffee = 500q ml
The volume of r extra-large coffee = 800r ml
As the total volume of coffee sold was 19,100mi, so
300p+500q+800r=19100
By using equation (i)
300p+500q+800(37-p-q)=19100
300p+500q+800 x 37 - 800p - 800q=19100
-500p-300q=19100-29600
500p+300q=10500
500p=10500-300q
p=21-0.6q...(ii)
Now, the cost of p regular coffee=$2.25p
The cost of q large coffee=$3.25q
The cost of r extra-large coffee=$5.75r
As the amount of money made in coffee sales was $133.25, so
2.25p+3.25q+5.75r=133.25
By using equations (1) we have
2.25p+3.25q+5.75(37-p-q)=133.25
2.25p+3.25q+212.75-5.75p-5.75q=133.25
3.50p+2.50q=79.5
From equation (ii)
3.5(21-0.6q)+2.50q=79.5
73.5-2.1q+2.5q=79.5
0.4q=79.5-73.5=6
q=6/0.4
q=15
From equation (ii)
p=21-0.6(15)
p=12
From equation (i)
r= 37-12-15
r=10
Hence, the number of regular, large, and extra-large drinks are 12, 15, and 10 respectively.
and get a common denominator of 8 on the right:
. Now simplify and cross-multiply: 39x = -16 andx = -16/39