At carl's combine diner, there are three size of coffee drinks regular (300ml), large (500ml) and extra large (800mL), and they cost $2.25, $3.25, and $5.75, respectively. on a particular afternoon, the diner sold at total of 37 coffees. The total volume of coffee sold was 19,100mL and the amount of money made in coffee sales was $133.25. How many of each size of drink did customers buy that afternoon?

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.

chad should not have included the values (1, 2) and (3, 6) because that would mean he only had half a healing card.

idk

step-by-step explanation: