How many pounds in regular coffee ( selling $4 per pound) and how many pounds of kona coffee (selling $11.50 per pound) must be combined to get 20 pounds of a mixture worth $6 per pound? be sure to write a system of two equations in two variables to solve the proplem.

Selling price of regular coffee = $4 per pound

Selling price of kona coffee = $11.5 per pound

Total mixture required = 20 pounds

Selling price of mixture = $6 per pound

Let quantity of regular coffee be 'a' and that of kona coffee be 'b'

⇒ Total mixture required = a + b

⇒ 20 = a + b ..................... (i)

Selling price of one pound of mixture = Price of total mixture ÷ Total pounds of mixture

⇒ 6 = [(4*a) + (11.5*b)] ÷ (a+b)

⇒ 6*a + 6*b = 4*a + 11.5*b

⇒ 2a = 5.5b

⇒ 20a = 55b

⇒ 4a = 11b

⇒ a = b ....................... (ii)

Substituting the value of (ii) in (i)

b + b = 20

⇒ 11b + 4b = 20 * 4

⇒ 15b = 80

⇒ 3b = 16

⇒ b =

and a = *

⇒ a =

So, to make the mixture we would require of regular coffee and of kona coffee.

d is the correct answer

step-by-step explanation:

x= 8 because 3x = 3 * 8, and 3*8 = 24