Question

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.

Answer 1

Let 'x' pounds of regular coffee and 'y' pounds of kona coffee be combined together to get a 20 pound mixture.

So, $x+y = 20$

x = 20-y (Equation 1)

Since,  regular coffee is to be sold at $4 per pound and kona coffee at $11.50 per pound to get 20 pounds of a mixture worth $6 per pound.

So, $(4 \times x) +(11.50 \times y) = (20 \times 6)$

$4x + 11.50y = 120$

Substituting the value of 'x' from equation 1, we get

4(20-y) + 11.50y = 120

$80-4y + 11.50y = 120$

$80+7.5y = 120$

$7.5y = 40$

y = 5.3

So, x = 20- y

x = 20-5.3

x = 14.7

So, 14.7 pounds of regular coffee and 5.3 pounds of kona coffee are mixed to get a 20 pound mixture.

Answer 2

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 = $\frac{11}{4}$b ....................... (ii)

Substituting the value of (ii) in (i)

$\frac{11}{4}$b + b = 20

⇒ 11b + 4b = 20 * 4

⇒ 15b = 80

⇒ 3b = 16

⇒ b = $\frac{16}{3}$

and a = $\frac{11}{4}$ * $\frac{16}{3}$

⇒ a = $\frac{44}{3}$

So, to make the mixture we would require $\frac{44}{3}$ of regular coffee and $\frac{16}{3}$ of kona coffee.