Question

A coffee mixture has beans that sell for 0.20 a pound and beans that sell for 0.68. If 120 pounds of beans create a mixture worth 0.54 a pound, how much of each bean is used ? Model the scenario then solve it. Then, in two or more sentences explain whether your solution is or isn't reasonable.

Answer 1

x=pounds of coffee bean  ($0.20 per pound)
y=pounds of coffee bean ($0.68 per pound )

We can suggest this system of equations:
x+y=120
(0.20x+0.68 y) / (x+y)=0.54 ⇒  (0.2x+0.68y)=0.54(x+y)

We can solve this system by substitution method.
x+y=120      ⇒  y=120-x

0.2x+0.68(120-x)=0.54[x+(120-x)]
0.2x+81.6-0.68x=0.54(120)
-0.48x+81.6=64.8
-0.48x=64.8-81.6
-0.48x=-16.8
x=-16.8/-0.48
x=35

y=120-x=120-35=85

Answer: the coffee mixture has 35 pounds of coffee beans sold to $0.2 a pound, and 85 pounds of coffee beans sold to $0.68 a pound, the solutions is reasonable because the price of a coffee mixture ($0.54 a pound) is greater  than $0.2 and smaller than $0.68.
             

Answer 2

Answer:

The coffee mixture has 35 pounds of coffee beans sold at $0.20 a pound and 85 pounds of coffee beans sold at $0.68 a pound. This is reasonable because if you add 35 and 85 it will give you 120 pounds.

Step-by-step explanation:

0.2x + 0.68(120 - x) = 0.54[x + (120 - x)]

 0.2x + 81.6 - 0.68x = 0.54(120)

          -0.48x + 81.6 = 64.8

                     -0.48x = 64.8 - 81.6

                      -0.48x = -16.8

                      /-0.48       /-0.48

                               x = 35

y = 120 - x

y = 120 - 35

y = 85