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 is not reasonable.

Answer 1

Let x be the amount of beans in mixture for $0.20 a pound  and y be the amount of beans in mixture for $0.68 a pound.  If 120 pounds of beans create a mixture, then

x+y=120.

x pounds beans for $0.20 a pound  cost $0.20x and y pounds beans for $0.68 a pound cost $0.68y. If mixture worth $0.54 a pound, then it costs $0.54·120=$64.8 for 120 pounds. Then

0.20x+0.68y=64.8

You get the system of equations:

$\left\{\begin{array}{l}x+y=120\\0.20x+0.68y=64.8.\end{array}\right.$

Solve it.

1. Multiply second equation by 100:

$\left\{\begin{array}{l}x+y=120\\20x+68y=6480.\end{array}\right.$

2. Express x from first equation and substitute it in second:

$\left\{\begin{array}{l}x=120-y\\20(120-y)+68y=6480.\end{array}\right.$

3. The equation 20(120-y)+68y=6480 contains only y and is easy to solve:

2400-20y+68y=6480,

48y=6480-2400,

48y=4080,

y=85.

4. Then x=120-y=120-85=35.

Answer: you should take 35 pounds of beans for $0.20 a pound  and 85 pounds of beans for $0.68 a pound.

Answer 2

Modeling an equation would be like this;

P = 0.2*X + 0.58*Y (i)

with the condition X + Y = 120 (ii)
and P = 0.54 then

from (ii) we have
X =120 - Y (iii)

from (i) and (iii)

0.54 = 0.2(120 -Y) +0.58Y

0.54 = 24 -0.2Y +0.58Y
Y = -61.73 pounds
which will mean that X = 181.73 pounds

The solutions is mathematically correct but does not make sense as there is no such thing as a negative pound.