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.

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.

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frika frika · Answer 2 · 2018-08-31T17:36:46+02:00

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:

[tex]\left\{\begin{array}{l}x+y=120\\0.20x+0.68y=64.8.\end{array}\right.[/tex]

Solve it.

1. Multiply second equation by 100:

[tex]\left\{\begin{array}{l}x+y=120\\20x+68y=6480.\end{array}\right.[/tex]

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

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

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.