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 the

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:

To model the scenario, let’s denote:

• Let x be the amount (in pounds) of beans that sell for $0.20 per pound.
• Let y be the amount (in pounds) of beans that sell for $0.68 per pound.

Given:

1. The total weight of the mixture is 120 pounds: $x+y=120$
2. The mixture is worth $0.54 per pound: $0.20x+0.68y=0.54\times 120$

Now, we have a system of two equations: {�+�=1200.20�+0.68�=0.54×120{x+y=1200.20x+0.68y=0.54×120​

We can solve this system of equations to find the values of x and y.

{�+�=1200.20�+0.68�=64.8{x+y=1200.20x+0.68y=64.8​

We can solve the first equation for x: $x=120-y$

Substituting this expression for x into the second equation:

$0.20(120-y)+0.68y=64.8$

$24-0.20y+0.68y=64.8$

$0.48y=40.8$

�=40.80.48y=0.4840.8​

$y=85$

Now, we can find the value of x:

$x=120-y=120-85=35$

Therefore, 35 pounds of beans that sell for $0.20 per pound and 85 pounds of beans that sell for $0.68 per pound are used in the mixture.

Explanation of the solution: The solution is reasonable because it satisfies the given conditions of the problem. The sum of the weights of the two types of beans equals 120 pounds, as required. Additionally, the cost of the mixture, when considering the weights and prices of the two types of beans, equals the desired price of $0.54 per pound. Hence, the solution accurately represents the scenario described in the problem.