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.
if x lbs are cheap beans, then the rest (120-x) are expensive.
Now just add up the value of the parts, and they must equal the value of the mix.
.20x + .68(120-x) = .54*120
To model the scenario, let's assume x pounds of the $0.20 beans are used, and 120 - x pounds of the $0.68 beans are used. The total cost of the mixture can be found by multiplying the cost per pound of each type of bean by the pounds used, and setting it equal to the cost per pound of the mixture multiplied by the total pounds used.
The equation can be written as:
0.20x + 0.68(120 - x) = 0.54 * 120
Now, let's solve this equation to find the values of x and (120 - x):
0.20x + 81.6 - 0.68x = 64.8
Combining like terms:
0.20x - 0.68x = 64.8 - 81.6
-0.48x = -16.8
Dividing both sides by -0.48:
x = -16.8 / -0.48
x = 35
Therefore, 35 pounds of the $0.20 beans are used, and (120 - 35) = 85 pounds of the $0.68 beans are used.
Explanation of the solution:
The solution suggests that 35 pounds of the $0.20 beans and 85 pounds of the $0.68 beans are used to create a 120-pound mixture worth $0.54 per pound. This means that a larger portion of the higher-priced beans is used in the mixture. The solution seems reasonable as the higher-priced beans contribute more to the total cost per pound, resulting in a price closer to the $0.68 per pound value.