She needs to prepare 160 pounds of blended coffee beans selling for $3.69per pound. She plans to do this by blending together a high-quality bean costing $5.00 per pound and a cheaper bean at $2.00 per pound. To the nearest pound, find how much high-quality coffee bean and how much cheaper coffee bean she should blend.
h + c = 160
... 200 h + 200 c = 200 * 160
500 h + 200 c = 369 * 160
subtract equations to eliminate c
solve for h, then substitute back to find c
Thank you so much Scott!
To find the amounts of high-quality beans and cheaper beans that need to be blended, we can use a system of equations.
Let's assume the amount of high-quality beans needed is x pounds, and the amount of cheaper beans needed is y pounds.
According to the given information, the total weight of the blended coffee beans is 160 pounds. So the first equation is:
x + y = 160 ...(Equation 1)
The cost per pound of the high-quality beans is $5.00, and the cost per pound of the cheaper beans is $2.00. The average price per pound of the blended coffee beans is $3.69. Therefore, the second equation is:
(5x + 2y) / (x + y) = 3.69 ...(Equation 2)
To solve this system of equations, we can use substitution or elimination.
Let's solve it using the substitution method:
From Equation 1, we have x = 160 - y. We can substitute this value of x into Equation 2:
(5(160 - y) + 2y) / (160 - y + y) = 3.69
(800 - 5y + 2y) / 160 = 3.69
(800 - 3y) / 160 = 3.69
Now, we can cross-multiply and solve for y:
800 - 3y = 160 * 3.69
800 - 3y = 590.4
Subtract 800 from both sides:
-3y = 590.4 - 800
-3y = -209.6
Divide by -3:
y = -209.6 / -3
y ≈ 69.87
Since we cannot have a fraction of a pound, we should round y to the nearest whole number:
y ≈ 70
Now, substitute the value of y back into Equation 1 to find x:
x + 70 = 160
x = 160 - 70
x = 90
Therefore, she should blend approximately 90 pounds of high-quality coffee beans and 70 pounds of cheaper coffee beans.