Sarah Meeham blends coffee for Tasti-Delight. She needs to prepare 150 pounds of blended coffee beans selling for $4.45 per pound. She plans to do this by blending together a high-quality bean costing $5.25per pound and a cheaper bean at $3.25 per pound. To the nearest pound, find how much high-quality coffee bean and how much cheaper coffee bean she should blend.
c + h = 150
... 325 c + 325 h = 325 * 150
325 c + 525 h = 445 * 150
subtract equations to eliminate c
... then substitute back to find h
A chemist wants to create a mixture that is 40% (0.40) of a certain chemical. She wants to create 50 liters. She has a stock of 25% (0.25) concentration and a stock of 60% (0.60) concentration. How much of each type should she use?
same problem:
If there are x liters of 25%, then the rest (50-x) is 60%. So,
.25x + .60(50-x) = .40 * 50
To find the amount of high-quality coffee beans and the amount of cheaper coffee beans that Sarah should blend, we can set up a system of equations.
Let's assume Sarah needs to blend x pounds of high-quality coffee beans and y pounds of cheaper coffee beans.
The total weight of coffee beans should add up to 150 pounds:
x + y = 150 .................(Equation 1)
The total cost of the blended coffee beans can be calculated by multiplying the weight of each type of coffee bean by its respective cost per pound and adding them together.
The cost equation is given by:
5.25x + 3.25y = 4.45(150) ...........(Equation 2)
Now, we can solve this system of equations using substitution or elimination.
Using substitution:
From Equation 1, we can rewrite it as: y = 150 - x
Now substitute the value of y in Equation 2:
5.25x + 3.25(150 - x) = 4.45(150)
Simplifying the equation:
5.25x + 487.5 - 3.25x = 667.5
Combining like terms:
2x + 487.5 = 667.5
Subtracting 487.5 from both sides:
2x = 180
Dividing by 2:
x = 90
Now substituting the value of x in Equation 1:
90 + y = 150
Subtracting 90 from both sides:
y = 60
Therefore, Sarah needs to blend 90 pounds of high-quality coffee beans and 60 pounds of cheaper coffee beans.