Candy mixture problem

Question

You have one type of candy that sells for $2.20/lb and another type of candy that sells for $9.60/lb. You would like to have 14.8 lbs of a candy mixture that sells for $6.00/lb. How much of each candy will you need to obtain the desired mixture?

Answer 1

c represents the amount of the cheaper

2.2c + (9.6)(14.8 - c) = (6)(14.8)

Colve for c

Answer 2

We have two variables or unknowns that we need to solve for, so we need two equations. First, we know that the total pounds of candy mixture is 14.8lbs which is equal to the sum of the pounds of candy x and candy y:

x + y = 14.8

Next, we know that the total cost of the candy mixture is equal to 14.8lbs times $6 per pound which is equal to the sum of the cost of candy x and candy y. Since we used x and y for pounds of candy x and y, the cost of candy x and y is x times $2.2 per pound and y times $9.6 per pound:

2.2x + 9.6y = 6(14.8)

Now that we have two equations, we can solve one of the equations for one of the variables in terms of the other variable:

x + y = 14.8

x = 14.8 - y

Then, we can substitute (14.8 - y) for x in the other equation:

2.2x + 9.6y = 6(14.8)

2.2(14.8 - y) + 9.6y = 6(14.8)

Solve for y in this equation. Then, to solve for pounds of candy x, substitute the value you got for y in the equation: x = 14.8 - y and solve for x.