solving a value mixture problem using a linear equation

Question

Miguel's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Miguel $5.65 per pound, and type B coffee costs $4.35 per pound. This month's blend used three times as many pounds of type B coffee as type A, for a total cost of $374.00. How many pounds of type A coffee were used?

Answer 1

To start, denote x as the number of lbs utilized per each coffee type (type A and type B). Further, let ax be the quantity of lbs for type A coffee and bx be the quantity of lbs for type B coffee. So, ax and bx are in units of quantity*lbs.

The cost of type A and type B coffee can be equated as follows:

Cost (A) = 5.65/lb and Cost (B) = 4.35/lb

The key information given ("This month's blend used three times as many pounds of type B coffee as type A, for a total cost of $374.00") can be represented algebraically as follows:

Cost (Blend) = Cost(A)*ax + Cost(B)*bx = $374.00. Since the quantity of type B coffee is 3 times the quantity of type A coffee, it follows that a = 1 and b = 3.

Therefore, 5.65/lb*(x) + 4.35/lb*(3x) = 374--> 5.65/lb*(x) + 13.05/lb*(x) = 374 or (5.65+13.05)(x) = 374.

Therefore, quantity (lbs) = 374/(13.05+5.65) = 374/(18.7) = 20.

Using the previous relation,we know that 3x + x = 20 or 4x = 20. So, the number of lbs for type A coffee utilized in the blend equates to the following:

x = 5 lbs.

Answer 2

Hi Alyssa,

To solve this problem we need to set up a set of linear equations for both cost and weight.

Setup:

I'm going to define:

C_A: Cost of A

C_B: Cost of B

C_T: Total Cost

W_A: Weight of A

W_B: Weight of B

Cost Equations:

The costs for each coffee are the product of their cost/weight in $/lb and their weight in lb:

C_A $ = (5.65 $/lb) * (W_A lb)

C_B $ = (4.35 $/lb) * (W_B lb)

C_T = C_A + C_B = 5.65*W_A + 4.35*W_B

Weight Equations:

The proportion of the weight of the objects was given to us:

3W_B = W_A

For any problem where you are given a proportion rewriting it as a ratio will save you a boatload of time:

W_B/W_A = 3/1

This ratio allows us to use a neat trick. Multiply the ratio by x/x:

W_B/W_A = 3x/1x

Thus,

W_B = 3x and W_A = x

Solving the linear equations.

We can now plug W_A and W_B into our formula for cost that we created above:

C_T = 5.65*W_A + 4.35*W_B = 5.65*x + 4.35*3x

Simplifying gives us

C_T = 10x

We know that C_T = $374

374 = 10x

x = 37.4

Now that we know x we can solve for W_A and W_B.

W_B = 3*37.4 = 112.2 lb

W_A = x = 37.4 lb

Had we not used the ratio method this would have potentially been much more complex. If you get used to using this technique now, it will save you a large amount of time in the future.

Best of luck,

Bryce