Algebra question

Question

A coffee distributor needs to mix a(n) Queen City coffee blend that normally sells for $8.30 per pound with a House coffee blend that normally sells for $9.00 per pound to create 70 pounds of a coffee that can sell for $8.93 per pound. How many pounds of each kind of coffee should they mix?

Answer 1

Let us consider, a = pounds of $8.30 Queen city coffee, and b = pounds of $9.00 House coffee,

then a + b = 70......(1), and another equation will be (8.30*a + 9.00*b)/70 = 8.93........(2).

Now, we have to solve these equations to find the values of a and b. Lets do it!!

From equation (1), we get that a = 70 - b, let us put this value in equation (2) and solve for b, we get

(8.30(70-b) +9.00b)/70 = 8.93

=> 581 - 8.30b + 9.00b = 8.93*70 = 625.10

=> 0.7b = 625.10 - 581 = 44.1

=> b = 44.1/0.7 = 63.

Put the value of b in a = 70 - b to get the value of a, we get a = 70 - 63 = 7.

Therefore, a = 7 and b = 63. Therefore, we get 7 pounds of Queen City coffee and 63 pounds of House coffee. This is the required answer for this problem.

Are we correct or wrong? Let us check it by plug-in the values of a and b obtained in equations (1) and (2),

a + b = 7 + 63 = 70 (So we are correct), and (8.30*a + 9.00*b)/70 = (58.1 + 567)/70 = 8.93 (Therefore, this also matches and therefore we are good).

Answer 2

to find how many pounds of each coffee they should buy, we first need to work with the variables that are all given. that means working with $. So the total amount of dollars for the end goal is $8.93 times 70lbs = $625.10 worth of coffee. So the equation to find out how much of each coffee to buy looks like:

8.30x+9.00y=625.10. And since we know how much the end weight needs to be, x+y=70. From here, you just substitute and plug in. So we can say y=70-x, and plug that it to the other equation to be 8.30x+9.00(70-x)=625.10 . Solve and you get 8.30x+630.00-9.00x=625.10 . solve this out and you get x=7. plug this in to get what y is, and y=63.