Asked by peggy
A coffee merchant has coffee beans that sell for $9 per
pound and $12 per pound. The two types are to be mixed to create 100 lb of a mixture
that will sell for $11.25 per pound. How much of each type of bean should be
used in the mixture?
My problem is that I can't figure out the correct way to set this up.
This is how I am setting the problem up...then I am stuck.
900x+1200y=1125
-900(900x+1200)y=(-900)(1125)
-1,080,000y=-1,012,500
y=.9375
The final product will be 100 lbs selling for $1125. So 1125 = 9x + 12 (100-x). Where x is the number of pounds of the cheaper brand. You will then need (100 - x) pounds of the other brand to fill the mixture b/c x + (100-x) = 100. So solving 1125 = 9x + 12(100-x) leaves you with x = 25. So y (or 100-x) is 75.
pound and $12 per pound. The two types are to be mixed to create 100 lb of a mixture
that will sell for $11.25 per pound. How much of each type of bean should be
used in the mixture?
My problem is that I can't figure out the correct way to set this up.
This is how I am setting the problem up...then I am stuck.
900x+1200y=1125
-900(900x+1200)y=(-900)(1125)
-1,080,000y=-1,012,500
y=.9375
The final product will be 100 lbs selling for $1125. So 1125 = 9x + 12 (100-x). Where x is the number of pounds of the cheaper brand. You will then need (100 - x) pounds of the other brand to fill the mixture b/c x + (100-x) = 100. So solving 1125 = 9x + 12(100-x) leaves you with x = 25. So y (or 100-x) is 75.
Answers
There are no human answers yet.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.