SOLUTION: Hi! I would like to know if you can help me with the following problem. Three kinds of coffee are to be mixed to get a blend. One kind costs $0.40 per pound, the second $0.80

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Hi! I would like to know if you can help me with the following problem. Three kinds of coffee are to be mixed to get a blend. One kind costs $0.40 per pound, the second $0.80       Log On


   

Question 26222: Hi!
I would like to know if you can help me with the following problem.
Three kinds of coffee are to be mixed to get a blend. One kind costs $0.40 per pound, the second $0.80 per pound and the third $1.00 per pound. 110 pounds of the mix is wanted and it should cost $0.60 per pound. The amount of coffee costing $0.80 per pound should be 2/3 of the amount at $0.40 per pound. How much of each kind of coffee should be used?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Let amount of 40 cent coffee be "x"
The amount of 80 cent coffee is (2/3)x
And amount of 100 cent coffee is (110-x-(2/3)x)=(110-(5/3)x)
Value of 40 cent coffee is 40x
Value of 80 cent coffee is 80(2/3)x
Value of 100cent coffee is 100(110-(5/3)x)
EQUATION:
Value of 40 + Value of 80 + Value of 100=60(110)cents
40x+80(2/3)x+100(110-(5/3)x)=60(110) cents
2x+4(2/3)x+5(110x)-(5/3))x=3(110) cents
6x+8x+1650-25x=990
-11x=-660
x=60 lbs. of 40cent coffee
(2/3)x=40 lb. of 80cent coffee
110-(5/3)x=110-(5/3)(60)=10 lbs. of $1.00 coffee
Cheers,
Stan H.