SOLUTION: A dealer mixed coffee worth 85 cents per pound with cofee worth 55 cents per pound. How many pounds of each kind did he use to make a mixture of 120 pounds to sell at 75 cents per
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: A dealer mixed coffee worth 85 cents per pound with cofee worth 55 cents per pound. How many pounds of each kind did he use to make a mixture of 120 pounds to sell at 75 cents per
Log On
Question 539518: A dealer mixed coffee worth 85 cents per pound with cofee worth 55 cents per pound. How many pounds of each kind did he use to make a mixture of 120 pounds to sell at 75 cents per pound?
please include explanation (step by step) im so confused! Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A dealer mixed coffee worth 85 cents per pound with coffee worth 55 cents per pound.
How many pounds of each kind did he use to make a mixture of 120 pounds to sell at 75 cents per pound?
:
Let x = amt of 85 cent coffee required
the total is to be 120, therefore
(120-x) = amt of 55 cent coffee
:
A typical mixture equation, we can solve it using cents
:
85x + 55(120-x) = 75(120)
85x + 6600 - 55x = 9000
85x - 55x = 9000 - 6600
30x = 2400
x = 2400/30
x = 80 lb of 85 cent coffee
then
120 - 80 = 40 lb of 55 cent coffee
:
:
Confirm our solution in the original mixture equation
85(80) + 55(40) = 75(120)
6800 + 2200 = 9000
:
Was this step-by-step enough for you, any questions on this?
