Question 211342: Juan has coffee that sells for $9 a pound and a coffee selling for $4 a pound. How many pounds of each must be mixed to get 20 pounds of a coffee worth $8.25 a pound?
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let x = the number of pounds of coffee at $9.00 per pound, then (20-x) = the number of pounds of coffee at $4.00 per pound. The sum of these two amounts is to equal 20 pounds at $8.25 per pound.
The cost of the x pounds of coffee can be expressed as:
$9(x) and, similarly, the cost of the (20-x) pounds can be expressed as:
$4(20-x) and the sum of these will equal $8.25(20).
Here's the equation:
9x+4(20-x) = 8.25(20)
9x+80-4x = 165
5x+80 = 165
5x = 85
x= = 17 and 20-x = 3.
Juan will to mix 17 pounds of coffee at $9.00 per pound with 3 pounds of coffee at $4.00 per pound to obtain a 20-pound mixture at $8.25 pound.
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