Question 958961: I cant figure out how to work this problem out and how they came up with the answer of 60lbs. A merchant has coffee worth $40 a pound that she wishes to mix with 20 pounds of coffee worth $80 a pound to get a mixture that can be sold for $50 a pound. How many pounds of the $40 coffee should be used?
Found 3 solutions by josgarithmetic, macston, MathTherapy:
Answer by josgarithmetic(39615)   (Show Source):
Answer by macston(5194)  (Show Source):
You can put this solution on YOUR website! F=pounds of $40:
$40F+$80(20lbs)=$50(20lbs+F)
$40F+$1600=$1000+$50F Subtract $40F from each side.
$1600=$1000+$10F Subtract $1000 from each side.
$600=$10F Divide each side by $10
60=F ANSWER: She needs 60 pounds of the $40 coffee.
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website!
I cant figure out how to work this problem out and how they came up with the answer of 60lbs. A merchant has coffee worth $40 a pound that she wishes to mix with 20 pounds of coffee worth $80 a pound to get a mixture that can be sold for $50 a pound. How many pounds of the $40 coffee should be used?
Let amount of $40-per lb coffee to mix, be F
Then total cost of the $40-per lb coffee = 40F
Total cost of the $80-per lb coffee: 20(80), or $1,600
Total cost of entire mixture: 40F + 1,600, and total weight is: F + 20
We then get: 
50(F + 20) = 40F + 1.600 ------ Cross-multiplying
50F + 1,000 = 40F + 1,600
50F - 40F = 1,600 - 1,000
10F = 600
F, or amount of $40-per lb coffee to mix =  , or  lbs
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