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Question 1164056: A merchant has coffee worth $20 a pound that she wishes to mix with 90 pounds of coffee worth $80 a pound to get a mixture that can be sold for $40 a pound . How many pounds of the $20 coffee should be used ?
Found 3 solutions by Boreal, greenestamps, ikleyn:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! x pounds of $20 coffee and 90 pounds of $80 coffee=20x+$7200 . units are pounds*dollars/pound=dollars
20x+7200=(90+x)*$40, the sum of the first two equals the product of the mixture
20x+7200=3600+40x
3600=20x
x=180 pounds of $20 coffee ANSWER
the mixture will be 270 pounds *$40/pound=$10800
180*$20=$3600 and $80*90=$7200. That sum is $10800.
Answer by greenestamps(13216)  (Show Source):
You can put this solution on YOUR website!
A formal algebraic setup....
x pounds of coffee at $20 per pound, plus 90 pounds of coffee at $80 per pound, makes (90+x) pounds of coffee at $40 per pound:
Solve using basic algebra.
Informally, if formal algebra is not required....
The $40 per pound price of the mixture is "twice as close to $20" as it is to $80. (The difference between 20 and 40 is 20; the difference between 40 and 80 is 40. 20 is half of 40, so 40 is twice as close to 20 as it is to 80.)
Therefore, the mixture must contain twice as much of the $20 per pound coffee as it does the $80 per pound coffee. Since there are 90 pounds of the more expensive coffee, you need 180 pounds of the less expensive coffee.
ANSWER: 180 pounds
Obviously, that is the answer you should get if you finish the problem by the formal algebraic method.
Answer by ikleyn(52915)  (Show Source):
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