SOLUTION: Please help me solve this equation:
A merchant wishes to mix two grades of coffee, one of which sells for $0.80 per pound and one of which sells for $1.20 per pound. He wa
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Question 165650: Please help me solve this equation:
A merchant wishes to mix two grades of coffee, one of which sells for $0.80 per pound and one of which sells for $1.20 per pound. He wants to sell the mixture $1.10 per pound. If he has 25 pounds of the $0.80 coffee, how much of the $1.20 per pound coffee must he add so that the value of the final mixture is equal to the total value of the other two?
Found 2 solutions by stanbon, checkley77:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A merchant wishes to mix two grades of coffee, one of which sells for $0.80 per pound and one of which sells for $1.20 per pound. He wants to sell the mixture $1.10 per pound. If he has 25 pounds of the $0.80 coffee, how much of the $1.20 per pound coffee must he add so that the value of the final mixture is equal to the total value of the other two?
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80 cent coffee DATA:
Amt. = 25 lbs. ; value = 80*25 = 2000 cents
--------------------------
120 cent coffee DATA:
Amt = x lbs. ; value = 120x cents
----------------------------
110 cent coffee DATA:
Amt = 25+x lbs value = 110(25+x)
-----------------------------
EQUATION:
value + value = value
2000 + 120x = 110(25+x)
2000 + 120x = 2750 + 110x
10x = 750
x = 75 lbs (amount of $1.20 coffee needed for the mixture)
==============================================================
Cheers,
Stan H.
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
.8*25+1.20x=1.10(25+x)
20+1.20x=27.5+1.10x
1.20x-1.10x=27.5-20
.10x=7.5
x=7.5/.1
x=75 pounds of $1.20 coffee is used.
Proof:
20+1.20*75=1.10(25+75)
20+90=1.10*100
110=110
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