Question 105670This question is from textbook
: A coffee wholesaler blends together three types of coffee that sell for $2.20, $2.30, and $2.60 per pound, so as to obtain 100 lb of coffee worth $2.40 per pound. If the wholesaler uses the same amount of the two higher priced coffees, how much of each type must be used in the blend?
This question is from textbook
Answer by fmo(8)   (Show Source):
You can put this solution on YOUR website! Let a represent the quantity of the $2.20 type of coffee.
Let b represent the quantity of the $2.30 type of coffee.
Let c represent the quantity of the $2.60 type of coffee.
From the problem we know that the wholesaler is making 100 lbs of coffee, so:
a + b + c = 100
We also know that he's using the same amount of the $2.30 type as of the $2.60 type, or b = c. That being the case we can substitute b in for c in the above equation and we get:
a + b + b = 100
Then by combining like terms:
a + 2b = 100
We also know from the problem that by mixing these coffee types, the wholesaler arrives at a price of $2.40 per pound of the blend. He's making 100 lbs, so:
2.40 * 100 = 240
is the amount of money used to make the blend. The cost of each type of coffee used then, would be its price per pound multiplied by the quantity used or:
2.20a + 2.30b + 2.60b = 240
2.20a + 4.90b = 240 (after adding the like terms)
Again, we didn't write 2.60c because the quantity of types b and c are equal. Now we have a system of equations that we can use to eliminate one of the variables:
2.20a + 4.90b = 240 eq1
a + 2b = 100 eq2
Multiplying eq1 by 2.20 so that we can eliminate variable a:
2.20a + 4.90b = 240 eq1
2.20a + 4.40b = 220 eq2*2.20
Subtracting eq2*2.20 from eq1 we get:
0.50b = 20
b = 10 (after multiplying both sides of the equation by 0.50)
Substituting b = 10 back into the original equation eq2:
a + 2(10) = 100
a + 20 = 100
a = 80
Therefore, the wholesaler used 80 lbs of the $2.20 coffee, 10 lbs of $2.30 coffee, and 10 lbs of the $2.60 coffee.
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