SOLUTION: Avis Proctor works at a gourmet delicatessen. She is preparing a cheese tray for a large reception. She is using some cheeses that sell for $8 per lb and others that sell for $12 p

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Question 184857: Avis Proctor works at a gourmet delicatessen. She is preparing a cheese tray for a large reception. She is using some cheeses that sell for $8 per lb and others that sell for $12 per lb. How many pounds of cheese at each price should she use in the oreder for the mixed cheeses on the tray to wigh of 56 lb and sell for $10.50 per lb?
Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Let x be the amount of the cheap cheese. Then 8x is the price of all the cheap cheese.

Since there are 56 pounds of cheese total, the amount of expensive cheese must be 56 - x. The price of all the expensive cheese is then .

The price of the entire 56 pounds of mixture is

Since there is nothing else in the mixture, the price of all the cheap cheese plus the price of all the expensive cheese must equal the price of the mixture:



Solve for x to get the amount of $8 cheese, and then subtract that amount from 56 to get the amount of $12 cheese.

I'll have a glass of wine with that, thank you.

John

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Avis Proctor works at a gourmet delicatessen. She is preparing a cheese tray for a large reception. She is using some cheeses that sell for $8 per lb and others that sell for $12 per lb. How many pounds of cheese at each price should she use in the order for the mixed cheeses on the tray to weigh 56 lb and sell for $10.50 per lb?
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Quantity Equation: x + y = 56
Value Equation...:8x + 12y = 10.5*56
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Modify to prepare for elimination:
8x + 8y = 8*56
8x + 12y = 10.5*56
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Subtract 1st from 2nd to solve for "y":
4y = 2.5*56
y = 2.5*14
y = 35 (# of lbs of $12 cheese in the mixture)
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Substitute into x+y = 56 to solve for "x":
x + 35 = 56
x = 21 (# of lbs of $8 cheese in the mixtue)
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cheers,
Stan H.