SOLUTION: I am having difficulty setting up this problem. I have tried several different ways but can not get the same answer as the book. Here is the problem.
A Merchant blends tea t
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A Merchant blends tea t
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Question 354203: I am having difficulty setting up this problem. I have tried several different ways but can not get the same answer as the book. Here is the problem.
A Merchant blends tea that sells for $3.00 a pound with tea that sells for $2.75 a pound to produce 80 lbs of a mixture that sells for $2.90 a pound. How many pounds of each type of tea does the merchant use in the blend.
I thought that if i multiplied $2.90 by 80 that would give me the price for the total amount of 80 lbs. I used the x variable to represent the unknown lbs.
Here is my equation 232(price of total 80 lbs)= $3.00x + $2.75x I thought this would work but it didn't. Here are the answers the book has 48 lbs of the $3.00 tea and 32 lbs of the $2.75 tea
I would greatly appreciate it if you could explain this thank you. Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website! You are on the right track...sort of!
Let x = the number of lbs of the $3.00-tea, then (80-x) = the number of lbs of the $2.75-tea. The sum of these two amounts (x+(80-x)) are to equal 80 lbs of $2.90-tea blend. So we can set up the necessary equation to find x.
$3.00(x)+$2.75(80-x) = $2.90(80) We'll drop the $ sign for convenience.
3x+2.75(80-x) = 2.9(80) Simplify.
3x+220-2.75x = 232 Combine like-terms.
0.25x+220 = 232 Subtract 220 from both sides.
0.25x = 12 Divide both sides by 0.25
x = 48 and 80-x = 32
The merchant will need to mix 48 lbs of the $3.00-tea with 32 lbs of the $2.75-tea to obtain 80 lbs of $2.90 tea blend.
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