SOLUTION: Trying to solve Algebra-1 Mixture problems. Some of the simple ones I understand, but I am stumped on this one:
Steven has 3 pounds of Brand 121 nuts that sell for $2.23 a pound
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Steven has 3 pounds of Brand 121 nuts that sell for $2.23 a pound
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Question 171548: Trying to solve Algebra-1 Mixture problems. Some of the simple ones I understand, but I am stumped on this one:
Steven has 3 pounds of Brand 121 nuts that sell for $2.23 a pound. He also has Brand 330 nuts that sell for $1.75 per pound. If he wants a mixture of nuts that sell for $1.93 a pound, how many pounds of Brand 330 should he add?
In following the previous examples they have given me, I build a table such as:
........$$$........LBS.......TOTAL
121.....2.23........3.........6.69
330.....1.75........X.........1.75X
MIX.....1.93.......3+X......1.93(3+X)
The answer is suppose to be 5 pounds, but I don't understand how to take the data from the table and get 5 pounds.
HELP!
and thanks! Found 2 solutions by stanbon, josmiceli:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Steven has 3 pounds of Brand 121 nuts that sell for $2.23 a pound. He also has Brand 330 nuts that sell for $1.75 per pound. If he wants a mixture of nuts that sell for $1.93 a pound, how many pounds of Brand 330 should he add?
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Value equation: 223*3 + 175x = 193(3+x)
669 + 175x = 579 + 193x
Subtract 579 from both sides;
Also, subtract 175x from both sides to get:
18x = 90
x = 5 lbs. (amt of Brand 330 nuts needed for the mixture)
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Cheers,
stan H.
You can put this solution on YOUR website! I take a little different approach. First I define the
cost per pound of the ingredients. Then I use those
definitions to define the cost of the mixture.
In words:
(price/pound of 121 nuts)(pounds of 121 nuts) +
(price/pound of 330 nuts)(pounds of 330 nuts) = price of mixture
Let = pounds of 330 nuts needed = pounds of mixture
5 pounds of 330 nuts are needed
If you understand this formula, you can always make
a table out of it
(