SOLUTION: A farmer wants to mix a liquid fertilizer that contains 2% nitrogen with one that contains 10% nitrogen to obtain 40 gallons of a fertilizer that contains 8% nitrogen. How many gal

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Question 283127: A farmer wants to mix a liquid fertilizer that contains 2% nitrogen with one that contains 10% nitrogen to obtain 40 gallons of a fertilizer that contains 8% nitrogen. How many gallons of each fertilizer should be used?
Found 2 solutions by richwmiller, oberobic:
Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website!
.02x+.10y=40*.08
x+y=40
x=10 y=30

Answer by oberobic(2304) (Show Source):
You can put this solution on YOUR website!
When solving solution problems, you need to keep track of how much 'pure' stuff you need. In this case the problem states you need 40 gal. of 8% nitrogen fertilizer: 40*.08 = 3.2 gal. 'pure' nitrogen.
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The materials on hand include 2% and 10% nitrogen in liquid form, so we assume we have as much of these two as we may need.
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Since the final amount needed is 40 gal., we can define the two mixtures in terms of the variable 'x'.
x = amount of 2% nitrogen fertilizer
40 - x = amount of 10% fertilizer.
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The objective equation we need to solve thus becomes:
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.02x + .10(40-x) = .08*40 = 3.20 gal.
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multiplying through by 100 will remove the decimals
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2x + 10(40-x) = 320
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expanding
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2x + 400 - 10x = 320
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collecting
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-8x = 320 - 400 = -80
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divide both sides by -8
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x = -80/-8 = 80/8 = 10
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Since we need 10 gal. of 2% solution, we need 40-10 = 30 gal of 10%.
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40-x = 30
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Therefore, we think that we need 10 gal. of 2% and 30 gal. of 10%.
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But we always need to check our answer...
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How much 'pure' stuff do we have in the 40 gallons we propose?
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10 gal. @ 2% = .02*10 = 0.2 gal.
30 gal. @ 10% = .10*30 = 3 gal.
0.2+3 = 3.2 gal.
Correct
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Answer:
Mix 10 gallons of 2% and 30 gallons of 10% nitrogen fertilizer to produce 40 gallons of 8% solution.
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Done.