Question 283127
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.