Question 505697
With mixture problems you always have to keep track of how much 'pure' stuff you have.
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At present you have 100 gal of 2% solution.
2%(100) = .02*100 = 2 gallons of pure bleach added to 98 gallons of water.
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What you need is 5 gallons of bleach added to 95 gallons of water.
But you cannot do that because you have a tank full.
So, you have to drain the same amount of solution as you add.
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Let 'x' be the amount you add, which = the amount you drain.
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2%(100) + 100%(x) - 2%(x) = 5%(100)
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.02(100) + 1(x) - .02(x) = .05(100)
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2 + x - .02x = 5
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.98x = 3
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98x = 300
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x = 300/98
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x = 3.0612
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Always check your solution to see if it is truly the answer.
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You plan to drain 3.06 gallons of the 2% solution and add 3.06 gallons of pure bleach.
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Would that be a 5% solution?
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The amount you drain would take some of the pure stuff with it.
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.02*(3.0612) = .0612 gal of pure stuff would be removed from the 2.0 gallons you had in total.
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So you would still have 1.9388 gal of pure stuff in the tank.
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Then you add 3.0612 gallons of pure stuff, which fills the tank again.
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1.9388 + 3.0612 = 5 gallons of pure stuff in the 100 gal tank.
That's a 5% solution.
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Answer:
Drain 3.0612 of 2% solution from the 100 gallon tank and add 3.0612 gallons of pure bleach to produce 100 gallons of 5% solution.
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Done.