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Problem Solving Problem

  1. Jul 19, 2005 #1
    Problem Solving.... Problem

    For the love of God I need some help.... Here is the question:

    How much water should be added to 20 gallons of a solution that is 50% antifreeze in order to get a mixture that is 30% antifreeze?


    I came up with two answers but am not sure if either are correct: 7.5 gallons and 12 gallons.

    Can anyone help me?
     
  2. jcsd
  3. Jul 19, 2005 #2

    Pengwuino

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    Gold Member

    Well think about it, dont be fooled by numbers and percentages :D.

    If theres 50% antifreeze in 20 gallons of solution... how many gallons of antifreeze is there? Then, how many gallons overall need to exist for that # of gallons to equal 30%?
     
  4. Jul 19, 2005 #3

    Hmm.... still having trouble... :confused:
     
  5. Jul 19, 2005 #4
    wow here is one where i actually think i can help

    Part A: heres a quick tip for figuring it out.

    (10gallons of antifreeze)/(20gallons of water-antifreeze solution) = a 50% solution.

    so if i wanted to make that 10 gallons of anti freeze only 30% of the solution i would have to add more water to the mixture. So you might have to add X number of gallons to the mixture. Make an equation that shows adding X number of gallons to the mixture and then make that equation equal to X.
     
  6. Jul 19, 2005 #5

    EnumaElish

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    Science Advisor
    Homework Helper

    Denote antifr. as AF and water as WR. So you have 10 AF + 10 WR = 20gal. mixture with 50% AF

    You want 10 AF + X WR = 33.3 gal mixture with 30% AF (we know this because 10/33.3 = 0.3)

    X = 33.3 - 10 = 23.3

    Need to add extra WR = 23.3 - 10 = 13.3 gal
     
  7. Jul 20, 2005 #6
    (10/0.3) - 20

    that's the answer!
     
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