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Math Help - Mixture/Solution Problem.

  1. #1
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    Mixture/Solution Problem.

    Q. How many litres of a solution that is 15% salt must be added to 5 litres of a solution that is 8% salt so that the resulting solution is 10% Salt.

    ok so we have 0.15x + 5(0.08) = 0.08 (x+5).

    I completely understand all the above, until the (x + 5). Why do we need to multiply the 8% by x + 5?

    Thanks a million.
    Last edited by dumluck; January 13th 2011 at 01:58 AM.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Well, in chemistry, you deal with moles which maybe makes it easier to understand, but I'll talk about mass here as an analogy to help you.

    Let 1 L of solution has a mass of 1 kg.

    Let's say the concentrations given are concentrations by mass, meaning that 15% is similar to saying 15% by mass of the 1L solution is salt. (ie, 0.15 kg of 1 kg of solution is salt)

    0.15x will give the total mass of salt in x litres of 15% solution.

    Similarly,

    8% is similar to saying 8% by mass of the 1L solution is salt. (ie, 0.08 kg of 1 kg of solution is salt)

    0.08(5) will give the total mass of salt in 5 litres of 8% solution.

    As such,

    10% is similar to saying 10% by mass of the 1L solution is salt. (ie, 0.1 kg of 1 kg of solution is salt)

    0.1(y) will give the total mass of salt in y litres of 10% solution.

    where y is the sum of volumes of both solutions.


    This gives a sum of masses:

    0.15x + 0.08(5) = 0.1(x + 5)

    Note that it's not 0.18. This must have been a mistake, because two more dilute solutions cannot give a more concentrated one.
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  3. #3
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    Thanks,
    So to convert this to laymans terms....

    1: Sol A: is 15x because we don't know how many litres
    2. Sol B: is 5(0.08) because we know that Sol B is 8% salt in 5 litres
    3. 0.10: Is the resulting amount of salt we want
    4. (x + 5) is the number of litres that contain the 10% of salt

    Is that correct?

    So to take another example (bare with me ).

    Q. Mixture A is 15% alcohol, Mixture B is 50% Alcohol. If the two are poured together to create a 4-gallon mixture that contains 30% alcohol. How may gallons of mixture A are in the mixture.

    1. 15x ( As we have 15% alcogol but don't know how many gallons of this mixture will be needed)
    2. 50 (As we know Mixture B is 50% Alcohol).
    3. (4-X): As we want the 50 to only account for Mix Bs solution, so we need to remove A's which is the total 4 minus the volume of Mix As.
    4. = 30*4 (the alcohol to volume ratio was want).

    Leaving: 15x + 50(4-x) = 30*4.

    Is my reasoning correct here?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    1. Yes, that's it

    2. Yes, good job! Well done
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