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    #1
    How much pure disinfectant must be added to 30 gallons of an 8% solution to increase its strength by 25%?

    Let x = pure disinfectant to be added

    The word PURE tells me that 100 will be in the equation somewhere.

    30 gallons of 8 percent = 0.08(30)

    Must be added to some unknown = plus x

    My equation is 0.08(30) + x = 0.25(x + 100)

    Correct? If not, can someone break this mixture problem step by step leading to the right equation?

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    #2
    The LHS of your equation is correct, however for the RHS, we want the concentration to be increase BY 25%, not TO 25%, which means instead of 8%, we want 1.25*8% = 10%. Also, the amount of the final solution will be x + 30, not x + 100...so the correct equation, at least for the way I am interpreting the problem, is:

    0.08(30) + x = 0.1(x + 30)

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    #3 Thread Author
    0.08(30) + x = 0.1(x + 30)

    2.4 + x = 0.1x + 3

    x - 0.1x = 2.4 + 3

    0.9x = 5.4

    x = 5.4/0.9

    x = 6

    So, 6 pure disinfectant must be added.

    Correct?

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    #4
    Quote Originally Posted by RTCNTC View Post
    0.08(30) + x = 0.1(x + 30)

    2.4 + x = 0.1x + 3

    x - 0.1x = 2.4 + 3
    You've subtracted 2.4 from the LHS, but added 2.4 to the RHS...

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    #5 Thread Author
    0.08(30) + x = 0.1(x + 30)

    2.4 + x = 0.1x + 3

    x - 0.1x = -2.4 + 3

    0.9x = 0.6

    x = 0.6/0.9

    x = 0.66

    Right?

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    #6
    Quote Originally Posted by RTCNTC View Post
    0.08(30) + x = 0.1(x + 30)

    2.4 + x = 0.1x + 3

    x - 0.1x = -2.4 + 3

    0.9x = 0.6

    x = 0.6/0.9

    x = 0.66

    Right?
    If you are going to round to 2 decimal places then x ≈ 0.67 gal., otherwise x = 2/3 gal.

    x = 0.6/0.9 = 6/9 = 2/3

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    #7 Thread Author
    Quote Originally Posted by MarkFL View Post
    If you are going to round to 2 decimal places then x ≈ 0.67 gal., otherwise x = 2/3 gal.

    x = 0.6/0.9 = 6/9 = 2/3
    Great as always.

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