Help with a math problem

Discussion in 'Basic Math' started by Rozza, Nov 2, 2021.

  1. Rozza

    Rozza

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    Sorry if this is in the wrong place. Just trying to find an answer somewhere.


    I’m having a problem with scaling up ratios, well I think that's the problem. The problem is regarding the dosage of medicine.


    The medicine is 1ml = 300mg, I would like to know the formula to work out how much i need to increase or decrease the amount of ml to get a different amount of mg from a dosage. For example if I want 500mg or 200mg what millilitres of the medicine do I need?

    Any help will be greatly appreciated.
     
    Rozza, Nov 2, 2021
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  2. Rozza

    MathLover1

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    medicine is 1ml = 300mg

    proportion

    1ml .................300mg
    xml..................200mg

    1/xml=300/200
    1/xml=3/2
    3xml=2
    xml=2/3 (200mg is (2/3)(300mg))



    1ml .................300mg
    xml..................500mg

    1/xml=300/500
    1/xml=3/5
    3xml=5
    xml=5/3 (500mg is (5/3)(300mg))
     
    MathLover1, Nov 2, 2021
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  3. Rozza

    Rozza

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    Thank you very much for replying!

    Honestly from your message I didn’t understand it, however I googled proportion and that was the key word I needed.
    Now I believe I understand it.

    Direct proportion

    1ml = 1000 (Cubic Millimeter)

    X is 300(mg) - y is 1000 (Cubic Millimeter)

    X 500 - y ?

    500x1000 = 500,000

    500,000 divide by 300 = 1666.66667 back into ml will be 1.666.

    500mg - 1.66ml.

    I think that's right.
     
    Rozza, Nov 2, 2021
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    nycmathguy likes this.
  4. Rozza

    MathLover1

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    yes, (5/3) =1.666666666666667 or 1.67
     
    MathLover1, Nov 2, 2021
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    Rozza and nycmathguy like this.
  5. Rozza

    nycmathguy

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    I like the way you reasoned your way to the answer.
     
    nycmathguy, Nov 2, 2021
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  6. Rozza

    nycmathguy

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    Cool. I like the way Rozza reasoned his way or her way to the answer. This is what I want to see---more interaction between members.
     
    nycmathguy, Nov 2, 2021
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  7. Rozza

    Rozza

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    Thank you very much for your help! I greatly appreciated it.
     
    Rozza, Nov 3, 2021
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    MathLover1 likes this.
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