Who is good at math? HELPPP!!!

Sarie_Bronish
Sarie_Bronish Posts: 255 Member
in Health and Weight Loss
My friend weighs 187 pounds and is 38% body fat currently.

I know that 38% of 187 pounds = 71 pounds.
This would leave 116 pounds as LBM

She wants to know what she would weigh at 33% body fat. - but I apparently skipped class those days in math class because I cant figure this out after 3 sheets of paper, 100+ equations, and more answers than I care to admit. Help me please!!!

Replies

  • Sarie_Bronish
    Sarie_Bronish Posts: 255 Member
    cool points if you show your work :P
  • j6o4
    j6o4 Posts: 871 Member
    She wants to lose 5%, 38% - 33% = 5%
    187lbs x 0.05 = 9.35lb, this is how much she needs to lose.
    187lbs - 9.35lbs = 177.65lbs, this is how much she would be at 33% if she loses no muscle in the process.
  • Aaron_K123
    Aaron_K123 Posts: 7,122 Member
    If her lean mass is 116 pounds and she is at 33% bodyfat and you want to determine her weight X then:

    116 + 0.33 * X = X


    Solve for X. 57 pounds of fat, 173 pounds would be her bodyweight assuming no loss of lean mass.
  • EvanKeel
    EvanKeel Posts: 1,903 Member
    are you assuming she only lost fat? If so, it should be fairly easy to figure out a 5% reduction in weight..unless I'm missing something.
  • Aaron_K123
    Aaron_K123 Posts: 7,122 Member
    She wants to lose 5%, 38% - 33% = 5%
    187lbs x 0.05 = 9.35lb, this is how much she needs to lose.
    187lbs - 9.35lbs = 177.65lbs, this is how much she would be at 33% if she loses no muscle in the process.

    But if she is 177.65 pounds at 33% bodyfat her lean mass would be 177.65 - (.33 * 177.6) = 119 which would mean she would have put on 3 pounds of lean mass.

    Answer is 173
  • EvanKeel
    EvanKeel Posts: 1,903 Member
    She wants to lose 5%, 38% - 33% = 5%
    187lbs x 0.05 = 9.35lb, this is how much she needs to lose.
    187lbs - 9.35lbs = 177.65lbs, this is how much she would be at 33% if she loses no muscle in the process.

    But if she is 177.65 pounds at 33% bodyfat her lean mass would be 177.65 - (.33 * 177.6) = 119 which would mean she would have put on 3 pounds of lean mass.

    Fair

    EDIT: I guess alternatively one could say X= (116/0.67) if assume 116 is constant.
  • j6o4
    j6o4 Posts: 871 Member
    She wants to lose 5%, 38% - 33% = 5%
    187lbs x 0.05 = 9.35lb, this is how much she needs to lose.
    187lbs - 9.35lbs = 177.65lbs, this is how much she would be at 33% if she loses no muscle in the process.

    But if she is 177.65 pounds at 33% bodyfat her lean mass would be 177.65 - (.33 * 177.6) = 119 which would mean she would have put on 3 pounds of lean mass.

    Answer is 173

    hmm, guess i made a mistake.
  • brisingr86
    brisingr86 Posts: 1,789 Member
    If her lean mass is 116 pounds and she is at 33% bodyfat and you want to determine her weight X then:

    116 + 0.33 * X = X


    Solve for X. 57 pounds of fat, 173 pounds would be her bodyweight assuming no loss of lean mass.

    I did the math differently, but came up with the same answer.
    1. Assuming LBM should remain constant: lean body mass=weight * (1-%BF).
    2. 187*(1-0.38)=X* (1-0.33)-->solve for X
    3. 187*0.62=0.67X
    4. 115.94/0.67=X
    5. 173.04=X=weight at 33% body fat
  • yewbic
    yewbic Posts: 37
    x - 116 = x(.33)
    x - .33x = 116
    .67x = 116
    x = 173
  • Sarie_Bronish
    Sarie_Bronish Posts: 255 Member
    thank you - yes, I WAYYYYYYYYYYYYYYYYYYYY over thought this, and i shouldve stuck to my gut. I got all twisted up and *kitten* deep in numbers that didnt matter while trying to be accurate.
  • Aaron_K123
    Aaron_K123 Posts: 7,122 Member

    EDIT: I guess alternatively one could say X= (116/0.67) if assume 116 is constant.

    Yeah that would be the easy way really then you don't even have to solve for X in the formula I layed out. Of course if you do do the algebra you end up with X = (116/.67) as laid out in what yewbic did. Guess I never showed my work ;-)

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