Results 1 to 4 of 4

Math Help - Need help with percent word problems.

  1. #1
    Newbie
    Joined
    Feb 2006
    Posts
    2

    Question Need help with percent word problems.

    Help! This problem has me in a muddle.

    In the Chamber of Commerce, 66 2/3% (sixty-six and two thirds percent) of the members are women and 200 of the members are men. How many Chamber of Commerce members are there in all?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by Ekdog
    Help! This problem has me in a muddle.

    In the Chamber of Commerce, 66 2/3% (sixty-six and two thirds percent) of the members are women and 200 of the members are men. How many Chamber of Commerce members are there in all?
    66 2/3 percent of x = Women
    So, (As long as all non-women are men),
    100% -(66 2/3)5% = (33 1/3)% of x are Men.

    Then,
    (33 1/3)% of x = 200
    So,
    (0.3333333...)x = 200
    x = 200 /(0.33333...)
    x = 600 members ---------------answer.

    Check,
    If 200 of the 600 members are Men, then (assuming again that all non-men are women) the women must be 600-200 = 400 in numbers.
    Is 400/600 equal to 66 2/3 percent?

    What is 66 2/3 percent?
    Make that improper fraction into proper fraction.
    66 2/3
    = (66*3 +2)/3
    = (198 +2)/3
    = 200/3

    So, 66 2/3 percent is (200/3)/100 = 2/3

    Then is 400/600 = 22 2/3 percent = 2/3?
    400/600
    = 4/6
    = 2/3 ---------yes, so, OK.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2006
    Posts
    2


    "66 2/3 percent of x = Women
    So, (As long as all non-women are men),
    100% -(66 2/3)5% = (33 1/3)% of x are Men."

    Thanks for helping, Ticbol, but I'm afraid I don't understand where you got the 5%.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by Ekdog


    "66 2/3 percent of x = Women
    So, (As long as all non-women are men),
    100% -(66 2/3)5% = (33 1/3)% of x are Men."

    Thanks for helping, Ticbol, but I'm afraid I don't understand where you got the 5%.
    Oh, Zeez, that 5% after the (66 2/3). It is a typo. I am good at that---making typographical errors. And most of ther time I fail to correct them.

    Okay, now, suppose that "5" were not there, that the (66 2/3)5% were (66 2/3)% only, could understand my solution?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Percent word problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 15th 2010, 02:27 AM
  2. percent word problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 28th 2009, 03:10 PM
  3. Percent word problem help
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 1st 2009, 04:00 PM
  4. Percent Word Problems
    Posted in the Algebra Forum
    Replies: 7
    Last Post: April 11th 2008, 04:00 PM
  5. Percent word problem
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 25th 2007, 03:17 AM

Search Tags


/mathhelpforum @mathhelpforum