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?

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- February 28th 2006, 12:24 AMEkdogNeed 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? - February 28th 2006, 02:02 AMticbolQuote:

Originally Posted by**Ekdog**

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. - February 28th 2006, 02:54 AMEkdog
:confused:

"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%. - February 28th 2006, 03:08 AMticbolQuote:

Originally Posted by**Ekdog**

Okay, now, suppose that "5" were not there, that the (66 2/3)5% were (66 2/3)% only, could understand my solution?