Need help with percent word problems.

• Feb 28th 2006, 12:24 AM
Ekdog
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?
• Feb 28th 2006, 02:02 AM
ticbol
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...)

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.
• Feb 28th 2006, 02:54 AM
Ekdog
: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%.
• Feb 28th 2006, 03:08 AM
ticbol
Quote:

Originally Posted by Ekdog
: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%.

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?