Thread: 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?

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.

"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%.

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?