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Question 1123204: A committee is to have 3 more women than men and the number on the committee must be at least 7 but no more than 15. What are the possible numbers of women on the committee?
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Start with 5W 2M
6W3M
7W4M
8W5M
9W6M
x=women, y=men
7<=x+y<=15, x=y+3
7<=2y+3<=15
4<=2y<=12
2<=y<=6
therefore, the women (y+3) can be 5,6,7,8,9 members.
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
(1) Algebraically....
We are asked to find the possible numbers of women on the committee, so use that as our variable; the number of men will be 3 less:
let x = number of women
then x-3 = number of men
The total number of people on the committee -- x, plus (x-3) -- is to be between 7 and 15:
7 <= x+(x-3) <= 15
7 <= 2x-3 <= 15
10 <= 2x <= 18
5 <= x <= 9
The number of women can be between 5 and 9.
Using logical reasoning....
Not counting the 3 "extra" women, the committee consists of equal numbers of men and women, and the total number on the committee is between 4 and 12; that means the numbers of men and women are between 4/2=2 and 12/2=6 each. And then counting the other 3 women, the number of women is between 2+3=5 and 6+3=9.
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