3/5 of the men in a chemistry class have beards and 2/3 of the women have long hair; if there are 120 in the class and 46 are not in the above groups, how many men and how many women are there in the class?
This is how I solved it. x = number of men, 120 - x = number of women.
since 46 is not in the above group, the above group must equal 120-46 or 74.
So my equation is, 3/5(x) + 2/3(120-x) = 74
3/5(x)+ 80 - 2/3(x) = 74
-1/15(x) +80 = 74
6 = 1/15(x)
x = 90
120-x = 30
Is that the right answer and did I solve it the correct way? Also what type of word problem would this be considered as?
thanks
This is how I solved it. x = number of men, 120 - x = number of women.
since 46 is not in the above group, the above group must equal 120-46 or 74.
So my equation is, 3/5(x) + 2/3(120-x) = 74
3/5(x)+ 80 - 2/3(x) = 74
-1/15(x) +80 = 74
6 = 1/15(x)
x = 90
120-x = 30
Is that the right answer and did I solve it the correct way? Also what type of word problem would this be considered as?
thanks