okay so the ratio of men to women is 5:2

(1) the ratio or children to men is 5:11
(2) the number of women<30

ratio problems are all about factors, common denominators and multiples. Looking at the ratio of men to women, 5:2, we know that for every 5 men, there are 2 women. another way to put it is that out of every 7 adults (men and women), 5 will be men and 2 will be women.

now (1) tells us that the ratio of C to M is 5:11. Now, because we can't have half of a child (that would look bad for the captain of the cruise ship), we need to find a nice whole number.

so our combined info for (1) are the two ratios

M:W = 5:2
C:M = 5:11

we can also write these in fraction terms as

m/w = 5/2

c/m = 5/11

what we want to do is find the LCD (lowest common denominator) between the men in each ratio. because 5 and 11 share no factors, then that means that their lowest common multiple (i am fuzzy on the terminology, so forgive me is I misspeak) will be 55, or 5x11.

so the next step is..

m/w = 5(11)/2(11) = 55/22
c/m = 5(5)/11(5) = 25/55

This means that the # of people = 55m + 22w + 25c

because we don't know what c,m, or w are,(1) is insufficient.

(2). we know that w<30
this by itself is insufficient, because there could be 20 women and 50 men, or 10 women and 25 men.

so we look at our equation, and see that there is only one value of w that would give us a # of w<30. so both together will be sufficient.

As per the ques...

W:C=5:2

from 1-
c:m=5:11 clearly this alone cannot suffice

therefore we go further
w:m= (w:c)*(c:m)= (5:2)*(5:11)=25:22
that means if x=m+w

w=(25/47)x which means x=(47w)/25

and m= (22/47)x= (22/47)*(47w/25)=(22w/25)
for m to be an integer(no of men)w=multiples of 25

Now from 2
w<30 hence w=25 and m=22

ahh, i wrongly copied the problem down as M:W, my explanation is still sufficient in theory though...