Is my answer correct to this homework involving sets? Of a group of people, each person is wearing green, blue, or both. One-fifth of those wearing green are also wearing blue. One-eighth of those wearing blue are also wearing green. Are more than one-third of this group wearing green?
I decided to solve it by picking an arbitrary number of people wearing only green. Let $g$ and $b$ be the number of people wearing green and blue respectively.
Let $G$ and $B$ be the sets of people wearing green, and people wearing blue respectively.
$|G \cap B|$ = $\frac{g}{5} = \frac{b}{8}$
$g = \frac{5b}{8}$
Does this show that the number of people wearing green is less than the number of people wearing blue? And if more than a third of people are wearing green then that implies $g > \frac{b}{2}$ which is true, therefore more than one-third of this group is wearing green?
 A: Think of it this way. Suppose one person is wearing both blue and green. How many people are wearing blue then? And how many people are wearing green? So how many total are there?
Well, if one person is wearing both, then five are wearing green. Why? one-fifth of those wearing green are also wearing blue. In other words, one-fifth of those wearing green are one person. Similarly, eight are wearing blue. So in all there are 5+8-1=12 people. Finishing the question is straightforward from here but I trust you've figured it out by now.
A: Let $G$ be the number wearing green, and $B$ the number wearing blue.  We could also define $g$ and $b$ as you did, but that may be introducing an unnecessarily large number of variables.
The number wearing green and blue is $G/5$. It is also $B/8$. So $8G=5B$.  Let $G=5k$. Then $B=8k$, and the number wearing both is $k$, It follows that the group has $12k$ people.   Now it should be easy to answer the question.
Another way: Let the number wearing both be $x$. Then $4x$ wear green only, and $7x$ wear blue only, for a total of $12x$, of whom $5x$ wear green.
