There are a couple of ways to approach the problem; I’ll start with a very straightforward one from first principles. First, here’s a picture:
The blue part of the diagram represents those students who do not participate in soccer, basketball, or track. We’re interested in $x$, the percentage who play all three sports. One way to solve the problem is to fill in the Venn diagram in terms of $x$. I’ve started: we know, for instance, that $2\%$ play soccer and basketball, and $x\%$ play all three sports, so $2-x$ must be the percentage playing soccer and basketball but not track. The figures $6-x$ and $9-x$ were obtained by similar reasoning. Now go after the percentage playing just soccer, just basketball, and just track. We know that $21\%$ play soccer, and $(2-x)+x+(6-x)$ percent play soccer and at least one other of the three sports, so what percentage (in terms of $x$) play just soccer? Repeat this procedure with basketball and track to fill in the last three regions of the Venn diagram. Finally, add up the percentages of all of the white regions; that total, which will involve $x$, must be the $100-36$ percent who play at least one of the three sports, and you’ll have a simple equation to solve for $x$.
If you know something about calculating the cardinalities of unions and intersections of sets, you can short-circuit this calculation. Let $S,B$, and $T$ be the sets of students playing soccer, basketball, and track, respectively. Pretend that there are $100$ students altogether, so that percentages are simply numbers of students. Then you want $|S\cap B\cap T|$, and you know $|S|$, $|B|$, $|T|$, $|S\cap B|$, $|S\cap T|$, $|B\cap T|$, and the cardinality of the complement of $S\cup B\cup T$. That last figure gives you $|S\cup B\cup T|$. Now, do you know a formula that relates all of these quantities? If so, you’re in business, because you know all but one of them.
Added: However, the problem is faulty, because if you actually carry out the computations, which I unfortunately did not bother to do when I originally posted the answer, you’ll find that $x$ is negative!
You can express the various overlaps with $3$ circles, since there are three states for each of the companies (elements of $C$):
- Either it is in retail (an element of $R$) or it is not.
- Either it is quoted in the stock exchange (an element of $Q$) or it is not.
- Either it failed (is an element of $F$) or it is not.
Added: To start with, we have the following diagram:
We want to find that the maximum and minimum for that unknown value $x$. Since $5$ of the $90$ failed companies were not in retail, then $90-5=85$ were in retail. That is, of the studied companies, there were $85$ that failed and were retail companies. This tells us that $0\le x\le 85$ and that:
The two unfilled regions of $F$ must add up to $5$ (since $5$ of the failed companies were not in retail). Likewise, since $85$ of the $140$ retail companies failed, then $140-85=55$ of them did not, so the two unfilled regions of $R$ must add up to $55$. Hence:
where $0\le y\le 5$ and $0\le z\le 55.$ Now, all told, there are $120$ quoted companies, and we've accounted for $85-x+z+y$ of them so far. Thus:
Now, our total is $180+x-y-z,$ so:
Now, remember that $0\le x\le 85.$ We can refine this just a bit. Of the $200$ studied companies, $85$ of them were retail companies that failed and $120$ of them were companies quoted on the stock exchange. In order to make sure that there is "enough room" for these two sets of companies, we will need at least $120+85-200=5$ companies to be retail companies that failed and were quoted on the stock exchange, meaning $85-x\ge 5,$ meaning $80\ge x.$ Hence, our bounds on $x$ are $0\le x\le 80.$ At this point, we're all set to go. We know that $0\le x\le 80,$ so we have upper and lower bounds. To see whether they are the maximum and minimum, we must see if it is possible for those bounds to be attained (that is, see if it is possible for $x=80$ or $x=0$). The substitution $x=80$ gives us:
Recall that $0\le z\le 55.$ Of the $200$ companies, $140$ are in retail and $120$ were quoted, so to make sure there is "enough room" for those sets of companies, we need at least $140+120-200=60$ retail companies that were quoted. That is, $z+5\ge 60,$ so $z\ge 55.$ Hence, we need $z=55,$ and so:
Similarly, we need at least $90+120-200=10$ companies to be quoted companies that failed, meaning $5+y\ge10,$ so $y\ge 5,$ and since $0\le y\le 5,$ then $y=5,$ so:
Therefore, $80$ is the maximum number attainable for $x$.
Now, let's check and see if $x=0$ can be obtained. Substitution gives us:
Since $y\ge 0$ and $35-y-z\ge 0,$ then $35-z\ge 35-y-z\ge 0,$ so $z\le 35,$ so our new bounds of $z$ (having specified $x$) are $0\le z\le 35.$ Likewise, $y\le 35,$ but this doesn't refine the constraints $0\le y\le 5$ that we already had. Let's just pick a value of $z$ and see what happens. Say $z=10.$ Then:
That didn't introduce any issues, either, nor did it narrow our constraints for $y$ at all, so we can still pick any $0\le y\le 5$ and get an appropriate Venn diagram. Thus, $0$ is the minimum number attainable for $x$.
Best Answer
Illustrating the ideas already indicated by the comments:
Consider the following chart:
\begin{array}{| r | r | r | r | r |} \hline \text{Variable} & \text{Tea} & \text{Milk} & \text{Coffee} & \text{Value} \\ \hline v_1 & t & t & t & 3 \\ \hline v_2 & t & t & f & 2 \\ \hline v_3 & t & f & t & \\ \hline v_4 & t & f & f & \\ \hline v_5 & f & t & t & 4 \\ \hline v_6 & f & t & f & 6 \\ \hline v_7 & f & f & t & \\ \hline v_8 & f & f & f & \\ \hline \end{array}
From the constraints:
$v_1 + v_2 + v_3 + v_4 = 9$.
$v_1 + v_3 + v_5 + v_7 = 9$.
$v_1 + v_2 + v_5 + v_6 = 15$.
$v_1 + v_2 = 5$.
$v_1 + v_5 = 7$.
$v_1 = 3$.
It is required to solve for $v_3$.
From the above constraints:
The constraint that $v_7 + v_8 = 1$ clearly has two possibilities, that lead to two separate solutions for $v_3$.
Either $v_7 = 1 \implies v_3 = 1$
or $v_7 = 0 \implies v_3 = 2.$