<p>From the first practice test. page 427 Section 9 # 16</p>
<li>Set X has x members and set Y has y members. Set Z consists of all members that are in either set X or set Y with the exception of the k common members (k>0). Which of the following represents the number of members in Set Z?</li>
</ol>
<p>A. x+y+k
B. x+y-k
C. x+y+2k
D. x+y-2k
E. 2x+2y-2k</p>
<p>I am sure most of you have done this problem by now, and the answer is D. Can anyone of you explain how you came up with the right answer? What approaches did you use, because i found this one to be challenging. It seems like a logic question tho.</p>
<p>Well...think of it this way. Let's say there are 15 students taking Math, and 12 students taking Biology, but 5 of the students are taking both Math and Biology. To figure out the total number of students, you would add the students taking math, those tatking Bio, and then SUBTRACT the ones taking both. This is to prevent counting students twice. SO it would be 15 + 12 - 5 = 22.
Now, for this particular problem, we not only want to prevent the members in both sets from being counted twice, but also stop them from being counted AT ALL. So instead of subtracting just k, you have to subtract 2k.
Let's call the numbers in set X that are not common with those in set Y a. The numbers in set Y that are exclusive of set X are b.
The common terms from sets X and Y are called k.
Set X:
a
k</p>
<p>Set Y:
b
k</p>
<p>Now it is easy to see why you must subtract 2k when forming set Z.
Z = a +b = X + Y -2k</p>