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If you’re one of those people who goes crazy over overlapping sets problems, this post is for you. Oftentimes, Venn Diagrams are the way to go on these problems, and I highly recommend that you master them in our GMAT course. However, if you’re more formula-oriented, there’s a handy shortcut you can use.

Let’s say we consider all the athletes at a school. We consider those on the football team and those on the hockey team. Some athletes are on both teams, and some athletes are on neither team. Let’s assign the following variables:

F = number of athletes on the football team (including those who also play hockey)
H = number of athletes on the hockey team (including those who also play football)
B = number of athletes who play both football and hockey
N = number of athletes who play neither football nor hockey
T = total number of athletes at the school.

The following formula would apply:

F + H – B + N = T

This applies to any pair of overlapping sets. You add the total number in each group (including the overlap in both cases), then you subtract the number in the overlap, add the number in the neither group, and that gets you the overall total.

In general, you could think of it as:

(Total in first group) + (Total in second group) – (overlap) + neither = Overall total

Now, it’s important to keep in mind that the total for each group includes the overlap and not just those elements that are only football or only hockey.

“All that is great,” you might say. But then you’d add, “What about the really annoying problems that involve not two but three overlapping sets?” Well, there’s even a formula for that situation. In general, if you have three overlapping groups, the formula would be:

(Total in Group 1) + (Total in Group 2) + (Total in Group 3) – (Overlap of 1 and 2) – (Overlap of 1 and 3) – (Overlap of 2 and 3) – [2 * (Overlap of 1, 2, and 3)] + (total in none of the groups) = Overall total

Very important reminder: When we say “Total in Group 1″, we mean everything in Group 1, including the elements in the overlaps. It does not mean those elements in Group 1 only!

If you’re curious about why we subtract the overlaps, it’s essentially because in adding up the totals of each group, we count the overlaps more than once. For example, in adding the totals of Groups 1 and 2, the overlap between them gets counted twice, so we must subtract out one of them. When we add the totals of all three groups, the overlap of all three groups gets counted 3 times, whereas it should only be counted once; that’s why we subtract that overlap twice.

Now, the formula is a great tool, but as you probably suspected, having it memorized is just a first step. The difficulty of GMAT problems lies not in complex formulas and calculations, but in the quirkiness of their constructions and setups. Try your hand at these two problems to see what I mean. The first is an Official Guide problem, and the second is one of my own. As usual, feel free to post your step-by-step solution in the comments. Good luck!

1. Of the 200 students at college T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from

A) 20 to 50
B) 40 to 70
C) 50 to 130
D) 110 to 130
E) 110 to 150

2. In a particular neighborhood of 150 households, some households have no electronic devices. The rest have some combination of televisions, laptops, and stereos. 75 households have televisions, 50 have laptops, and 20 have all three devices. If 45 households have exactly two of the three devices, and the number of households that have stereos is four times the number of households that have none of the three devices, how many households have stereos?

A) 84
B) 88
C) 92
D) 96
E) 100

Posted in GMAT, Quant Guide | 8 comments



  • http://rolipolli.wordpress.com/ Rolipolli

     1. Answer:

     T= 200
     Ch = 130
     Bio = 150
     N = 30 

    Both Ch and Bio = ?  
    T = Ch + Bio – B + N
    200 = 130 + 150 – B + 30

    B = 110
    Hence, we are down to answer choices D & E.

    The range can calculated from the given number of students in Ch and Bio as maximum number of students who can major in both is the students of Chemistry. 

    So, the answer is D
      

    • http://rolipolli.wordpress.com/ Rolipolli

      I got confused with question no. 2.  Can you please explain?  

  • Visitashi

    For 1st question,
    i m getting option D
    here ,given that atleast 30 are not majoring in either.
    let X are majoring in both.then ,13-X +150-X +X=170
    =>X=110
    now,Max no. of people who do not major in either would be when those who major in Chemistry overlap fully with the ones who major in Biology.
    so,atmost 50 people do not major in any subject.
    with this info,i get X=130
    so the range will be 110<=Xthose who have stereo =4X
    so equation would be like,
    75+4X-a-20+50-20-c-b+X=150
    putting a+b+c=45,we get
    X=22
    =>4X=88,so option B.
    sir, i would request if u please enlighten us with finding Maximum /Minimum values in this same genre of questions.

    • http://rolipolli.wordpress.com/ Rolipolli

      It helped! 

    • Visitashi

      in the second question above,we must note that a,b,c are the numbers for exactly 2 region,they do not include the all 3 region.

  • Rich Zwelling (Knewton)

    Nice work guys!

    As you pointed out, the first problem can be narrowed down to D or E:  The formula would look like this:

    Total = Chem + Bio – Both + Neither

    200 = 130 + 150 – B + N

    We know that N must be at least 30, so let’s see what happens when N=30:

    200 = 130 + 150 – B + 30

    200 = 310 – B

    B = 110

    Even if you’re not sure whether 110 is the highest or lowest B could possibly be, it doesn’t matter, because you know it’s a boundary point of the range, which means only answers D or E could be correct.

    Then, the trick is to realize that B could not be greater than 130, because even if every chem major was also a bio major, there are only 130 chem majors total.  D is the winner.

    Regarding the second problem, if T represents televisions, L represents laptops, S represents stereos, and N represents none of them, the formula would look like this:

    T + L + S – (Overlap of T and L) – (Overlap of T and S) – (Overlap of L and S) – [2 * (Overlap of all three)] + N =  total

    75 + 50 + S -  (Overlap of T and L) – (Overlap of T and S) – (Overlap of L and S) – 2 * 20 + N = 150

    There are two catches here:  1.  The remaining overlaps can be combined, because we’re told that “45 households have exactly two of the three devices”;  2.  We’re told that “the number of households that have stereos is four times the number of households that have none of the three devices”, so we know that N=S/4.  (You could also substitute 4N for S, but keep in mind that you’re looking for S in the end).

    Substitute:

    75 + 50 + S – 45 - 2*20 + S/4 = 150

    125 + S – 45 - 40 + S/4 = 150

    40 + S + S/4 = 150

    5S/4 = 110

    5S=440

    S=88

    • http://rolipolli.wordpress.com/ Rolipolli

      Thank you Rich!

    • Arjun

      Hi Rich,
      In your post where you state the formula for 3 sets, does overlap of 1&2 include overlap of 1&2&3 also?