**Learn this technique to master set questions of GMAT Quant word problem. **

## Practice questions

First, try these challenging practice questions.

1) Of the 80 houses in a development, 50 have a two-car garage, 40 have an in-the-ground swimming pool, and 35 have both a two-car garage and an in-the-ground swimming pool. How many houses in the development have neither a two-car garage nor an in-the-ground swimming pool?

A. 10

B. 15

C. 20

D. 25

E. 30

2) A certain school has three performing arts extracurricular activities: Band, Chorus, or Drama. Students must participate in at least one, and may participate in two or even in all three. There are 120 students in the school. There are 70 students in Band, 73 in the Chorus, and 45 in the Drama. Furthermore, 37 students are in both the Band and Chorus, 20 are in both the Band and the Drama, and 8 students are in all three groups. Twenty-five students are just in the chorus, not in anything else. How many students participate in only the drama?

A. 11

B. 12

C. 14

D. 17

E. 21

## Introduction to sets

The idea of a set is, in some sense, the most fundamental idea in all of mathematics. Nonetheless, it’s a very simple idea. A set is simply a collection of objects or elements. When the members of a set are all numbers, then we use “set notation”, which consists of brackets. For example:

A = {2, 3, 3, 5, 7}

That notation denotes set A with five numerical members. When all the members of the set are numbers, typical questions involve computations like the mean or the median: here, the mean (or average) of set A is 4, and the median of A is 3. You can read more about those calculations at this post.

## Overlapping sets

Numerical sets can be handled with statistical calculations. Non-numerical sets, sets in which the members are people or cars or companies, are the stuff of tricky word problems. Ordinarily, there’s nothing particular challenging if there’s only one set: some of the people are in that set, whatever it is, and the rest aren’t. No challenge.

Things get more interesting if there are two or more overlapping sets. For example, in the SF Bay Area, many adult residents were born out of state — many, but not all; many adult residents have a college education — many, but not all; and many adult residents are SF Giants fans — many, but not all. Those are three overlapping sets —- any particular adult resident of the SF Bay Area many be a member of none, one, two, or all three of those three categories. As it happens, I am a member of exactly two of those categories. This is exactly the situation of the practice questions posed above.

## Venn diagrams

Venn diagrams are the best method for untangling overlapping sets. If you have two overlapping sets, you need a two-circle Venn diagram:

This diagram contains three discrete regions:

A = those elements in just the left circle

B = those element in both categories, in the overlap

C = those elements in just the right circle

There may also be a fourth discrete region, those elements that are not members of any set. Typically, the problem will only give us information about totals — the total number of elements altogether, the total number in each circle, and the overlap. If you are told there are 70 members in the right circle, and 20 members in the overlap, then you would know B + C = 70 and B = 20, so from that you could deduce C = 50, the number of elements that are just in the portion of the circle labeled C. In general, you work from the center outward, figuring out one discrete region after another. Once you know the value of each individual discrete region, you will be able to answer any question about the number in any particular grouping.

If there are three overlapping categories, we use a three-circle Venn diagram:

This diagram has at least seven discrete regions

A = members of all three circles

B = members of the green and blue circles, but not the red circle

C = members of the green and red circles, but not the blue circle

D = members of the blue and red circles, but not the green circle

E = members of the green circle but of neither the blue nor the red circles

F = members of the blue circle but of neither the green nor the red circles

G = members of the red circle but of neither the green nor the blue circles

Depending on context, there may also be a eighth discrete region, those elements that are not members of any of the three set. Typically, the problem will only give us information about totals. This gets very tricky. If we are told the total in any one circle, that includes four discrete regions; for example, the green circle includes A + B + C + E. Similarly, the overlap of two circles contains two discrete regions: for example, the overlap of the blue and red circles includes A + D. The problem will always tell you how many elements are in the central region (A), and will often tell you how many are in each circle, and how many in each overlap of two circles. In general, you work from the center outward, figuring out one discrete region after another. Once you know the value of each individual discrete region, you will be able to answer any question about the number in any particular grouping.

With these strategies, you may find the practice problems at the beginning somewhat more approachable. Try them again, before reading the explanations below.

In the next post, we will look at set problems in which each element is categorized according to two different variables at once.

## Practice question explanations

1) Here, we have two categories: (a) with or without two-car garage, and (b) with or without an in-the-ground pool. Houses can be members of either, both, or neither category. We will use a two circle Venn diagram:

We know the total of the group is 80 —– A + B + C + D = 80. We know the green circle, two-car garages, has 50 members, so A + B = 50. We know the blue circle, in-the-ground pool, has 40 members, so B + C = 40. We also know the crucial overlap region, B = 35. If B = 35, in the green circle, we can deduce that A = 15, and in the blue circle, we can deduce that C = 5. Then

A + B + C + D = 15 + 35 + 5 + D = 80

D = 25

Thus, 25 houses in this development have neither a two-car garage nor an in-the-ground swimming pool. Answer = **D**.

2) Here, we have three categories, so we need three circles. Every student must take at least one of these three performing arts extracurricular activities, so there will be no one outside the three circles.

The sum of all seven = 120 (we never use this number in this question)

The totals for the band (70), the chorus (73), and the drama (45) each involve the sum of four discrete regions. We will have to find other information before we can employ them.

“8 students are in all three groups”

**N = 8**.

“37 students are in both the Band and Chorus”

37 = K + N = K + 8 —> **K = 29**

“20 are in both the Band and the Drama”

20 = M + N = M + 8 —> **M = 12**

“twenty-five students are just in the chorus, not in anything else”

**L = 25**

We now have identified three of the regions in the Chorus circle, so we can solve for P.

chorus = 73 = K + L + N + P

73 = 29 + 25 + 8 + P

**P = 11**

Now, we have identified three of the regions in the Drama circle, so we can solve for Q.

drama = 45 = M + N + P + Q

45 = 12 + 8 + 11 + Q

**Q = 14**

This is precisely what the question was asking: how many students are only in drama? There are 14 students who take only drama.

Answer = **C**

### Most Popular Resources

Thanks. it’s very nice

Hi,

I solved the second question without the information that C only is 25. I used instead the fact that total is 120 and the following general formula for 3 way venn diagrams:

Total = Neither + A + B + C – (A and B) – (A and C) – (B and C) + (A and B and C).

In that way, C only = 25 is in actually redundant information.

So D only = D – (B and D) – (C and D) + (C and B and D).

However, I do see that given C only = 25 you can figure it out faster.

Nice work! Thanks for sharing your solution 🙂

@kai Aeberli what you’re saying would have worked if we had “only” A and B. But here A and B is inclusive of A, B and C too.

Questions in GMAT do not normally have redundant information.

Hi,

I have a problem with this question.

In a class of 60 students, the number of students who passed biology is 6 more than those who passed chemistry. Every student passed at least one of the two subjects and 8 students passed both subjects.

How many passed;

1. Biology

2. Chemistry

3. Only one subject.

Thanks

Hi there 🙂

We’d be happy to help you with this question, but could you let us know where it’s from? Please note that we generally don’t answer questions from materials other than Magoosh and GMAC Official resources. If the question’s from the official materials, we’ll get back to you as soon as I can!

If the question is from another source, I’d recommend checking to see if the question has already been posted on one or more of the following sites: Urch’s forums, GMAT Club’s forum, or Beat the GMAT’s forum.

Also, it’s a great help to us in giving explanations if you tell us more about how you approached the problem and where you got stuck. That way, we can give you more specific, better explanations. 🙂

Thanks!

It is from neither actually

Then we unfortunately cannot help you here, but there are many vibrant student communities where someone should be able to help. 🙂

G. Banku qtn reply::

total students = 60

B=C+6

Common in B&C are 8

Only B = C+6-8

Only C = C-8

Total = 60 = (C+6-8)+8+(C-8)

Solve for C we get C = 31

So Only B are 29

Only C are 23

1. Biology… 29

2. Chemistry…23

3. Only one subject…29+23=52

but bro question did not ask only biology or only chemistry . so you should add 8 in the first to answers i.e. 29+8=37 and 23+8=31.

Thanks, but calculating everything together is more than 120, how do we arrive at 120 which is the total of the students? Thanks

Hi Mariam,

I have put the associated values for each region here in the diagram to show that we can add them all up to be the 120 we should have. You can see it here.

Remember, just because we know the totals for the band (70), the chorus (73), and the drama (45) doesn’t mean we just add those numbers. If a student is in drama and band, we have counted that same student twice in each of these figures, so we have to use algebra, like our explanation shows, to find the actual number in each region of the diagram.

I hope this helps. 🙂

Thanks Mike! 🙂

how did you find L=25 ?

Its mentioned in the question.

This value is given in the statement

Hi Peter. Thanks for the super clear explanations. I always get stumped with 3 category questions. Hopefully, I can push my Quant score up a few points with this review

Thanks you so much Mike for the wonderful explanation

Dear Vivek,

You are quite welcome, my friend! I’m glad you found it helpful! Best of luck to you!

Mike 🙂

Thanks a lot

Dear Faisal,

You are quite welcome! 🙂 Best of luck to you!

Mike 🙂

Hey Mike,

Long time reader, first time poster –

I recently took a practice GMAT from GMAT PREP (From GMAC itself) that did, in fact, have a 3 variable venn-diagram – essentially the question provided a table along the lines of – 3 people like A and B, 5 people like B and C, 8 people like A and C, 3 people like all three, etc. and there are 30 total people asked you to calculate how many liked 0 or 1 thing. I used your technique above and was able to solve in under two minutes.

-Eric

Dear Eric,

I’m glad to hear that this helped you, my friend. Best of luck to you.

Mike 🙂

Mike from where can we get questions with 3 variables for Venn diagram?

Dear Kumar,

That’s a good question. These questions are relatively rare, so most collections of GMAT Math problems have few, if any, of these. In fact, I think you could sit for the GMAT a 100 times, a different test for a 100 days in a row, and never see a problem of this sort. It

couldappear, but it’s very rare. Practicing it will make you skilled with what is more likely to appear.In this blog:

https://magoosh.com/gmat/2013/best-gmat-books-and-resources-2014/

We recommend the NOVA GMAT Math Bible. It’s a big set of practice problems. They definitely have problems that involve Venn Diagrams, but it’s not clear to me that any are three-variable problems.

I also found a few sites, each with one or two practice problems of this sort:

http://www.math.fsu.edu/~wooland/hm2ed/Part1Module3/Part1Module3.pdf

http://www.regentsprep.org/regents/math/algebra/ap2/lvenn.htm

http://www.purplemath.com/modules/venndiag4.htm

Once again, do not make the mistake of believing you need to master the three-variable case for the GMAT. Think of the three-variable case as something that will build your problem-solving skills for the two-variable case that is much much more likely to appear on the GMAT.

Does all this make sense?

Mike 🙂

Mike,

Thanks for explaining Double Matrix and Venn Diagram approaches.

Two questions

1) When we see a word problem, what is the best way to know whether given word problem can be solved using either Veen Diagram approach or by using Double Matrix approach ?

e.g. Let’s say I spend first 30 seconds to try Veen Diagram approach and then later figured out (after vesting 30 seconds) that problem cannot be solved using Veen diagram but instead Double Matrix approach should be used.

Any suggestion for avoiding such kind of this mistake?

2) If in case given word problem can be solved using both methods, then how do we know which approach would be quicker (to save time)?

Thanks.

Peter,

Let’s draw a distinction of variables vs. categories. In both of these problems, there is one “variable” — say, in the second, the variable is “extracurricular activity”, and then all the groups are categories of that one variable — that’s a Venn Diagram scenario. Individuals are classified according to one or more categories of the single variable.

In the Double-Matrix case, there are two separate variables, and each individual is classified into exactly one case of one variable and one case of the other scenario.

Some problems can be solved by both methods, if it just yes-or-no on A and yes-or-no on B. Thinking in terms of variables vs. categories will help you see the differences.

Mike 🙂

Thanks Mike.

So do you think that it would be good idea to always try out Venn Diagram approach if question describes “one variable”(where this one variable has multiple categories) and Double-Matrix approach if there are “multiple variables” (where each of the variable is having multiple categories )

About two examples that you have described above, I was able to solve first example using Double-Matrix approach even though it has one variable. However, for second example, I couldn’t figure out a way to solve it using Double-Matrix.

So it looks like

1) “one” variable type problems can “always” be solved using Veen Diagram approach

However, sometime they can also be solved using Double-Matrix approach.

2) “two or more variable” type problems can only be solved using Veen Diagram.

Is my analysis correct? Appreciate your reply.

Thanks

Peter,

For one variable problems, I would say the Venn Diagrams method is almost always the method of choice — I can’t think of an exception off the top of my head.

For two variables — if each one simply has the categories “yes” & “no”, then you could use a two circle Venn diagram or a Double Matrix method, whichever you prefer. If, for two variables, the categories are more complicated, and especially if either variable has more than two categories, then you must use a Double Matrix method.

For three variables, you would need a 3D array of numbers to solve it — that already is well beyond the GMAT. Don’t worry about three or more variables.

Mike 🙂

Thanks a lot Mike.

Peter,

You are quite welcome. Best of luck to you.

Mike 🙂