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GRE Quant - Overlapping Set Theory

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GRE Quant - Overlapping Set Theory [#permalink] New post   10 Jul 2020, 10:30
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This post is a part of [GRE MATH BOOK]

Frequency of the concepts tested: Medium


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Re: GRE Quant - Overlapping Set Theory [#permalink] New post   10 Jul 2020, 23:54
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Overlapping Sets Made Easy!



A rare question involves an overlapping set for group members, These questions a fairly easy but they could result intimidating to those students who have never seen them before.

Say, there are a total of 100 people in a housing society. There are two clubs close to the society – A and B. You are given that of the 100 people of the housing society, 60 people are members of club A and 50 people are members of club B.

Image

Question 1: How many people are members of both the clubs?

We are looking for the number of people in the green region. The answer here is not 10. It is ‘cannot be determined’ i.e. you cannot say how many people are members of both the clubs. The reason is that you do not know how many people belong to neither club.

Say, for all future questions (unless mentioned otherwise), you are given that 20 people belong to neither club. What can you say about the number of people who belong to both the clubs? Now, out of the pool of 100, 20 are out. Only 80 people are club members. Since 60 are members of club A and 50 people are members of club B which gives us a total of 110, there must be an overlap of 30 people i.e. 30 people must belong to both the clubs (80 = 60 + 50 – Both)

Question 2: How many people belong to only one club?

We found above that 30 people belong to both the clubs. So out of the 60 people of club A, 30 belong to only club A. Out of the 50 people of club B, 20 belong to only club B. So a total of 30+20 = 50 people belong to only one club, either A or B but not both. (60 – Both + 50 – Both = 30 + 20 = 50)

Question 3: Say, you don’t know the number of people who belong to neither club. What is the minimum number of people who must belong to both the clubs?

We know that there are a total of 100 people. 60 belong to club A and 50 belong to club B which adds up to 110. Therefore, AT LEAST 10 people must have membership of both the clubs. Now if you increase the number of people who do not belong to either club, the number of people who belong to both will increase by the same number. Think in terms of the Venn diagram. If the ‘Neither’ number increases, the number of people who are members decreases. Hence, the overlap increases to keep A = 60 and B = 50.

Let’s look at the promised question which will make this concept clear.

Question: Of the 400 members at a health club, 260 use the weight room and 300 use the pool. If at least 60 of the members do not use either, then the number of members using both the weight room and the pool must be between:

(A) 40 to 100
(B) 80 to 140
(C) 160 to 260
(D) 220 to 260
(E) 220 to 300

Solution: When we minimize “number of members who do not use either”, we are minimizing the “number of members who use both” as well.

Look at the equation:

Total = A + B – Both + Neither



Eventually, this is the only formula you may need to memorize to tackle the overlapping sets on the GRE

Since the total sum 400 is constant, if we increase the ‘Neither’ i.e. 60, we will have to increase the ‘Both’ term too to maintain the sum of 400 (Assuming A and B are constant which they are since they are given to us).
Least value of ‘number of members who use neither’ is 60. We will get the least value of ‘number of members who use both’ when we put ‘Neither’ = 60.
400 = 260 + 300 – ‘Minimum value of both’ + 60

Minimum value of both = 220

On the same lines, if we maximize “number of members who use neither”, we are maximizing the “number of members who use both” as well.

What is the maximum number of people who use neither? Out of a total of 400 people, 300 people use the pool. Hence at least 300 people use at least one of the two facilities. This means that there can be AT MOST 100 people (total 400 – 300 who use pool) who use neither facility.

400 = 260 + 300 – ‘Maximum value of both’ + 100

Maximum value of both = 260

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Re: GRE Quant - Overlapping Set Theory [#permalink] New post   11 Jul 2020, 00:34
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Three Overlapping Sets



How a the three overlapping sets diagram looks like.

Image

Notice that the total comprises of the elements that do not fall in any of the three sets and the elements that are a part of at least one of the three sets.

The elements falling in the red, yellow or green region (region a, b or c) fall in only one set. The elements falling in region d, e or f fall in exactly two sets and the elements falling in region g fall in all three sets.

Now some quick questions to get a clear picture:

Question 1: Which regions represent the elements that belong to at least 2 sets?
Answer 1: d + e + f + g

Question 2: Which regions represent the elements that belong to at least 1 set?
Answer 2: a + b + c + d + e + f + g = Total – None

Question 3: Which regions represent the elements that belong to at most 2 sets?
Answer 3: None + a + b + c + d + e + f = Total – g

Hope there are no doubts up till now. Let’s look at a question to see how to apply these concepts.

Question: Three table runners have a combined area of 200 square inches. By overlapping the runners to cover 80% of a table of area 175 square inches, the area that is covered by exactly two layers of runner is 24 square inches. What is the area of the table that is covered with three layers of runner?

(A) 18 square inches
(B) 20 square inches
(C) 24 square inches
(D) 28 square inches
(E) 30 square inches

Solution: Let’s first try to understand what exactly is given to us. The area of all the runners is equal to 200 square inches.

Runner 1 + Runner 2 + Runner 3 = 200

In our diagram, this area is represented by
(a + d + g + e) + (b + d + g + f) + (c + e + g + f) = 200

(We need to find the value of g i.e. the area of the table that is covered with three layers of runner.)

Area of table covered is only 80% of 175 i.e. only 140 square inches. This means that if each section is counted only once, the total area covered is 140 square inches.
a + b + c + d + e + f + g = 140

So the overlapping regions are obtained by subtracting second equation from the first. We get d + e + f + 2g = 60

But d + e + f (area with exactly two layers of runner) = 24
So 2g = 60 – 24 = 36

g = 18 square inches.

Note that you don’t need to make all these equations and can directly jump to d + e + f + 2g = 60. We wrote these equations down only for clarity. It is a matter of thinking vs solving. If we think more, we have to solve less. Let’s see how.

Combined area of runners is 200 square inches while area of table they cover is only 140 square inches. So what does the extra 60 square inches of runner do? It covers another runner!

Wherever there are two runners overlapping, one runner is not covering the table but just another runner. Wherever there are three runners overlapping, two runners are not covering the table but just the third runner at the bottom.

So can we say that (d + e + f) represents the area where one runner is covering another runner and g is the area where two runners are covering another runner?

Put another way, can we say d + e + f + 2g = 60?

We know that d + e + f = 24 giving us g = 18 square inches

This entire ‘thinking process’ takes ten seconds once you are comfortable with it and your answer would be out in about 30 sec!



In the end, it's not that difficult of a concept to memorize if you simply extrapolate the formula for overlapping sets as follows:

Total = Group1 + Group2 + Group3 - Those in Exactly Two Groups - 2(Those in All Three Groups) + None

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Re: GRE Quant - Overlapping Set Theory [#permalink] New post   11 Jul 2020, 00:34
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How to Use a Double Matrix on the GRE



Using a Venn diagram when dealing with most Sets questions is correct and effective, there are some questions in which a double-matrix is necessary (and much more powerful than a wimpy Venn). This little guy will make even the scariest-looking Sets question into a simple set of rows and columns, and its ability to help us determine whether a statement is sufficient in DS is unmatched!

To make a double-matrix, simply create a chart with rows and columns. The rows are assigned to one variable, and the columns to another. At the bottom of each column and at the far-right of each row, place a box for the column-total and row-total. Let’s check out a sample GMAT question to see what this might look like!

Question: 33 out of the 47 students in an advanced degree program have a higher than average GPA. How many students in the program are receiving some form of academic scholarship?

(1) More students do not have a scholarship than have a scholarship.
(2) The same number of students have a higher than average GPA and are receiving some form of academic scholarship as have neither a higher than average GPA nor an academic scholarship.

Solution: From the question-stem, we know 33 students have a high GPA, while 14 do not. We need more information about which of these students have scholarships to be able to answer this value question. Statement (1) is insufficient because it does not give us information to find the exact numerical value of the students receiving some form of scholarship.

Statement (2) tells us that the number of students who fit “both” is equal to the number of students who fit “neither.” Let’s set up a chart to visualize the four possible categories for the students. Since “both” = “neither,” let’s fill in “x” for those boxes.

Image

Since each column and row must total, if there are “x” students receiving no scholarship and not a higher GPA, and the total students who don’t have a higher GPA is 14, then 14-x students must not have a higher GPA and have a scholarship. The total we are looking for is represented by the red “?,” and we can set up an equation to solve: x + (14 – x) = 14. Sufficient.

The correct response is (B).


To wrap up, keep in mind that it’s possible (although highly unlikely) that you might see a Sets question on the GRE involving three variables instead of two. In that case, you’d have 5 columns instead of 4, and 5 rows instead of 4, but the same rules of totaling apply!
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Re: GRE Quant - Overlapping Set Theory [#permalink] New post   11 Jul 2020, 00:34
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How to draw a Venn Diagram




At a certain school, each of the 150 students takes between 1 and 3 classes. The 3 classes available are Math, Chemistry and English. 53 students study math, 88 study chemistry and 58 study english. If 6 students take all 3 classes, how many take exactly 2 classes?
A. 37
B. 43
C. 45
D. 60

Step 1: Deconstruction

This is where you extract the given information from a problem. Use a symbol for each section given to you. I usually use the first letter in caps. Please remember at this point that any number specified without explicitly mentioning that it's "exactly" or "only" for a certain subject is not to be taken so.

For instance, in this case, it says 53 math students. This doesn't mean students who are taking only 1 subject (Math). This could include students who are taking Math and Chemistry or Math and English or all three too. So now break down the numbers given to us.

Total = 150
M (Total) = 53
C (Total) = 88 [Why the hell are people studying more Chemistry than Math?]
E (Total) = 58
MCE = 6

We are asked to find MC + CE + EM.

Step 2: Drawing the diagram

ALWAYS start from the center of the Venn Diagram wherever information is available. This will make life infinitely easier.

In this case, the center is the intersection of all three circles, i.e MCE = 6

So fill that in to the diagram you've drawn.

Image

Now that you've gotten that, let's start filling in a variable for each section not known to us. Here, consider each letter to represent only that specific section and not the entire circle or a larger portion.

Image

So now that you have the parts filled in, what you need to do is write down what you have in the diagram in terms of numbers. So we are given the totals for each subject.

Look at math first. There are four types of people taking math (each group of these people mutually exclusive, and not in common with any other group)
1. Only math: a
2. Math and Chemistry: y
3. Math and English: x
4. All three: 6

So now represent this as a sum and you get

\(a+x+y+6 = 53\) and hence \(a+x+y = 47\)

Similarly for the other subjects you get:
\(b+x+z+6 = 58\) and hence \(b+x+z = 52\)

\(c+y+z+6\) = 88 and hence \(c+y+z = 82\)

And then you have the total:

\(x+y+z+a+b+c+6=150\) and hence \(x+y+z+a+b+c = 144\)

Step 3: Solution

This is perhaps the most intuitive part, but in my experience the first part of this step is the same in all overlapping sets problem. It's only what's asked for that's different.

Add all the individual equations together to get a combined equation with all the variables and a number.

So you get:

\(2(x+y+z) + a+b+c = 47+52+82 = 181\)

Rearranging this to get \(a+b+c\)we get \(a+b+c = 181 - 2(x+y+z)\)

Substitute this into the total equation we derived earlier saying \(x+y+z+a+b+c = 144\) so you get:

\(x+y+z+181 - 2(x+y+z)= 144\)

Upon rearranging this you get:

\(x+y+z = 181-144 = 37 \) which is option A the right answer.
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Re: GRE Quant - Overlapping Set Theory [#permalink] New post   19 Jul 2020, 08:10
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Overlapping Set Questions by Tag



Quantitative Comparison Questions



Multiple-choice Questions — Select One Answer Choice




Multiple-choice Questions — Select One or More Answer Choices



Numeric Entry Question



Data Interpretation Sets

  • HARD Questions
  • MEDIUM Questions
  • EASY Questions
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Re: GRE Quant - Overlapping Set Theory [#permalink] New post   19 Jul 2020, 10:17
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1) GRE Lesson: Double Matrix Method



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Re: GRE Quant - Overlapping Set Theory [#permalink] New post   26 Apr 2022, 07:41
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