Overlapping Sets Made Easy! : Quantitative Questions

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Bunuel

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16 Sep 2015, 07:10

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Three Overlapping Sets

BY KARISHMA, VERITAS PREP

Here, let’s take a look at how to use Venn diagrams to solve questions involving three overlapping sets.

First, let me show you what the three overlapping sets diagram looks like.

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. This question is discussed HERE.

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!

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SetsThree_1_23Sept.jpg [ 20.19 KiB | Viewed 120871 times ]

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Re: Overlapping Sets Made Easy! [#permalink]

16 Sep 2015, 07:02

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A Sets Question That Upsets Many

BY KARISHMA, VERITAS PREP

Example: A group of people were given 2 puzzles. 79% people solved puzzle X and 89% people solved puzzle Y. What is the maximum and minimum percentage of people who could have solved both the puzzles?

(A) 11%, 0%
(B) 49%, 33%
(C) 68%, 57%
(D) 79%, 68%
(E) 89%, 79%

Solution: The first thing to note here is that we do not know the % of people who could not solve either puzzle. All we know is that puzzle X was solved by 79% of the people and puzzle Y was solved by 89% of the people.

Let’s first try to maximize the % of people who solved both the puzzles. We want to make these two sets overlap as much as possible i.e. we need to get them as close to each other as possible. Region of overlap can be 79% at most since we know that only 79% people solved puzzle X. In this case, the venn diagram will look something like this.

Hence, the maximum % of people who could have solved both the puzzles is 79%.

Now, let’s try to minimize the % of people who solved both the puzzles. We want the sets to be as far apart as possible. In this case, the % of people who solved neither puzzle must be 0. Only then will the overlap of the sets be as little as possible.

In this case, 68% people must have solved both the puzzles.

Hence, the answer is (D) This question is discussed HERE.

Note: If the question instead gave you the % of people who did not solve either puzzle (e.g. by giving you that everyone solved at least one puzzle), then there is no question of maximizing/minimizing the % of people who solved both the puzzles. Consider this:

(For ease, let’s drop the percentage and work with just numbers.)

Total no. of people = No. of people who solved X + No. of people who solved Y – No. of people who solved both + No. of people who solved neither

100 = 79 + 89 – No. of people who solved both + No. of people who solved neither

No. of people who solved both – No. of people who solved neither = 68

We can maximize/minimize the two numbers by adjusting them against each other. If one increases, the other increases too. If one decreases, the other decreases too.

If no. of people who solve both = 68, no. of people who solve neither = 0

If no. of people who solve both = 69, no. of people who solve neither = 1

If no. of people who solve both = 79, no. of people who solve neither = 11

If you are given the number of people who solved neither, you have a fixed number of people who solved both. Hence, maximizing or minimizing becomes pointless.

You can also work on this concept logically –

Say, every solution of every puzzle was written down on a separate sheet of paper. Then there would be 168 sheets of paper in all (79 + 89). If everyone solved 1 puzzle, then we have accounted for 100 sheets. The other 68 sheets were made by 68 people who have already solved one puzzle each. Hence, 68 people would have solved both the puzzles. If instead, 99 people solve at least one puzzle and one person solves no puzzle, then 69 (obtained by 168 – 99) sheets would have been made by people who have already solved one. Hence 69 people would have solved both the puzzles. Note here that the number of people who solved neither and the number of people who solved both are not independent of each other. One number depends on the other. It will be good if you make a note of this in your log book.

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Re: Overlapping Sets Made Easy! [#permalink]

16 Sep 2015, 07:17

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Rocking a Venn Diagram

BY Vivian Kerr, VERITAS PREP

Set theory is no one’s favorite GMAT concept (unless you’re a masochist), but since nearly all test-takers will see at least one overlapping-sets question on the Quantitative section of the GMAT, it’s certainly important. And take solace in this – becoming confident with this challenging type of word problem can be as simple as learning how to rock a Venn diagram.

To refresh a bit on Venn diagrams, we have to recall two definitions: union and intersection. The union of sets is all elements from all sets. The intersection of sets is only those elements common to all sets. Let’s look at how we can take apart a complex question without any messy set theory formulas.

Question: This year, x people won an Olympic medal for water competitions. One-third of the winners earned a medal for swimming and one-fourth of those who earned a medal for swimming also earned a medal for diving. How many people won an Olympic medal for water competitions but did not both receive a medal for swimming and a medal for diving?

(A) 11x/12
(B) 7x/12
(C) 5x/12
(D) 6x/7
(E) x/7

Solution: Let’s work on a Venn diagram and fill in what we do know – an important first step as many of these problems come down in large part to “getting organized”. We know there will be some medal winners who swam but did not dive, some who dove but did not swim, and some who swam AND dove.

“x” here will be at the top of our Venn because it is the total for ALL PARTS of the Venn. That is, all three categories will sum to x. x/3 represents the 1/3 of the total (“x”) who swam, including those who swam only AND those who swam and dove.

We can make up a variable, let’s say “y,” to represent the total number of divers. The key to understanding this question lies in the last sentence and the phrase “not both.”

We need to know the people who ONLY swam but did NOT dive, and the people who ONLY dove but did NOT swim. I made up variables for these two groups: “a” and “z.”

Let’s use the answer choices to our advantage! Since they have the denominators of 12 and 7, let’s use one of those and work backwards! 12 appears more often, so we can start there.

If x = 12, there are 12/3 = 4 swimmers total, (12/3)/4 = 1 of whom swam and dove. That means a = 3. If 4 people swam, then 12-4 = 8 dove, so z = 8.

The two categories we’re looking for (a + z) are 3 + 8 = 11. We are looking for an answer choice that gives us 11 when x = 12.

The answer is (A). This question is discussed HERE.

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Re: Overlapping Sets Made Easy! [#permalink]

16 Sep 2015, 07:24

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How to Use a Double Matrix on the GMAT

BY Vivian Kerr, VERITAS PREP

While my personal preference is to use a Venn diagram when dealing with most Sets questions, 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.

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). This question is discussed HERE.

To wrap up, keep in mind that it’s possible (although highly unlikely) that you might see a Sets question on the GMAT 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: Overlapping Sets Made Easy! [#permalink]

16 Sep 2015, 07:36

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Other Resources on Overlapping Sets

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Re: Overlapping Sets Made Easy! [#permalink]

08 Mar 2016, 05:07

Bunuel wrote:

Three Overlapping Sets

BY KARISHMA, VERITAS PREP

Here, let’s take a look at how to use Venn diagrams to solve questions involving three overlapping sets.

First, let me show you what the three overlapping sets diagram looks like.

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. This question is discussed HERE.

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!

Show Spoiler

Attachment:

SetsThree_1_23Sept.jpg

Hi

I am somehow not able to apply the logic of sets here. Conceptually, n (AUBUC)=nA+nB+nC-(taken two at a time)+(taken three at a time). But this is not giving the result. Pls advice.

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Re: Overlapping Sets Made Easy! [#permalink]

08 Mar 2016, 05:29

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sinhap07 wrote:

Bunuel wrote:

Three Overlapping Sets

BY KARISHMA, VERITAS PREP

Here, let’s take a look at how to use Venn diagrams to solve questions involving three overlapping sets.

First, let me show you what the three overlapping sets diagram looks like.

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. This question is discussed HERE.

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!

Show Spoiler

Attachment:

SetsThree_1_23Sept.jpg

Hi

I am somehow not able to apply the logic of sets here. Conceptually, n (AUBUC)=nA+nB+nC-(taken two at a time)+(taken three at a time). But this is not giving the result. Pls advice.

There are 2 different formulas for three overlapping sets. Please go through the link below to know when to use each of them.

ADVANCED OVERLAPPING SETS PROBLEMS

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Re: Overlapping Sets Made Easy! [#permalink]

03 Sep 2016, 10:33

I don't understand the derivation of 7th formula for 3 overlapping sets.

So lets cal a , b and c the regions of "Only" (i.e. the non overlapping regions) and A, B, C the full region of the sets (with the overlapping) So:

a = A - AnB - AnC - AnBnC
b = B - BnA - BnC - AnBnC
c = C - CnA - CnB - AnBnC

Since AnB = BnA or AnC = CnA etc. the Σum of a, b and c will be

n(only one) = n(exactly one) = a + b + c = = A + B + C -2AnB - 2AnC - 2BnC - 3AnBnC .... I dont undertstand how the "+" comes up for the "+ 3AnBnC.

Can somebody explain it slowly?

I think also I noticed a pattern for the formulas. Whenever you want to find the n(EXACTLY of "something") = n(LEAST of something) - n(LEAST something+1) Hence EXACTLY = LEAST - LEAST

Also for the two group case is it safe to say: n(EXACTLY ONE) = A + B - 2*Both ?

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Re: Overlapping Sets Made Easy! [#permalink]

26 Sep 2016, 10:43

Ndkms wrote:

I don't understand the derivation of 7th formula for 3 overlapping sets.

So lets cal a , b and c the regions of "Only" (i.e. the non overlapping regions) and A, B, C the full region of the sets (with the overlapping) So:

a = A - AnB - AnC - AnBnC
b = B - BnA - BnC - AnBnC
c = C - CnA - CnB - AnBnC

Since AnB = BnA or AnC = CnA etc. the Σum of a, b and c will be

n(only one) = n(exactly one) = a + b + c = = A + B + C -2AnB - 2AnC - 2BnC - 3AnBnC .... I dont undertstand how the "+" comes up for the "+ 3AnBnC.

Can somebody explain it slowly?

I think also I noticed a pattern for the formulas. Whenever you want to find the n(EXACTLY of "something") = n(LEAST of something) - n(LEAST something+1) Hence EXACTLY = LEAST - LEAST

Also for the two group case is it safe to say: n(EXACTLY ONE) = A + B - 2*Both ?

Let us try to understand the 7th formula in this way:

You want the area which has no overlapping, so you have to add the individual area A + B + C and remove the overlapping part.

When you added A + B + C, you have Added -- AnB, BnC and CnA -- twice, so to remove the overlapping part you have to remove these overlapping part two times.

So Step 1 -> A + B + C - 2AnB - 2BnC - 2CnA

I think you have already understood this part.

Now removing the area of 3 overlap.

While adding A+ B + C , you have added AnBnC Thrice, so we have to subtract that:

Step 2 -> A + B + C - 2AnB - 2BnC - 2CnA - 3AnBnC

But in step 1, while removing AnB you have removed AnBnC as well, so removing 2AnB implies that 2AnBnC was removed as well
So - 2AnB - 2BnC - 2CnA implies that AnBnC was removed 6 extra times, so we have to add that back

Step 3 -> A + B + C - 2AnB - 2BnC - 2CnA - 3AnBnC + 6AnBnC which is the required formula.

Hope that makes sense.

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Re: Overlapping Sets Made Easy! [#permalink]

18 Mar 2018, 19:33

Bunuel chetan2u VeritasPrepKarishma

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.

Only for two overlapping sets, can you confirm below equations:

Total = A + B - A (inter) B + Neither

OR

Total = Only A + Only B + A (inter) B + Neither

The one mentioned on post (below)while solving the problem seems incorrect:
Total = A + B – Both + Neither (In diagram we have considered BLUE as only A)
but I think A - BLUE + GREEN. ie Only A + A (intersection) B

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Re: Overlapping Sets Made Easy! [#permalink]

18 Mar 2018, 20:59

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adkikani wrote:

Bunuel chetan2u VeritasPrepKarishma

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.

Only for two overlapping sets, can you confirm below equations:

Total = A + B - A (inter) B + Neither

OR

Total = Only A + Only B + A (inter) B + Neither

The one mentioned on post (below)while solving the problem seems incorrect:
Total = A + B – Both + Neither (In diagram we have considered BLUE as only A)
but I think A - BLUE + GREEN. ie Only A + A (intersection) B

Both formulas below are correct:

Total = A + B – Both + Neither (A here is the whole blue circle, so Only A + Both, the same for B)

Total = Only A + Only B + Both + Neither

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Re: Overlapping Sets Made Easy! [#permalink]

18 Mar 2018, 21:09

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Bunuel wrote:

adkikani wrote:

Bunuel chetan2u VeritasPrepKarishma

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.

Only for two overlapping sets, can you confirm below equations:

Total = A + B - A (inter) B + Neither

OR

Total = Only A + Only B + A (inter) B + Neither

The one mentioned on post (below)while solving the problem seems incorrect:
Total = A + B – Both + Neither (In diagram we have considered BLUE as only A)
but I think A - BLUE + GREEN. ie Only A + A (intersection) B

Both formulas below are correct:

Total = A + B – Both + Neither (A here is the whole blue circle, so Only A + Both, the same for B)

Total = Only A + Only B + Both + Neither

Total = A + B - A (inter) B + Neither
A is A only + both
B is B only and both..
substitute above..
Total = A only + both+B only + both - both + neither = A only + B only+both + neither
This is SAME below....

OR

Total = Only A + Only B + A (inter) B + Neither

so both mean same. you have to choose on basis of info given in question

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Re: Overlapping Sets Made Easy! [#permalink]

18 Mar 2018, 22:39

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adkikani wrote:

Bunuel chetan2u VeritasPrepKarishma

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.

Only for two overlapping sets, can you confirm below equations:

Total = A + B - A (inter) B + Neither

OR

Total = Only A + Only B + A (inter) B + Neither

The one mentioned on post (below)while solving the problem seems incorrect:
Total = A + B – Both + Neither (In diagram we have considered BLUE as only A)
but I think A - BLUE + GREEN. ie Only A + A (intersection) B

One last thing, Blue is "only A" and Green is "A and B".
So A is "Blue + Green".

B

Loop123

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Re: Overlapping Sets Made Easy! [#permalink]

26 Apr 2020, 13:51

Sounds great,this useful information !

G

RahulJain293

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Re: Overlapping Sets Made Easy! [#permalink]

01 Jan 2024, 20:09

Bunuel wrote:

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.

Hi Bunuel,

I did not understand the third sentence.. Is it not correct?

L

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Re: Overlapping Sets Made Easy! [#permalink]

02 Jan 2024, 00:13

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Expert Reply

RahulJain293 wrote:

Bunuel wrote:

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.

Hi Bunuel,

I did not understand the third sentence.. Is it not correct?

Given:

That sentence implies that increasing {Neither} by a certain quantity would result increasing {Both} by the same quantity to balance the sum to the same number of 100. For example, if {Neither} = 20, then {Both} = 30:

60 + 50 - 30 + 20 = 100
However, if {Neither} is 30 more, so 50, then {Both} is also 30 more, so 60.

60 + 50 - 60 + 50 = 100
Hope it's clear.

G

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Re: Overlapping Sets Made Easy! [#permalink]

02 Jan 2024, 00:57

Bunuel wrote:

RahulJain293 wrote:

Bunuel wrote:

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.

Hi Bunuel,

I did not understand the third sentence.. Is it not correct?

Given:

That sentence implies that increasing {Neither} by a certain quantity would result increasing {Both} by the same quantity to balance the sum to the same number of 100. For example, if {Neither} = 20, then {Both} = 30:

60 + 50 - 30 + 20 = 100
However, if {Neither} is 30 more, so 50, then {Both} is also 30 more, so 60.

60 + 50 - 60 + 50 = 100
Hope it's clear.

Yup thanks! The wording 'either' and 'neither' threw me off

gmatclubot

Re: Overlapping Sets Made Easy! [#permalink]

02 Jan 2024, 00:57