Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 65808

Overlapping Sets Made Easy!
[#permalink]
Show Tags
16 Sep 2015, 05:58
Overlapping Sets Made Easy!Nuances of Sets We will start with sets today. Your Veritas Prep GMAT book explains you the basics of sets very well so I am not going to get into those. If you have gone through the concepts, you know that we can use Venn diagrams to solve the sets questions. First, let’s look at why we should focus on terminology in sets question. Thereafter, we will put up a very nice question from our very own book which is simple but takes down many people (just like a typical GMAT question): 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.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 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 constants 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 ‘a 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 The maximum value of both = 260 Answer (D) This question is discussed HERE. Attachment:
Ques3.jpg [ 12.46 KiB  Viewed 60789 times ]
_________________




Math Expert
Joined: 02 Sep 2009
Posts: 65808

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
16 Sep 2015, 06:29
Solving InferenceBased 700+ Level Official GMAT Questions Sometimes, to solve some tough questions, we need to make inferences. Those inferences may not be apparent at first but once you practice, they do become intuitive. Today we will discuss one such inference based high level question of an official GMAT practice test. Question: In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:(A) 65 (B) 55 (C) 45 (D) 35 (E) 25 Solution: We need to find the value of x – y What is x? It is the greatest possible number of households that have all three devices What is y? It is the lowest possible number of households that have all three devices Say there are 100 households and we have three sets: Set DVD including 75 households Set Cell including 80 households Set MP3 including 55 households We need to find the values of x and y to get x – y. We need to maximise the overlap of all three sets to get the value of x and we need to minimise the overlap of all three sets to get the value of y. Maximum number of households that have all three devices: We want to bring the circles to overlap as much as possible. The smallest set is the MP3 set which has 55 households. Let’s make it overlap with both DVD set and Cell set. These 55 households are the maximum that can have all 3 things. The rest of the 45 households will definitely not have an MP3 player. Hence the value of x must be 55. Note here that the number of households having no device may or may not be 0 (it doesn’t concern us anyway but confuses people sometimes). There are 75 – 55 = 20 households that have DVD but no MP3 player. There are 80 – 55 = 25 households that have Cell phone but no MP3 player. So they could make up the rest of the 45 households (20 + 25) such that these 45 households have exactly one device or there could be an overlap in them and hence there may be some households with no device. In the figure we show the case where none = 0. Now, let’s focus on the value of y i.e. minimum number of households with all three devices: How will we do that? Before we delve into it, let us consider a simpler example: Say you have 3 siblings (A,B and C) and 5 chocolates which you want to distribute among them in any way you wish. Now you want to minimise the number of your siblings who get 3 chocolates. No one gets more than 3. What do you do? Will you leave out one sibling without any chocolates (even if he did rat you out to your folks!)? No. Because if one sibling gets no chocolates, the other siblings get more chocolates and then more of them will get 3 chocolates. So instead you give 1 to each and then give the leftover 2 to 2 of them (one each). This way, no sibling gets 3 chocolates and you have successfully minimised the number of siblings who get 3 chocolates. Basically, you spread out the goodies to ensure that minimum people get too many of them. This is the same concept. When you want to minimise the overlap, you basically want to spread the goodies around. You want minimum people to have all three. So you give at least one to all of them. Here there will be no household which has no device. Every household will have at least one device. So you have 80 households which have cell phone. The rest of the 20 households say, have a DVD player so the leftover 55 households (75 – 20) with DVD player will have both a cell phone and a DVD player. There are 55 households who already have two devices and 45 households with just one device. Now how will you distribute the MP3 players such that the overlap between all three is minimum? Give the MP3 players to the households which have just one device so 45 MP3 player households are accounted for. But we still need to distribute 10 more MP3 players. These 10 will fall on the 55 overlap of the previous two sets. Hence there are a minimum of 10 households which will have all three devices. This means y = 10 x – y = 55 – 10 = 45 Answer (C) This question is discussed HERE.
_________________




Math Expert
Joined: 02 Sep 2009
Posts: 65808

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
16 Sep 2015, 06:02
A Sets Question That Upsets Many 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.
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 65808

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
16 Sep 2015, 06:10
Three Overlapping Sets 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! Attachment:
SetsThree_1_23Sept.jpg [ 20.19 KiB  Viewed 62545 times ]
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 65808

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
16 Sep 2015, 06:17
Rocking a Venn Diagram Set theory is no one’s favorite GMAT concept (unless you’re a masochist), but since nearly all testtakers will see at least one overlappingsets 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. Onethird of the winners earned a medal for swimming and onefourth 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 124 = 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. Attachment:
VKerrVenn.jpg [ 7.85 KiB  Viewed 61737 times ]
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 65808

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
16 Sep 2015, 06:24
How to Use a Double Matrix on the GMAT While my personal preference is to use a Venn diagram when dealing with most Sets questions, there are some questions in which a doublematrix is necessary (and much more powerful than a wimpy Venn). This little guy will make even the scariestlooking 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 doublematrix, 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 farright of each row, place a box for the columntotal and rowtotal. 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 questionstem, 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 14x 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! Attachment:
VK1.jpg [ 13.6 KiB  Viewed 61140 times ]
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 65808

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
16 Sep 2015, 06:36
Other Resources on Overlapping Sets
_________________



Manager
Joined: 27 Aug 2014
Posts: 63

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
08 Mar 2016, 04:07
Bunuel wrote: Three Overlapping Sets 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! 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.



Math Expert
Joined: 02 Sep 2009
Posts: 65808

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
08 Mar 2016, 04:29
sinhap07 wrote: Bunuel wrote: Three Overlapping Sets 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! 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
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 65808

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
30 Mar 2016, 10:55
7 Formulas for Tackling Three Overlapping Sets on the GMAT In the previous posts, we saw how to solve three overlapping sets questions using venn diagrams. Here, we will look at all of the various formulas floating around on three overlapping sets. Most of these are self explanatory but we will look into the details of some of them. There are two basic formulas that we already know: 1) Total = n(No Set) + n(Exactly one set) + n(Exactly two sets) + n(Exactly three sets)2) Total = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C) + n(No Set)From these two formulas, we can derive all others. n(Exactly one set) + n(Exactly two sets) + n(Exactly three sets) gives us n(At least one set). So we get: 3) Total = n(No Set) + n(At least one set)From (3), we get n(At least one set) = Total – n(No Set) Plugging this into (2), we then get: 4) n(At least one set) = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C)Now let’s see how we can calculate the number of people in exactly two sets. There is a reason we jumped to n(Exactly two sets) instead of following the more logical next step of figuring out n(At least two sets) – it will be more intuitive to get n(At least two sets) after we find n(Exactly two sets). n(A and B) includes people who are in both A and B and it also includes people who are in A, B and C. Because of this, we should remove n(A and B and C) from n(A and B) to get n(A and B only). Similarly, you get n(B and C only) and n(C and A only), so adding all these three will give us number of people in exactly 2 sets. n(Exactly two sets) = n(A and B) – n(A and B and C) + n(B and C) – n(A and B and C) + n(C and A) – n(A and B and C). Therefore: 5) n(Exactly two sets) = n(A and B) + n(B and C) + n(C and A) – 3*n(A and B and C)Now we can easily get n(At least two sets): 6) n(At least two sets) = n(A and B) + n(B and C) + n(C and A) – 2*n(A and B and C)This is just n(A and B and C) more than n(Exactly two sets). That makes sense, doesn’t it? Here, you include the people who are in all three sets once and n(Exactly two sets) converts to n(At least two sets)! Now, we go on to find n(Exactly one set). From n(At least one set), let’s subtract n(At least two sets); i.e. we subtract (6) from (4) n(Exactly one set) = n(At least one set) – n(At least two sets), therefore: 7) n(Exactly one set) = n(A) + n(B) + n(C) – 2*n(A and B) – 2*n(B and C) – 2*n(C and A) + 3*n(A and B and C)You don’t need to learn all these formulas. Just focus on first two and know how you can arrive at the others if required. Let’s try this in an example problem: Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?(A) 185 (B) 180 (C) 175 (D) 190 (E) 195 You are given that: n(At least one channel) = 250 n(Exactly two channels) = 50 So we know that n(At least one channel) = n(Exactly 1 channel) + n(Exactly 2 channels) + n(Exactly 3 channels) = 250 250 = n(Exactly 1 channel) + 50 + n(Exactly 3 channels) Let’s find the value of n(Exactly 3 channels) = x We also know that n(At least one channel) = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C) = 250 Also, n(Exactly two channels) = n(A and B) + n(B and C) + n(C and A) – 3*n(A and B and C) So n(A and B) + n(B and C) + n(C and A) = n(Exactly two channels) + 3*n(A and B and C) Plugging this into the equation above: 250 = n(A) + n(B) + n(C) – n(Exactly two channels) – 3*x + x 250 = 116 + 127 + 107 – 50 – 2x x = 25 250 = n(Exactly 1 channel) + 50 + 25 n(Exactly 1 channel) = 175, so your answer is C. This question is discussed HERE.
_________________



Manager
Joined: 09 Aug 2016
Posts: 59

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
03 Sep 2016, 09:33
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 ?



Intern
Joined: 13 Jul 2016
Posts: 36

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
26 Sep 2016, 09:43
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.



IIMA, IIMC School Moderator
Joined: 04 Sep 2016
Posts: 1428
Location: India
WE: Engineering (Other)

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
18 Mar 2018, 18:33
Bunuel chetan2u VeritasPrepKarishmaSay, 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
_________________
It's the journey that brings us happiness not the destination. Feeling stressed, you are not alone!!



Math Expert
Joined: 02 Sep 2009
Posts: 65808

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
18 Mar 2018, 19:59
adkikani wrote: Bunuel chetan2u VeritasPrepKarishmaSay, 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
_________________



Math Expert
Joined: 02 Aug 2009
Posts: 8794

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
18 Mar 2018, 20:09
Bunuel wrote: adkikani wrote: Bunuel chetan2u VeritasPrepKarishmaSay, 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 + Neitherso both mean same. you have to choose on basis of info given in question
_________________



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10785
Location: Pune, India

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
18 Mar 2018, 21:39
adkikani wrote: Bunuel chetan2u VeritasPrepKarishmaSay, 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".
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Status: Undergraduate
Joined: 03 Nov 2019
Posts: 23
Location: Cyprus
Concentration: Finance, Economics
GPA: 3

Re: Overlapping Sets Made Easy!
[#permalink]
Show Tags
26 Apr 2020, 12:51
Sounds great,this useful information !
_________________




Re: Overlapping Sets Made Easy!
[#permalink]
26 Apr 2020, 12:51




