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Trouble with set theory!

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Trouble with set theory!

by papgust » Sun Feb 07, 2010 5:53 am
I'm having some trouble with understanding the concept of Overlapping sets. I'm fine with basic overlapping sets such as two-set diagrams. But when it comes to three-set diagrams with questions such as "exactly 2 of the sets" or "atleast 3 of the sets" etc, i'm finding it difficult to understand the concepts behind it.

Further, i want to know when and how we must discount double-counting. I'm pretty overwhelmed by the different kinds of formulae for the scenarios mentioned above. I would like to understand the concept behind each formulae. Can someone explain the concepts in detail or point to a good resource where i can find whatever i'm looking for?
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by harsh.champ » Sun Feb 07, 2010 6:17 am
papgust wrote:I'm having some trouble with understanding the concept of Overlapping sets. I'm fine with basic overlapping sets such as two-set diagrams. But when it comes to three-set diagrams with questions such as "exactly 2 of the sets" or "atleast 3 of the sets" etc, i'm finding it difficult to understand the concepts behind it.

Further, i want to know when and how we must discount double-counting. I'm pretty overwhelmed by the different kinds of formulae for the scenarios mentioned above. I would like to understand the concept behind each formulae. Can someone explain the concepts in detail or point to a good resource where i can find whatever i'm looking for?

Now, in 3 sets diagrams,

The main formula that is to be kept in mind is:-
P(A U B U C) = P(A) + P(B) + P(C) - [P(A n B) + P(B n C) + P(C n A) ] + P(A n B n C)

where "U" denotes union and "n" denotes intersection.

"Exactly two of the sets means(though can you clarify this point) " means P(A) + P(B) + P(C) - [P(A n B) + P(B n C) + P(C n A) ]
"atleast 3 of the sets(though can you also clarify this point) " means P(A U B U C)
Just plot out a Venn diagram and you will know why this formula exists.

As for the problem of double-counting,start practising solving the question by the formal and the proper approach rather the hit-and-trial or plug-in technique.Initially it will be time taking and take up a lot of time but once you become conversant with the method you will automatically start improving.[This is from my personal experience-Initially I also had this problem of double counting when I used to adopt shortcut techniques]
Last edited by harsh.champ on Sun Feb 07, 2010 6:52 am, edited 2 times in total.
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by papgust » Sun Feb 07, 2010 6:24 am
I have the 3rd edition book. I felt that the content is not enough for me to understand the concepts better.
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by harsh.champ » Sun Feb 07, 2010 6:59 am
papgust wrote:I'm having some trouble with understanding the concept of Overlapping sets. I'm fine with basic overlapping sets such as two-set diagrams. But when it comes to three-set diagrams with questions such as "exactly 2 of the sets" or "atleast 3 of the sets" etc, i'm finding it difficult to understand the concepts behind it.
Can you post any question in the same thread regarding "exactly two of the sets" and "atleast 3 of the sets"...
Though I am sure that my post above is sufficient,but still I would like to look at a question and justify it.
I hope you are acquainted with terms "union" and "intersection".

In exactly two of the sets , just keep in mind not to count the intersection of all the 3 sets i.e. the term P(A n B n C)
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by papgust » Sun Feb 07, 2010 9:26 pm
Hey Harsh,

Firstly, thanks a lot for showing interest to help me. As you requested, i searched for a couple of problems in this forum and picked them. Here you go!

Q1 :
In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?

OA 20

Q2 :
There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German.

OA 5


I need a traditional method of solving this problem without any shortcuts. If i see any problem of this sort, i must be able to solve them with ease.

Thanks for your help once again!
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by harsh.champ » Mon Feb 08, 2010 2:03 am
papgust wrote:Hey Harsh,

Firstly, thanks a lot for showing interest to help me. As you requested, i searched for a couple of problems in this forum and picked them. Here you go!

Q1 :
In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?

OA 20

I need a traditional method of solving this problem without any shortcuts. If i see any problem of this sort, i must be able to solve them with ease.

Thanks for your help once again!
Well,thanks for thanking me .Can you please also click on the thank button . :):P

Now, taking one question at a time :_
Q1) P(1 U 2 U 3) = 0.85 [85% of those surveyed liked at least one of three products]
P(1) = 0.5 [50% of those asked liked product 1]
P(2)=0.3 [30% liked product 2 ]
P(3)=0.2 [20% liked product 3 ]
P(1 n 2 n 3 ) = 0.05

Applying the main formula I gave above,
P(A U B U C) = P(A) + P(B) + P(C) - [P(A n B) + P(B n C) + P(C n A) ] + P(A n B n C) - (X)
From,equation (X) , 0.85 = 0.5 + 0.3 + 0.2 - [P(1 n 2) + P(2 n 3) + P(3 n 1 ] + 0.05
=>[P(1 n 2) + P(2 n 3) + P(3 n 1 ] = 0.2

% of survey participants that liked more than 1 product = [P(1 n 2) + P(2 n 3) + P(3 n 1 ] - 2P(1 n 2 n 3) {as P(1 n 2 n 3 ) is counted 3 times) -(Y)

From eqn. (X) and (Y), [spoiler]we get the answer as 0.2 - 2(0.05) = 0.1.
I dont know how why answer is coming to be 0.2.[/spoiler]
I have rechecked all the equations also.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by harsh.champ » Mon Feb 08, 2010 2:29 am
papgust wrote:
Q2 :
There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German.

OA 5


I need a traditional method of solving this problem without any shortcuts. If i see any problem of this sort, i must be able to solve them with ease.

Thanks for your help once again!
P(M U E U G)=70
P(M) -P(M n E) - P(M n G) + P(M n E n G) = 40 -(1)
P(G) -P(G n E) - P(G n M) + P(M n E n G) = 30 - (2)
P(E) -P(G n E) - P(M n E) + P(M n E n G) = 35 - (3)

Now, P(M n E n G) = 15
Applying master formula:-
P(A U B U C) = P(A) + P(B) + P(C) - [P(A n B) + P(B n C) + P(C n A) ] + P(A n B n C)
=> 70 = P(M) + P(G) + P(E) -P(G n E) - P(M n E) - P(M n G) + P(M n E n G) - (4)

Now,adding (1),(2) and (3) we get
P(M) + P(G) + P(E) - 2[P(G n E) + P(M n E) + P(M n G)] + 3P(M n E n G) = 105
Now, P(M n E n G) = 15
=>P(M) + P(G) + P(E) - 2[P(G n E) + P(M n E) + P(M n G)] + 3x15 = 105
=>P(M) + P(G) + P(E) - 2[P(G n E) + P(M n E) + P(M n G)] = 60 -(5)
Using (4) and (5), we get
70 = 60 + [P(G n E) + P(M n E) + P(M n G)] + 15
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by Stuart@KaplanGMAT » Mon Feb 08, 2010 12:26 pm
For 3 set questions, there are 2 formulae that you can use.

For the sake of simplicity, I'm going to call our 3 characteristics A, B and C.

Picture a Venn diagram; the first formula is just the sum of all of the various parts:

True # of objects = (# only A) + (# only B) + (# only C) + (# only AB) + (# only AC) + (# only BC) + (# only ABC)

The second formula is the one we use more often:

True # of objects = (total # A) + (total # B) + (total #C) - (# only AB) - (# only AC) - (# only BC) - 2(# ABC)

[Note that, technically, we should add a "+ (# with none of ABC)" to the end of each equation, but I've never seen a 3 set question on the GMAT that had a "none" component.]

We can simplify the second equation to:

True # of objects = (sum of total characteristics) - (sum of doubles) - 2(triples)

To understand why we have to subtract the doubles once and the triples twice, again picture a Venn diagram.

If an object is in the AB portion of the diagram, it's already been counted in the A circle and the B circle. In other words, it's been counted twice. To get a true count, therefore, we must subtract it once.

If an object is in the ABC portion of the diagram, it's already been counted in the A circle, the B circle AND the C circle. In other words, it's been counted three times. To get a true count, therefore, we must subtract it twice.

Here's the primary principle we're using:

every object should be counted exactly once.
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by Stuart@KaplanGMAT » Mon Feb 08, 2010 12:32 pm
papgust wrote: Q1 :
In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?

OA 20
Let's say there's 100 people, just to use numbers instead of percents. Since 85% like at least one of 3 products, we'll use 85 as our base number.

Since we have "at least one" information, we need to use the second formula:

True # of objects = (sum of total characteristics) - (sum of doubles) - 2(triples)

plugging in what we know:

85 = 50 + 30 + 20 - doubles - 2(5)

85 = 100 - 10 - doubles

85 = 90 - doubles

doubles = 5

The question is what % like more than 1, so we want to solve for:

doubles + triples

= 5 + 5 = 10

So, 10% is the final answer.

(The posted OA is wrong.)
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by Stuart@KaplanGMAT » Mon Feb 08, 2010 12:34 pm
papgust wrote:
Q2 :
There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German.

OA 5
We use the exact same formula:

True # of objects = (sum of total characteristics) - (sum of doubles) - 2(triples)

70 = (40 + 30 + 35) - (doubles) - 2(15)

70 = 105 - doubles - 30

70 = 75 - doubles

doubles = 5
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by harsh.champ » Mon Feb 08, 2010 2:01 pm
Stuart Kovinsky wrote:
papgust wrote: Q1 :
In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?

OA 20
Let's say there's 100 people, just to use numbers instead of percents. Since 85% like at least one of 3 products, we'll use 85 as our base number.

Since we have "at least one" information, we need to use the second formula:

True # of objects = (sum of total characteristics) - (sum of doubles) - 2(triples)

plugging in what we know:

85 = 50 + 30 + 20 - doubles - 2(5)

85 = 100 - 10 - doubles

85 = 90 - doubles

doubles = 5

The question is what % like more than 1, so we want to solve for:

doubles + triples

= 5 + 5 = 10

So, 10% is the final answer.

(The posted OA is wrong.)
Thats true Stuart.
I also got the answer as 0.1.
Can you check my soln. approach above and tell me if my solving method was right or not??
P(A U B U C) = P(A) + P(B) + P(C) - [P(A n B) + P(B n C) + P(C n A) ] + P(A n B n C) - (X)
From,equation (X) , 0.85 = 0.5 + 0.3 + 0.2 - [P(1 n 2) + P(2 n 3) + P(3 n 1 ] + 0.05
=>[P(1 n 2) + P(2 n 3) + P(3 n 1 ] = 0.2

% of survey participants that liked more than 1 product = [P(1 n 2) + P(2 n 3) + P(3 n 1 ] - 2P(1 n 2 n 3) {as P(1 n 2 n 3 ) is counted 3 times) -(Y)

From eqn. (X) and (Y), we get the answer as 0.2 - 2(0.05) = 0.1.
I dont know how why answer is coming to be 0.2.

Also,in the 2nd question,I seem to get a -ve value.Is that question right or am I missing something?
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by papgust » Mon Feb 08, 2010 8:51 pm
Thanks Stuart! That was great.. You've given every concept of Venn diagrams in a nutshell. This was what i expected.
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by papgust » Mon Feb 08, 2010 11:29 pm
Stuart, I've seen some problems with intersection of all sets being subtracted "thrice" i.e. - 3(triples). When and why do we need to do that?
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by Stuart@KaplanGMAT » Tue Feb 09, 2010 8:34 pm
papgust wrote:Stuart, I've seen some problems with intersection of all sets being subtracted "thrice" i.e. - 3(triples). When and why do we need to do that?
I've never seen a question on which we'd subtract 3 times... do you have an example?
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by papgust » Tue Feb 09, 2010 9:09 pm
Stuart,

Thanks for your prompt response. I've seen some problems of that type. I tried to search for those type of questions but couldn't find any. However, i came across a post here.

https://www.beatthegmat.com/formulas-for ... 16479.html

This post tells that for "exactly 2 of the sets" problems, intersection of all must be subtracted thrice.

I'm actually overwhelmed by seeing so many forumale in this link. I've a couple of questions for you.

1. Do we really need to know these for GMAT?
2. Can we actually answer any type of questions (as posted in the link) with the 2 formulae you shared with us in this thread?
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