least value venn diagram

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by pnvpratik » Mon May 17, 2010 10:49 pm
Stuart Kovinsky wrote: The second equation is derived from another version of the triple-overlapping set equation:

True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)

We set the "total in exactly 1 group" to 0, so we get:

100 = 0 + AB + AC + BC + ABC
Hi
I know this thread is quite old , but I was having a look at the triple-overlapping concept
and i am confused as to how the second formula mentioned by Stuart was derived . And why we substituted 0 for "total in exactly 1 group"
Any explanations ?

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by Stuart@KaplanGMAT » Mon May 17, 2010 11:55 pm
pnvpratik wrote:
Stuart Kovinsky wrote: The second equation is derived from another version of the triple-overlapping set equation:

True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)

We set the "total in exactly 1 group" to 0, so we get:

100 = 0 + AB + AC + BC + ABC
Hi
I know this thread is quite old , but I was having a look at the triple-overlapping concept
and i am confused as to how the second formula mentioned by Stuart was derived . And why we substituted 0 for "total in exactly 1 group"
Any explanations ?
Hi!

Let's start with the easier question: we substituted 0 for "total in exactly 1 group" because our goal is to minimize the people in all 3 groups, and to do so we need to maximize the number of people in exactly 2 groups (as explained earlier in my original post).

The formula is derived from a triple circle Venn diagram.

If you look at all the regions in the diagram, you'll see that there are 3 areas for singles (1 per circle), 3 areas for doubles (each circle overlaps with 2 other circles) and 1 area for triples (the area in the middle is for the people in all 3 groups). Those 7 areas add up to everyone in the system (assuming that everyone is in at least one group, which is the case in the vast vast majority of GMAT 3 group overlapping set questions).

If we call the 3 circles A, B and C, we have:

Singles: A, B, C
Doubles: AB, AC, BC
Triple: ABC

and the formula:

Total # = A + B + C + AB + AC + BC + ABC

which we can rewrite as:

Total = (A + B + C) + (AB + AC + BC) + (ABC)

and then simplify as:

Total = (singles) + (doubles) + (triples)

which is the final version derived above.
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by pnvpratik » Tue May 18, 2010 12:53 am
Stuart Kovinsky wrote:
pnvpratik wrote:
Stuart Kovinsky wrote: The second equation is derived from another version of the triple-overlapping set equation:

True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)

We set the "total in exactly 1 group" to 0, so we get:

100 = 0 + AB + AC + BC + ABC
Hi
I know this thread is quite old , but I was having a look at the triple-overlapping concept
and i am confused as to how the second formula mentioned by Stuart was derived . And why we substituted 0 for "total in exactly 1 group"
Any explanations ?
Hi!

Let's start with the easier question: we substituted 0 for "total in exactly 1 group" because our goal is to minimize the people in all 3 groups, and to do so we need to maximize the number of people in exactly 2 groups (as explained earlier in my original post).

The formula is derived from a triple circle Venn diagram.

If you look at all the regions in the diagram, you'll see that there are 3 areas for singles (1 per circle), 3 areas for doubles (each circle overlaps with 2 other circles) and 1 area for triples (the area in the middle is for the people in all 3 groups). Those 7 areas add up to everyone in the system (assuming that everyone is in at least one group, which is the case in the vast vast majority of GMAT 3 group overlapping set questions).

If we call the 3 circles A, B and C, we have:

Singles: A, B, C
Doubles: AB, AC, BC
Triple: ABC

and the formula:

Total # = A + B + C + AB + AC + BC + ABC

which we can rewrite as:

Total = (A + B + C) + (AB + AC + BC) + (ABC)

and then simplify as:

Total = (singles) + (doubles) + (triples)

which is the final version derived above.
Ok.. Actually had a Edison moment while reading your reply.
So in summary even though the 2 formula's are for triple overlapping , they approach the problem in a different way.
The first one when it says
(total # in group 1) it is the total # of people who at least love Apple. ( means AB AC ABC are included)

While in the second one
(total in exactly 1 group) it means people who love just apple.

Also if i understood correctly
Since there are 125 more "fruit loves" than there are people there will always be people who love more than 1 fruit.

As i am writing this I understand why we wanted to minimize only 1 fruit lovers.

Thanks Stuart. Your replies help a lot.

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by waltz2salsa » Mon Oct 18, 2010 10:07 am
even after going through all the wonderful explanations here, i am still unsure about when to use the following formulas :

1. True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in exactly 2 groups) - 2(# in all 3 groups) as quoted here and

2. P(AuBuC) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC) as quoted in this forum as well as lot other places such as https://gmat-maths.blocked/2008/05/ ... mulas.html.

Please explain the different scenarios when they must be applied.....

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by GMATGuruNY » Mon Oct 18, 2010 11:02 am
agganitk wrote:According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the minimum percentage of people who like all three?

A. 15%
B. 20%
C. 25%
D. 0%
E. 35%

Ans??
We can plug in the answers choices, which represent the minimum percentage who like all 3 fruits.

The total of all the percentages given is 70+75+80 = 225%. When we subtract all the overlaps (those who like all 3 fruits and those who like exactly 2), the total of all the remaining percentages must be 100% or less. We can eliminate D because the maximum percentage who like exactly 2 of the fruits can't be more than 100%, and we need to subtract from 225 at least 125, so some of the people must like all 3 fruits.

Answer choice A: 15% like all 3 fruits
70-15 = 55% who like apples and perhaps 1 other fruit
75-15 = 60% who like bananas and perhaps 1 other fruit
80-15 = 65% who like cherries and perhaps 1 other fruit
New total is 55+60+65 = 180%
We need to subtract another 80%. Let's try to maximize the percentage who like exactly 2 of the fruits.
People who like cherries and bananas = 60 (the maximum overlap since only 60% like bananas)
5 cherry lovers left. Let's say that they also like apples.
180-60-5 = 115. Doesn't work.

Answer choice B: 20% like all 3 fruits
70-20 = 50% who like apples and perhaps 1 other fruit
75-20 = 55% who like bananas and perhaps 1 other fruit
80-20 = 60% who like cherries and perhaps 1 other fruit
New total is 50+55+60 = 165%
We need to subtract another 65%. Let's try to maximize the percentage who like exactly 2 of the fruits.
People who like cherries and bananas = 55 (the maximum overlap since only 55% like bananas)
5 cherry lovers left. Let's say that they also like apples.
165-55-5 = 105. Doesn't work.

105 is so close to 100 that the correct answer must be C, the next largest.

The correct answer is C.

For the skeptical among us:

Answer choice C: 25% like all 3 fruits
70-25 = 45% who like apples and perhaps 1 other fruit
75-25 = 50% who like bananas and perhaps 1 other fruit
80-25 = 55% who like cherries and perhaps 1 other fruit
New total is 45+50+55 = 150%
We need to subtract another 50%. Let's try to maximize the percentage who like exactly 2 of the fruits.
People who like cherries and bananas = 50% (the maximum overlap since only 50% like bananas)
150-50 = 100. Success!
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by fskilnik@GMATH » Tue Oct 19, 2010 9:19 am
heshamelaziry wrote:In a group of 68 students, each student is registered for at least one of three classes - History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?
Hi, heshamelaziry!

The problem discussed in this post is much harder than the one you ask (quoted above)!

Please note that one of the equations they were dealing with was:

(reunion of A, B and C) = (total in A) + (total in B) + (total in C) - (total in exactly 2 groups) - 2(total in all 3 groups)

Therefore just plug in the numbers given (in "your" exercise) at the proper places to find the answer you are looking for.

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by pzazz12 » Fri Oct 22, 2010 3:59 am
life is a test wrote:My attempt at an explanation in the attached diag.

hope that helps.
thanks... for u r simple process in diagrammatically way..........

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by pzazz12 » Mon Oct 25, 2010 3:52 am
Stuart Kovinsky wrote:
agganitk wrote:According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the minimum percentage of people who like all three?

A. 15%
B. 20%
C. 25%
D. 0%
E. 35%

Ans??
Let's say we have 100 people to make things simple. We want to minimize the triple group, so let's minimize how many people like each kind of fruit, giving us:

70 apple lovers, 75 banana lovers and 80 cherry lovers.

Now, 70 + 75 + 80 = 225, so we have 225 "fruit loves" spread out among 100 people.

Therefore, there are 125 more "fruit loves" than there are people.

Our job is to minimize the number of people who love all 3; to do so, we want to maximize the number of people who love exactly 2 of the 3 and minimize the number of people who love exactly 1 of the 3.

So, we can come up with two equations:

AB + AC + BC + 2ABC = 125

and

AB + AC + BC + ABC = 100

AB = number who like just apple/banana
AC = number who like just apple/cherry
BC = number who like just banana/cherry
ABC = number who like all 3

The first equation is derived from the triple-overlapping set equation:

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in exactly 2 groups) - 2(# in all 3 groups)

100 = 70 + 75 + 80 - AB - AC - BC - 2(ABC)

and when we rearrange to get all variables on one side:

AB + AC + BC + 2ABC = 125

The second equation is derived from another version of the triple-overlapping set equation:

True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)

We set the "total in exactly 1 group" to 0, so we get:

100 = 0 + AB + AC + BC + ABC

So, back to our equations:

AB + AC + BC + 2ABC = 125
AB + AC + BC + ABC = 100

If we subtract the second from the first, we get:

ABC = 25... done!

Now, at this point you may be saying, "umm.. ok.. but I asked for a simple way to solve, that seemed super complicated and time consuming!"

However, if you understand the concepts behind triple-overlap (or double-overlap) questions, it's fairly intuitive; the complicated part is getting to the stage at which you have that deeper understanding.

Of course, this exact question won't appear on the GMAT. So, as always, after you do a question you ask yourself: "what did I learn from this question that's going to help me on future questions?"

Here's our takeaways:

1) there are multiple ways to solve overlapping sets questions. The more you familiarize yourself with the 3 major approaches (equations/venn diagrams/matrices(the last only works when there are 2 overlapping sets, unless you're really good at drawing a 3-dimensional matrix)), the more likely it is that the quickest approach will jump out at you on test day.

2) if you're shooting for a 600+, learn the two equations noted above.

3) whenever you're asked to minimize something, think "what do I need to maximize to achieve that result?"
thank you...is there any simple method to solve it in short period..........

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by gmatmachoman » Sat Oct 30, 2010 4:48 am
@Stuart/Guru/fskil

If the problem asks for Maximum percentage of people who like all three?


Is it 62.5 %??

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by nipunkathuria » Thu Nov 18, 2010 10:30 am
Stuart Kovinsky wrote:
agganitk wrote:According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the minimum percentage of people who like all three?

A. 15%
B. 20%
C. 25%
D. 0%
E. 35%

Ans??
Let's say we have 100 people to make things simple. We want to minimize the triple group, so let's minimize how many people like each kind of fruit, giving us:

70 apple lovers, 75 banana lovers and 80 cherry lovers.

Now, 70 + 75 + 80 = 225, so we have 225 "fruit loves" spread out among 100 people.

Therefore, there are 125 more "fruit loves" than there are people.

Our job is to minimize the number of people who love all 3; to do so, we want to maximize the number of people who love exactly 2 of the 3 and minimize the number of people who love exactly 1 of the 3.

So, we can come up with two equations:

AB + AC + BC + 2ABC = 125

and

AB + AC + BC + ABC = 100

AB = number who like just apple/banana
AC = number who like just apple/cherry
BC = number who like just banana/cherry
ABC = number who like all 3

The first equation is derived from the triple-overlapping set equation:

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in exactly 2 groups) - 2(# in all 3 groups)

100 = 70 + 75 + 80 - AB - AC - BC - 2(ABC)

and when we rearrange to get all variables on one side:

AB + AC + BC + 2ABC = 125

The second equation is derived from another version of the triple-overlapping set equation:

True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)

We set the "total in exactly 1 group" to 0, so we get:

100 = 0 + AB + AC + BC + ABC

So, back to our equations:

AB + AC + BC + 2ABC = 125
AB + AC + BC + ABC = 100

If we subtract the second from the first, we get:

ABC = 25... done!

Now, at this point you may be saying, "umm.. ok.. but I asked for a simple way to solve, that seemed super complicated and time consuming!"

However, if you understand the concepts behind triple-overlap (or double-overlap) questions, it's fairly intuitive; the complicated part is getting to the stage at which you have that deeper understanding.

Of course, this exact question won't appear on the GMAT. So, as always, after you do a question you ask yourself: "what did I learn from this question that's going to help me on future questions?"

Here's our takeaways:

1) there are multiple ways to solve overlapping sets questions. The more you familiarize yourself with the 3 major approaches (equations/venn diagrams/matrices(the last only works when there are 2 overlapping sets, unless you're really good at drawing a 3-dimensional matrix)), the more likely it is that the quickest approach will jump out at you on test day.

2) if you're shooting for a 600+, learn the two equations noted above.

3) whenever you're asked to minimize something, think "what do I need to maximize to achieve that result?"

Hi...can u plz explain how did u arrive at the following relationship:

AB + AC + BC + 2ABC = 125

and

AB + AC + BC + ABC = 100
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by Stuart@KaplanGMAT » Thu Nov 18, 2010 11:00 am
nipunkathuria wrote:Hi...can u plz explain how did u arrive at the following relationship:

AB + AC + BC + 2ABC = 125

and

AB + AC + BC + ABC = 100
Hi,


We have 100 people and 225 "fruit loves", so we know that there are 125 (225-100) duplicates out there. AB, AC and BC represent the people who love exactly two types of fruit and ABC represents the people who love all three types.

The two types people get counted once in our equation, because each of those people provides 1 extra "love". The three types people get counted twice in our equation, because each of those people provides 2 extra "loves".

The second equation is derived from the Venn diagram.

My original post goes through the derivation of both in quite a bit of detail, so if there's a specific part of the explanation on which you'd like me to elaborate, let me know and I'm happy to do so.
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by kashefian » Thu Nov 18, 2010 11:24 pm
I solved this problem like this:

P (A & B & C) = P(A) + P(B) + P(C) - P(A or B or C) or
(A & B & C) = (A) + (B) + (C) - (A or B or C)
x = 300 - 225 = 75

Therefore the probability is 75/300 = 25%

Note that A + B + C = 300 because we have a total of 300 people. It took me around 30 seconds to answer the question.

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by jd792104 » Fri Dec 03, 2010 4:03 pm
All

It seems that the solution assumes that there are no people that do not like either apple, banana or cherries? Am I missing something?

If that is the case then can such an assumption always be made in problems like this?

look fwd to responses.

jd

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by djiddish98 » Thu Feb 03, 2011 8:03 am
I've approached problems like these in a very simple manner (although I do enjoy the underlying logic that has been presented in the discussion)

I first looked at the overlap between the bananas and apples

70%+75% = 145%.
145% - 100% (the maximum) = 45%, which means there is 45% overlap between the groups.

Now factor in cherries

45% + 80% = 125%

125% - 100% = 25% overlap for all 3.

Is there anything wrong with taking a step-by-step approach like this?

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by Namritha » Mon Mar 07, 2011 4:49 am
To get equation 2
100= AB+BC+CA+ ABC
why do we set the total of group1 to be '0'

thanks