least value venn diagram

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by GMATGuruNY » Mon Jun 22, 2015 7:30 am
kulsim wrote:Hi! Apologies if this sounds trivial but just trying to make sure I am getting this fully. I now understood why we got 25% and just trying to think about this problem in reverse. If we were asked to maximize the ABC, would the answer be simply 75%?
Not quite.
To make the reasoning easier to see, let's assume that everyone in the survey likes at least one of the 3 fruits.
In this case, we can use the following formula:

T = A + B + C - (AB + AC + BC) - 2(ABC)

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in A, everyone in B, and everyone in C:
Those in exactly 2 of the groups (AB+AC+BC) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (ABC) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.

In the problem above:
T = 100%.
A = 70%.
B = 75%.
C = 80%.

Plugging these values into the formula, we get:
100 = 70 + 75 + 80 - (AB + AC + BC) - 2(ABC).
100 = 225 - (AB + AC + BC) - 2(ABC)
(AB + AC + BC) + 2(ABC) = 125.

MINIMUM value of ABC:
To MINIMIZE the value of ABC, we must MAXIMIZE the value of (AB + AC + BC).
Here, the maximum value of (AB + AC + BC) = 75, as shown in my post above:
75 + 2(ABC) = 125
2(ABC) = 50
ABC = 25%.
Thus, the least possible value of ABC = 25%.

MAXIMUM value of ABC:
To MAXIMIZE the value of ABC, we must MINIMIZE the value of (AB + AC + BC).
If AB + AC + BC = 0, we get:
0 + 2(ABC) = 125
2(ABC) = 125
ABC = 62.5%.
Thus, the greatest possible value of ABC = 62.5%.

If it's possible that some people like NONE of the 3 fruits, we can use a variation of the formula above:

T = A + B + C - (AB + AC + BC) - 2(ABC) + None.

In this case -- since the percentage who like all 3 fruits cannot exceed the percentage who like cherries -- the greatest possible value of ABC is 70% (the percentage attributed to C).
The least possible for (AB + AC + BC) is still 0.
Plugging all of the percentages into the amended formula, we get:

100 = 70 + 75 + 80 - 0 - 2(70) + None
100 = 85 + None
None = 15%.
Implication:
If 15% like none of the 3 fruits, then it's possible that 70% like all 3 fruits.

For more practice with triple-overlapping groups, check my post here:
https://www.beatthegmat.com/og-13-178-vi ... 11188.html
Last edited by GMATGuruNY on Tue Jun 23, 2015 9:54 am, edited 1 time in total.
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by kulsim » Tue Jun 23, 2015 7:55 am
Mitch

Thanks for your reply. I now understand how the formula works and your explanation is very helpful. Trying to think about this problem for a different angle I drew down the overlaps to visualise them.I am able to get the correct answer of 25% when I try to minimise the overlap (minimum overlap between A and B is 45% and then when we add C it is 25%). However, when I do the same I get the maximum overlap of 70%. Which constraints am I not taking into account?

Thanks!

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by GMATGuruNY » Tue Jun 23, 2015 10:00 am
kulsim wrote:Mitch

Thanks for your reply. I now understand how the formula works and your explanation is very helpful. Trying to think about this problem for a different angle I drew down the overlaps to visualise them.I am able to get the correct answer of 25% when I try to minimise the overlap (minimum overlap between A and B is 45% and then when we add C it is 25%). However, when I do the same I get the maximum overlap of 70%. Which constraints am I not taking into account?

Thanks!
Case 1:
If everyone must like at least one of the 3 fruits, then the greatest possible value of ABC is 62.5%.
Case 2:
If it's possible that some people like none of the 3 fruits, then the greatest possible value of ABC is 70%.
Please revisit my post above.
I've amended it to include both cases.
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by E.T. » Sat Nov 19, 2016 1:34 am
Ok so I haven't read all 7 pages so I don't know if my method is going to add anything to this thread. I'm also curious to see whether this would work for all problems of this type, so please do correct me/comment if I'm wrong/whether you know any situations that this won't work in!

I took a group of n=20 'cause that got me nice round numbers with the percentages.

So if 70% likes Apples, 1-70%=30% doesn't like them --> 14 people like and 6 people don't like (so they can at most like 2 fruits).
75% (15 people) like bananas, 25% (5) doesn't like (so can only like at most 2 fruits).
80% (16) likes cherries, 20% (4) doesn't (can only like at most 2 fruits).
We're looking at the minimum possible number that has to like 3 fruits, so we can just pretend all those people that like at most 2 fruits are different people --> we can just add them all up. That gets you 6+5+4=15. That means that 5 people have to like all 3 fruits. 5/20=25%.

Hope this could be of any help to anyone!

Cheers!

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by Lakshya05 » Wed Apr 12, 2017 5:06 pm
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?"

This question has been posted long time ago but can anyone advise why we can equate A+B+C to 0 and not (AB+BC+CA) in below equation

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

100 = 0 + AB + AC + BC + ABC