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least value venn diagram

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Master | Next Rank: 500 Posts Default Avatar
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kashefian wrote:
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.
Hi!

I did not understand 300??

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Hey finance,

This may sound complex like others but is logically correct. Please don't hate me if you don't understand Smile
Lets try to solve this problem by assuming that there are 20 people.

Now to find the minimum number of people who like all the three,

1. Those who love apples - 70% = 14 (lets assume remaining six love cherries dislike bananas)

2. Those who love bananas - 75% = 15 (remaining 5 love cherries and dislike apples )

3. Those who love cherries - 80% = 16 (assuming remaining 4 just don't like all the three).

subtract those who dislike bananas and apples out of this 16.
16- (remaining from bananas) - (remaining from apples) = who like all the three fruits
16- 5-4 = 5
So out of 20, 5 like all three which is 25%.


Thanks
Krishna.


Lets say

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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?"
Can we solve the problem as stated below -
Lets assume 100% of ppl like all the 3 fruits -
Now according to data given -
100 - 30(who dont like app) = 70
70 - 25(who dont like cherries) = 45
45 - 20(who dont like banana) = 25
So our answer is 25 %.
Am i correct or there is some flaw in my logic ??

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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?"
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 Stuart,
Thanks for the vivid description. It was really helpful.
But I have some problems
I couldn't understand this equation and I have not seen it anywhere. Can you plz explain and also how come you assume the total in exactly 1 group to be 0.

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nitin9003 wrote:
Stuart Kovinsky wrote:
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 Stuart,
Thanks for the vivid description. It was really helpful.
But I have some problems
I couldn't understand this equation and I have not seen it anywhere. Can you plz explain and also how come you assume the total in exactly 1 group to be 0.
Hi Nitin,

here's a good general rule for GMAT mini/maxi questions: whenever you're asked to minimize or maximize one part of a problem, do so by maximizing or minimizing other parts of the problem.

Here's a simpler example:
Quote:
One quarter of the cars on a parking lot have air conditioning and one third of the trucks on the same lot have air conditioning. If there is at least one car and at least one truck on the lot, no other vehicles on the lot, and a total of 72 vehicles on the lot, what's the maximum number of cars that could have air conditioning?
We think: we want to maximize the number of cars with AC, so we need to minimize the number of trucks with AC.

I'll let you try that problem on your own for practice - the answer is 15.

For the question in this thread the reasoning is similar, but more complex.

We want to minimize the percentage of people who like all three types of fruit, but we have some overlap among the groups (since 70+75+80>100); so, we need to maximize the percentage of people who like exactly two types of fruit.

To maximize the percentage of people who like exactly two types of fruit, we need to minimize the percentage of people who like exactly one type of fruit.

Accordingly, to solve, we set "total in exactly one group" to zero.

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[quote="Stuart Kovinsky"]
agganitk wrote:
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.
Hey Stuart thanks a lot for your elaborate explanation but,

I have a question regarding the second equation y we minimize number of ppl who love exactly one shud nt we also maximize them to get minimum love all three exactly types

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wontheGMAT,

Good approach but I just don't understand one thing:
we know that 70 out of 100 eat apples right,
you say that remaining 30 should eat bananas and cherries, both.
why do you assume that those 30 eat both, maybe some of them eat only bananas or cherries?

Maybe I am missing something, so it would be great if you explain.
Thank you

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thanks for the question and the explanation

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And what if there are people that don't eat apples, bananas or cherries in the survey?

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Quote:
And what if there are people that don't eat apples, bananas or cherries in the survey?
The question specifies that there are 80, 75 and 70 cherry/banana/apple lovers respectively.



------

here's an easy way to solve it by plugging in numbers.

you have 225 fruit loves total.
to min the number of 3 fruit lovers, you max the number of two fruit lovers while keeping within 100 people.


since we're minimizing the number of three fruit lovers, start off by letting it be zero

AB=35
AC=35
BC=30
ABC=0

that equals 100 people and 200 fruit loves.
shit, we need 25 more fruit loves while keeping the number of people constant.


AB=30
AC=30
BC=30
ABC=10

that gives us 100 people and 210 fruit loves. not quite there.
what will the next guess be? well, you can see adding 10 3-fruit-lovers gave us an additional 10 fruit loves. we need 15 more fruit loves, so try increasing that number by 15 up to 25.


ABC=25
AB=25
AC=25
BC=25

So we have 100 people. We have 225 fruit loves. What's wrong though? It doesn't meet all the criteria mentioned. there have to be 70 apple lovers, whereas we have 75. there have to be 80 cherry lovers, and we have 75. it's ultimately irrelevent how they're laid out, but for the sake of completeness - take 5 people out of AB and place it into BC.


now we have
ABC=25
AB=20
AC=25
BC=30

100 people
225 fruit loves
A=70
B=75
C=80

winnar

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I have this method .

The Blocks/lines can be used to represent the quantiites.
This is just lika Venn Diagram ,but with lines.
The biggest advantage is you can use it for quantities more >3 very easily.


First lets us assume 100 people

So people who like Apples =70 atleast
I have represented this as the green bar.

Now we have ppl who like bananas=75 atleast
Our Job is to minimize the overlap (so the the ppl who like all three fruits is minimised)
So we have to start the bar from the below so the overlap is minimum
which in this case comes to 15.


Now for the third category of ppl liking cherrieds is 80.

To draw with minimising the overlap ,drawing continuos bar from below or top will
not suffice.
So I divided into 40-40 with gives us a minimun overlap of 25

Hence OA is 25%


I hope the method is clear .
We can take (50-30) or (60-20) split for eth third bar,the answer can be verified to be minimum

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My Solution ::

at least 70% of people like apples, at least 75% like bananas so at least 45 % will like both, call it X. Now take the third fruit 80% like cherries.

Now we have to find percentage of people who will like at least of x and cherries, it will be 25 %

so the answer is 25%

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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?"
Hey Stuart,

I think there is much simpler approach. We are looking for the number of people who like all three and a minimum no at that. Let's try and find out the number of people who are not in A or B or C. That would equate to people who are not in ABC (a set comprising of people who like all three). Simplest way to find this out is to add up all the people who do not like A or B or C. Since we want a maximum no to this figure lets take maximums. 70% like Apples, so 30% do not. 25% do not like bananas and 20% do no like cherries. Add them all up:

30+25+20 = 75 this should give us the "MAXIMUM POSSIBLE" no of people who do not like any particular fruit of the 3. Subtract that from the total no:

100 - 75 = 25% and that should be the answer. The key is in realising that the question asks for the Minimum Possible people who like A, B, and C.
Fining the maximum possible people who do not like any of A, B or C should give us the minimum possible for ABC.

Is there anything flawed with the approach above?

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20% no cherries - possible minimum of people who might like all three fruits

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I got 25% but only out of an intuitive guess!

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