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Set Theory challenge

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by Rabeea » Tue Mar 01, 2011 11:02 am
why we have taken cheese the largest possible value for the range?
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by fskilnik@GMATH » Tue Mar 01, 2011 11:30 am
Rabeea wrote:why we have taken cheese the largest possible value for the range?
Because we were looking for Xmax!

What I mean? If you have 20 people that like magazine A and 30 people that like magazine B, what is the maximum number of people that may like both magazines? 20. Why? Because 21 or more people could not be (why?) and 19 or less people is not the maximum possible, because ALL people that like mag A could also like mag B.

I hope you got it.

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Fabio.
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by Rabeea » Tue Mar 01, 2011 11:43 am
sir there is still confusion,
For example we have three sets,
say it set A , set B and Set C
20 people in A, 20 in B and 26 in C
How can we find the maximum number of people that is AuBuC
Rj
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by fskilnik@GMATH » Tue Mar 01, 2011 1:11 pm
Rabeea wrote:sir there is still confusion,
For example we have three sets,
say it set A , set B and Set C
20 people in A, 20 in B and 26 in C
How can we find the maximum number of people that is AuBuC
Hi there, Rabeea.

You must understand (and memorize) the following very-useful "formula":

AuBuC = A+B+C - (Sum of Exactly 2 of them) - 2* (Exactly 3 of them)

(The reason for why this is true is REALLY easy. Try to find by yourself "counting" the regions of the Venn diagram...)

Therefore AuBuC = 20+20+26 - (sum of exactly 2) - 2(exactly 3).

From the fact that (sum of exactly 2) and (exactly 3) are positive or zero (always), to maximize AuBuC the "ideal" would be to put zero in each one of them! The question is: is it possible to have A = 20, B = 20, C = 26 and all intersections mentioned equal to zero?

The answer is: it is! Please note that 20+20+26 is less or equal than 100 (percent), therefore we could have A = 20 = only A, B = 20 = only B and C = 26 = only C and the rest would go to neither A, nor B, nor C.

In this scenario AuBuC = 20+20+26 = 66 and this would be the maximum value for AuBuC (with the numbers you provided).

Is this what you were looking for (as an answer) ?

Regards,
Fabio.
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by Rabeea » Tue Mar 01, 2011 1:46 pm
Actual question was
Each of the 49 students in the class understands equations,excersiceregularly or loves litreture.20 understand equations, 20 excercise regularly and 26 loves litreture.
36 understands equations or excersice regularly
38 understands equation or loves litreture
40 excersice regularly or loves litreture
Suppose you dont know the no.of students in the class,What is the largest possible no.of students in class?
Rj
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by fskilnik@GMATH » Tue Mar 01, 2011 2:06 pm
Rabeea wrote:Actual question was
Each of 49 students in the class understands equations,exercises regularly or loves literature.
20 understand equations, 20 excercise regularly and 26 loves litreture.
36 understands equations or excersice regularly
38 understands equation or loves litreture
40 excersice regularly or loves litreture
Suppose you dont know the no.of students in the class,What is the largest possible no.of students in class?
Total = (at least one thing) + (None), that is, Total = AuBuC + Remainder.

A = understand equations
B = exercise regularly
C = loves literature

From the question stem we know that:

AuBuC = 49
A = 20
B = 20
C = 26

AuB = 36
AuC = 38
BuC = 40

Imagine (Exactly 3) = x students

From what I mentioned we have:

49 = AuBuC = 20 + 20 + 26 - [(36-x) + (38-x) + (40-x)] - 2x therefore x = 97 (check that).

That means ABSURD. The data given in the question stem is impossible to be satisfied!

The red number means that I guess you typed it wrong.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Rabeea » Tue Mar 01, 2011 11:54 pm
this 49 was in my question may be absurd.
But we have to suppose that this 49 is not mentioned now we have to find the largest possible number of students.

from your concept if i get AuBuC 66
then by applying inclusion exclusion principle i can find the largest possible number of students.
that is to say
A inter B inter C
Rj
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