In a multiplex 3 shows A, B, and C run in 3 auditoriums. One day, 1000

good question. But in your case,

only A = 170
ABC, which is B = 250
only C = 200

So, adding all = 620
Which means neither = 380, but per question neither = 0.
Therefore, this case is not possible.

Just a question crossed in my mind... generally in such questions.... we have a tendency to pick the the minimum of A,B,C which is B=250 as maximum. The venn diagram will look as attached
Why we did not take this approach here?

if the maximum is indeed 60.... how will the venn diagram look like?

rsrighosh

When the Union of the overlapping sets is defined (I.e., we are old that every one or the 1,000 people watched at least 1 movie, meaning there aren’t any people who didn’t watch a movie) the method I find the easiest to use is the following:

(1st) We are maximizing the amount of ppl who saw all 3 movies. That number can never exceed the smallest of the 3 sets - in this case, the MAX can not exceed the 250 ppl who watched B

(2nd) Total up all the recorded members who saw a movie (the “entire circle” for A , B , and C)

(420) + (250) + (450) = 1,120

So we have 1,120 recorded views of any 1 movie. Yet we have only 1,000 unique people.

This surplus of +120 must be accommodated by the fact that some people watched 2 movies or 3 movies.

(2nd) imagine we have 1,120 pieces of candy to hand out. Each piece of candy represents 1 viewing out of the 1,120 total viewings of any particular movie that we Summed up above (let us say that a person gets one piece of candy every time he or she goes to view a movie).

We have to account for all 1,120 pieces of candy among the 1,000 unique people.

(3rd) we need to Take Care of the Minimum Requirement:

since all 1,000 people saw at least one movie, we have to give all 1,000 people (1) piece of candy when they see their first movie.

Right now at this point we will have:

1,000 people = saw 1 movie

+120 pieces of candy yet to hand out to some of these people when some go view another movie.

(4th) The Goal is to MAX the number of people who see all 3 movies. As of right now, every person has (1) candy.

If we were to hand (+2) more candies to any one person, then that person would have (3) candies in total—- i.e., this means he or she will have seen all 3 movies - A, B, and C

And no person can see more than 3 movies - this is obvious because there is only 3 movies to see.

We have 120 candies left. At most, we can hand out (+2) more candies to:

120 / 2 = 60 people

These 60 people will get (+2) more candies for seeing 2 more movies - bringing their total up to (3) candies

Summary:
940 people = see only 1 movie

60 people = see All 3 movies


(Answer)
MAX amount of ppl who could have seen all 3 movies ———-> 60 people


60



If an Algebra approach makes more sense:

Let

I = # of ppl who watch exactly 1 movie

II = # of ppl who watch exactly 2 movies

III = # of ppl who watch exactly 3 movies

Since everyone sees at least one movie (there is no “Neither/nor” region), all 1,000 people will fall in one of the 3 categories above.

1,000 = (I) + (II) + (III)


Now if we add up the total who saw movie A + total who save movie B + total who saw movie C, we are accounting for all the “viewings” of movies

(450) + (250) + (420) = 1,120 **(equation 1)


For instance, all the people who saw movie A will include:

(Ppl who saw ONLY movie A)

(Some Ppl who saw Exactly 2, A and B)

(Some other ppl who saw Exactly 2, A and C)

(Ppl who saw All 3 movies)


Rather then categorize the people according to which specific movie group they are part of (A, B, or C), you can just characterize them as:

(A person who saw 1 movie) = I

(A person who saw exactly 2 movies) = II

(A person who saw all 3 movies) = III

If a person is part of the group of people who saw exactly 2 movies, then that one person will be responsible for 2 of the viewings among the 1,120 viewings totaled above.

Similarly, if a person is part of the group of people who saw all 3 movies, then that person will be responsible for 3 of the viewings among the 1,120 viewings totaled above.

We can set up a second equation to account for these 1,120 viewings based on the number of people in each group

1,120 = 1*(I) + 2*(II) + 3*(III)
**(equation 2)


The question asks us what is the maximum possible no. of ppl who could be part of (III), meaning they saw all 3 movies.

(Step 1) subtract equation 1 from equation 2

1.120 = 1*(I) + 2*(II) + 3*(III)

- (1,000 = I + II + III)
________________________

120 = (II) + 2 * (III)


Again, we want to maximize the number of people who are part of (III), since this represents the no. of ppl who saw all 3 movies.

By making (II) = 0 (MINIMIZING this variable) ———- we MAXIMIZE the (III) Variable

120 = 0 + 2*(III)

III = 60

Answer: 60 is the MAX





Posted from my mobile device

A simple way to approach these type of questions is as follows:
Let T be total number of people of watched the movie i.e. 1000
Let O be the number of overlaps i.e. 1120
Point to note here is that the number of overlaps include people who watched one, two or even all three movies.

Now all 1000 people watched at least one movie, so 1120-1000 = 120
This means that there are more 120 shows of the movies that need to be watched. So to maximize the number of people who watched all movies, we will assume the fact that these 120 shows were watched by the people who will in total watch 3 movies. Now they have already watched one movie, so (120/2) will give the maximum number if people that can watch the remaining 120 shows such that they will have watched all three movies.
IMO

Pont to note : Minimum number of people to watch all three movies would be zero as the 120 shows can be all seen by the people who decide to ONLY watch two movies.

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