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Well its 'could have watched' which can be interpreted as maximum number of people who could have watched both the movie[IMO],so i'll go with X = 68%. ( 79 + 89 - 100).

But yes the questions seems inadequate as it open to differing interpretation.
_________________

There are 2 movies. 79% watch movie X and 89% watch movie Y. How many COULD have watched both movies ?

No OA guys

Responding to a pm:

'COULD' have watched implies the maximum number of people who could have watched both movies. What we need to remember here is that they do not say that everyone watched at least one movie. If the question setter wanted to tell you that everyone watched at least one movie, he/she would have specifically said it. We want to make these two sets overlap as much as possible. Region of overlap can be 79% at the most since we know that only 79% people watched movie X. Hence, here the answer will be 79%.

If the question was modified and it mentioned that everyone watched at least one movie, then there is no question of maximizing/minimizing the number of people who watched both the movies. Consider this:

Total no of people = people who watch no movie + people who watch X + people who watch Y - people who watch both

100 = [highlight]people who watch no movie[/highlight] + 79 + 89 - [highlight]people who watch both[/highlight]

[highlight]people who watch both[/highlight] - [highlight]people who watch no movie[/highlight] = 68

We can maximize/minimize the highlighted parts by adjusting them with each other. If one increases, the other increases too. If people who watch both = 68, people who watch neither = 0 If people who watch both = 69, people who watch neither = 1 If people who watch both = 79, people who watch neither = 11

or think logically - there were 168 screenings. If everyone watched 1 movie, that's 100 screenings. 68 must have been on 68 people who have already watched one movie. If instead, 99 people watch a movie, then 69 screenings can be for people who have already watched one. I hope the logic is clear.

In case we know that everyone watches at least one movie, people who watch neither = 0 and people who watch both = 68. There is no maximizing/minimizing people who watch both. The question will not be 'how many could have watched both?'. It will be 'how many watched both?'

This reminds me of a great little sets question in the Veritas book which is simple but takes down many people (just like a typical GMAT question). I will put it up in some time.
_________________

There are 2 movies. 79% watch movie X and 89% watch movie Y. How many COULD have watched both movies ?

No OA guys

Responding to a pm:

'COULD' have watched implies the maximum number of people who could have watched both movies. What we need to remember here is that they do not say that everyone watched at least one movie. If the question setter wanted to tell you that everyone watched at least one movie, he/she would have specifically said it. We want to make these two sets overlap as much as possible. Region of overlap can be 79% at the most since we know that only 79% people watched movie X. Hence, here the answer will be 79%.

If the question was modified and it mentioned that everyone watched at least one movie, then there is no question of maximizing/minimizing the number of people who watched both the movies. Consider this:

Total no of people = people who watch no movie + people who watch X + people who watch Y - people who watch both

100 = [highlight]people who watch no movie[/highlight] + 79 + 89 - [highlight]people who watch both[/highlight]

[highlight]people who watch both[/highlight] - [highlight]people who watch no movie[/highlight] = 68

We can maximize/minimize the highlighted parts by adjusting them with each other. If one increases, the other increases too. If people who watch both = 68, people who watch neither = 0 If people who watch both = 69, people who watch neither = 1 If people who watch both = 79, people who watch neither = 11

or think logically - there were 168 screenings. If everyone watched 1 movie, that's 100 screenings. 68 must have been on 68 people who have already watched one movie. If instead, 99 people watch a movie, then 69 screenings can be for people who have already watched one. I hope the logic is clear.

In case we know that everyone watches at least one movie, people who watch neither = 0 and people who watch both = 68. There is no maximizing/minimizing people who watch both. The question will not be 'how many could have watched both?'. It will be 'how many watched both?'

This reminds me of a great little sets question in the Veritas book which is simple but takes down many people (just like a typical GMAT question). I will put it up in some time.

We can maximize/minimize the highlighted parts by adjusting them with each other. If one increases, the other increases too. If people who watch both = 68, people who watch neither = 0 If people who watch both = 69, people who watch neither = 1 If people who watch both = 79, people who watch neither = 11

How do you know when to stop increasing? Why not 80 people who watch both and 12 that watch neither? (I know that this combination is not possible but what is the restriction that you stop at 79?

How would you deal with a min, max question with 3 overlapping sets like this one:

"In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. Every village has at least one of these three devices. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is?

How do you know when to stop increasing? Why not 80 people who watch both and 12 that watch neither? (I know that this combination is not possible but what is the restriction that you stop at 79?

Since only 79 people watch X, you cannot go beyond 79. Both the movies cannot be watched by 80 people. 79 (the lower of the two - 79 and 89) makes the upper limit of intersection. You have to be mindful of it.

How would you deal with a min, max question with 3 overlapping sets like this one:

"In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. Every village has at least one of these three devices. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is?[/quote]

There are 100 households and a total of 75+80+55 = 210 devices. You have to distribute these 210 devices among 100 people such that each person gets at least one device. Maximize the overlap of all 3: You give 100 devices to 100 people. (Say, give 75 DVD players to 75 people and out of 80 cell phones, give 25 to the remaining people.) You have 110 devices (55 MP3 and 55 cell phone) remaining. Of the 75 people who got DVD, you select 55 and give them the 55 MP3s and the 55 cell phones. Overlap of all 3 is 55 which is the maximum possible since there are only 55 MP3s.

Minimize: You have 210 devices. You want to minimize the overlap of all 3. So you give 200 devices to 100 people (2 each). You still have 10 devices left. You will have to give these to 10 people and they will have all 3 devices. So minimum overlap of all 3 is 10.

How do you know when to stop increasing? Why not 80 people who watch both and 12 that watch neither? (I know that this combination is not possible but what is the restriction that you stop at 79?

Since only 79 people watch X, you cannot go beyond 79. Both the movies cannot be watched by 80 people. 79 (the lower of the two - 79 and 89) makes the upper limit of intersection. You have to be mindful of it.

How would you deal with a min, max question with 3 overlapping sets like this one:

"In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. Every village has at least one of these three devices. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is?

There are 100 households and a total of 75+80+55 = 210 devices. You have to distribute these 210 devices among 100 people such that each person gets at least one device. Maximize the overlap of all 3: You give 100 devices to 100 people. (Say, give 75 DVD players to 75 people and out of 80 cell phones, give 25 to the remaining people.) You have 110 devices (55 MP3 and 55 cell phone) remaining. Of the 75 people who got DVD, you select 55 and give them the 55 MP3s and the 55 cell phones. Overlap of all 3 is 55 which is the maximum possible since there are only 55 MP3s.

Minimize: You have 210 devices. You want to minimize the overlap of all 3. So you give 200 devices to 100 people (2 each). You still have 10 devices left. You will have to give these to 10 people and they will have all 3 devices. So minimum overlap of all 3 is 10.

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