In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S

In order to have the minimum number of students who watch at least one of the
given channels, we must have a maximum number of people watching all 3 channels.
Using Venn diagram, we can solve this problem as follows


Here, P(Watching all 3 channels) = 25
But P(Watching ABC) = 40+25+35 > 90. This scenario is not possible!


Here, P(Watching all 3 channels) = 30
P(Watching ABC) = 35+30+30 > 90. Again, this scenario is not possible!


Here, P(Watching all 3 channels) = 35
P(Watching ABC) = 30+35+25 = 90.
P(Watching SkySports) = 15+35+25+5 = 80 P(Only Watching SkySports) = 5
P(Watching Fox) = 15+35+30+40 = 120 P(Only Watching Fox) = 40

Therefore, the minimum number of students who watch at least one of the channels is 40+5+30+35+25+15 = 150(Option D)

Thanks for the solution! Could you please explain the logic behind this statement?

Hey Krotishka1

The logic is simple: If 1 student watches all 3 channels, it reduces 1 student from each
of the individual channel's students count, so there is a net loss of 2 students from the
overall number of students watching the channels. The more the number of students
watching all 3 channels the minimum the number of students watching TV.

I'll try and explain by means of an example.

With zero students watching all 3 channels

Here, the total number of students watching TV is 60 + 30 + 20 + 30 + 30 + 30 = 200

With 20 students watching all 3 channels

Here, the total number of students watching TV is 40 + 10 + 0 + 30 + 30 + 30 + 20 = 160
If you observe because of the inclusion of 20 students watching all the 3 channels, the number
of students watching TV reduces by 40(2*20)

Hope this helps you!

Hi

I understand your logic but the examples you gave don't satisfy the conditions of intersection of 2 of them.
n (Fox and Sky sport) =50
n( Sky Sport and ABC) =60
n( ABC and Fox) =65.

Also please point out the flaw in my explanation.
As per my approach "None of them = 135 - n(all of them). "
hence, to get min value of union of three, we should have maximum value of None of them which asks for min value of n(all of them).

Hello Scott / All - just to confirm that we can not set N equal to zero because that implies there are 135 students who watch 3 movies and this is impossible. Correct ?

You are right, n can not be zero in this equation.

Moreover , since we need to find the minimum number of students who watch atleast one of the given channels. We are looking for the maximum value of n , So that the number of students who watch atleast one of the given channels (Union of all news watchers) is minimum.

OA: D
As gmatbusters has already explained

To maximixe N , we have to make value of \(x\) as small as possible.
lets take \(x=0\), then Number of student who watch Sky Sports alone and Number of student who watch ABC alone become \(-30\) and \(-35\) respectively, which is not possible.
Now lets take \(x=30\),then Number of student who watch Sky Sports alone and Number of student who watch ABC alone become \(0\) and \(-5\) respectively, which is also not possible.
Finally lets take \(x=35\),then Number of student who watch Sky Sports alone and Number of student who watch ABC alone become \(5\) and \(0\) respectively,
all other terms in venn diagram are also non negative, so \(35\) is minimum value of \(x\) that can be possible.

Using \(x = 35\), we get \(None(N)=135-35=100\)
Minimum No of student,who watch atleast one tv channel \(= 250- N=250-100 =150\)

 
In above sentence, why have we taken only the overlaps with ABC, why not overlapps with Skysports or fox. could you please explain the logic behind this?


Thnks in advance!


­

Hi, the section where minimum value of t is ascertained uses the Set C as the anchor to get 35. However, why pick set C as the anchor? I tried with set A (t min comes out to be 0) and set B (t min comes out to be 30). So, I'd like to ask how you zoned in on Set C as the target. Thanks.

Bunuel KarishmaB, why cannot we use this approach? Technically - if I want to minimise 'student watches at least one channel' then isn't the best way to do that is to ensure that the centre part where all 3 channels overlap is maximised? That way, we would need the least amount to populate the other 'exactly 1 channel' and 'exactly 2 channel' areas.

Also, say the Q asked, maximise the number of 'students who watch all 3 channels', then shouldn't the answer be 50? Least number of students who watch 2 or more channels is Fox and Sky = 50. If we assume all of them watch ABC too, then all students who watch all 3 becomes 50.
Students who watch exactly Fox and Sky = 50 - 50 = 0.
Students who watch exactly Sky and ABC = 60 - 50 = 10.
Students who watch exactly Fox and ABC = 65 - 50 = 15.

Students who watch only Fox = 120 - 50 - 15 - 0 = 55
Students who watch only Sky = 80 - 50 - 20 - 0 = 10
Students who watch only ABC = 90 - 50 - 15 - 10 = 25

Minimum students who watch at least 1 channel = who watch exactly 1 channel + who watch exactly 2 channels + who watch exactly 3 channels
At least 1 channel = (55 + 10 + 25) + (0 + 10 + 15) + 50 = 90 + 25 + 50 = 165.

I get that it is wrong, but could you point out the error for both: (1) Why can't this way be used to solve this question? (2) What can be the maximum value of 'students watch all 3 channels'? Thank you.

Student watches at least one channel = Total students (T) - students watching none of the channels (n)

So to minimize students watching at least one channel, you've to maximize n

Post solving, you end up with, 135 - n = Students watching all three channels (k) => n = 135 - k

Now since you want to maximize n, you need to minimize k i.e. minimize students who watch all 3 channels