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

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In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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New post 11 May 2018, 01:09
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In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky Sports, and 90 watch ABC. 50 students watch both Fox News and Sky Sports, 60 students watch both Sky sports and ABC and 65 students watch both Fox News and ABC. What is the minimum number of students who watch atleast one of the given channels?
A.35
B.50
C.135
D.150
E.165

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Re: In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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New post 12 May 2018, 05:25
3
2
One formula for 3 overlapping sets is as follows:

Total = Group 1 + Group 2 + Group 3 - (at least 2 of the groups) + (all 3 groups) + none.

In the problem above:
Total = 250.
Group 1 = 120.
Group 2 = 80.
Group 3 = 90.
(at least 2 of the groups) = 50+60+65 = 175.
Let x = all 3.
Let N = none.

Plugging these values into the equation above, we get:
250 = 120 + 80 + 90 - 175 + x + N.
250 = 115 + x + N
N = 250 - 115 - x.
N = 135 - x.

To MINIMIZE the number who watch at least 1 channel, we must MAXIMIZE the value of N: the number who watch NONE of the channels.
In the equation above, the value of N will be maximized if x (all 3) is as small as possible.

Of the given overlaps -- 50, 60, 65 -- the two greatest overlaps both include A:
(at least S and A) + (at least F and A) = 60+65 = 125.
Now this value ((at least S and A) + (at least F and A)) is 35 greater than the total number who watch A (90).
Implication:
At least 35 students must watch all 3 channels.
Thus, the minimum value of x = 35.

Plugging x=35 into N = 135 - x, we get:
N = 135 - 35 = 100.
Since the maximum number who watch none of the channels = 100, the minimum number who watch at least 1 channel = 250-100 = 150.

Krotishka1 wrote:
pushpitkc wrote:
In order to have the minimum number of students who watch at least one of the
given channels, we must have a maximum Minimum number of people watching all 3 channels.


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

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Re: In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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New post 11 May 2018, 05:00
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gmatbusters wrote:
In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky Sports, and 90 watch ABC. 50 students watch both Fox News and Sky Sports, 60 students watch both Sky sports and ABC and 65 students watch both Fox News and ABC. What is the minimum number of students who watch atleast one of the given channels?
A.35
B.50
C.135
D.150
E.165


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

Attachment:
Play.png
Play.png [ 6.96 KiB | Viewed 3034 times ]


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

Attachment:
Play1.png
Play1.png [ 8.29 KiB | Viewed 3034 times ]


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

Attachment:
Play2.png
Play2.png [ 7.18 KiB | Viewed 3031 times ]


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)
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Re: In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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New post 12 May 2018, 04:49
pushpitkc wrote:
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.


Thanks for the solution! Could you please explain the logic behind this statement?
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In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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New post 12 May 2018, 05:37
Krotishka1 wrote:
pushpitkc wrote:
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.


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
Attachment:
Play1.png
Play1.png [ 8.73 KiB | Viewed 2890 times ]


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

With 20 students watching all 3 channels
Attachment:
Play.png
Play.png [ 8.62 KiB | Viewed 2894 times ]


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!
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In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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New post 12 May 2018, 06:51
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).

pushpitkc wrote:
Krotishka1 wrote:
pushpitkc wrote:
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.


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
Attachment:
Play1.png


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

With 20 students watching all 3 channels
Attachment:
Play.png


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!

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Re: In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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New post 14 May 2018, 16:42
gmatbusters wrote:
In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky Sports, and 90 watch ABC. 50 students watch both Fox News and Sky Sports, 60 students watch both Sky sports and ABC and 65 students watch both Fox News and ABC. What is the minimum number of students who watch atleast one of the given channels?
A.35
B.50
C.135
D.150
E.165


To minimize the number of students who watch at least one channel, we want to maximize the number of the students who watch none of these channels.

We can create the equation:

250 = 120 + 80 + 90 - (50 + 60 + 65) + triple + none

250 = 290 - 175 + t + n

250 = 115 + t + n

135 - t = n

Since we want to maximize the value of n, we need to minimize the value of t. Let’s let a = the number of students who watch ABC only. Recall that t is the number of students who watch all 3 channels. Thus we have 65 - t students watching ABC and Fox (but not Sky) and 60 - t students watching ABC and Sky (but not Fox) and we can create the equation for all the students who watch ABC:

(65 - t) + (60 - t) + t + a = 90

125 - t + a = 90

35 = t - a

Since we want to minimize the value of t, so a must be 0. Thus t = 35

Therefore, n = 135 - t = 135 - 35 = 100.

Since there could be a maximum 100 students who watch none of the 3 channels, there must be a minimum of 250 - 100 = 150 who watch at least one channel.

Answer: D
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Re: In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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New post 19 May 2018, 01:21
ScottTargetTestPrep wrote:
gmatbusters wrote:
In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky Sports, and 90 watch ABC. 50 students watch both Fox News and Sky Sports, 60 students watch both Sky sports and ABC and 65 students watch both Fox News and ABC. What is the minimum number of students who watch atleast one of the given channels?
A.35
B.50
C.135
D.150
E.165


To minimize the number of students who watch at least one channel, we want to maximize the number of the students who watch none of these channels.

We can create the equation:

250 = 120 + 80 + 90 - (50 + 60 + 65) + triple + none

250 = 290 - 175 + t + n

250 = 115 + t + n

135 - t = n

Since we want to maximize the value of n, we need to minimize the value of t. Let’s let a = the number of students who watch ABC only. Recall that t is the number of students who watch all 3 channels. Thus we have 65 - t students watching ABC and Fox (but not Sky) and 60 - t students watching ABC and Sky (but not Fox) and we can create the equation for all the students who watch ABC:

(65 - t) + (60 - t) + t + a = 90

125 - t + a = 90

35 = t - a

Since we want to minimize the value of t, so a must be 0. Thus t = 35

Therefore, n = 135 - t = 135 - 35 = 100.

Since there could be a maximum 100 students who watch none of the 3 channels, there must be a minimum of 250 - 100 = 150 who watch at least one channel.

Answer: D


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 ?
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Re: In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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New post 19 May 2018, 03:21
1
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.

tixan wrote:
ScottTargetTestPrep wrote:
gmatbusters wrote:
In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky Sports, and 90 watch ABC. 50 students watch both Fox News and Sky Sports, 60 students watch both Sky sports and ABC and 65 students watch both Fox News and ABC. What is the minimum number of students who watch atleast one of the given channels?
A.35
B.50
C.135
D.150
E.165


To minimize the number of students who watch at least one channel, we want to maximize the number of the students who watch none of these channels.

We can create the equation:

250 = 120 + 80 + 90 - (50 + 60 + 65) + triple + none

250 = 290 - 175 + t + n

250 = 115 + t + n

135 - t = n

Since we want to maximize the value of n, we need to minimize the value of t. Let’s let a = the number of students who watch ABC only. Recall that t is the number of students who watch all 3 channels. Thus we have 65 - t students watching ABC and Fox (but not Sky) and 60 - t students watching ABC and Sky (but not Fox) and we can create the equation for all the students who watch ABC:

(65 - t) + (60 - t) + t + a = 90

125 - t + a = 90

35 = t - a

Since we want to minimize the value of t, so a must be 0. Thus t = 35

Therefore, n = 135 - t = 135 - 35 = 100.

Since there could be a maximum 100 students who watch none of the 3 channels, there must be a minimum of 250 - 100 = 150 who watch at least one channel.

Answer: D


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 ?

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Re: In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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New post 20 May 2018, 05:59
Thanks gmatbusters! :thumbup: :-)
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Re: In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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New post 24 May 2018, 11:10
OA: D
As gmatbusters has already explained
Quote:
One formula for 3 overlapping sets is as follows:

Total = Group 1 + Group 2 + Group 3 - (at least 2 of the groups) + (all 3 groups) + none.

In the problem above:
Total = 250.
Group 1 = 120.
Group 2 = 80.
Group 3 = 90.
(at least 2 of the groups) = 50+60+65 = 175.
Let x = all 3.
Let N = none.

Plugging these values into the equation above, we get:
250 = 120 + 80 + 90 - 175 + x + N.
250 = 115 + x + N
N = 250 - 115 - x.
N = 135 - x.

To MINIMIZE the number who watch at least 1 channel, we must MAXIMIZE the value of N: the number who watch NONE of the channels.
In the equation above, the value of N will be maximized if x (all 3) is as small as possible.

Attachment:
set.PNG
set.PNG [ 37.81 KiB | Viewed 2361 times ]


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 \(= 250- N=250-100 =150\)
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Re: In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S  [#permalink]

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Re: In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S   [#permalink] 11 Aug 2019, 08:28
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