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

Question

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 at least one of the given channels?

A. 35
B. 50
C. 135
D. 150
E. 165

A. 35 
B. 50 
C. 135 
D. 150 
E. 165

GMATBusters · Accepted Answer

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.

ScottTargetTestPrep · Answer

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

Princ · Answer

OA: D
As gmatbusters has already explained

Attachment:

set.PNG

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 \(= 250- N=250-100 =150\)

pushpitkc · Answer

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 Play.png Here, P(Watching all 3 channels) = 25 But P(Watching ABC) = 40+25+35 > 90. This scenario is not possible! Play1.png Here, P(Watching all 3 channels) = 30 P(Watching ABC) = 35+30+30 > 90. Again, this scenario is not possible! Play2.png 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)

Krotishka1 · Answer

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

pushpitkc · Answer

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 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 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!

GMATBusters · Answer

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) .

tixan · Answer

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 ?

GMATBusters · Answer

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.

SudsMeister · Answer

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.

Krunaal · Answer

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

consistentprep · Answer

You can  solve by options  as well.
Once you reach till here:
Total(T) - None(N) = (120+80+90)-(50+65+60)+x
T-N= 115+x

a) 35 = 115+x will give negative value of x , eliminate
b) 50 = 115+x , again eliminate
c) 135= 115+x , gives x=20 , but SkySports will be 30+20+40 which exceeds the value of 80, eliminate
d) 150= 115+x , gives x=35 , satisfies the value for all the sets
e) No need to check the next higher value

Adit_ · Answer

Plenty of layers to this problem. I will solve it using my approach for this:

Drawing the venn diagram is a first.
Second thing is realising that there are number of people who have potentially not watched any channel and considering that we are asked the question of the minimum number who watched atleast one.

Basic equations:

290 - Twice - 2  = Atleast 1.

Thus to minimize Atleast 1, we need to maximum the expression Twice+2

Now Lets keep it as x+2y for simplicity.

Now consider the expressions:
b+a=50,
c+a=65,
d+a=60.

These are the equations for atleast 2 items at a time.

Now Twice+2  = b+c+d-2a

Lets add up all the atleast 2 items equations:

b+c+d+3a = 175.
b+c+d+2a = 175-a 
This is the most important part, we have managed to get to our maximizing equation using those 2 items at a time equations added up.
Now it becomes clear that 175-a is our key equation. Here a needs to be MINIMIZED.
Why? Because we need to maximise the entire expression 175-a, so "a" need to be minimized.

What is "a"? It is 3 items.

Again one can solve using options, but I am doing the methodical approach.

Here question is what is a?
Now here we need the venn diagram. It is important to note that take any random value like 1 then atleast one of the summations will exceed.
One needs to realise that making the ABC part of venn diagram 0, is the only way we can get the most minimum value.

We get 35 as value of a here. And solving the remaining variables working back we find that total adds up to  150.

135 is not possible here because if 135 then a=20. That is not possible.
You will realise that ABC will exceed 90 on keeping a=20. Basically the values of a range from 35 to 50. 50 because that is the minimum atleast 2 items value.

Sunsh · Answer

Hi, Can you tell me why you took ABC to get the t(min)... and not Sky sports.. as doing the same with Sky Sports giving me t(min) = 30..I know I am wrong, want to know the reason.

egmat · Answer

Hi Sunsh,

Your work is actually correct; the only thing to fix is how you read the two results.

Look at what each channel gives you. When you write the "watches only this channel" region and force it to be non-negative:

-  Sky Sports only  = t -  30  >= 0 -> t >=  30 
-  ABC only  = t -  35  >= 0 -> t >=  35

Neither of these is "the" answer on its own. They are both  conditions that t must satisfy at the same time . t has to make  every  region non-negative - Sky Sports' region, ABC's region, all of them at once.

So ask: which value of t keeps  both  alive?

- If t =  30 : Sky Sports only =  0  (fine), but ABC only = 30 - 35 =  -5 . Impossible - you can't have  -5  students.
- If t =  35 : Sky Sports only =  5  (fine) and ABC only =  0  (fine). Everything works.

That's the whole point. When you have several lower bounds (t >=  30and  t >=  35 ), the true minimum is the  largest  of them - because that's the only value big enough to clear all the conditions. ABC isn't "the special group"; it just happens to give the  tightest  requirement. Your Sky Sports bound of  30  is real, but it's too small to also satisfy ABC, so it gets overruled.

Quick way to lock it in:  suppose one rule says t >=  4  and another says t >=  7 . What's the smallest t that obeys both? It's  7 , not  4  - the bigger floor wins every time. Same logic here: the most demanding constraint (ABC, t >=  35 ) sets the minimum.

Answer: D

SwiftMonk · Answer

it is not required to make N=0.As we need to make as max as possible to keep the watchers at minimum

bharathhr · Answer

thanks for a diagramitic solution

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

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