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

Question

Competition Mode Question

In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000 visited the multiplex and watched at least one of 3 shows. If 420 people watched A, 250 watched B, 450 watched C, what is the maximum number of people who watched all the 3 shows ?

(A) 0
(B) 50
(C) 60
(D) 100
(E) 250

(A) 0
(B) 50
(C) 60
(D) 100
(E) 250

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CrackverbalGMAT · Accepted Answer

This is a fairly straightforward question on maximization of regions of a Venn diagram. In all maximization and minimization questions, remember that it’s not about taking the largest or the smallest value. It’s about taking the largest of the smallest value that does not make any region of the Venn diagram negative.

Let us go ahead and draw the Venn diagram now:

Attachment:

22nd Nov 2019 - Reply 4.jpg [ 50.16 KiB | Viewed 13668 times ]

Since 1000 people visited the multiplex, the rectangle has 1000 people in total. The question also says that everyone watched at least one of the shows. This means everyone who visited the multiplex will be inside one of the circles. The area inside the rectangle but outside the circles will be ZERO.

In order to maximize the number of people who watched all 3, we will have to minimize the other regions. The minimum possible value of any region in a Venn diagram is 0.

From the diagram, we see that the sum of the sets = 420 + 250 + 450 = 1120. But, the sum of the sets contains the regions a, b and c twice and the region x thrice. If we remove a, b and c once and x twice from 1120, we get 1000 since this is the actual number of people who visited the multiplex (without repetition of any region). This means,
1120 –(a+b+c+2x) = 1000 which, on simplification, gives us,
a + b + c + 2x = 120. To maximize x, we take a+b+c = 0 and so, 2x = 120 or x = 60.

The correct answer option is C.
Hope this helps!

Archit3110 · Answer

1000 - 0 = 420 + 250 + 450 - In Exactly two - 2(all 3)
120 = Exactly two + 2(all 3)
 exactly three we have to take exactly two as Zero
120 = 2(all 3)
60 = all 3
IMO C

In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000 visited the multiplex and watched at least one of 3 shows. If 420 people watched A, 250 watched B, 450 watched C, what is the maximum number of people who watched all the 3 shows ?

(A) 0
(B) 50
(C) 60
(D) 100
(E) 250

(A) 0
(B) 50
(C) 60
(D) 100
(E) 250

CareerGeek · Answer

Let the number of people who watched all the 3 shows = X and those who watched exactly 2 shows = Y

Formula: Total - Neither = A + B + C - (Exactly 2 categories) - 2(Exactly 3 categories)

—> 1000 - 0 = 450 + 250 + 420 - Y - 2X
—> 2X = 1120 - 1000 - Y
—> 2X = 120 - Y
—> X = 60 - Y/2

For X to be maximum, Y has to be minimum = 0

—> Maximum value if X = 60 - 0 = 60

IMO Option C

exc4libur · Answer

T=A+B+C-both-2(mid)+neither
1000=420+250+450-both-2mid+0
-120=-both-2mid
2mid=120-both
mid=60-both/2
both=0; mid=60

Ans (C)

madgmat2019 · Answer

From image:
A=a+b+d+e=420
B=c+b+e+f=250
C=d+e+f+g=450

Total=a+b+c+d+e+f+g=1000=a+b+d+e+c+b+e+f+d+e+f+g - (d+b+f) - 2e

1000 = 420 + 250 + 450 - In Exactly two - 2(In Exactly 3)
120 = Exactly two + 2(In Exactly 3)
to maximize exactly three we have to take exactly two as Zero
120 = 2(In Exactly 3)
Exactly 3=60

OA:C

lacktutor · Answer

Total= 1000
A= 420. B= 250. C= 450
Neither —0(zero)

The maximum number of people—Z(3 shows) —???

A= a+ a1+b1 +Z= 420
B= b+b1+ c1+Z= 250
C= c+c1+ a1+ Z= 450
——————————
a+b+c+a1+b1+c1+Z= 1000
a+b+c+ 2(a1+b1+c1)+ 3Z= 1120
—> (a1+b1+c1)+ 2Z= 120

—> if (a1+b1+c1)= 0, then 2Z= 120
—> Z= 60

The answer is C

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eakabuah · Answer

We are given that a total of 1000 people patronized at least one of three shows, A, B, and C. This implies none =0.
We are also given that the number that attended show A=420, the number that attended show B=250, and the number that attended show C=450.

Let m, n, and z be the number who attend both A and B only, both A and C only, and both B and C only respectively.
Likewise let a, b, and c be the number who attend only show A, only B, and only C respectively.
Let x be the number who attend all three shows.
a+b+c+m+n+z+x=1000
but a=420-m-n-x, b=250-m-z-x, and c=450-n-z-x
so, (420-m-n-x)+(250-m-z-x)+(450-n-z-x)+m+n+z+x=1000
1120-(m+n+z)-2x=1000
120-(m+n+z)=2x
hence x=60-(m+n+z)/2
For maximum x, then (m+n+z)=0
Therefore the maximum possible people who attended all three shows is 60.

The answer is therefore option C.

ayushkumar22941 · Answer

answer should be d, we get (450+420+250)=1120 , now get maximum ppl who watched all show substitute value from given answer . it gets solved in 30 sec max .

srik410 · Answer

There is some subtlety to this type of question that other answers haven't acknowledged. It seems at first glance(from the Venn diagrams) that A ∩B cannot be zero with A∩B∩C being non zero(because it is impossible to draw Venn diagram with that description).

But that is perfectly reasonable by wording. In this case, say a movie goer goes to shows A and B and C ie he went to all three. So this data point belongs to A∩B∩C set. In fact, you can say because he went to all three, he does not belong to the A∩B set. As a result, the pictures look a little different as shown below.

DonBosco7 · Answer

Hi  KarishmaB 
I have a doubt on this question as to why the answer for Maximum value is not E = 250.
See there is this one rather similar question i came across earlier on min max and overlapping sets, which had values for one circle as 70, other as 100 and third circle as 110 and the max value answer to that question was 70, and the reasoning for it was that if we draw the overlapping circles the other 2 big circle will encompass 70 and hence the maximum value will be 70. 
My question is why that same reasoning is not valid for this question ?

KarishmaB · Answer

This question is a little different. We are given that none cannot be 0 here. If we place one circle inside another, then there would be 550 people who would have watched no show. When none is not 0, the question is best done thinking about instances. 
There are total 420 + 250 + 450 = 1120 instances 
Distribute 1000 among 1000 people and you are left with 120 extra. Of these extra, to maximize the overlap of all 3, give away 2 each to 60 people so that we have a maximum overlap of all 3 of 60.

ManishaPrepMinds · Answer

We are given:
 
 Total people = 1000 
 Watched A = 420 
 Watched B = 250 
 Watched C = 450 
  Goal : Find the maximum number of people who watched all three shows

Let

x = number of people who watched  all 3 shows .

Use the Formula :
Total = A + B + C 
 - ( exactly two - shows ) 
 - 2×(all three viewers: x)
 + neither
To  maximize x , we  minimize the number of people  who watched  exactly two shows  → assume it is  0 .
As all watched at least one of 3 shows ->  neither =0

-> Total = A + B + C - 2x
-> 1000 = 420 + 250 + 450 - 2x
->1000 = 1120 - 2x
-> 2x = 120
-> x = 60

The  maximum  number of people who could have watched all three shows is 60.

https://drive.google.com/file/d/1JTVdlmZ0gXMRad2qxnU3uxZZYM1kfHrA/view?usp=sharing

udbhavtewari · Answer

What if the question asked to minimize the number of people who watched all the 3 shows ? How would we be using this formula then ?

Mannisha

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

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

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