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

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Joined: 02 Sep 2009
Posts: 64322
In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000  [#permalink]

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22 Nov 2019, 01:40
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Difficulty:

55% (hard)

Question Stats:

59% (02:05) correct 41% (02:02) wrong based on 80 sessions

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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

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Re: In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000  [#permalink]

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22 Nov 2019, 04:00
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1
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 1127 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!
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Re: In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000  [#permalink]

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22 Nov 2019, 02:24
1
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
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Re: In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000  [#permalink]

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Updated on: 23 Nov 2019, 03:40
1
1
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

Originally posted by Dillesh4096 on 22 Nov 2019, 03:11.
Last edited by Dillesh4096 on 23 Nov 2019, 03:40, edited 2 times in total.
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Re: In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000  [#permalink]

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22 Nov 2019, 03:27
1
Quote:
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

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)
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Re: In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000  [#permalink]

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22 Nov 2019, 05:04
1
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
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Re: In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000  [#permalink]

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22 Nov 2019, 13:10
1
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

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Re: In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000  [#permalink]

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23 Nov 2019, 05:36
1
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.
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Re: In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000  [#permalink]

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23 Nov 2019, 05:53
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 .
Re: In a multiplex 3 shows A, B and C run in 3 auditoriums. One day, 1000   [#permalink] 23 Nov 2019, 05:53