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Forty students watched films A, B and C over a week. Each student watched either only one film or all three. Thirteen students watched film A, sixteen students watched film B and nineteen students watched film C. How many students watched all three films?

  1. 0
  2. 2
  3. 4
  4. 8
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Migrated from GO Mechanical 4 years ago by Arjun

3 Answers

Best answer
5 votes
5 votes

Total Students who watched Film $= 40$

Let

  • $A =$ Students who watched Film "A" alone
  • $B =$ Students who watched Film "B" alone
  • $C =$ Students who watched Film "C" alone
  • $X =$ Students who watched Both Film "A" and "B" but not Film "C"
  • $Y =$ Students who watched Both Film "A" and "C" but not Film "B"
  • $Z =$ Students who watched Both Film "B" and "C" but not Film "A"
  • $O =$ Students who watched all the 3 Films

By Principle of Inclusion and Exclusion, we have 

Total Students who watched Film $= A + B + C + X + Y + Z + O$

Now from the Venn diagram it is clear that $13 = A + X + Y + O$

No student watched exactly $2$ films. Hence, $X = Y = Z = 0$ 

Therefore, $13 = A + 0 + 0 + O$

$\implies 13 - O = A$ 

Similarly, $B = 16 - O$ and $C = 19 - O$

So, Total Students who watched Film $= A + B + C + X + Y + Z + O$

$\implies 40 = A + B + C + 0 + 0 + 0 + O$

$\implies 40 = A + B + C + O$

$\implies 40 = (13 - O) + (16 - O) + (19 - O) + O$

$\implies 40 = 48 - 2O$

$\implies -8 = -2O$

$\implies O = 4.$

Hence, C is correct.

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8 votes
8 votes

$|AUBUC|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|A\cap C|+|A\cap B\cap C|$

$40=13+16+19-x-x-x+x$

$2x=8$

$x=4$

$Note: Why\ |A\cap B|,|B\cap C|,|A\cap C|\ is\ x\ ?$

Each student watched either only one film or all three which means $A\ \&\ B\ but\ not\ C=0$

1 votes
1 votes

Since students watch either 1 or all 3:

let a = students who watched only A,

b = students who watched only B,

c = students who watched only C,

n = students who watched A, B and C.

Given data implies

a + n = 16 ……….[1]

b + n = 13 ……….[2]

c + n = 19 ……….[3]

and since total strength of class is 40,

a+b+c+n = 40 ……….[4]

solving for n ( [1] + [2] + [3] – [4] )

2n = 8

=> n=4

Answer:

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