PROBLEMS ON INTERSECTION OF THREE SETS

Question 1 :

For any three sets A,B and C if n(A) = 17, n(B) = 17, n(C) =  17, n(A∩B)  =  7, n(B∩C)  =  6, (A∩C)  =  5 and n(A∩B∩C) = 2, find n (AUBUC).

Solution :

n(A)  =  17, n(B)  =  17, n(C)  =  17, n(A∩B)  =  7

n(B∩C)  =  6, (A∩C)  =  5 

n(AUBUC)

= n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(A∩C)+n (A∩B∩C)

  =  17 + 17 + 17 - 7 - 6 - 5 + 2

  =  53 - 18 + 2

  =  55 - 20

n (AUBUC)  =  35

Question 2 :

verify

n(AUBUC)

  =  n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B ∩C)

(i) A  =  {4, 5, 6}, B  =  {5, 6, 7, 8} and C  =  {6, 7, 8, 9}

Solution :

A  =  {4, 5, 6}

B  =  {5, 6, 7, 8}

C  =  {6, 7, 8, 9}

n(A)  =  3, n(B)  =  4, n(C)  =  4

A∩B∩C  =  {6}, n(A∩B∩C)  =  1

n(AUBUC)

  =  n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(A∩C)+n(A∩B∩C)

  =  3 + 4 + 4 - 2 - 3 - 1 + 1

  =  11 - 6 + 1

  =  12 - 6

  =  6

(ii) A  =  {a, b, c, d, e} B  =  {x, y, z} and C = {a, e, x}

Solution :

A  =  {a, b, c, d, e} B  =  {x, y, z} and C  =  {a, e, x}

n(A)  =  5 , n(B)  =  3, n(C)  =  3

n (A∩B∩C)  =  0

n(AUBUC)

  =  n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(A∩C)+n(A∩B∩C)

  =  5 + 3 + 3 - 0 - 1 - 2 + 0

  =  11 - 3

 n (AUBUC) = 8

Question 3 :

If the sets A and B are given by

A = {1, 2, 3, 4}

B = {2, 4, 6, 8, 10}

and the universal set

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

then :

(a) (A 𝖴 B)' = {5, 7, 9}      (b) (A ∩ B)'= {1, 3, 5, 6, 7}

(c) (A ∩ B)' = {1, 3, 5, 6, 7, 8}     (d) None of these

Solution :

Option a :

A u B = {1, 2, 3, 4, 6, 8, 10}

(AuB)' = {5, 7, 9}

Option a is true

Option b :

AnB = {2, 4}

(A ∩ B)'= {1, 3, 5, 6, 7, 8, 9, 10}

So, option b is False.

Option c is also false.

Option d is false.

Question 4 :

If A = {1, 2, 3, 4}, B = {2, 3, 5, 6} and C = {3, 4, 6, 7}, then

(a) A – (B ∩ C) = {1, 3, 4}        (b) A – (B ∩ C) = {1, 2, 4}

(c) A – (B 𝖴 C) = {2, 3}       (d) A – (B 𝖴 C) = {Ø}

Solution :

Option a :

B n C = {3, 6}

A – (B ∩ C) = {1, 2, 3, 4} - {3, 6}

= {1, 2, 4}

So, option a is incorrect.

Option b :

B n C = {3, 6}

A – (B ∩ C) = {1, 2, 3, 4} - {3, 6}

= {1, 2, 4}

So, option b is correct.

Question 5 :

Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}. Which of the following may be considered as universal set for all the three sets A, B and C:

(a) {0, 1, 2, 3, 4, 5, 6}           (b) Ø

(c) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(d) {1, 2, 3, 4, 5, 6, 7, 8}

Solution :

A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}

Universal set will contain all the elements in the sets A, B and C.

AuBuC = {0, 1, 2, 3, 4, 5, 6, 8} = U

So, the universal set will have the elements of A, B and C. Option c is correct.

Question 6 :

From 50 students taking examination in Mathematics,Physics and Chemistry, each of the students has passed in at least one of the subject, 37 passed Mathematics, 24 Physics and 43 Chemistry. Atmost 19 passed Mathematics and Physics, atmost 29 Mathematics and Chemistry and atmost 20 Physics and Chemistry. Then, the largest numbers that could have passed all three examinations, are:

(a) 12      (b) 14     (c) 15      (d) 16

Solution :

Total number of students = 50

Number of students who passes all three exams = n(A∩BnC)

Let x be the number of students who passes in all three.

n(AUBUC)

50 = n(A) + n(B) + n(C) - [n(A∩B)+n(B∩C)+n(A∩C)] + x

50 = 37 + 24 + 43 - [19 + 20 + 29] + x

50 = 104 - 68 + x

50 = 36 + x 

x = 50 - 36

= 14

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