# CLASS-8PROBLEM & SOLUTION OF SETS

PROBLEM & SOLUTION OF SETS

Example.A)

Let A = { letters of the word BADMINTON } and B = { letters of the word CRICKET }

Find  1) A B,  2) A U B,  3) B U A,  4)  B A,  5) A – B,  6) B – A

1)   A B

As per the given condition, we have got A = { letters of the word BADMINTON } and  B = { letters of the word CRICKET }

A = { letters of the word BADMINTON } = { B, A, D, M, I, N, T, O }

B = { letters of the word CRICKET } =  { C, R, I, K, E, T }

As per the given instruction, we have to find out

A B = { B, A, D, M, I, N, T, O } { C, R, I, K, E, T }                                                                                                        =  { I, T }

2)   A U B

Ans.) As per the given condition, we have got A = { letters of the word BADMINTON } and  B = { letters of the word CRICKET }

A = { letters of the word BADMINTON }

= { B, A, D, M, I, N, T, O }

B  = { letters of the word CRICKET } =  { C, R, I, K, E, T }

As per the given instruction, we have to find out

A U B =  { B, A, D, M, I, N, T, O }  U { C, R, I, K, E, T }                                                                                                       =   { B, A, D, M, I, N, T, O, C, R, K, E }

3)   B U A

Ans.) As per the given condition, we have got A = { letters of the word BADMINTON } and B = { letters of the word CRICKET }

A = { letters of the word BADMINTON }

=  { B, A, D, M, I, N, T, O }

B  = { letters of the word CRICKET }

=  { C, R, I, K, E, T }

As per the given instruction, we have to find out

B U A =  { C, R, I, K, E, T } { B, A, D, M, I, N, T, O }                                                                                                        =  { C, R, I, K, E, T, B, A, D, M, N, O }

4)    B A

Answer) As per the given condition, we have got A = { letters of the word BADMINTON } and  B = { letters of the word CRICKET }

A = { letters of the word BADMINTON }

= { B, A, D, M, I, N, T, O }

B  = { letters of the word CRICKET }

=  { C, R, I, K, E, T }

As per the given instruction, we have to find out

B  A = { C, R, I, K, E, T }  { B, A, D, M, I, N, T, O }                                                                                                         = { I, T }

5)    A – B

Answer) As per the given condition, we have got A = { letters of the word BADMINTON } and B = { letters of the word CRICKET }

A = { letters of the word BADMINTON }

=  { B, A, D, M, I, N, T, O }

B  = { letters of the word CRICKET }  =  { C, R, I, K, E, T }

As per the given instruction, we have to find out  A – B

A – B  =  { B, A, D, M, I, N, T, O } – { C, R, I, K, E, T }

=  { B, A, D, M, N, O }

6)    B – A

Answer) As per the given condition, we have got A = { letters of the word BADMINTON } and B = { letters of the word CRICKET }

A = { letters of the word BADMINTON } = { B, A, D, M, I, N, T, O }

B = { letters of the word CRICKET } =  { C, R, I, K, E, T }

As per the given instruction, we have to find out  A – B

B – A  =  { C, R, I, K, E, T } – { B, A, D, M, I, N, T, O }

=  { C, R, K, E }

Example.B)

If U = { x : x W and  6 ≤ x ≤ 12 } , A = { 6, 8, 9 } , B = { 7, 8, 11 } and C = { 6 } find the following sets.

1)  A’ ,  2)   B’ ,  3)  C’ ,  4)  A – B ,   5)  (B U C )’ ,  6)  ( A B )’ ,  7)  A – ( B U C ),   8)   A – ( B C ).

1)    A’

Answer) As per the given condition,

U = { x : x W and  6 ≤ x ≤ 12 } ,

U = { 6, 7, 8, 9, 10, 11, 12 } and  A = { 6, 8, 9 }

As per the rule of complement of set

A’ =  U – A  =  { 6, 7, 8, 9, 10, 11, 12 } –  { 6, 8, 9 }

=  { 7, 10, 11, 12 }

2)   B’

Answer) As per the given condition,

U = { x : x W and  6 ≤ x ≤ 12 },

U = { 6, 7, 8, 9, 10, 11, 12 } and  B = { 7, 8, 11 }

As per the rule of complement of set

A’ =  U – A  =  { 6, 7, 8, 9, 10, 11, 12 }  –  { 7, 8, 11 }                                                                                                            =  { 6, 9, 10, 12 }

3)   C’

Answer) As per the given condition,

U = { x : x W and  6 ≤ x ≤ 12 },

U = { 6, 7, 8, 9, 10, 11, 12 } and  C = { 6 }

As per the rule of complement of set

A’ =  U – A  =  { 6, 7, 8, 9, 10, 11, 12 } –  { 6 }

=  { 7, 8, 9, 10, 11, 12 }

4)  A – B

Answer) As per the given condition, A = { 6, 8, 9 } , B = { 7, 8, 11 },

we have to find out A – B

So,  A – B = { 6, 8, 9 } – { 7, 8, 11 }  =  { 6, 9 }

5)  (B U C )’

Answer) As per the given condition,

U = { x : x W and  6 ≤ x ≤ 12 },

U = { 6, 7, 8, 9, 10, 11, 12 }, B = { 7, 8, 11 }, C = { 6 }

First, we have to find , B U C

So,  B U C = { 7, 8, 11 } U { 6 }  =  { 6, 7, 8, 11 }                                                           [ As per laws of Union Set ]

Now, (B U C )’ = U - (B U C )

= { 6, 7, 8, 9, 10, 11, 12 } - { 6, 7, 8, 11 }                                         [ As per Laws of Complement Set ]

=  { 9, 10, 12 }

So, (B U C )’  =  { 9, 10, 12 }

6)  ( A B )’

Answer) As per the given condition,

U = { x : x W and  6 ≤ x ≤ 12 },

U = { 6, 7, 8, 9, 10, 11, 12 }, B = { 7, 8, 11 }, C = { 6 }

First, we have to find, B U C

So,  B C  =  { 7, 8, 11 } { 6 }

=  φ                 [ Overlapping rules ]

Now,  ( A B )’  =  U - ( A B )

=  { 6, 7, 8, 9, 10, 11, 12 }  - φ

=  { 6, 7, 8, 9, 10, 11, 12 }  =  U

So, ( A B )’   =   U

7)  A – ( B U C ),

Answer) As per the given condition,

U = { x : x W and  6 ≤ x ≤ 12 },

A = { 6, 8, 9 }, B = { 7, 8, 11 } and C = { 6 }

We have to find out =  ( B U C )

( B U C ) =  { 7, 8, 11 } U { 6 }  =  { 6, 7, 8, 11 }

Now,  A – ( B U C )  =  { 6, 8, 9 } - { 6, 7, 8, 11 }

=  { 9 }

8)  A – ( B C )

Answer ) As per the given condition,

U = { x : x W and  6 ≤ x ≤ 12 },

A = { 6, 8, 9 }, B = { 7, 8, 11 } and C = { 6 }

B C  =  { 7, 8, 11 } { 6 }  =  φ

Now,  A – ( B C )  =  { 6, 8, 9 } – φ

=  { 6, 8, 9 }

=   A

So, A – ( B C )  =  A

Example. C)

Let U = { x : x N and 2 ≤ x ≤ 12 }, A = { x : x is an even number }, B = { x : x is a multiple of 3}, and C = { x : x is a multiple of 4 }, verify the following  1) ( A U B )’ = A’ B’ ,  2) ( A B )’ = A’ U  B’ ,  3) A U ( B C ) = ( A U B ) ( A U C ),  4) A ( B U C ) = ( A B ) U ( A C )

1)   ( A U B )’ = A’ B’

Answer) As per the given condition, we can find,

U = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 },

A = { 2, 4, 6, 8, 10, 12 },

B = { 3, 6, 9, 12 },

C = { 4, 8, 12 }

Here we consider ( A U B )’ = 1st Part  and  A’ B’ = 2nd Part

1st Part

So, ( A U B )  =  { 2, 4, 6, 8, 10, 12 }  U  { 3, 6, 9, 12 }

=  { 2, 3, 4, 6, 8, 9, 10, 12 }

( A U B )’  =  U -  ( A U B )

=   { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 2, 3, 4, 6, 8, 9, 10, 12 }

=   { 5, 7, 11 }

2nd Part

So,  A’ B’  =  ( U – A ) ∩  ( U – B )

{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 2, 4, 6, 8, 10, 12 } ∩ U – B )

= { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 2, 4, 6, 8, 10, 12 } { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 3, 6, 9, 12 }

=  { 3, 5, 7, 9, 11 }  ∩  { 2, 4, 5, 7, 8, 10, 11 }

=  { 5, 7, 11 }

So, we can observe that  1st Part = 2nd Part

( A U B )’ =  A’ B’    (Proven)

2)  ( A B )’ = A’ U  B’

Answer) As per the given condition, we can find , U = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }, A = { 2, 4, 6, 8, 10, 12 }, B = { 3, 6, 9, 12 }, C = { 4, 8, 12 }

Here we consider ( A B )’ = 1st Part  and  A’ U B’ = 2nd Part

1st Part

( A B ) = ( A B ) =  { 2, 4, 6, 8, 10, 12 } { 3, 6, 9, 12 }

=  { 6, 12 }

( A B )’  =  U - ( A B )

=  { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 6, 12 }

=  { 2, 3, 4, 5, 7, 8, 9, 10, 11 }

2nd Part

A’ U  B’  = ( U – A )  U ( U – B )

=  { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 2, 4, 6, 8, 10, 12 } U { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } – { 3, 6, 9, 12 }

=  { 3, 5, 7, 9, 11 }  U  { 2, 4, 5, 7, 8, 10, 11 }

=   { 2, 3, 4, 5, 7, 8, 9, 10, 11 }

So, we can observe that  1st Part = 2nd Part

( A B )’ = A’ U  B’      (Proven)

3)  A U ( B C )  =  ( A U B ) ( A U C )

Answer) As per the given condition, we can find, U = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }, A = { 2, 4, 6, 8, 10, 12 }, B = { 3, 6, 9, 12 }, C = { 4, 8, 12 }

Here we consider  A U ( B C ) = 1st Part

and  ( A U B ) ( A U C ) = 2nd Part

1st Part

A U ( B C )  =  A U { 3, 6, 9, 12 } ∩  { 4, 8, 12 }

=  A U { 12 }

=  { 2, 4, 6, 8, 10, 12 } U { 12 }

=  { 2, 4, 6, 8, 10, 12 }

So,  A U ( B C ) =  { 2, 4, 6, 8, 10, 12 }

2nd Part

( A U B ) ( A U C )

=  { 2, 4, 6, 8, 10, 12 } U { 3, 6, 9, 12 } ∩ { 2, 4, 6, 8, 10, 12 } U { 4, 8, 12 }

=  { 2, 3, 4, 6, 8, 9, 10, 12 } { 2, 4, 6, 8, 10, 12 }

=   { 2, 4, 6, 8, 10, 12 }

So, we can see that, 1st Part = 2nd Part

So, now we can say -

A U ( B C ) =  ( A U B ) ( A U C )    (Proven)

4)   A ( B U C ) =  ( A B ) U ( A C )

Answer) As per the given condition, we can find , U = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }, A = { 2, 4, 6, 8, 10, 12 }, B = { 3, 6, 9, 12 }, C = { 4, 8, 12 }

Here we consider  A ( B U C ) = 1st Part

and   ( A B ) U ( A C )  = 2nd Part

1st Part

A ( B U C )  =   A { 3, 6, 9, 12 } U { 4, 8, 12 }

=   A { 3, 4, 6, 8, 9, 12 }

=   { 2, 4, 6, 8, 10, 12 } { 3, 4, 6, 8, 9, 12 }

=   { 4, 6, 8, 12 }

So,  A ( B U C ) =  { 4, 6, 8, 12 }

2nd Part

( A B ) U ( A C )

=  { 2, 4, 6, 8, 10, 12 } { 3, 6, 9, 12 }  U  ( A C )

=  { 6, 12 } U { 2, 4, 6, 8, 10, 12 }  { 4, 8, 12 }

{ 6, 12 } U { 4, 8, 12 }

=  { 4, 6, 8, 12 }

So, we can observe that, 1st Part = 2nd Part

So,  A ( B U C )  =  ( A B ) U ( A C )     (Proven)