Question

6. IfA 3,5, 7,9, 11 ), B 17,9, 11, 13], C 111, 13, 15and D(ii) BnC(v) BnDA(15, 17); find(i) An Biv) An CVi1(x) (Au D)n Bu C)(ii) AnCn D(vi) An(BU C)V1(vii) AnD(viii)(BUD)(i)(AB)n(BUC)Vill

Answer 1

Given, A={ 3,5,7,9,11 } , B={ 7,9,11,13 } , C={ 11,13,15 } and D={ 15,17 }

(i)

The intersection between sets A and B is denoted as A∩B .

So, the intersection between sets A and B is,

A∩B={ 3,5,7,9,11 }∩{ 7,9,11,13 } ={ 7,9,11 }

Hence, A∩B={ 7,9,11 } .

(ii)

The intersection of the sets B and C is denoted as B∩C .

Given, B={ 7,9,11,13 } and C={ 11,13,15 }

So, the intersection between sets B and C is,

B∩C={ 7,9,11,13 }∩{ 11,13,15 } ={ 11,13 }

Hence, B∩C={ 11,13 } .

(iii)

The intersection of the sets A , C and D is denoted as A∩C∩D .

Given, A={ 3,5,7,9,11 } , C={ 11,13,15 } and D={ 15,17 }

So, A∩C∩D can be calculated as

A∩C∩D={ 3,5,7,9,11 }∩{ 11,13,15 }∩{ 15,17 } ={ 11 }∩{ 15 } =ϕ

Hence, A∩C∩D=ϕ .

(iv)

The intersection of the sets A and C is denoted as A∩C .

Given, A={ 3,5,7,9,11 } and C={ 11,13,15 }

So, A∩C can be calculated as

A∩C={ 2,3,4,5,6 }∩{ 9,11,13,15 } =ϕ

Hence, A∩C=ϕ .

(v)

The intersection of the sets B and D is denoted as B∩D .

Given, B={ 7,9,11,13 } and D={ 15,17 }

So, the intersection between B and D is

B∩D={ 7,9,11,13 }∩{ 15,17 } =ϕ

Hence, B∩D=ϕ .

(vi)

Given, A={ 3,5,7,9,11 } , B={ 7,9,11,13 } and C={ 11,13,15 }

So, A∩( B∪C ) can be calculated as

A∩( B∪C )={ 3,5,7,9,11 }∩( { 7,9,11,13 }∪{ 11,13,15 } ) ={ 3,5,7,9,11 }∩{ 7,9,11,13,15 } ={ 7,9,11 }

Hence, A∩( B∪C )={ 7,9,11 } .

(vii)

The intersection of the sets A and D is denoted as A∩D .

Given, A={ 3,5,7,9,11 } and D={ 15,17 }

So, A∩D can be calculated as

A∩D={ 2,3,4,5,6 }∩{ 15,17 } =ϕ

Hence, A∩D=ϕ .

(viii)

Given, A={ 3,5,7,9,11 } , B={ 7,9,11,13 } and D={ 15,17 }

So, A∩( B∪D ) can be calculated as

A∩( B∪D )={ 3,5,7,9,11 }∩( { 7,9,11,13 }∪{ 15,17 } ) ={ 3,5,7,9,11 }∩{ 7,9,11,13,15,17 } ={ 7,9,11 }

Hence, A∩( B∪D )={ 7,9,11 } .

(ix)

Given, A={ 3,5,7,9,11 } , B={ 7,9,11,13 } and C={ 11,13,15 }

So, ( A∩B )∩( B∪C ) can be calculated as

( A∩B )∩( B∪C )=( { 3,5,7,9,11 }∩{ 7,9,11,15 } )∩( { 7,9,11,13 }∪{ 11,13,15, } ) ={ 7,9,11 }∩{ 7,9,11,13,15 } ={ 7,9,11 }

Hence, ( A∩B )∩( B∪C )={ 7,9,11 } .

(x)

Given, A={ 3,5,7,9,11 } , B={ 7,9,11,13 } , C={ 11,13,15 } and D={ 15,17 }

So, ( A∪D )∩( B∪C ) can be calculated as

( A∪D )∩( B∪C )=( { 3,5,7,9,11 }∪{ 15,17 } )∩( { 7,9,11,13 }∪{ 11,13,15, } ) ={ 3,5,7,9,11,15,17 }∩{ 7,9,11,13,15 } ={ 7,9,11,15 }

Hence, ( A∪D )∩( B∪C )={ 7,9,11,15 } .