## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 6 Operation on sets Venn Diagrams Ex 6.1

Question 1.

If A = {0, 1, 2, 3, …….., 8}, B = {3, 5, 7, 9, 11} and C = {0, 5, 10, 20}, find

(i) A ∪ B

(ii) A ∪ C

(iii) B ∪ C

(iv) A ∩ B

(v) A ∩ C

(vi) B ∩ C

Also find the cardinal number of the sets B ∪ C, A ∪ B, A ∩ C and B ∩ C.

Solution:

A = {0, 1, 2, 3, …, 8}

B = {3, 5, 7, 9, 11}

C = {0, 5, 10, 20}

(i) A ∪ B = {0,1, 2, 3, 4, 5, 6, 7, 8, 9,11},

n(A ∪ B) = 11

(ii) A ∪ C = {0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 20},

n(A ∪ C) = 11

(iii) B ∪ C = {0, 3, 5, 7, 9, 10, 11, 20},

n(B ∪ C) = 8

(iv) A ∩ B = {3, 5, 7}, n(A ∩ B) = 3

(v) A ∩ C = (0, 5), n(A ∩ C) = 2

(vi) B ∩ C = {5}, n(B ∩ C) = 1

Question 2.

Find A’ when

(i) A= {0, 1, 4, 7} and E, = {x | x ϵ W, x ≤ 10}

(ii) A = {consonants} and ξ = {alphabets of English}

(iii) A = boys in class VIII of all schools in Bengaluru} and ξ = {students in class VIII of all schools in Bengaluru}

(iv) A = {letters of KALKA} and ξ = {letters of KOLKATA}

(v) A = {odd natural numbers} and ξ = {whole numbers}.

Solution:

Find A’

(i) A = {0, 1,4,7} and ξ= {x|x ϵ W, x ≤10}

ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

∴ A’ = {2, 3, 5, 6, 8, 9, 10}

(ii) A = {consonants} and ξ = {alphabets of English}

∴ A’ = {Vowels}

(iii) A = boys in class VIII of all schools in Bengaluru} and

ξ = {students in class VIII of all schools in Bengaluru}

∴ A’ = {Girls in class VIII of all schools in Bengaluru}

(iv) A = {letters of KALKA}

ξ = {letters of KOLKATA}

= {KAL} and ξ = {KOLAT}

∴ A’ = {O, T}

(v) A = {odd natural numbers}

ξ = {whole numbers}

∴ A’ = {0, even whole numbers}

Question 3.

If A {x : x ϵ N and 3 < x < 1} and B = {x : x ϵ Wand x ≤ 4}, find

(i) A ∪ B

(ii) A ∩ B

(iii) A – B

(iv) B – A

Solution:

A = {x : x ϵ N and 3 < x < 7}

= {4, 5, 6}

B = {x: x ϵ W and x ≤ 4}

= {0, 1, 2, 3, 4}

(i) A ∪ B = {0, 1, 2, 3, 4, 5, 6}

or {x : x ϵ W and x ≤ 4}

(ii) A ∩ B = {4}

(iii) A – B = {5, 6}

(iv) B – A = {0, 1, 2, 3}

Question 4.

If P = {x : x ϵ W and x < 6} and Q = {x : x ϵ N and 4 ≤ x ≤ 9}, find

(i) P ∪ Q

(ii) P ∩ Q

(iii) P – Q

(iv) Q – P

Is P ∪ Q a proper superset of P ∩ Q ?

Solution:

P = {x: x ϵ W and x < 6}

= {0, 1, 2, 3, 4, 5}

Q = {x : x ϵ N and 4 ≤ x ≤ 9}

= (4, 5, 6, 7, 8}

(i) P ∪ Q = {0, 1, 2, 3, 4, 5, 6, 7, 8}

{x : x ϵ W and x ≤ 8}

(ii) P ∩ Q = {4, 5}

(iii) P – Q = {0, 1, 2, 3}

(iv) Q – P = {6, 7, 8}

Yes, P ∪ Q is the super set of P ∩ Q

because P ∩ Q is subset of P ∪ Q.

Question 5.

If A = (letters of word INTEGRITY) and B = (letters of word RECKONING), find

(i) A ∪ B

(ii) A ∩ B

(iii) A – B

(iv) B – A

Also verify that:

(a) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

(b) n(A – B) = n(A ∪ B) – n(B)

= n(A) – n(A ∪ B)

(c) n(B – A) = n(A ∪ B) – n(A)

= n(B) – n(A ∩ B)

(d) n(A ∪ B) = n(A – B) + n(B – A) + n(A ∩ B).

Solution:

A = (I, N, T, E, G, R, Y}

B = {R, E, C, K, O, N, I, G}

(i) A ∪ B = {I, N, T, E, G, R, Y} ∪ {R, E, C, K, O, N, I, G}

= {I, N, T, E, G, R, Y, C, K, 0}

(ii) A ∩ B = {I, N, T, E, G, R, Y} ∩ {R, E, C, K, O, N, I, G}

= {I, N, E, R, G}

(iii) A – B = {R, E, C, K, O, N, I, G} – {1, N, T, E, G, R, Y}

= {T, Y}

(iv) B – A = – {R, E, C, K, O, N, I, G} {I, N, T, E, G, R, Y}

= {C, K, O}

Verification:

Here n(A ∪ B) = 10

n(A) = 7

n(B) = 8

n(A ∩ B) = 5

n(A – B) = 2

n(B – A) = 3

(a) n(A ∪ B) = 10

and n(A) + n(B) – n(A ∩ B)

= 7 + 8 – 5 = 15 – 5 = 10

∴ n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

(b) n(A – B) = 2

n(A ∪ B) – n(B) = 10 – 8 = 2

and n(A) – n(A ∩ B) = 7 – 5 = 2

∴ n(A – B) = n(A ∪ B) – n(B)

= n(A) – n(A ∩ B)

(c) n(B – A) = 3

n(A ∪ B) – n(A) = 10 – 7 = 3

and n(B) – n(A ∩ B) = 8 – 5 = 3

∴ n(B – A) = n(A ∪ B) – n(A)

= n(B) – n(A ∩ B)

(d) n(A ∪ B) = 10

and n(A – B) + n(B – A) + n(A ∩ B)

= 2 + 3 + 5 = 10

∴ n(A ∪ B) = n(A – B) + n(B – A) + n(A ∩ B).

Question 6.

If ξ = {natural numbers between 10 and 40}

A = {multiples of 5} and

B = {multiples of 6}, then

(i) find A ∪ B and A ∩ B

(ii) verify that

n(A ∪ B) = B (A) + n(B) – n(A ∩ B).

Solution:

Here, ξ = {11, 12, 13, ……., 39}

It is understood that A and B are subsets of ξ {Universal set)

So, the elements of A and B are to be taken only from ξ.

A= {multiples of 5}

= multiples of 5 which belong to ξ

= {15, 20, 25, 30, 35}

B = {multiples of 6}

= multiples of 6 which belong to ξ

= {12, 18, 24, 30, 36}

(i) A ∪ B = {15, 20, 25, 30, 35, 12, 18, 24, 36}

A ∩ B = {30}

(ii) n(A) = 5, n(B) = 5

n(A ∪ B) = 9 and n(A ∩ B) = 1

∴ n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

9 = 5 + 5 – 1 = 10 – 1 = 9

Hence, verified.

Question 7.

If ξ ={1,2, 3, …. 9}, A = {1, 2, 3, 4, 6, 7, 8} and B = {4, 6, 8}, then find.

(i) A’

(ii) B’

(iii) A ∪ B

(iv) A ∩ B

(v) A – B

(vi) B – A

(vii) (A ∩ B)’

(viii) A’ ∪ B’

Also verify that:

(a) (A ∩ B)’ = A’ ∪ B’

(b) n(A) + n(A’) = n(ξ)

(c) n(A ∩ B) + n((A ∩ B)’) = n(ξ)

(d) n(A – B) + n(B – A) + n(A ∩ B)

= n(A ∪ B).

Solution:

The given sets in the roster form are:

ξ = {1,2, 3, …, 9}

A = {1,2, 3, 4, 6, 7,8} and B = {4, 6, 8}

(i) A’ = ξ – A

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

= {5, 8, 9}

(ii) B’ = ξ – B

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

= {1,2,3, 5, 7, 9}

(iii) A ∪ B = {1, 2, 3, 4, 6, 7, 8} ∪ {4, 6, 8}

= {1,2, 3, 4, 6, 7, 8}

(iv) A ∩ B = {1, 2, 3, 4, 6, 7, 8} ∩ {4, 6, 8}

= {4, 6, 8}

(v) A – B = {1, 2, 3, 4, 6, 7, 8} – {4, 6, 8}

= {1, 2, 3, 7}

(vi) B – A = {4, 6, 8} – {1, 2, 3, 4, 6, 7, 8}

= {}

(vii) (A ∩ B)’ = ξ – (A ∩ B)

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

= {1,2, 3, 5, 7, 9}

(viii) A’ ∪ B’ = {5, 8, 9} ∪ {1, 2, 3, 5, 7, 9}

= {1,2, 3, 5, 7, 8, 9}

Verification :

Here n(A) = 6

n(A’) = 3

n(ξ) = 9

n(A ∩ B) = 3

n((A ∩ B)’) = 6

n(A – B) = 4

n(B – A) = 0

n(A ∩ B)’ = 6

n( A’ ∩ B’) = 7

n(A ∪ B) = 8

(a) (A ∩ B)’ = ξ – (A ∩ B)

= {1,2, 3, 4, 5, 6, 7, 8, 9} – (4, 6, 8}

= {1,2, 3, 5, 7, 9}

A’ ∪ B’ = {5, 7, 9} ∪ {1, 2, 3, 5, 7, 9}

= {1, 2, 3, 5, 7, 9}

(A ∩ B)’ = A’ ∪ B’ (Verified)

(b) n(A) + n(A’) = 6 + 3 = 9

n(ξ) = 9

∴ n(A) + n(A’) = n(ξ) (Verified)

(c) n(A ∩ B) + n((A ∩ B)’) = 3 + 6 = 9

n(ξ) = 9

∴ n(A ∩ B) + n((A ∩ B)’) = n(ξ) (Verified)

(d) n(A – B) + n(B – A) + n(A ∩ B)

= 4 + 0 + 3 = 7

n(A ∪ B) = 7

∴ n(A – B) + n(B – A) + n(A ∩ B)

= n(A ∪ B) (Verified)

Question 8.

If 4 = {x : x ϵ W, x ≤ 10}, A. = {x : x ≥ 5} and B = {x : 3 ≤ x < 8}, then verify that:

(i) (A ∪ B)’ = A’ ∩ B’

(ii) (A ∩ B)’= A’ ∪ B’

(iii) A – B = A ∩ B’

(iv) B – A = B ∩ A’

Solution:

The given sets in the roster form are :

ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {5, 6, 7, 8, 9, 10}

B = {3, 4, 5, 6, 7}

(i) L.H.S. = (A ∪ B)’

A ∪ B = {5, 6, 7, 8, 9, 10} ∪ {3, 4, 5, 6, 7}

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

∴ (A ∪ B)’ = ξ – (A ∪ B)

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7, 8, 9, 10}

= {0, 1, 2}

R.H.S. = A’ ∩ B’

A’ = ξ – A

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}

= {0, 1, 2, 3, 4}

B’ = ξ – B

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}

= {0, 1, 2, 8, 9, 10}

A’ ∩ B’ = {0, 1, 2, 3, 4} ∩ {0, 1, 2, 8, 9, 10}

= {0, 1, 2}

Hence (A ∪ B)’ = A’ ∩ B’ (Verified)

(ii) L.H.S. = (A ∩ B)’

A n B = {5, 6,7, 8,9, 10} n {3,4, 5,6, 7} = {5, 6, 7}

(A ∩ B)’ = ξ – (A ∩ B)

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

– {5, 6, 7}

= {0, 1, 2, 3, 4, 8, 9, 10}

R.H.S. = A’ u B’

A’ = ξ – A

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}

= {0, 1, 2, 3, 4}

B’ = ξ – B

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}

= {0, 1, 2, 8, 9, 10}

∴ A’ ∪ B’ = {0, 1, 2, 3, 4} ∪ {0, 1, 2, 8, 9, 10}

= {0, 1, 2, 3, 4, 8, 9, 10}

Hence (A ∩ B)’ =A’ ∪ B’ (Verified)

(iii) L.H.S. = A – B

∴ A – B = {5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}

= {8, 9, 10}

R.H.S. = A ∩ B’

B’ = ξ – B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}

= {0, 1, 2, 8, 9, 10}

∴ A ∩ B’ = {5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 8, 9, 10}

= {8, 9, 10}

Hence A – B = A ∩ B’

(iv) L.H.S. = B – A

∴ B – A = {3, 4, 5, 6, 7} – {5, 6, 7, 8, 9, 10}

= {3, 4}

R.H.S. = B ∩ A’

A’ = ξ – A

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}

= (0, 1,2, 3, 4}

B ∩ A’ = {3, 4, 5, 6, 7} ∩ {0, 1, 2, 3, 4} = {3, 4}

Hence B – A = B ∩ A’.

Question 9.

If n(A) = 20, n(B) = 16 and n(A ∪ B) = 30, find n(A ∩ B).

Solution:

Given that,

n(A) = 20, n(B) = 16 and

n(A ∪ B) = 30 Then n(A ∩ B) = ?

We know that,

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

⇒ 30 = 20 + 16 – n(A ∩ B)

⇒ 30 = 36 – n(A ∩ B)

⇒ n(A ∩ B) = 36 – 30

⇒ n(A ∩ B) = 6

Hence n(A ∩ B) = 6.

Question 10.

If n ξ = 20 and n(A’) = 7, then find n(A).

Solution:

We know that,

n(A’) = n ξ – n(A)

7 = 20 – n(A)

7 – 20 = – n(A)

-13 = -n(A) ⇒ n(A) = 13.

Question 11.

If n(ξ) = 40, n(A) = 20, n(B’) = 16 and n(A ∪ B) = 32, then find n(B) and n(A ∩ B).

Solution:

We know that,

n(B’) = n(ξ) – n(B)

16 = 40 – n(B)

16 – 40 = – n(B)

-24 = -n(B)

n(B) = 24

and also, we know that,

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

32 = 20 + 24 – n(A ∩ B)

32 = 44 – n(A ∩ B)

32 – 44 = – n(A ∩ B)

⇒ -12 = -n(A ∩ B)

⇒ n(A ∩ B) = 12

Question 12.

If n(ξ) = 32, n(A) = 20, n(B) = 16 and n((A ∪ B)’) = 4, find :

(i) n(A ∪ B)

(ii) n(A ∩ B)

(iii) n(A – B)

Solution:

Given that,

n(ξ) = 32, n(A) = 20, n(B) = 16,

n((A ∪ B)’) = 4

(i) n(A ∪ B) = n(ξ) – n((A ∪ B)’)

= 32 – 4 = 28

(ii) We know that,

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

⇒ 28 = 20 + 16 – n(A ∩ B)

⇒ 28 = 36 – n(A ∩ B)

⇒ n(A ∩ B) = 36 – 28 ⇒ n(A ∩ B) = 8

Hence n(A ∩ B) = 8.

(iii) n(A – B) = n(A) – n(A ∩ B)

= 20 – 8 = 12.

Question 13.

If n(ξ) = 40, n(A’) = 15, n(B) = 12 and n((A ∩ B)’) = 32, find :

(i) n(A)

(ii) n(B’)

(iii) n(A ∩ B)

(iv) n(A ∪ B)

(v) n(A – B)

(vi) n(B – A)

Solution:

Given that,

n(ξ) = 40

n(A’)= 15

n(B) = 12

n((A ∩ B)’) = 32

(i) n(A) = n(ξ) – n(A’)

= 40 – 15 = 25

(ii) n(B’) = n(ξ) – n(B) = 40 – 12 = 28

(iii) n(A ∩ B) = n(ξ) – n((A ∩ B)’)

= 40 – 32 = 8

(iv) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

= 25 + 12 – 8 = 25 + 4 = 29

(v) n(A – B) = n(A) – n(A∩B)

= 25 – 8 = 17

(vi) n(B – A) = n(B) – n(A ∩ B)

= 12 – 8 = 4

Question 14.

If n(A – B) = 12, n(B – A) = 16 and n(A ∩ B) = 5, find:

(i) n(A)

(ii) n(B)

(iii) n(A ∪ B)

Solution:

Given that n(A – B) = 12,

n(B – A)= 16 and n(A ∩ B) = 5

(i) n(A) = n(A – B) + n(A ∩ B)

{∵ n(A – B) = n(A) – n(A ∩ B)}

= 12 + 5 = 17

(ii) n(B) = n(B – A) + n(A ∩ B)

{∵ n(B – A) = n(B) – n(A ∩ B)}

= 16 + 5 = 21

(iii) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

= 17 + 21 – 5 = 38 – 5 =33