## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 6 Operation on sets Venn Diagrams Check Your Progress

Question 1.

If ξ={x : x ϵ N, r < 25} and A = {x : x is a composite number}, then find A’ in set builder from and also in roster form.

Solution:

ξ = {x : x ϵ N, x < 25}

A = {x : x is a composite number},

then find A’ in set builder form and also in roster form

ξ = {0, 2, 3, 4, ……., 24}

A = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24}

∴ A’ = (1, 2, 3, 5, 7, 11, 13, 17, 19, 23}

= {x : x ϵ N, x is a prime number less than 25}

Question 2.

If ξ = {x | x ϵ I, -5 ≤ x ≤ 5},

A= {-5, -3, 0, 3, 5} and B = {-4, -2, 0, 2, 4}, then

(i) Find A ∪ B and A ∩ B

(ii) Verify that n(A ∪ B) = n(A) + n(B) – n(A ∩ B).

(iii) Find A’ and verify that n(A) + n(A’) = n(ξ).

(iv) Find B’ and verify that n(B’) = n(ξ) – n(B)

(v) Find A ∪ A’, A ∩ A’, B ∪ B’ and B ∩ B’.

(vi) Find (A ∪ B)’ and A’ ∩ B’. Are they equal ?

(vii) Find (A ∩ B)’ and A’ ∪ B’. Are they equal ?

Solution:

ξ = {x | x ϵ 1, – 5 ≤ x ≤ 5}

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

n(ξ) = 11

A = {-5, -3, 0, 3, 5} ⇒ n(A) = 5

B = {-4,-2, 0, 2, 4} ⇒ n(B) = 5

(i) A ∪ B = {-5, -4, -3, -2, 0, 2, 3, 4, 5}

⇒ n(A ∪ B) = 9

A ∩ B = {0}

⇒ n(A n B) = 1

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

L.H.S. = n(A ∪ B) = 9

R.H.S. = n(A) + n(B) – n(A ∩ B)

= 5 + 5 – 1 = 9

L.H.S. = R.H.S.

(iii) A’ = ξ – A = {-4, -2, -1, 1, 2, 4}

⇒ n(A’) = 6

Now, n(A) + n(A’) = 5 + 6 = 11 = n(ξ)

Hence verified

(iv) B’ = ξ – B = {-5, -3, -1, 1, 3, 5}

n(B’) = 6

n(ξ) – n(B) = 11 – 5 = 6 = n(B’)

(v) A ∪ A’ = ξ

A ∩ A’ = ϕ

B ∪ B’ = ξ, B ∩ B’ = ϕ

(vi) (A ∪ B)’ = {-1, 1} and A’ ∩ B’= {-1, 1}

Yes both are equal.

(vii) (A ∩ B)’ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5}

A’ ∪ B’ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5}

Yes, (A ∩ B)’ = A’ ∪ B’

Question 3.

If ξ = {x | x ϵ N, x ≤ 12}, A = {prime numbers} and B = {odd numbers}, then

(i) Find A ∪ B and A ∩ B.

(ii) Verify that n(A ∪ B) + n(A ∩ B) = n(A) + n(B).

(iii) Find A’ and B’.

(iv) Find (A ∪ B)’ and verify that n(A ∪ B) + n(A ∪ B)’ = n(ξ).

(v) Find (A ∩ B)’ and A’ ∪ B’. Are they equal ?

Solution:

Here,

ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

A = {2, 3, 5, 7, 11}

B = {1, 3, 5, 7, 9, 11}

(i) A ∪ B = {1, 2, 3, 5, 7, 9,11}

and A ∩ B = {3, 5, 7, 11}

(ii) n(A ∪ B) + n(A ∩ B) = n(A) + n(B)

7 + 4 = 5 + 6

11 = 11

Hence, verified.

(iii) A’ = ξ – A = {1, 4, 6, 8, 9, 10, 12}

and B’ = ξ – B = {2, 4, 6, 8, 10, 12}

(iv) (A ∪ B)’ = ξ – (A ∪ B)= {4, 6, 8, 10, 12}

To show,

n(A ∪ B) + n(A ∪ B)’ = n(ξ)

7 + 5 = 12

12 = 12

Hence verified

(v) (A ∩ B)’ = ξ – (A ∪ B)

= {1, 2, 4, 6, 8, 9, 10, 12}

A’ ∪ B’ = {1, 2, 4, 6, 8, 9, 10, 12}

Yes, the given sets are equal

because they have the same elements.

Question 4.

If ξ = {x : x ϵ N, x ≤ 12}, A= {x : x ≥ 7} and B = {x : 4 < x < 10}. 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:

(i) (A ∪ B)’ = A’ ∩ B’

(ii) A – B = A ∩ B’

(iii) n(A ∪ B) + n((A ∪ B)’) = n(ξ)

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

Solution:

The given sets in the roster form are :

ξ = {1,2, 3, …, 12}

A = {7, 8, 9, 10, 11, 12}

B = {5, 6, 7, 8, 9}

(i) A’ = ξ – A

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} – {7,8,9,10,11,12}

= {1,2, 3, 4, 5, 6}

(ii) B’ = ξ – B

= {1, 2, 3, …, 12} – {5, 6, 7, 8, 9}

= {1, 2, 3, 4, 10, 11, 12}

(iii) A ∪ B = {7, 8, 9, 10, 11, 12} ∪ {5, 6, 7, 8, 9}

= {5, 6, 7, 8, 9, 10, 11, 12}

(iv) A ∩ B = {7, 8, 9, 10, 11, 12} ∩ {5, 6, 7, 8, 9}

= {7, 8, 9}

(v) A – B = {7, 8, 9, 10, 11, 12} – {5, 6, 7, 8, 9}

= {10, 11, 12}

(vi) B – A = {5, 6, 7, 8, 9} – {7, 8, 9, 10, 11, 12}

= {5, 6}

(vii) (A ∪ B)’ = ξ – A ∪ B

= {1,2,3,…, 12} – {5, 6, 7, 8, 9, 10, 11, 12}

= {1,2,3, 4}

(viii) A’ ∩ B’= {1,2, 3, 4, 5, 6} ∩ {1,2, 3, 4, 10, 11, 12}

= {1, 2, 3, 4}

Verification :

(i) (A ∪ B)’ = {1, 2, 3, 4} (By VII part)

and A’ ∩ B’ = {1, 2, 3, 4} (By VIII part)

Hence (A ∪ B)’ = A’ ∩ B’ (Verified)

(ii) A – B = {10, 11, 12} (By (v) part)

and A ∩ B’ = {7, 8, 9, 10, 11, 12} ∩ {1, 2, 3, 4, 10, 11, 12} (By (ii) part)

= {10, 11, 12}

Hence A – B = A ∩ B’ (Verified)

(iii) n(A ∪ B) + n((A ∪ B)’) = 8 + 4=12

n(ξ) = 12

Hence n(A ∪ B) + n((A ∪ B)’) = n(ξ) (Verified)

(iv) n(A ∪ B) = 8

and n(A – B) + n(B – A) + n(A ∩ B)

= 3 + 2 + 3 = 8

Hence n(A ∪ B) = n(A – B) + n(B – A) + n(A ∩ B) (Verified)

Question 5.

Given A = {students who like cricket} and B = {students who like tennis}; n(A) = 20, n(B) = 15 and n(A ∩ B)= 5.

Illustrate this through a Venn diagram. Hence find n(A ∪ B).

Solution:

Here, n(A) = 20

n(B) = 15

n(A ∩ B) = 5

Now, only A i.e. students who like cricket only

= n(A) – n(A ∩ B) = 20 – 5 = 15

Only B i.e. students who like tennis only

= n(B) – n(A ∩ B) = 15 – 5 = 10

n(A ∪ B)= 15 + 5 + 10 (from Venn diagram)

Using formula

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

= 20 + 15 – 5 = 35 – 5 = 30.

Question 6.

If n(ξ) = 50, n(A) = 15, n(B) = 13 and n(A ∩ B) = 10. Find n(A’), n(B’) and n(A ∪ B).

Solution:

Here, n(ξ) = 50

n(A) = 15

n(B) = 13

n(A ∩ B) = 10

Using formula,

n(A’) = n(ξ) – n(A)

= 50 – 15 = 35

n(B’) = n(ξ) – n(B)

= 50 – 13 = 37

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

= 15 + 13 – 10 = 18.

Question 7.

If n(ξ) = 60, n(A) = 35, n(B’) = 36 and n((A ∩ B)’) = 51, find :

(i) n(B)

(ii) n(A ∩ B)

(iii) n(A ∪ B)

(iv) n(A – B)

Solution:

Given n(ξ) = 60

n(A) = 35

n(B’) = 36

and n((A ∩ B)’) = 51

(i) n(B) = n(ξ) – n(B’) = 60 – 36 = 24

(ii) n(A ∩ B) = n(ξ) – n((A ∩ B)’) = 60 – 51 = 9

(iii) n(A ∪ B) = n(A) + n(B) – n(A ∩ B) = 35 + 24 – 9 = 59 – 9 = 50

(iv) n(A – B) = n(A) – n(A ∩ B) = 35 – 9 = 26

Question 8.

In a city, 50% people read newspaper A, 45% read newspaper B, and 25% read neither A nor B. What percentage of people read both the newspapers A as well as B ?

Solution:

Let A = Those people who read newspaper A

B = Those people who read newspaper B

Here. n(ξ) = 100%

n(A) = 50%

n(B) = 45%

n(A ∪ B)’ = 25%

∴ n(A ∪ B) = n(ξ) – n((A ∪ B)’)

= 100% – 25% = 75%

We know that,

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

⇒ 75% = 50% + 45% – n(A ∩ B)

⇒ n(A ∩ B) = 95% – 75%

⇒ n(A ∩ B) = 20%