Sets Class- 7th RS Aggarwal Exe-6 C Goyal Brothers ICSE Maths Solution

Sets Class- 7th RS Aggarwal Exe-6 C Goyal Brothers ICSE Math Solution . We provide step by step Solutions of lesson-6 Sets for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan . Our Solutions contain all type Questions of Exe-6 C to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics. We would learn in this article about operations on sets in details.

Sets Class- 7th RS Aggarwal Exe-6 C Goyal Brothers ICSE Maths Solution

Board ICSE
Publications Goyal brothers Prakashan
Subject Maths
Class 7th
Chapter-6 Sets
Writer RS Aggrawal
Book Name Foundation
Topics Solution of Exe-6 C
Academic Session 2023 – 2024

operations on sets

Set operations can be defined as the operations that are performed on two or more sets to obtain a single set containing a combination of elements from both all the sets being operated upon.

Type of operation on set

There are basically three types of operation on sets in Mathematics; they are: The Union of Sets (∪) The Intersection of Sets (∩).

The Union of Sets (∪)

The union of two sets A and B is defined as the set of all the elements which lie in set A and set B or both the elements in A and B altogether. The union of the set is denoted by the symbol ‘∪’.

The Intersection of Sets (∩)

The intersection of sets A and B is the set of all elements which are common to both A and B.

Exercise – 6 C

Sets Class- 7th RS Aggarwal Goyal Brothers ICSE Maths Solution

1. Let A = {2, 4, 6, 8}, B = {6, 8, 10, 12} and C = {7, 8, 9, 10}. Find :

(i) A ∪ B

Answer: A ∪ B = {2, 4, 6, 8, 10, 12}

(ii) A ∪ C

Answer: A ∪ C = {2, 4, 6, 7, 8, 9, 10}

(iii) B ∪ C

Answer: B ∪ C = {6, 7, 8, 9, 10, 12}

(iv) A ∩ B

Answer: A ∩ B = {6, 8}

(v) A ∩ C

Answer: A ∩ C = {8}

(vi) B ∩ C

Answer: B ∩ C = {8, 10}

2. Let P = {x : x is a factor of 18} and Q = {x : x is a factor of 24}.

(i) Write each one of P and Q in Roster form.

Answer: (i) P = {1, 2, 3, 6, 9, 18}, Q = {1, 2, 3, 4, 6, 8, 12, 24}

(ii) Find : (a) P ∪ Q                                     (b) P ∩ Q

Answer: (ii) (a) P ∪ Q = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24}

(b) P ∩ Q = {1, 2, 3, 6}

3. Let A = {a, b, c}, B = {b, d, e} and C = {e, f, g}, verify that :

(i) A ∪ B = B ∪ A

Answer: A ∪ B = {a, b, c, d, e}………..(i)

B ∪ A = {a, b, c, d, e}………….(ii)

By (i) and (ii), we get

A ∪ B = B ∪ A

(ii) (A ∪ B) ∪ C = A ∪ (B ∪ C)

Answer: B ∪ C = {b, d, e, f, g}

A ∪ (B ∪ C) = {a, b, c} ∪ {b, d, e, f, g}

= {a, b, c, d, e, f, g}……(i)

(A ∪ B) ∪ C = {a, b, c, d, e} ∪ {e, f, g}

= {a, b, c, d, e, f, g}…….(ii)

From (i) and (ii) we get

(A ∪ B) ∪ C = A ∪ (B ∪ C)

(iii) A ∩ B = B ∩ A

Answer: A ∩ B = {b}……(i)

B ∩ A = {b}…..(ii)

From (i) and (ii) we get

A ∩ B = B ∩ A

(iv) (A ∩ B) ∩ C = A ∩ (B ∩ C)

Answer: (A ∩ B) = {b}

(A ∩ B) ∩ C = {b} ∩ {e, f, g}

= { } or ∅……(i)

(B ∩ C) = {e}

A ∩ (B ∩ C) = {a, b, c} ∩ {e}

= { } or ∅……(ii)

From (i) and (ii) we get

(A ∩ B) ∩ C = A ∩ (B ∩ C)

4. Let A = {x : x is a multiple of 2, x < 15}, B = {x : x is a multiple of 3, x < 20}, C = {x : x is a prime, x < 20}.

(i) Write each one of the sets A, B, C in Roster form.

Answer: (i) A = {2, 4, 6, 8, 10, 12, 14}

B = {3, 6, 9, 12, 15, 18}

C = {2, 3, 5, 7, 11, 13, 17, 19}

(ii) Find : (a) A ∪ B       (b) A ∪ C       (c) B ∪ C       (d) A ∩ B        (e) A ∩ C      (f) B ∩ C

Answer: (a) A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 17, 19}

(b) A ∪ C = {2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19}

(c) B ∪ C  = {2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19}

(d) A ∩ B = {6, 12}

(e) A ∩ C = {2}

(f) B ∩ C = {3}

5. Let A = {b, d, e, f}, B = {c, d, g, h} and C = {e, f, g, h}. Find :

(i) A – B          (ii) B – C       (iii) C – A        (iv) (A – B) ∪ (B – A)      (v) (B – C) ∪ (C – B)

Answer: A = {b, d, e, f}, B = {c, d, g, h}, C = {e, f, g, h}

(i) A – B = {b, e, f}

(ii) B – C = {c, d}

(iii) C – A  = {g, h}

(iv) (A – B) ∪ (B – A) = {b, c, e, f, g, h}

(v) (B – C) ∪ (C – B) = {c, d, e, f}

6. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} be the universal set and let A = {2, 3, 4, 5, 6} and B = {3, 5, 7, 8} be its subsets .

Find : (i) A′                        (ii) B′                          (iii) A′ ∩ B′                               (iv) A′ ∪ B′
Verify that : (v) (A ∪ B)′ = (A′ ∩ B′)                  (vi) (A ∩ B′) = (A′ ∪ B)′

Answer: Hance § = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

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

B = {3, 5, 7, 8}

∴ A′ = § – A = {1, 7, 8, 9, 10}

B′ = § – B = {1, 2, 4, 6, 9, 10}

(i) A′ 

Answer: A′ = {1, 7, 8, 9, 18}

(ii) B′ 

Answer: B′ = {1, 2, 4, 6, 9, 10}

(iii) A′ ∩ B′ 

Answer: A′ ∩ B′ = {1, 9, 10}

(iv) A′ ∪ B′ 

Answer: A′ ∪ B′ = {1, 2, 4, 6, 7, 8, 9, 10}

(v) (A ∪ B)′ = (A′ ∩ B′)

Answer: (A ∪ B) = {2, 3, 4, 5, 6, 7, 8,}, (A ∪ B)′ = U – (A ∪ B) = {1, 9, 10}………(i)

Now (A′ ∩ B′) = {1, 7, 8, 9, 10} ∩ {1, 2, 4, 6, 9, 10}

From (i) and (ii) we get

= {1, 9, 10}………(ii)

∴ (A ∪ B)′ = (A′ ∩ B′)

(vi) (A ∩ B′) = (A′ ∪ B)′

Answer: (A ∩ B) = {3, 5}

(A ∩ B)′ = U – (A ∩ B)

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

= {1, 2, 3, 4, 6, 7, 8, 9, 10}……..(i)

Now, (A′ ∪ B)′ = {1, 7, 8, 9, 10} ∪ {1, 7, 4, 6, 9, 10}

= {1, 2, 3, 4, 6, 7, 8, 9, 10}………(ii)

From (i) and (ii), we get

(A ∩ B′) = (A′ ∪ B)′

7. Let U = {a, b, c, d, e, f, g} be the universal set and let its subsets be A = {a, b, d, e} and B = {b, e, g}.

Verify that : (i) (A ∪ B)′ = (A′ ∩ B′)                (ii) (A ∩ B′) = (A′ ∪ B)′

Answer: (i) (A ∪ B)′ = (A′ ∩ B′)

U = {a, b, c, d, e, f, g}

A = {a, b, d, e}

B = {b, e, g}

∴ A′ = {c, f, g}

B′ = {a, c, d, f}

A ∪ B = {a, b, d, e} ∪ {b, e, g}

= {a, b, d, e, g}

(A ∪ B)′= {c, f}…….(i)

(A′ ∩ B′) = {c, f, g} ∩ {a, c, d, f}

= {c, f}…….(ii)

(i) and (ii), we get

(A ∪ B)′ = (A′ ∩ B′)

Answer: (ii) (A ∩ B′) = (A′ ∪ B)′

(A ∩ B) = {a, b, d, e} ∩ {b, e, g}

= {b, e}

(A ∩ B′) = {a, c, d, f, g}…….(i)

(A′ ∪ B)′ = {c, f, g} ∪ {a, c, d, f}

= {a, c, d, f, g}……..(ii)

From (i) and (ii), we get

(A ∩ B′) = (A′ ∪ B)′

8. Let U = {3, 6, 9, 12, 15, 18, 21, 24} be the universal set and let A = {6, 12, 18, 24} be its subset.

Verify that : (i) A ∪ A = A                    (ii) A ∩ A = A                  (iii) A ∩ A′ = ∅
(iv) A ∪ A′ = U                   (v) (A′)′ = A

Answer: U = {3, 6, 9, 12, 15, 18, 21, 24} and A = {6, 12, 18, 24}

Answer: (i) A ∪ A = A

∴ A ∪ A = {6, 12, 18, 24} ∪ {6, 12, 18, 24}

A ∪ A = {6, 12, 18, 24} = A

Answer: (ii) A ∩ A = A

∴ A ∩ A = {6, 12, 18, 24} ∩ {6, 12, 18, 24}

A ∩ A = {6, 12, 18, 24} = A

Answer: (iii) A ∩ A′ = ∅

∴ A′= ∪ – A

= {3, 9, 15, 21}

Now, A ∩ A′ = {6,12, 18, 24} ∩ {3, 9, 15, 21}

= ∅

Answer: (iv) A ∪ A′ = U

A ∪ A′ = {6,12, 18, 24} ∪ {3, 9, 15, 21}

= {3, 6, 9, 12, 15, 18, 21, 24} = U

Answer: (v) (A′)′ = A

∴ (A′)′ = U – A′

= {3, 6, 9, 12, 15, 18, 21, 24} – {6,12, 18, 24}

= {3, 9, 12, 21}

— : end of Sets Class- 7th RS Aggarwal Exe-6 C Goyal Brothers ICSE Maths Solution:–

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