# 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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