ICSE Board: Suggested Books ICSE Board: Foundation Mathematics Class 8: Reference Books Class 8: NTSE Preparation --------------------------------------------------------------

Q1. Let A – {*a, b, c, d*}, B = {*b, c, e*} and C = {*a, b, e*}. Find:

- A ∪ B = {
*a, b, c, d, e*} - B ∪ C = {
*a, b, c, e*} - A ∪ C = {
*a, b, c, d, e*} - A ∩ B = {
*b, c*} - B ∩ C = {
*b, e*} - A ∩ C = {
*a, b*}

Note:

The union of sets A and B, denoted by A ∪ B, is the set of all those elements, each one of which is either in A or in B or in both A and B

If there is a set A = {2, 3} and B = {*a, b*}, then A ∪ B = {2, 3, *a, b*}

- So if A ∪ B = {
*x*|*x*∈ A or x ∈ B} then*x*∈ A ∪ B which means*x*∈ A or*x*∈ B- And if
*x*∉ A ∪ B which means*x*∉ A or*x*∉ B

- The intersection of sets A and B is denoted by A ∩ B, and is a set of all elements that are common in sets A and B.
- If A = {1, 2, 3} and B = {2, 4, 5}, then A ∩ B = {2} as 2 is the only common element.
- Thus A ∩ B = {
*x*:*x*∈ A and*x*∈ B}- then
*x*∈ A ∩ B i.e.*x*∈ A and*x*∈ B - And if
*x*∉ A ∩ B i.e.*x*∉ A and*x*∉ B

- then

Q2. Let A = {2, 3, 4, 6}, B = {4, 6, 8, 9} and C = {2, 7, 8, 9}. Find:

- A ∪ B = {2, 3, 4, 6, 8, 9}
- B ∪ C = {2, 4, 6, 7, 8, 9}
- A ∪ C = {2, 3, 4, 7, 8, 9}
- A ∩ B = {4, 6}
- B ∩ C = {8, 9}
- A ∩ C = {2}

Q3. Let A = {1, 4, 7, 8} and B = {4, 6, 8, 9}. Find i) A – B and ii) B – A

- A – B = {1, 4, 7, 8} – {4, 6, 8, 9} = {1, 7}
- B – A = {4, 6, 8, 9} – {1, 4, 7, 8} = {6, 9}

Note:

- For any two sets A and B, the difference A – B is a set of all those elements of A which are not in B.
- i.e. if A = {1, 2, 3, 4, 5} and B = {4, 5, 6}
- Then A – B = {1, 2, 3} and B – A = {6}

- Therefore, A – B = {
*x*|*x*∈ A and*x*∉ B}, then*x*∈ A – B then*x*∈ A but x ∉ B

Q4. Let ξ = {13, 14, 15, 16, 17, 18, 19, 20, 21}, A = {13, 17, 19} and B = {14, 16, 18, 20}. Find i) A′ and ii) B′

- A′ = {14, 15, 16, 18, 20, 21}
- B′ = {13, 15, 17, 19, 21}

Note:

- Let
*x*be the universal set and let A⊆*x*. Then the complement of A, denoted by A′ is the set of all those elements of*x*which are not in A.- i.e. let ξ = {1, 2, 3, 4, 5,6 ,7 ,8} and A = {2, 3, 4}, then A′ = {1, 5, 6, 7, 8}
- Thus A′ = {
*x*|*x*∈ ξ and*x*∉ A} clearly*x*∈ A′ and*x*∉ A - Please note
- ϕ′ = ξ and ϕ′= ξ
- A ∪ A′ = ξ and A ∩ A′ = ϕ

Q5. Let ξ = {*x* | *x* ∈ Z, -4 ≤ x ≤ 4}, A = {*x* | *x* ∈ W, *x* <4} and B = {*x* | *x* ∈ N, 2 <x ≤ 4}. Find i) A′ and ii) B′

Note: First find out the elements of the three given sets.

ξ = {-4, -3, -2, -1, 0, 1, 2, 3, 4}, A = {0, 1, 2, 3} and B = {3, 4}

A′ = {-4, -3, -2, -1, 4}

B′ = {-4, -3, -2, -1, 0, 1, 2}

Q6. Let ξ = {*x* | *x* ∈ N, *x* is a factor of 144}, A = {*x* | *x* ∈ N, *x* is a factor of 24}, B = {*x* | *x* ∈ N, *x* is a factor of 36} and C = {*x* | *x* ∈ N, *x* is a factor of 48}. Find:

- A′
- B′
- C′
- A ∪ B
- B ∪ C
- A ∪ C
- A ∩ B′
- B ∩ C′
- C – A
- A – (B ∩ C)

Note: First calculate what the elements of each of these sets are. Therefore, we have the following:

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}. This is the universal set.

A = {1, 2, 3, 4, 6, 8, 12, 24}

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

Now we can easily calculate.

- A′ = {9, 16, 18, 36, 48, 72, 144}
- B′ = {8, 16, 24, 48, 72, 144}
- C′ = {9, 18, 36, 72, 144}
- A ∪ B = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}
- B ∪ C = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 26, 48}
- A ∩ C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
- A ∩ B′ = {8, 24}
- B ∩ C′ = {9, 18, 36}
- C – A = {16, 48}
- A – (B ∩ C)
- First calculate (B ∩ C) = {1, 2, 3, 4, 6, 12}
- Now Calculate A – (B ∩ C) = {8, 24}

Q7. Considering the sets given in Q.6 state whether the following statements are True or False.

- A ∩ (B ∪ C) = A
- A ⊂ C
- B ⊆ C
- A ∩ C′ = ϕ

*Note:* First calculate what the elements of each of these sets are. Therefore, we have the following:

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}. This is the universal set.

A = {1, 2, 3, 4, 6, 8, 12, 24}

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

A ∩ (B ∪ C) = A

A = {1, 2, 3, 4, 6, 8, 12, 24}

B ∪ C = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}

Therefore, A ∩ (B ∪ C) = {1, 2, 3, 4, 6, 8, 12, 24} = A

Hence Proved. TRUE

A ⊂ C

A = {1, 2, 3, 4, 6, 8, 12, 24}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} Every element of A is in C and there are two elements (16 and 18) which are in C but not in A. A is a proper subset of B.

Hence, TRUE.

B ⊆ C

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

Here every element of B is not in C. Hence it is not a subset of C. Hence FALSE.

A ∩ C′ = ϕ

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}. This is the universal set.

A = {1, 2, 3, 4, 6, 8, 12, 24}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} Therefore C′ = {9, 18, 36, 72, 144}

If we look at the two sets A and C’, we see that there are no common elements and hence

A ∩ C′ = f or null set. Hence TRUE

Q8. Let A = {*a, b, c, d, e*}, B = {*a, c, e, g*} and C = {*b, e, f, g*}. Then verify the following identities:

B ∪ C = C ∪ B

Answer:

B = {*a, c, e, g*} & C = {*b, e, f, g*}. Therefore, B ∪ C = {*a, b, c, e, f, g*}.

Similarly, C ∪ B = {*a, b, c, e, f, g*}.

Hence B ∪ C = C ∪ B

B ∩ C = C ∩ B

Answer:

B = {*a, c, e, g*} & C = {*b, e, f, g*}. Therefore, B ∩ C = {*a, b, c, e, f, g*}.

Similarly, C ∪ B = {*a, b, c, e, f, g*}.

A ∪ (B ∪ C) = (A ∪ B) ∪ C

Answer:

A = {*a, b, c, d, e*}, B = {*a, c, e, g*}, C = {*b, e, f, g*

Therefore:

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

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

Therefore, LHS = RHS. Hence Proved.

A ∩ (B ∩ C) = (A ∩ B) ∩ C

Answer:

A = {*a, b, c, d, e*}, B = {*a, c, e, g*}, C = {*b, e, f, g*}

Therefore:

LHS: A ∩ (B ∩ C) = {*a, b, c, d, e*} ∩ {*e, g*} = {*e*}

RHS: (A ∩ B) ∩ C = {*a, c, e*} ∩ {*b, e, f, g*} = {*e*}

Therefore, LHS = RHS. Hence Proved.

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Answer:

A = {*a, b, c, d, e*}, B = {*a, c, e, g*}, C = {*b, e, f, g*} Therefore:

LHS: A ∪ (B ∩ C) = {*a, b, c, d, e*} ∪ {*e, g*} = {*a, b, c, d, e, g*}

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

Therefore, LHS = RHS. Hence Proved.

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Answer:

A = {*a, b, c, d, e*}, B = {*a, c, e, g*}, C = {*b, e, f, g*} Therefore:

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

RHS: (A ∩ B) ∪ (A ∩ C) = {*a, c, e*} ∩ {*b, e*} = {*a, b, c, e*}

Q9. Let A = {*b, c, d, e*} and B = {*d, e, f, g*} be the two sub sets of the universal set ξ = {*b, c, d, e, f, g*}. Then verify the following:

(A ∪ B)′ = (A′∩ B′)

Answer:

LHS: A ∪ B = {*b, c, d, e*} ∪ {*d, e, f, g*} = {*b, c, d, e, f, g*}

Therefore, (A ∪ B)′ = ϕ

RHS: A′ = {*f, g*} and B′ = {*b, c*}

Therefore, A′∩ B′ = ϕ

Therefore, LHS = RHS. Hence proved.

(A ∩ B)′ = (A′∪ B′)

Answer:

x = {*b, c, d, e, f, g*}

LHS: A ∩ B = {*b, c, d, e*} ∩ {*d, e, f, g*} = {*d, e*}

Therefore, (A ∩ B)′ = {*b, c, f, g*}

RHS: A′ = {*f, g*} and B′ = {*b, c*}

Therefore, A′∪ B′ = {*b, c, f, g*}

Therefore, LHS = RHS. Hence proved.

Q10. Fill in the blanks

- A ∪ A = A
- A ∩ A = A
- A ∪ f = A
- A ∩ f = A
- (A ∪ B)′ = (A′∩ B′)
- (A ∩ B)′ = (A′∪ B′)

Q11. Let ξ ={x | x ∈ N 4 ≤ x < 18} and A, B, and C are subsets of ξ. Given A = {x | x is a multiple of 2}, B = {x | x is a multiple of 3}, C = {x | x ∈ N x < 11}. Then verify, the following:

Note: First identify the elements of all the 4 sets in this question. That would make the solution easier.

ξ= {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

A = {4, 6, 8, 10, 12, 14, 16}

B = {6, 9, 12, 15}

C = {4, 5, 6, 7, 8, 9, 10}.

(A ∪ B)′ = (A′∩ B′)

Answer:

LHS: (A ∪ B) = {4, 6, 8, 9, 10, 12, 14, 15, 16}

(A ∪ B)′ = {5, 7, 11, 13, 17}

RHS: (A′∩ B′) = {5, 7, 9, 11, 13, 15, 17} ∩ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}

(A′∩ B′) = {5, 7, 11, 13, 17}

Therefore, LHS = RHS. Hence proved.

(A ∩ B)′ = (A′∪ B′)

Answer:

LHS: (A ∩ B)′ = {4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17}

RHS: (A′∪ B′) = {5, 7, 9, 11, 13, 15, 17} ∪ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}

= {4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17}

Therefore, LHS = RHS. Hence proved.

A – B = A ∩ B′

Answer:

A – B = {4, 6, 8, 10, 12, 14, 16} – {6, 9, 12, 15} = {4, 8, 10, 14, 16}

A ∩ B′ = {4, 6, 8, 10, 12, 14, 16} ∩ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17} = {4, 8, 10, 14, 16}

Therefore, LHS = RHS. Hence proved.

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Answer:

A = {4, 6, 8, 10, 12, 14, 16}

B = {6, 9, 12, 15}

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

A ∪ (B ∩ C) = {4, 6, 8, 9, 10, 12, 14, 16}

(A ∪ B) ∩ (A ∪ C) = {4, 6, 8, 9, 10, 12, 14, 15, 16} ∩ {4, 5, 6, 7, 8, 9, 10, 12, 14, 16} = {4, 6, 8, 9, 10, 12, 14, 16}

Therefore, LHS = RHS. Hence proved