Class 8: Sets – Exercise 1C – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1. Let $A = \{a, b, c, d \}$ , $B = \{b, c, e \}$ and $C = \{a, b, e \}$ . Find:

(i) $A \cup B$ (ii) $B \cup C$ (iii) $A \cup C$ (iv) $A \cap B$ (v) $B \cap C$ (vi) $A \cap C$

Answer:

(i) $A \cup B = \{a, b, c, d, e \}$ (ii) $B \cup C = \{a, b, c, e \}$ (iii) $A \cup C = \{a, b, c, d, e \}$

(iv) $A \cap B = \{b, c \}$ (v) $B \cap C = \{b, e \}$ (vi) $A \cap C = \{a, b \}$

Note: The union of sets $A$ and $B$ , denoted by $A \cup B$ , 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 $A \cup B = \{2, 3, a, b \}$

So if $A \cup B = \{x | x \in A \ or \ x \in B \}$ then

$x \in A \cup B$ which means $x \in A$ or $x \in B$

And if $x \notin A \cup B$ which means $x \notin A$ or $x \notin B$

The intersection of sets $A$ and $B$ is denoted by $A \cap B$ , 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 $A \cap B = \{2 \}$ as $2$ is the only common element.

Thus $A \cap B = \{x: x \in A \ and \ x \in B \}$

then $x \in A \cap B$ i.e. $x \in A$ and $x \in B$

And if $x \notin A \cap B$ i.e. $x \notin A$ and $x \notin B$

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Question 2. Let $A = \{2, 3, 4, 6 \}$ , $B = \{4, 6, 8, 9 \}$ and $C = \{2, 7, 8, 9 \}$ . Find: (i) $A \cup B$ (ii) $B \cup C$ (iii) $A \cup C$ (iv) $A \cap B$ (v) $B \cap C$ (vi) $A \cap C$

Answer:

(i) $A \cup B = \{2, 3, 4, 6, 8, 9 \}$ (ii) $B \cup C = \{2, 4, 6, 7, 8, 9 \}$

(iii) $A \cup C = \{2, 3, 4, 7, 8, 9 \}$ (iv) $A \cap B = \{4, 6 \}$

(v) $B \cap C = \{8, 9 \}$ (vi) $A \cap C = \{2 \}$

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Question 3. Let $A = \{1, 4, 7, 8 \}$ and $B = \{4, 6, 8, 9 \}$ . Find i) $A - B$ and ii) $B - A$

Answer:

(i) $A - B = \{1, 4, 7, 8 \} - \{4, 6, 8, 9 \} = \{1, 7 \}$

(ii) $B - A = \{4, 6, 8, 9 \} - \{1, 4, 7, 8 \} = \{6, 9 \}$

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, $A - B = \{x | x \in A \ and \ x \notin B \}$ , then

$x \in A - B$ then $x \in A$ but $x \notin B$

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Question 4. Let $\xi = \{13, 14, 15, 16, 17, 18, 19, 20, 21 \}$ , $A = \{13, 17, 19 \}$ and $B = \{14, 16, 18, 20 \}$ . Find (i) $A'$ and (ii) $B'$

Answer:

(i) $A' = \{14, 15, 16, 18, 20, 21 \}$

(ii) $B' = \{13, 15, 17, 19, 21 \}$

Note: Let $x$ be the universal set and let $A \subseteq x$ . 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 $\xi = \{1, 2, 3, 4, 5,6 ,7 ,8 \}$ and $A = \{2, 3, 4 \}$ , then $A' = \{1, 5, 6, 7, 8 \}$

Thus $A' = \{x | x \in \xi \text{ and } x \notin A \}$ clearly $x \in A'$ and $x \notin A$

Please note

$\phi^{'} = \xi$ and $\phi^{'}= \xi$

$A \cup A' = \xi$ and $A \cap A' = \phi$

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Question 5. Let $\xi = \{x | x \in Z, -4 \leq x \leq 4 \}$ , $A = \{x | x \in W, x <4 \}$ and $B = \{x | x \in N, 2 <x \leq 4 \}$ . Find (i) $A'$ and (ii) $B'$

Answer:

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

$\xi = \{-4, -3, -2, -1, 0, 1, 2, 3, 4 \}, A = \{0, 1, 2, 3 \}$ and $B = \{3, 4 \}$

(i) $A' = \{-4, -3, -2, -1, 4 \}$ (ii) $B' = \{-4, -3, -2, -1, 0, 1, 2 \}$

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Question 6. Let $\xi = \{x | x \in N, x \text{ is a factor of} 144 \}$ , $A = \{x | x \in N, x \text{ is a factor of } 24 \}$ , $B = \{x | x \in N, x \text{ is a factor of } 36 \}$ and $C = \{x | x \in N, \ x \text{ is a factor of } 48 \}$ . Find:

(i) $A'$ (ii) $B'$ (iii) $C'$ (iv) $A \cup B$ (v) $B \cup C$ (vi) $A \cup C$ (vii) $A \cap B'$ (viii) $B \cap C'$ (ix) $C - A$ (x) $A - (B \cap C)$

Answer:

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

$\xi = \{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144 \}$ . 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.

(i) $A' = \{9, 16, 18, 36, 48, 72, 144 \}$ (ii) $B' = \{8, 16, 24, 48, 72, 144 \}$

(iii) $C' = \{9, 18, 36, 72, 144 \}$ (iv) $A \cup B = \{1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 \}$

(v) $B \cup C = \{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 26, 48 \}$

(vi) $A \cap C = \{1, 2, 3, 4, 6, 8, 12, 16, 24, 48 \}$ (vii) $A \cap B' = \{8, 24 \}$

(viii) $B \cap C' = \{9, 18, 36 \}$ (ix) $C - A = \{16, 48 \}$ (x) $A - (B \cap C)$

First calculate $(B \cap C) = \{1, 2, 3, 4, 6, 12 \}$

Now Calculate $A - (B \cap C) = \{8, 24 \}$

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Question 7. Considering the sets given in Q.6 state whether the following statements are True or False.

(i) $A \cap (B \cup C) = A$ (ii) $A \subset C$ (iii) $B \subseteq C$ (iv) $A \cap C' = \phi$

Answer:

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

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

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

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

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

(i) $A \cap (B \cup C) = A$

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

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

Therefore, $A \cap (B \cup C) = \{1, 2, 3, 4, 6, 8, 12, 24 \} = A$

Hence Proved. TRUE

(ii) $A \subset C$

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

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

(iii) $B \subseteq C$

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

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

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

(iv) $A \cap C' = \phi$

$\xi = \{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144 \}$ . 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

$A \cap C' = \phi$ or null set. Hence TRUE

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Question 8. Let $A = \{a, b, c, d, e \}$ , $B = \{a, c, e, g \}$ and $C = \{b, e, f, g \}$ . Then verify the following identities:

(i) $B \cup C = C \cup B$ (ii) $B \cap C = C \cap B$ (iii) $A \cup (B \cup C) = (A \cup B) \cup C$

(iv) $A \cap (B \cap C) = (A \cap B) \cap C$ (v) $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

(vi) $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

Answer:

(i) $B \cup C = C \cup B$

$B = \{a, c, e, g \}$ & $C = \{b, e, f, g \}$ . Therefore, $B \cup C = \{a, b, c, e, f, g \}$ .

Similarly, $C \cup B = \{a, b, c, e, f, g \}$ .

Hence $B \cup C = C \cup B$

(ii) $B \cap C = C \cap B$

$B = \{a, c, e, g \}$ & $C = \{b, e, f, g \}$ . Therefore, $B \cap C = \{a, b, c, e, f, g \}$ .

Similarly, $C \cup B = \{a, b, c, e, f, g \}$ .

(iii) $A \cup (B \cup C) = (A \cup B) \cup C$

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

Therefore:

LHS: $A \cup (B \cup C) = \{a, b, c, d, e \} \cup \{a, b, c, e, f, g \} = \{a, b, c, d, e, f, g \}$

RHS: $(A \cup B) \cup C = \{a, b, c, d, e, g \} \cup \{b, e, f, g \} = \{a, b, c, d, e, f, g \}$

Therefore, LHS = RHS. Hence Proved.

(iv) $A \cap (B \cap C) = (A \cap B) \cap C$

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

Therefore:

LHS: $A \cap (B \cap C) = \{a, b, c, d, e \} \cap \{e, g \} = \{e \}$

RHS: $(A \cap B) \cap C = \{a, c, e \} \cap \{b, e, f, g \} = \{e \}$

Therefore, LHS = RHS. Hence Proved.

(v) $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

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

LHS: $A \cup (B \cap C) = \{a, b, c, d, e \} \cup \{e, g \} = \{a, b, c, d, e, g \}$

RHS: $(A \cup B) \cap (A \cup C) = \{a, b, c, d, e, g \} \cap \{a, b, c, d, e, f, g \} = \{a, b, c, d, e, g \}$

Therefore, LHS = RHS. Hence Proved.

(vi) $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

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

LHS: $A \cap (B \cup C) = \{a, b, c, d, e \} \cap \{a, b, c, e, f, g \} = \{a, b, c, e \}$

RHS: $(A \cap B) \cup (A \cap C) = \{a, c, e \} \cap \{b, e \} = \{a, b, c, e \}$

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Question 9. Let $A = \{b, c, d, e \}$ and $B = \{d, e, f, g \}$ be the two sub sets of the universal set $\xi = \{b, c, d, e, f, g \}$ . Then verify the following:

(i) $(A \cup B)' = (A' \cap B')$ (ii) $(A \cap B)' = (A' \cup B')$

Answer:

(i) $(A \cup B)' = (A' \cap B')$

LHS: $A \cup B = \{b, c, d, e \} \cup \{d, e, f, g \} = \{b, c, d, e, f, g \}$

Therefore, $(A \cup B)' = \phi$

RHS: $A' = \{f, g \}$ and $B' = \{b, c \}$

Therefore, $A' \cap B' = \phi$

Therefore, LHS = RHS. Hence proved.

(ii) $(A \cap B)' = (A' \cup B')$

$x = \{b, c, d, e, f, g \}$

LHS: $A \cap B = \{b, c, d, e \} \cap \{d, e, f, g \} = \{d, e \}$

Therefore, $(A \cap B)' = \{b, c, f, g \}$

RHS: $A' = \{ f, g \}$ and $B' = \{ b, c \}$

Therefore, $A' \cup B' = \{b, c, f, g \}$

Therefore, LHS = RHS. Hence proved.

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Question 10. Fill in the blanks

(i) $A \cup A = \ldots$ (ii) $A \cap A = \ldots$ (iii) $A \cup \phi = \ldots$

(iv) $A \cap \phi = \ldots$ (v) $(A \cup B)' = \ldots$ (vi) $(A \cap B)' = \ldots$

Answer:

(i) $A \cup A = A$ (ii) $A \cap A = A$ (iii) $A \cup \phi= A$ (iv) $A \cap \phi= A$

(v) $(A \cup B)' = (A' \cap B')$ (vi) $(A \cap B)' = (A' \cup B')$

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Question 11. Let $\xi = \{x | x \in N 4 \leq x < 18 \}$ and $A, B$ , and $C$ are subsets of $\xi$ . Given $A = \{x | x \text{ is a multiple of } 2 \}$ , $B = \{x | x \text{ is a multiple of } 3 \}$ , $C = \{x | x \in N x < 11 \}$ . Then verify, the following: