If $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, $A = \{1, 2, 3, 5\}$, $B = \{2, 4, 6, 7\}$ and $C = \{2, 3, 4, 8\}$. Then (i) $(B \cup C)'$ is ______________. (ii) $(C - A)'$ is ______________.

If U = \1, 2, 3, 4, 5, 6, 7, 8, 9, 10\, A = \1, 2, 3, 5\, B = \2, 4, 6, 7\ and C = \2, 3, 4, 8\. Then (i) (B ∪

NCERT Exemplar · Q50

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

(i)

(B \cup C)'

is ______________.

(ii)

(C - A)'

is ______________.

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Concept understanding — Set Operations

The idea in plain words

Think of your two favourite groups of friends — the ones who play cricket and the ones who play football. Some friends are in both groups, some in only one, and some in neither. Set operations are simply the mathematical ways to answer questions like: "Who's in at least one team?" or "Who's only in the cricket team?".

The whole secret? Each operation is just a different way of combining or comparing two collections — like sorting your friends into different buckets.

Why this works

Sets are just labelled buckets that hold distinct items. The universal set $U$ is the "whole world" of things we're talking about — say, all your friends. Then each operation picks out a specific bucket:

Operation	What it asks	Bucket contains
Union ( $A \cup B$ )	In either?	Everything from $A$ or $B$ (or both)
Intersection ( $A \cap B$ )	In both?	Only the overlap
Difference ( $A \setminus B$ )	In $A$ but not $B$ ?	Just the part of $A$ that doesn't touch $B$
Complement ( $A^c$ )	Not in $A$ ?	Everything outside $A$ (inside $U$ )
Symmetric Difference ( $A \triangle B$ )	In exactly one?	The two crescent-shaped parts, excluding the overlap

Step by step

Let's take two concrete sets so you can see each operation in action:

A = \{1, 2, 3\}, \quad B = \{3, 4, 5\}

Step 1: Union — gather everything from both, but don't repeat anything.

A \cup B = \{1, 2, 3, 4, 5\}

Step 2: Intersection — only what's common to both.

A \cap B = \{3\}

Step 3: Difference ( $A$ minus $B$ ) — start with $A$ , remove anything that's also in $B$ .

A \setminus B = \{1, 2\}

Step 4: Complement — needs a universal set. Let $U = \{1, 2, 3, 4, 5\}$ . Then:

A^c = \{4, 5\}

Step 5: Symmetric Difference — combine the two differences:

A \triangle B = (A \setminus B) \cup (B \setminus A) = \{1, 2\} \cup \{4, 5\} = \{1, 2, 4, 5\}

A common slip …

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Use complement and set-difference operations: $(B \cup C)'$ finds elements in $U$ but not in $B$ or $C$ ; $(C - A)'$ finds elements in $U$ but not in " $C$ but not $A$ ".

The complement of a set $X$ with respect to a universal set $U$ , written $X'$ , contains every element in $U$ that is not in $X$ . The set difference $C - A$ contains elements that belong to $C$ but do not belong to $A$ . These operations let us carve out subsets by exclusion.

Part (i): Finding $(B \cup C)'$

First we need $B \cup C$ , the union of $B$ and $C$ —every element that appears in at least one of the two sets.

Form the union $B \cup C$ .

$B = \{2, 4, 6, 7\}$ and $C = \{2, 3, 4, 8\}$ .

Collecting all distinct elements:

B \cup C = \{2, 3, 4, 6, 7, 8\}.

Take the complement with respect to $U$ .

The universal set is $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ .

$(B \cup C)'$ consists of elements in $U$ that are not in $B \cup C$ . Remove $\{2, 3, 4, 6, 7, 8\}$ from $U$ :

(B \cup C)' = \{1, 5, 9, 10\}.

Tip

A quick check: count the sizes. $|B \cup C| = 6$ and $|(B \cup C)'| = 4$ , and $6 + 4 = 10 = |U|$ . ✓

Part (ii): Finding $(C - A)'$

The set difference $C - A$ removes from $C$ any element that also belongs to $A$ .

Compute $C - A$ .

$C = \{2, 3, 4, 8\}$ and $A = \{1, 2, 3, 5\}$ . …

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Q.If $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ , $A = \{1, 2, 3, 5\}$ , $B = \{2, 4, 6, 7\}$ and $C = \{2, 3, 4, 8\}$ . Then

Concept understanding — Set Operations

The idea in plain words

Why this works

Step by step

A common slip …

Part (i): Finding $(B \cup C)'$

Part (ii): Finding $(C - A)'$

Step 1 — Compute the union $B\cup C$

Step 2 — Compute the difference $A-(B\cup C)$

Q.If U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}U={1,2,3,4,5,6,7,8,9,10}, A={1,2,3,5}A = \{1, 2, 3, 5\}A={1,2,3,5}, B={2,4,6,7}B = \{2, 4, 6, 7\}B={2,4,6,7} and C={2,3,4,8}C = \{2, 3, 4, 8\}C={2,3,4,8}. Then

Part (i): Finding (B∪C)′(B \cup C)'(B∪C)′

Part (ii): Finding (C−A)′(C - A)'(C−A)′

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Q.If $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ , $A = \{1, 2, 3, 5\}$ , $B = \{2, 4, 6, 7\}$ and $C = \{2, 3, 4, 8\}$ . Then

Part (i): Finding $(B \cup C)'$

Part (ii): Finding $(C - A)'$

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