Let $S = \{x \mid x \text{ is a positive multiple of 3 less than 100}\}$, $P = \{x \mid x \text{ is a prime number less than 20}\}$. Then $n(S) + n(P)$ is (A) $34$ (B) $31$ (C) $33$ (D) $30$

Let S = \x mid x is a positive multiple of 3 less than 100\, P = \x mid x is a prime number less than 20\. The

NCERT Exemplar · Q42

S = \{x \mid x \text{ is a positive multiple of 3 less than 100}\}

Rajasthan RbseMCQ· 1mImportance★★★★★

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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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$n(S)=33$ and $n(P)=8$ , so $n(S)+n(P)=41$ . This value does not match any of the four printed options (34, 31, 33, 30) -- the printed question appears to contain an error.

Step 1: Count $S$ .

$S$ is the set of positive multiples of 3 less than 100: $3, 6, 9, \dots, 99$ . Since $99 = 3\times 33$ , there are 33 such multiples. So $n(S) = 33$ .

Step 2: Count $P$ .

$P$ is the set of prime numbers less than 20: $2, 3, 5, 7, 11, 13, 17, 19$ . That is 8 primes. So $n(P) = 8$ .

Step 3: Add them.

n(S) + n(P) = 33 + 8 = 41

Step 4: Compare with the given options.

The options are (A) 34, (B) 31, (C) 33, (D) 30. None of these equals 41. …

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Q.Let $S = \{x \mid x \text{ is a positive multiple of 3 less than 100}\}$ , $P = \{x \mid x \text{ is a prime number less than 20}\}$ . Then $n(S) + n(P)$ is
(A) $34$
(B) $31$
(C) $33$
(D) $30$

Concept understanding — Set Operations

The idea in plain words

Why this works

Step by step

A common slip …

Step 1 — Compute the union $B\cup C$

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

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