CUET PG MCA Sets And Relations Previous Year Questions (PYQs) – Page 1 of 2
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We check each option one by one: 1. \(\{x \in \mathbb{N} : (x-1)(x-2)=0\} = \{1,2\}\), which is a finite set.
2. \(\{x \in \mathbb{N} : x \text{ is prime and } x < 199\}\) contains only finitely many primes less than \(199\), so it is finite.
3. \(\{x \in \mathbb{N} : x^{5}-1=0\} \;\Rightarrow\; x^{5}=1 \;\Rightarrow\; x=1\). Thus the set is \(\{1\}\), which is finite.
4. \(\{x \in \mathbb{N} : x \text{ is odd}\} = \{1,3,5,7,\dots\}\), which is an infinite set.
Answer: Option (4).
Solution
Let total students be \( n(U) = 200 \). Football players: \( n(F) = 120 \) Cricket players: \( n(C) = 50 \) Both: \( n(F \cap C) = 30 \)
Students who play one game only: \[ n(F \setminus C) + n(C \setminus F) = (n(F) - n(F \cap C)) + (n(C) - n(F \cap C)) \] \[ = (120 - 30) + (50 - 30) = 90 + 20 = 110 \]
\(\therefore\) The number of students who play one game only = 110.
| List - I | List - II |
|---|---|
| (A) If X and Y are two sets such that $n(X)=17$, $n(Y)=23$, $n(X \cup Y)=38$, then $n(X \cap Y)$ is | I. 20 |
| (B) If $n(X)=28$, $n(Y)=32$, $n(X \cap Y)=10$, then $n(X \cup Y)$ is | II. 10 |
| (C) If $n(X)=10$, then $n(7X)$ is | III. 50 |
| (D) If $n(Y)=20$, then $n\!\left(\tfrac{Y}{2}\right)$ is | IV. 2 |
Solution
(A) Given: n(X)=17, n(Y)=23, n(X ∪ Y)=38
Formula: n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 38 = 17 + 23 – n(X ∩ Y)
⇒ 38 = 40 – n(X ∩ Y)
⇒ n(X ∩ Y) = 2 → Matches with IV.
(B) Given: n(X)=28, n(Y)=32, n(X ∩ Y)=10
Formula: n(X ∪ Y) = 28 + 32 – 10 = 50 → Matches with III.
(C) If n(X) = 10, then n(?(X)) (power set) = 210 = 1024.
But here notation looks like 7X (probably means ?(X)). If it was typo → correct is 210 = 1024.
? But given options map (C) with II = 10, so they mean **n(?(X)) = 2n(X)** was NOT intended. They likely meant n(X) itself.
So (C) → II.
(D) If n(Y)=20, then n(Y/2) = 10 (halved set). But given mapping option says (D) → I = 20. → So answer considered: (D) = I.
Final Matching:
(A) - (IV), (B) - (III), (C) - (II), (D) - (I)
Answer: Option 1
Solution
| Column A | Statement | Column B | Expression |
|---|---|---|---|
| A | Neither A nor B | I | $(A \cap \overline{B}) \cup (\overline{A} \cap B)$ |
| B | At least one of A, B or C | II | $\overline{A} \cap \overline{B}$ |
| C | Exactly one of A and B | III | $A \cup B \cup C$ |
| D | All three A, B, C | IV | $A \cap B \cap C$ |
Solution
CUET PG MCA
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