NIMCET Previous Year Questions (PYQs) – Sets And Relations – Page 1 of 3

Explanation

• Total possible ordered pairs on a set of 10 elements = 10² = 100.
• A reflexive relation must include all pairs of the form (a,a) for all a ∈ A.
• Thus, at least 10 pairs must be in R ⇒ m ≥ 10.
• The maximum relation is the universal relation with all 100 pairs ⇒ m ≤ 100.

✅ Final Answer

The number of ordered pairs m can take any value in the range:
10 ≤ m ≤ 100

Let $|S|=m$.

Count incidences via the $A_i$: There are 30 sets each of size 5, so total memberships $=30\times 5=150$. Each element of $S$ lies in exactly 10 of the $A_i$, so also $=m\times 10$. Hence $m={\displaystyle \frac{150}{10}}=15$.

Count incidences via the $B_j$: There are $n$ sets each of size 3, so total memberships $=n\times 3$. Each element of $S$ lies in exactly 9 of the $B_j$, so also $=m\times 9=15\times 9=135$.

Thus $n\times 3=135\Rightarrow n={\displaystyle \frac{135}{3}}=\boxed{{\displaystyle 45}}.$

We are given:

$$\sin (4x)=\frac{1}{2},x\in (-9\pi ,3\pi )$$

General solutions for $\sin (\theta )=\frac{1}{2}$

$$\theta =\frac{\pi }{6}+2n\pi \text{or}\theta =\frac{5\pi }{6}+2n\pi$$

Let $\theta =4x$, so we get:

• $x=\frac{\pi }{24}+\frac{n\pi }{2}$
• $x=\frac{5\pi }{24}+\frac{n\pi }{2}$

Count how many such $x$ fall in the interval $(-9\pi ,3\pi )$

By checking all possible $n$ values, we find:

• For $x=\frac{\pi }{24}+\frac{n\pi }{2}$: 24 valid values
• For $x=\frac{5\pi }{24}+\frac{n\pi }{2}$: 24 valid values

Total distinct values = 24 + 24 = 48

✅ Final Answer: $\boxed{{\displaystyle 48}}$

Given: $$x^2+2y^2=3\text{ and }x>y\text{ with }x,y\in \mathbb{Z}$$

Solutions to the equation are: $$\{(1,1),(1,-1),(-1,1),(-1,-1)\}$$

Among them, only $(1,-1)$ satisfies $x>y$.

Answer: $$\boxed{{\displaystyle 1}}$$

Maximum Students Passing All Three Exams

Given:

• Total students = 50
• $|M|=37$, $|P|=24$, $|C|=43$
• $|M\cap P|\le 19$, $|M\cap C|\le 29$, $|P\cap C|\le 20$

We use the inclusion-exclusion principle:

$$|M\cup P\cup C|=|M|+|P|+|C|-|M\cap P|-|M\cap C|-|P\cap C|+|M\cap P\cap C|$$

Let $x=|M\cap P\cap C|$. Then:

$$50\ge 37+24+43-19-29-20+x\Rightarrow 50\ge 36+x\Rightarrow x\le 14$$

✅ Final Answer: $\boxed{{\displaystyle 14}}$