NIMCET 2027 2023 Previous Year Questions (PYQs) – Page 3 of 12

Given: Lines   $4ax+2ay+c=0$   and   $5bx+2by+d=0$.

Condition: Intersection lies in the 4th quadrant and is equidistant from the axes $\Rightarrow$ point is of the form $(t,-t)$ with $t>0$.

Substitute $y=-x$ in the two lines:
$4ax+2a(-x)+c=0 \;\Rightarrow\; 2x+\dfrac{c}{2a}=0 \;\Rightarrow\; x=-\dfrac{c}{2a}.$
$5bx+2b(-x)+d=0 \;\Rightarrow\; 3x+\dfrac{d}{b}=0 \;\Rightarrow\; x=-\dfrac{d}{3b}.$

Equate $x$ from both:   $-\dfrac{c}{2a}=-\dfrac{d}{3b} \;\Rightarrow\; \boxed{3bc=2ad}.$

(QIV check: $x>0,\ y<0 \;\Rightarrow\; \dfrac{c}{a}<0$ and $\dfrac{d}{b}<0$ which is consistent.)

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

Same Side of a Line — Geometric Condition

Line: \( x + y - 1 = 0 \)
Point 1: (3, 2) → Lies on the side where value is positive:
\[ f(3, 2) = 3 + 2 - 1 = 4 > 0 \]

Point 2: \( (\cos\theta, \sin\theta) \) lies on same side if: \[ \cos\theta + \sin\theta > 1 \] Using identity: \[ \cos\theta + \sin\theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) \Rightarrow \sin\left(\theta + \frac{\pi}{4}\right) > \frac{1}{\sqrt{2}} \] So: \[ \theta \in \left(0,\ \frac{\pi}{2}\right) \]

✅ Final Answer: \( \boxed{\theta \in \left(0,\ \frac{\pi}{2}\right)} \)

Vector Projection Problem

Given: A vector of magnitude 5 makes equal angles with x, y, and z axes.
To Find: Sum of magnitudes of projections on each axis.

Let angle with each axis be \( \alpha \). Then, from direction cosine identity: \[ \cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \Rightarrow 3\cos^2\alpha = 1 \Rightarrow \cos\alpha = \frac{1}{\sqrt{3}} \]

Projection on each axis: \( 5 \cdot \frac{1}{\sqrt{3}} \)
Sum = \( 3 \cdot \frac{5}{\sqrt{3}} = \frac{15}{\sqrt{3}} = \boxed{5\sqrt{3}} \)

✅ Final Answer: \( \boxed{5\sqrt{3}} \)

Conditional Probability – Bayes' Theorem

Given: One ball is transferred from Bag I to Bag II, and then a ball is drawn from Bag II and is black.

Goal: Find the probability that the transferred ball was red, given that a black ball was drawn.

Using Bayes' theorem: \[ P(R|A) = \frac{P(R \cap A)}{P(A)} = \frac{\frac{3}{10} \cdot \frac{5}{10}}{\frac{3}{10} \cdot \frac{5}{10} + \frac{4}{10} \cdot \frac{6}{10} + \frac{3}{10} \cdot \frac{5}{10}} = \frac{15}{54} = \boxed{\frac{5}{18}} \]

✅ Final Answer: \( \boxed{\frac{5}{18}} \)