NIMCET 2027 2024 Previous Year Questions (PYQs) – Page 2 of 12

Sequencing Logic: Arrange the Process

Words Given:

1. Software
2. Code
3. Data
4. Analysis
5. Report

Logical Order:

3 → 1 → 2 → 4 → 5

Explanation:

• Data is the raw input.
• Software handles and organizes the data.
• Code processes the data within the software.
• Analysis is performed based on the processed data.
• Report

Given: A one-one function from set $\{1,2,3\}$ to set $\{a,b,c,d,e\}$

Step 1: One-one (injective) function means no two elements map to the same output.

We choose 3 different elements from 5 and assign them to 3 inputs in order.

So, total one-one functions = $P(5,3) = 5 \times 4 \times 3 = 60$

✅ Final Answer: $\boxed{60}$

Given: Volume of a parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is 4 cubic units.

Vectors:

• $\vec{a} = m\hat{i} + \hat{j} + \hat{k}$
• $\vec{b} = \hat{i} - \hat{j} + \hat{k}$
• $\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$

Step 1: Volume = $|\vec{a} \cdot (\vec{b} \times \vec{c})|$

First compute $\vec{b} \times \vec{c}$:

$ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}((-1)(-1) - (1)(2)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(2) - (-1)(1)) \\ = \hat{i}(1 - 2) - \hat{j}(-1 - 1) + \hat{k}(2 + 1) = -\hat{i} + 2\hat{j} + 3\hat{k} $

Step 2: Compute dot product with $\vec{a}$:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = (m)(-1) + (1)(2) + (1)(3) = -m + 2 + 3 = -m + 5$

Step 3: Volume = $| -m + 5 | = 4$

So, $|-m + 5| = 4 \Rightarrow -m + 5 = \pm 4$

• Case 1: $-m + 5 = 4 \Rightarrow m = 1$
• Case 2: $-m + 5 = -4 \Rightarrow m = 9$

✅ Final Answer: $\boxed{m = 1 \text{ or } 9}$

Speed & Distance Problem: Car A vs Car B

Given:

• Car A speed = 60 km/h
• Car B speed = 80 km/h
• Time travelled = 1.5 hours

Calculations:

• Distance by Car A = 60 × 1.5 = 90 km
• Distance by Car B = 80 × 1.5 = 120 km

✅ Difference: 120 km – 90 km = 30 km

Given: Vectors:

• $\vec{a} = \lambda^2 \hat{i} + \hat{j} + \hat{k}$
• $\vec{b} = \hat{i} + \lambda^2 \hat{j} + \hat{k}$
• $\vec{c} = \hat{i} + \hat{j} + \lambda^2 \hat{k}$

Condition: Vectors are coplanar ⟹ Scalar triple product = 0

$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$

Step 1: Use determinant:

$ \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} \lambda^2 & 1 & 1 \\ 1 & \lambda^2 & 1 \\ 1 & 1 & \lambda^2 \end{vmatrix} $

Step 2: Expand the determinant:

$ = \lambda^2(\lambda^2 \cdot \lambda^2 - 1 \cdot 1) - 1(1 \cdot \lambda^2 - 1 \cdot 1) + 1(1 \cdot 1 - \lambda^2 \cdot 1) \\ = \lambda^2(\lambda^4 - 1) - (\lambda^2 - 1) + (1 - \lambda^2) $

Simplify:

$= \lambda^6 - \lambda^2 - \lambda^2 + 1 + 1 - \lambda^2 = \lambda^6 - 3\lambda^2 + 2$

Step 3: Set scalar triple product to 0:

$\lambda^6 - 3\lambda^2 + 2 = 0$

Step 4: Let $x = \lambda^2$, then:

$x^3 - 3x + 2 = 0$

Factor:

$x^3 - 3x + 2 = (x - 1)^2(x + 2)$

So, $\lambda^2 = 1$ (double root), or $\lambda^2 = -2$ (discard as it's not real)

Thus, real values of $\lambda$ are: $\lambda = \pm1$

✅ Final Answer: $\boxed{2}$ distinct real values

Profit & Discount Problem: Marked Price vs Cost Price

Given:

• Cash Discount = 20%
• Profit = 11.11% (i.e., 1/9 of Cost Price)

Step-by-Step:

• Assume Cost Price = ₹100
• Then Selling Price = ₹111.11
• SP = 80% of Marked Price ⇒ MP = (111.11 × 100) / 80 = ₹138.89
• Markup = ₹138.89 − ₹100 = ₹38.89

✅ Final Answer: The trader marks the goods 38.89% above the cost price.

Total bottles and boxes: 9 each, labeled 1 to 9.

We are asked to count permutations of bottles such that exactly 5 bottles go into their own numbered boxes.

Step 1: Choose 5 positions to be fixed points (i.e., bottle number matches box number).

Number of ways = $\binom{9}{5}$

Step 2: Remaining 4 positions must be a derangement (no bottle goes into its matching box).

Let $D_4$ be the number of derangements of 4 items.

$D_4 = 9$

Step 3: Total ways = $\binom{9}{5} \times D_4 = 126 \times 9 = 1134$

✅ Final Answer: $\boxed{1134}$