Numerical Computation & Estimation, Data Structures, Matrix Decompositions… Test for GATE DA: 21 Questions with Solutions & Analysis

Attempt the Numerical Computation & Estimation, Data Structures, Matrix Decompositions… test for GATE DA: 21 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

21 questions · 33 marks · 0.99 minutes. Probability and Statistics: 6 · Programming, Data Structures and Algorithms: 3 · Machine Learning: 3 · Analytical Aptitude: 3 · Linear Algebra: 2 · Quantitative Aptitude: 1 · Database Management and Warehousing: 1 · Calculus and Optimization: 1 · Artificial Intelligence: 1

Free sample questions from Numerical Computation & Estimation, Data Structures, Matrix Decompositions…

Question 1 · Probability and Statistics · 2024 MCQ

A fair six-sided die (with faces numbered 1, 2, 3, 4, 5, 6) is repeatedly thrown
independently.
What is the expected number of times the die is thrown until two consecutive throws
of even numbers are seen?

A.
2
B.
4
C.
6
D.
8

Correct Answer:

Step-by-Step Solution

Insight: This is a "wait for a pattern" question — two consecutive evens. The trials are independent, so track progress toward the pattern using states, and write one equation per state.

Exam route:

Let $p = 1/2$ (probability of even on a fair die).

States: $S_{0}$ (no progress / last was odd), $S_{1}$ (last throw was even), $S_{2}$ (done — two consecutive evens).

Let $E_{i}$ = expected additional throws from state $S_{i}$ .

From $S_{0}$ : throw once. With prob $p$ get even $\to S_{1}$ ; with prob $1 - p$ get odd $\to S_{0}$ .

$E_{0} = 1 + p E_{1} + (1 - p) E_{0}$

From $S_{1}$ : throw once. With prob $p$ get even $\to S_{2}$ (done, 0 more); with prob $1 - p$ get odd $\to S_{0}$ .

$E_{1} = 1 + p \cdot 0 + (1 - p) E_{0}$

Substitute $p = 1/2$ :

$E_{0} = 1 + \frac{1}{2} E_{1} + \frac{1}{2} E_{0} \Rightarrow \frac{1}{2} E_{0} = 1 + \frac{1}{2} E_{1} \Rightarrow E_{0} = 2 + E_{1}$

$E_{1} = 1 + \frac{1}{2} E_{0}$

Substitute $E_{0} = 2 + E_{1}$ into the second:

$E_{1} = 1 + \frac{1}{2} (2 + E_{1}) = 2 + \frac{1}{2} E_{1} \Rightarrow \frac{1}{2} E_{1} = 2 \Rightarrow E_{1} = 4$

$E_{0} = 2 + 4 = 6$ .

Answer: 6 (option C).

Learning route:

A fair die has 3 even faces (2, 4, 6) and 3 odd faces (1, 3, 5), so $p = P (even) = 1/2$ .

We want the expected number of throws until we see two evens in a row.

Define states by how much of the target pattern we have completed:

$S_{0}$ : no useful progress (we are at the start, or the last throw was odd).
$S_{1}$ : the last throw was even (we are one step into the pattern).
$S_{2}$ : we have just seen two consecutive evens (absorbing state, we stop).

Let $E_{i}$ be the expected number of additional throws needed to reach $S_{2}$ starting from $S_{i}$ .

From $S_{0}$ , we spend 1 throw. With probability $1/2$ we get even and move to $S_{1}$ ; with probability $1/2$ we get odd and stay in $S_{0}$ . So $E_{0} = 1 + \frac{1}{2} E_{1} + \frac{1}{2} E_{0}$ .

From $S_{1}$ , we spend 1 throw. With probability $1/2$ we get even and reach $S_{2}$ (done, 0 more throws); with probability $1/2$ we get odd and fall back to $S_{0}$ (all progress lost). So $E_{1} = 1 + \frac{1}{2} (0) + \frac{1}{2} E_{0}$ .

Solving the first equation: $E_{0} - \frac{1}{2} E_{0} = 1 + \frac{1}{2} E_{1} \Rightarrow E_{0} = 2 + E_{1}$ .

Substituting into the second: $E_{1} = 1 + \frac{1}{2} (2 + E_{1}) = 2 + \frac{1}{2} E_{1} \Rightarrow E_{1} = 4$ .

Therefore $E_{0} = 2 + 4 = 6$ .

The expected number of throws is 6.

Wrong paths:

Option A (2): uses $E [T] = 1/ p = 2$ . This is the expected wait for a SINGLE even, not two consecutive.
Option B (4): uses $E [T] = 1/ p^{2} = 4$ . This treats two consecutive evens as two independent events, ignoring the sequential overlap of attempts.
Option D (8): uses $2/ p^{2} = 8$ . This incorrectly doubles the $1/ p^{2}$ formula, perhaps thinking "two evens means multiply by 2".

Question 2 · Probability and Statistics · 2025 MCQ

A random variable

X

is said to be distributed as Bernoulli

(θ)

, denoted by

X \sim B er n o u l l i (θ)

, if

P (X = 1) = θ, P (X = 0) = 1 - θ

for

0 < θ < 1

. Let

Y = \sum_{i = 1}^{300} X_{i}

, where

X_{i} \sim B er n o u l l i (θ)

i = 1, 2, ..., 300

be independent and identically distributed random variables with

θ = 0.25

. The value of

P (60 \leq Y \leq 90)

, after approximation through Central Limit Theorem, is given by

Recall that

ϕ (x) = \frac{1}{2 π} \int_{- \infty}^{x} e^{- \frac{t ^{2}}{2}} d t

A.
$ϕ (2) - ϕ (- 2)$
B.
$ϕ (1) - ϕ (- 1)$
C.
$ϕ (3) - ϕ (- 3)$
D.
$ϕ (90) - ϕ (60)$

Correct Answer:

Step-by-Step Solution

Insight: Sum of 300 i.i.d. Bernoulli $⟹$ use CLT with mean $n p$ and SD $n p (1 - p)$ .

Exam route:

$Y = \sum_{i = 1}^{300} X_{i}$ , $X_{i} \sim Bernoulli (0.25)$ .
$E [Y] = 300 \times 0.25 = 75$ .
$Var (Y) = 300 \times 0.25 \times 0.75 = 56.25$ .
$SD (Y) = 56.25 = 7.5$ .
By CLT, $Y \approx N (75, 56.25)$ .
$P (60 \leq Y \leq 90) = P (\frac{60 - 75}{7.5} \leq Z \leq \frac{90 - 75}{7.5}) = P (- 2 \leq Z \leq 2)$ .
$= ϕ (2) - ϕ (- 2)$ .
Answer: A.

Learning route:

$Y$ is a sum of i.i.d. random variables, so by the Central Limit Theorem, $Y$ is approximately normal for large $n$ .

Mean: $μ = n p = 300 \times 0.25 = 75$ .

Variance: $σ^{2} = n p (1 - p) = 300 \times 0.25 \times 0.75 = 56.25$ .

Standard deviation: $σ = 7.5$ .

Standardize: $Z = \frac{Y - μ}{σ}$ .

$P (60 \leq Y \leq 90) = P (\frac{60 - 75}{7.5} \leq Z \leq \frac{90 - 75}{7.5}) = P (- 2 \leq Z \leq 2)$ .

Using the CDF $ϕ$ : $P (- 2 \leq Z \leq 2) = ϕ (2) - ϕ (- 2)$ .

Common trap: A student might forget to standardize and write $ϕ (90) - ϕ (60)$ , which is wrong because $ϕ$ is the CDF of the standard normal, not $Y$ .

Verification: $(90 - 75) /7.5 = 15/7.5 = 2$ . $(60 - 75) /7.5 = - 15/7.5 = - 2$ . The standardization is correct.

Question 3 · Probability and Statistics · 2025 MCQ

A random experiment consists of throwing $100$ fair dice, each die having six faces numbered $1$ to $6$ . An event $A$ represents the set of all outcomes where at least one of the dice shows a $1$ . Then, $P (A) =$

A.
$0$
B.
$1$
C.
$1 - (\frac{5}{6})^{100}$
D.
$(\frac{5}{6})^{100}$

Correct Answer:

Step-by-Step Solution

Insight: "At least one" is the classic complement trigger — compute $1 - P (none)$ .

Exam route:

Each die: $P (shows 1) = 1/6$ , $P (not 1) = 5/6$ .
100 independent dice. $P (none shows 1) = (5/6)^{100}$ .
$P (at least one 1) = 1 - (5/6)^{100}$ .
Answer: C.

Learning route:

Let $X$ = number of dice showing 1. $X \sim Binomial (100, 1/6)$ .

Event $A = {X \geq 1}$ .

$P (A) = 1 - P (X = 0) = 1 - (0 100) (1/6)^{0} (5/6)^{100} = 1 - (5/6)^{100}$ .

Common trap: A student might think $P (at least one) = 100 \times (1/6)$ by adding probabilities, which is wrong because events overlap. The complement method avoids this.

Verification: $(5/6)^{100}$ is extremely small ( $\approx 1.2 \times 1 0^{- 8}$ ), so $P (A) \approx 1$ , which makes sense — with 100 dice, it's almost certain at least one shows 1.

Question 4 · Programming, Data Structures and Algorithms · 2024 MCQ

Match the items in Column 1 with the items in Column 2 in the following table:

Column 1	Column 2
(p) First In First Out	(i) Stacks
(q) Lookup Operation	(ii) Queues
(r) Last In First Out	(iii) Hash Tables

A.
(p) − (ii), (q) − (iii), (r) − (i)
B.
(p) − (ii), (q) − (i), (r) − (iii)
C.
(p) − (i), (q) − (ii), (r) − (iii)
D.
(p) − (i), (q) − (iii), (r) − (ii)

Correct Answer:

Step-by-Step Solution

Key idea: This is an ADT property matching question, recognizable because it asks to pair fundamental access patterns (FIFO, LIFO, Lookup) with their corresponding data structures.

Step 1: Analyze "First In First Out" (p). This is the defining property of a Queue, where the first element added is the first to be removed. So, (p) matches with (ii).

Step 2: Analyze "Last In First Out" (r). This is the defining property of a Stack, where the most recently added element is the first to be removed. So, (r) matches with (i).

Step 3: Analyze "Lookup Operation" (q). Hash Tables are specifically designed to provide fast, average-case $O (1)$ key-based lookup operations. So, (q) matches with (iii).

Step 4: Combine the matches: (p) - (ii), (q) - (iii), (r) - (i).

Answer: Option A

Question 5 · Programming, Data Structures and Algorithms · 2024 NAT

The fundamental operations in a double-ended queue

D

are:
insertFirst(e) – Insert a new element e at the beginning of D.
insertLast(e) – Insert a new element e at the end of D.
removeFirst() – Remove and return the first element of D.
removeLast() – Remove and return the last element of D.
In an empty double-ended queue, the following operations are performed:
insertFirst(10)
insertLast(32)
a ←removeFirst()
insertLast(28)
insertLast(17)
a ←removeFirst()
a ← removeLast()
The value of a is ______.

Correct Answer:

Step-by-Step Solution

Key idea: This is a deque simulation question, recognizable because it provides a sequence of double-ended queue operations and asks for the final value of a variable.

Step 1: Start with an empty deque $D$ .

Step 2: insertFirst(10): $D = [10]$

Step 3: insertLast(32): $D = [10, 32]$

Step 4: a <- removeFirst(): Removes and returns the first element ($10$). $a = 10$ , $D = [32]$

Step 5: insertLast(28): $D = [32, 28]$

Step 6: insertLast(17): $D = [32, 28, 17]$

Step 7: a <- removeFirst(): Removes and returns the first element ($32$). $a = 32$ , $D = [28, 17]$

Step 8: a <- removeLast(): Removes and returns the last element ($17$). $a = 17$ , $D = [28]$

Step 9: The final value assigned to $a$ is $17$.

Answer: 17

Question 6 · Programming, Data Structures and Algorithms · 2025 NAT

Consider the following pseudocode.

Create empty stack S

Set x=0, flag=0, sum=0

Push x onto S

while (S is not empty){

    if (flag equals 0){

        Set x = x+1

        Push x onto S}

    if (x equals 8):

    Set flag=1

    if (flag equals 1){

    x = Pop(S)

    if (x is odd):

    Pop(S)

    Set sum = sum + x}

}

Output sum

The value of sum output by a program executing the above pseudocode is
(Answer in integer)

Correct Answer:

Step-by-Step Solution

Key idea: This is a stack simulation question, recognizable because it provides pseudocode with stack operations (Push, Pop) and conditional logic, asking for the final state of a variable.

Step 1: Initialize variables. $x = 0$ , $f l a g = 0$ , $s u m = 0$ . Stack $S$ is initialized with $0$, so $S = [0]$ .

Step 2: Trace the loop while $S$ is not empty.

Iterations 1 to 7: $f l a g$ is $0$. $x$ increments by $1$ each time and is pushed. $S$ becomes $[0, 1, 2, 3, 4, 5, 6, 7]$ . $x$ reaches $7$.
Iteration 8: $f l a g = 0$ , so $x$ becomes $8$ and is pushed. $S = [0, 1, 2, 3, 4, 5, 6, 7, 8]$ . Now $x = 8$ , so $f l a g$ becomes $1$. Since $f l a g = 1$ , we execute the pop block: $x = P o p (S) = 8$ . $8$ is even, so the inner pop is skipped. $s u m = 0 + 8 = 8$ .
Iteration 9: $f l a g = 1$ . $x = P o p (S) = 7$ . $7$ is odd, so we $P o p (S)$ again, removing $6$. $S$ is now $[0, 1, 2, 3, 4, 5]$ . $s u m = 8 + 7 = 15$ .
Iteration 10: $f l a g = 1$ . $x = P o p (S) = 5$ . $5$ is odd, so we $P o p (S)$ again, removing $4$. $S$ is now $[0, 1, 2, 3]$ . $s u m = 15 + 5 = 20$ .
Iteration 11: $f l a g = 1$ . $x = P o p (S) = 3$ . $3$ is odd, so we $P o p (S)$ again, removing $2$. $S$ is now $[0, 1]$ . $s u m = 20 + 3 = 23$ .
Iteration 12: $f l a g = 1$ . $x = P o p (S) = 1$ . $1$ is odd, so we $P o p (S)$ again, removing $0$. $S$ is now $[]$ . $s u m = 23 + 1 = 24$ .

Step 3: The stack is now empty, so the loop terminates. The final value of $s u m$ is $24$.

Answer: 24

Question 7 · Machine Learning · 2026 MCQ

Consider that you are training a classifier for a 10-class classification problem. Each input is represented as a 512-dimensional vector. There are 1000 samples, out of which first 100 will be used for testing. Let Leave-One-Out-Cross-Validation (LOOCV) be used for selection of the classifier model before testing.
Which of the following options is the correct number of validation splits that will be generated?

A.
10
B.
512
C.
900
D.
1000

Correct Answer:

Step-by-Step Solution

Insight: The test set is completely held out and does not participate in cross-validation. Only the remaining samples form the pool for LOOCV.

Exam route: Subtract test samples from total samples to get the training/validation pool size. For LOOCV, the number of splits equals this pool size. 1000 - 100 = 900.

Learning route:

Identify total samples: $N = 1000$ .
Identify test samples: $N_{t es t} = 100$ . These are strictly for final evaluation and are never used in cross-validation.
Compute the training/validation pool: $N_{t r ain} = N - N_{t es t} = 1000 - 100 = 900$ .
Apply LOOCV rule: In Leave-One-Out Cross-Validation, each sample in the training pool is used as a validation set exactly once.
Therefore, the number of validation splits generated is exactly $N_{t r ain} = 900$ .
The dimensionality (512) and number of classes (10) are distractors and do not affect the count of splits.

Question 8 · Machine Learning · 2024 MCQ

Consider the dataset with six datapoints:

{(x_{1}, y_{1}), (x_{2}, y_{2}), \dots, (x_{6}, y_{6})}

,
where

x_{1} = [10]

x_{2} = [01]

x_{3} = [0 - 1]

x_{4} = [- 1 0]

x_{5} = [22]

x_{6} = [- 2 - 2]

and the labels are given by

y_{1} = y_{2} = y_{5} = 1

, and

y_{3} = y_{4} = y_{6} = - 1

. A hard
margin linear support vector machine is trained on the above dataset.
Which ONE of the following sets is a possible set of support vectors?

A.
${x_{1}, x_{2}, x_{5}}$
B.
${x_{3}, x_{4}, x_{5}}$
C.
${x_{4}, x_{5}}$
D.
${x_{1}, x_{2}, x_{3}, x_{4}}$

Correct Answer:

Step-by-Step Solution

Insight: The dataset is perfectly symmetric around the origin, implying the optimal hard-margin hyperplane passes through the origin ( $b = 0$ ). The support vectors are the points closest to this boundary.

Exam route: Set up the hard-margin constraints $y_{i} (w^{T} x_{i} + b) \geq 1$ . Adding the constraints for symmetric points (e.g., $x_{1}$ and $x_{4}$ ) directly yields $w_{1} \geq 1$ and $w_{2} \geq 1$ . Minimizing $∥ w ∥^{2}$ gives $w = (1, 1)$ and $b = 0$ . Checking which points yield exactly 1 identifies the support vectors.

Learning route:

Write down the 6 constraints for hard-margin SVM: $y_{i} (w^{T} x_{i} + b) \geq 1$ .
For $x_{1}$ ( $y = 1$ ): $w_{1} + b \geq 1$
For $x_{4}$ ( $y = - 1$ ): $- (- w_{1} + b) \geq 1 ⟹ w_{1} - b \geq 1$
Adding these two gives $2 w_{1} \geq 2 ⟹ w_{1} \geq 1$ .
Similarly, for $x_{2}$ ( $y = 1$ ) and $x_{3}$ ( $y = - 1$ ), we get $w_{2} + b \geq 1$ and $w_{2} - b \geq 1$ , which sum to $2 w_{2} \geq 2 ⟹ w_{2} \geq 1$ .
The SVM objective is to minimize $\frac{1}{2} ∥ w ∥^{2} = \frac{1}{2} (w_{1}^{2} + w_{2}^{2})$ .
Given $w_{1} \geq 1$ and $w_{2} \geq 1$ , the minimum is achieved at $w_{1} = 1, w_{2} = 1$ .
Substituting back, $1 + b \geq 1 ⟹ b \geq 0$ and $1 - b \geq 1 ⟹ b \leq 0$ , so $b = 0$ .
Verify with $x_{5}$ and $x_{6}$ : $y_{5} (w^{T} x_{5} + b) = 1 (2 + 2 + 0) = 4 \geq 1$ . $y_{6} (w^{T} x_{6} + b) = - 1 (- 2 - 2 + 0) = 4 \geq 1$ . Both are strictly greater than 1, so they are not support vectors.
The points satisfying $y_{i} (w^{T} x_{i} + b) = 1$ are $x_{1}, x_{2}, x_{3}, x_{4}$ . Thus, they are the support vectors.

Question 9 · Machine Learning · 2025 MSQ

A.
$w = (44)$ and $b = 1$
B.
The number of support vectors is 3
C.
The margin is $2$
D.
Training accuracy is 98%

Correct Answer:

["B","C"]

Step-by-Step Solution

Insight: The support vectors are the closest points between the two classes. The optimal hard-margin hyperplane lies exactly midway between the parallel margin boundaries passing through these support vectors.

Exam route: Identify support vectors: $(0, 0)$ from Class-2, and $(2, 0), (0, 2)$ from Class-1. Total 3 support vectors. The margin boundaries are $x + y = 0$ and $x + y = 2$ . The margin width is $\frac{2}{1 ^{2} + 1 ^{2}} = 2$ .

Learning route:

Step 1: Identify the closest points between classes. Distance from $(0, 0)$ to $(2, 0)$ is 2, to $(0, 2)$ is 2, and to $(2, 2)$ is $22$ . Thus, $(2, 0)$ and $(0, 2)$ are the support vectors for Class-1, and $(0, 0)$ is the support vector for Class-2. Total = 3 support vectors. (Option B is correct).

Step 2: The line through $(2, 0)$ and $(0, 2)$ is $x + y = 2$ . The parallel line through $(0, 0)$ is $x + y = 0$ .

Step 3: The optimal separating hyperplane is exactly in the middle: $x + y = 1$ , which means $w = (11)$ and $b = - 1$ . Option A claims $w = (44)$ and $b = 1$ , which is incorrect.

Step 4: The margin width is the perpendicular distance between $x + y = 2$ and $x + y = 0$ , which is $\frac{∣2 - 0∣}{1 ^{2} + 1 ^{2}} = 2$ . (Option C is correct).

Step 5: Since the data is perfectly linearly separable, a hard-margin SVM achieves 100% training accuracy, not 98%. (Option D is incorrect).

Answer: B, C

Question 10 · Analytical Aptitude · 2025 MCQ

Let

p

and

q

be any two propositions. Consider the following propositional statements.

S_{1} : p \to q, S_{2} : \neg p \land q, S_{3} : \neg p \lor q, S_{4} : \neg p \lor \neg q,

where

\land

denotes conjunction (AND operation),

\lor

denotes disjunction (OR operation), and

\neg

denotes negation (NOT operation). Which one of the following options is correct?
(Note:

\equiv

denotes logical equivalence)

A.
$S_{1} \equiv S_{3}$
B.
$S_{2} \equiv S_{3}$
C.
$S_{2} \equiv S_{4}$
D.
$S_{1} \equiv S_{4}$

Correct Answer:

Step-by-Step Solution

Insight: $p \to q$ is logically equivalent to $\neg p \lor q$ by material implication.

Exam route: Recall the material implication law $p \to q \equiv \neg p \lor q$ . Compare $S_{1} = p \to q$ with $S_{3} = \neg p \lor q$ . They are identical in meaning.

Learning route:

Step 1: Write out each statement clearly.

$S_{1} : p \to q$

$S_{2} : \neg p \land q$

$S_{3} : \neg p \lor q$

$S_{4} : \neg p \lor \neg q$

Step 2: Apply the material implication equivalence to $S_{1}$ .

$p \to q \equiv \neg p \lor q$

Step 3: Compare the result with $S_{3}$ .

$S_{1} \equiv \neg p \lor q \equiv S_{3}$

Step 4: Verify with a truth table to be absolutely sure.

$p$ | $q$ | $S_{1}$ | $S_{3}$

T | T | T | T

T | F | F | F

F | T | T | T

F | F | T | T

The columns for $S_{1}$ and $S_{3}$ match exactly.

Answer: $S_{1} \equiv S_{3}$ , which is option A.

Common trap: Students sometimes confuse $\neg p \lor q$ with $\neg p \land q$ or $\neg p \lor \neg q$ . Remember that implication becomes a disjunction ( $\lor$ ), not a conjunction ( $\land$ ).

Question 11 · Analytical Aptitude · 2024 MSQ

Let

x

and

y

be two propositions. Which of the following statements is a tautology
/are tautologies?

A.
$(\neg x \land y) ⟹ (y ⟹ x)$
B.
$(x \land \neg y) ⟹ (\neg x ⟹ y)$
C.
$(\neg x \land y) ⟹ (\neg x ⟹ y)$
D.
$(x \land \neg y) ⟹ (y ⟹ x)$

Correct Answer:

["B","C","D"]

Step-by-Step Solution

Insight: Convert each implication to disjunction form and simplify. A tautology simplifies to $T$ .

Exam route: For each option, replace $A \to B$ with $\neg A \lor B$ , then simplify using De Morgan's and absorption. If the result is $T$ , it's a tautology.

Learning route:

Step 1: Recall $A \to B \equiv \neg A \lor B$ .

Step 2: Evaluate option A: $(\neg x \land y) \to (y \to x)$ .

$= \neg (\neg x \land y) \lor (\neg y \lor x)$

$= (x \lor \neg y) \lor (\neg y \lor x)$

$= x \lor \neg y$

This is NOT always true (false when $x = F, y = T$ ). So A is not a tautology.

Step 3: Evaluate option B: $(x \land \neg y) \to (\neg x \to y)$ .

$= \neg (x \land \neg y) \lor (x \lor y)$

$= (\neg x \lor y) \lor (x \lor y)$

$= \neg x \lor y \lor x \lor y$

$= (\neg x \lor x) \lor (y \lor y)$

$= T \lor y$

$= T$

This IS a tautology.

Step 4: Evaluate option C: $(\neg x \land y) \to (\neg x \to y)$ .

$= \neg (\neg x \land y) \lor (x \lor y)$

$= (x \lor \neg y) \lor (x \lor y)$

$= x \lor \neg y \lor x \lor y$

$= x \lor (\neg y \lor y)$

$= x \lor T$

$= T$

This IS a tautology.

Step 5: Evaluate option D: $(x \land \neg y) \to (y \to x)$ .

$= \neg (x \land \neg y) \lor (\neg y \lor x)$

$= (\neg x \lor y) \lor (\neg y \lor x)$

$= (\neg x \lor x) \lor (y \lor \neg y)$

$= T \lor T$

$= T$

This IS a tautology.

Answer: B, C, D are tautologies.

Question 12 · Analytical Aptitude · 2026 MCQ

Assume that a Creative

(C)

person will Succeed

(S)

if the person is also Disciplined

(D)

, but will not succeed otherwise. Now, consider the following statements:

(i)

C \land S \Leftrightarrow D

(ii)

C \Rightarrow (S ⟺ D)

(iii)

C \Leftrightarrow ((D \Rightarrow S) \lor \neg S)

Which of the following options is correct?

A.
Both (i) and (ii) are TRUE
B.
Only (ii) is TRUE
C.
Both (ii) and (iii) are TRUE
D.
Only (iii) is TRUE

Correct Answer:

Step-by-Step Solution

Insight: "A creative person will succeed if disciplined, but not otherwise" means: if creative, then (succeed iff disciplined).

Exam route: Translate the English sentence step by step. "Creative person will succeed if disciplined" gives $C \to (D \to S)$ . "But not otherwise" adds $C \to (\neg D \to \neg S)$ . Together: $C \to (S ⟺ D)$ . Check which option matches.

Learning route:

Step 1: Parse the sentence structure.

"A Creative (C) person will Succeed (S) if the person is also Disciplined (D), but will not succeed otherwise."

The phrase "but not otherwise" is the key. It means the condition is both necessary and sufficient.

Step 2: Translate the "if" part.

"will succeed if disciplined" $⟹ D \to S$ (within the context of being creative).

Step 3: Translate the "but not otherwise" part.

"will not succeed otherwise" $⟹ \neg D \to \neg S$ (within the context of being creative).

Step 4: Combine 2 and 3.

$(D \to S) \land (\neg D \to \neg S) \equiv S ⟺ D$ .

Step 5: Apply the context.

The whole rule applies to creative people, so: $C \to (S ⟺ D)$ .

Step 6: Evaluate the given statements.

(i) $C \land S ⟺ D$ : This means $(C \land S) ⟺ D$ , which is not equivalent. For example, if $C = F, D = F$ , the original gives $T$ (vacuously), but (i) gives $F ⟺ F = T$ . Wait, let me check $C = F, S = T, D = F$ : original gives $T$ , (i) gives $(F \land T) ⟺ F = F ⟺ F = T$ . Let me check $C = F, S = F, D = T$ : original gives $T$ , (i) gives $(F \land F) ⟺ T = F ⟺ T = F$ . Not equivalent. So (i) is FALSE.

(ii) $C \to (S ⟺ D)$ : This is exactly what we derived. TRUE.

(iii) $C ⟺ ((D \to S) \lor \neg S)$ : Simplify the right side. $(D \to S) \lor \neg S = (\neg D \lor S) \lor \neg S = \neg D \lor (S \lor \neg S) = \neg D \lor T = T$ . So (iii) becomes $C ⟺ T$ , which means $C$ must be true. This is not equivalent to $C \to (S ⟺ D)$ . FALSE.

Answer: Only (ii) is TRUE, which is option B.

Question 13 · Linear Algebra · 2024 NAT

Consider the

3 \times 3

matrix

M = 134213336

.
The determinant of

(M^{2} + 12 M)

is ______.

Correct Answer:

0.00

Step-by-Step Solution

Insight: The matrix polynomial $M^{2} + 12 M$ factors as $M (M + 12 I)$ . If $det (M) = 0$ , the whole product has determinant $0$ without any need to square the matrix or compute the second factor.

Exam route:

Factor the expression: $M^{2} + 12 M = M (M + 12 I)$ .
Use multiplicativity: $det (M^{2} + 12 M) = det (M) \cdot det (M + 12 I)$ .
Inspect $M$ for linear dependence. Row 3 $= (4, 3, 6) = (1, 2, 3) + (3, 1, 3) = R_{1} + R_{2}$ .
Therefore $det (M) = 0$ .
The product is $0 \times det (M + 12 I) = 0$ .

Learning route:

The determinant is multiplicative: $det (A B) = det (A) det (B)$ . Since $M$ and $M + 12 I$ are both polynomials in the same matrix $M$ , they commute, and the factorisation $M^{2} + 12 M = M (M + 12 I)$ is valid at the matrix level. Hence

$det (M^{2} + 12 M) = det (M) \cdot det (M + 12 I) .$

Compute $det (M)$ by inspection. The rows of $M$ are

$R_{1} = (1, 2, 3), R_{2} = (3, 1, 3), R_{3} = (4, 3, 6) .$

Observe that $R_{1} + R_{2} = (1 + 3, 2 + 1, 3 + 3) = (4, 3, 6) = R_{3}$ . The rows are linearly dependent, so $det (M) = 0$ .

(Equivalently, expanding along the first row: $det (M) = 1 (1 \cdot 6 - 3 \cdot 3) - 2 (3 \cdot 6 - 3 \cdot 4) + 3 (3 \cdot 3 - 1 \cdot 4) = 1 (- 3) - 2 (6) + 3 (5) = - 3 - 12 + 15 = 0$ .)

Since one factor has determinant $0$, the entire product has determinant $0$, regardless of the value of $det (M + 12 I)$ .

The common wrong path is to compute $M^{2}$ explicitly (a $3 \times 3$ matrix multiplication), then add $12 M$ , then expand a $3 \times 3$ determinant. That is legal but wasteful and invites arithmetic errors. The factorisation shortcut collapses the work to a single dependency check.

Verification: $det (M) = 0$ is confirmed by two independent routes (row dependence and cofactor expansion). Any product containing a singular factor is singular, so $det (M (M + 12 I)) = 0$ is exact.

Question 14 · Linear Algebra · 2024 NAT

Let

u = 12345

, and let

σ_{1}, σ_{2}, σ_{3}, σ_{4}, σ_{5}

be the singular values of the matrix

M = u u^{T}

(where

u^{T}

is the transpose of

u

). The value of

\sum_{i = 1}^{5} σ_{i}

is ______.

Correct Answer:

Step-by-Step Solution

Key idea: This is a rank-one matrix singular value problem. The matrix is given as an outer product $M = u u^{T}$ .

Step 1: Recognize that $M = u u^{T}$ is a symmetric, positive semi-definite matrix of rank 1.

Step 2: For a rank-one matrix of the form $u u^{T}$ , the only non-zero eigenvalue is given by $u^{T} u = ∥ u ∥^{2}$ .

Step 3: Calculate the squared Euclidean norm: $∥ u ∥^{2} = 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} = 1 + 4 + 9 + 16 + 25 = 55$ .

Step 4: Since $M$ is symmetric and positive semi-definite, its singular values are the absolute values of its eigenvalues. Thus, the singular values are $σ_{1} = 55$ and $σ_{2} = σ_{3} = σ_{4} = σ_{5} = 0$ .

Step 5: The sum of the singular values is $55 + 0 + 0 + 0 + 0 = 55$ .

Answer: 55

Question 15 · Quantitative Aptitude · 2025 MCQ

A rectangle has a length $L$ and a width $W$ , where $L > W$ . If the width, $W$ , is increased by $10%$ , which one of the following statements is correct for all values of $L$ and $W$ ?

A.
Perimeter increases by $10%$ .
B.
Length of the diagonals increases by $10%$ .
C.
Area increases by $10%$ .
D.
The rectangle becomes a square.

Correct Answer:

Step-by-Step Solution

Insight: Area is the product of length and width; a percentage change in only one dimension translates directly to the exact same percentage change in the area, regardless of the other dimension.

Exam route: Multiply the original area by the width multiplier (1.1) to get the new area. The factor is 1.1, meaning a 10% increase.

Learning route:

Identify the original dimensions: Length = $L$ , Width = $W$ . Original Area = $L \times W$ .
Identify the change: Width is increased by 10%, so the new width is $1.1 W$ . Length remains $L$ .
Calculate the new area: $New Area = L \times 1.1 W = 1.1 L W$ .
Calculate the percentage change in area:

$\frac{1.1 L W - L W}{L W} \times 100% = 0.1 \times 100% = 10%$

Verify other options to be absolutely sure:

Perimeter: Old = $2 L + 2 W$ , New = $2 L + 2.2 W$ . Increase is $0.2 W$ , which is not 10% of $2 L + 2 W$ .
Diagonal: Old = $L^{2} + W^{2}$ , New = $L^{2} + 1.21 W^{2}$ . The increase is not a flat 10%.
Square: $L > W$ , but $1.1 W$ is not guaranteed to equal $L$ .

The correct statement is that the area increases by 10%.

Question 16 · Database Management and Warehousing · 2026 NAT

Consider an ER model with the entities

E_{1} (A_{11}, A_{12}, A_{13})

and

E_{2} (A_{21}, A_{22}, A_{23})

, where,

A_{11}, A_{12}, A_{13}

are the attributes of

E_{1}

, and

A_{21}, A_{22}, A_{23}

are the attributes of

E_{2}

. Let

A_{22}

be a multi-valued attribute.

A_{11}

and

A_{21}

are the primary keys of

E_{1}

and

E_{2}

, respectively.

Let

R_{12}

be a many-to-many relationship between

E_{1}

and

E_{2}

. Participation of both

E_{1}

and

E_{2}

R_{12}

is total.

The minimum number of relations required to convert the ER model into relational model (assuming there is no other functional dependency) where each relation is in third normal form (3NF) is __________ . (Answer in integer)

Correct Answer:

4.00

Step-by-Step Solution

Insight: Multi-valued attributes require a separate relation, and M:N relationships require a separate relation to achieve 1NF and 3NF.

Exam route: Count relations for each component. E1 -> 1 relation. E2 with multi-valued A22 -> 2 relations (one for simple attributes, one for multi-valued). M:N relationship R12 -> 1 relation. Total = 1 + 2 + 1 = 4.

Learning route:

To convert an ER model to a relational model in 3NF, we follow standard mapping rules:

Strong Entity E1: Maps to 1 relation $E 1 (A_{11}, A_{12}, A_{13})$ with PK $A_{11}$ .
Strong Entity E2: Has a multi-valued attribute $A_{22}$ . To satisfy 1NF (and thus 3NF), multi-valued attributes must be separated. This creates 2 relations:

$E 2_s im pl e (A_{21}, A_{23})$ with PK $A_{21}$ .
$E 2_m u l t i (A_{21}, A_{22})$ with PK $(A_{21}, A_{22})$ .

M:N Relationship R12: Must map to a separate relation containing the primary keys of both participating entities: $R 12 (A_{11}, A_{21})$ with PK $(A_{11}, A_{21})$ .

Total number of relations = 1 (E1) + 2 (E2) + 1 (R12) = 4.

All resulting relations are in 3NF as there are no transitive dependencies.

Question 17 · Calculus and Optimization · 2025 MSQ

Consider two functions $f : R \to R$ and $g : R \to (1, \infty)$ . Both functions are differentiable at a point $c$ . Which of the following functions is/are ALWAYS differentiable at $c$ ? The symbol $\cdot$ denotes product and the symbol $\circ$ denotes composition of functions.

A.
$f \pm g$
B.
$f \cdot g$
C.
$\frac{f}{g}$
D.
$f \circ g + g \circ f$

Correct Answer:

["A","B","C"]

Step-by-Step Solution

Insight: Algebraic operations ( $\pm, \cdot, /$ ) only require differentiability at the point $c$ , but composition requires differentiability of the outer function at the inner function's value.

Exam route: Check if the outer function is guaranteed to be differentiable at the required point. For $f \circ g$ , $f$ must be diff at $g (c)$ , which is not guaranteed if $f$ is only diff at $c$ .

Learning route:

We are given $f$ and $g$ are differentiable at $c$ . This does not imply they are differentiable everywhere.
Option A ( $f \pm g$ ): Sum/difference rule requires $f, g$ diff at $c$ . Always true.
Option B ( $f \cdot g$ ): Product rule requires $f, g$ diff at $c$ . Always true.
Option C ( $f / g$ ): Quotient rule requires $f, g$ diff at $c$ and $g (c) \neq = 0$ . Since $g : R \to (1, \infty)$ , $g (c) > 1 \neq = 0$ . Always true.
Option D ( $f \circ g + g \circ f$ ): Chain rule for $f \circ g$ requires $f$ to be differentiable at $g (c)$ . Since $f$ is only guaranteed to be differentiable at $c$ , this fails if $g (c) \neq = c$ . Similarly, $g \circ f$ requires $g$ to be diff at $f (c)$ , which fails if $f (c) \neq = c$ . Not always true.

Correct options are A, B, C.

Question 18 · Artificial Intelligence · 2025 MCQ

Consider game trees Tree-1 and Tree-2 as shown. The first level is a MAX agent and the second level is a MIN agent. The value in the square node is the output of the utility function.

For what ranges of

x

and

y

, the right child of node

B

and the right child of node

E

will be pruned by alpha-beta pruning algorithm?

A.
$x \in [1, \infty)$ and $y \in (- \infty, 2]$
B.
$x \in (- \infty, 2]$ and $y \in (- \infty, 5]$
C.
$x \in (- \infty, 2]$ and $y \in [2, \infty)$
D.
$x \in [1, \infty)$ and $y \in (- \infty, 5]$

Correct Answer:

Step-by-Step Solution

Insight: Pruning occurs when $α \geq β$ . Track $α$ and $β$ carefully as they are passed down and updated at each node.

Exam route:

Tree 1: Root gets $α = 2$ from the left child. At node B (MIN), the left child is $x$ , so $β$ becomes $x$ . The right child is pruned if $α \geq β ⟹ 2 \geq x ⟹ x \in (- \infty, 2]$ .

Tree 2: Node D (MIN) returns $min (5, y)$ to the root. The root updates $α = min (5, y)$ . At node E (MIN), the left child is 2, so $β$ becomes 2. The right child is pruned if $α \geq β ⟹ min (5, y) \geq 2 ⟹ y \in [2, \infty)$ .

Learning route:

Step 1: Analyze Tree 1. Root A (MAX) starts with $α = - \infty, β = \infty$ . It evaluates the left child (value 2) and updates $α = max (- \infty, 2) = 2$ .

Step 2: Move to right child B (MIN). B inherits $α = 2, β = \infty$ . B evaluates its left child ( $x$ ) and updates $β = min (\infty, x) = x$ .

Step 3: For B's right child to be pruned, the condition $α \geq β$ must hold immediately after evaluating $x$ . Thus, $2 \geq x$ , meaning $x \in (- \infty, 2]$ .

Step 4: Analyze Tree 2. Root C (MAX) starts with $α = - \infty, β = \infty$ . It evaluates left child D (MIN). D evaluates its left child (5), so $β = 5$ . D evaluates its right child ( $y$ ), so $β = min (5, y)$ . D returns $min (5, y)$ to C.

Step 5: Root C updates $α = max (- \infty, min (5, y)) = min (5, y)$ .

Step 6: Move to right child E (MIN). E inherits $α = min (5, y)$ and $β = \infty$ . E evaluates its left child (2) and updates $β = min (\infty, 2) = 2$ .

Step 7: For E's right child to be pruned, we need $α \geq β ⟹ min (5, y) \geq 2$ . Since $5 \geq 2$ is always true, this condition simplifies to $y \geq 2$ , meaning $y \in [2, \infty)$ .

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