GATE DA 2026 Question Paper – Data Science and Artificial Intelligence

GATE 2026 Data Science and Artificial Intelligence (DA) – Exam Pattern & Question Paper

The Graduate Aptitude Test in Engineering (GATE) 2026 – Data Science and Artificial Intelligence (DA) is a national-level computer-based examination conducted to evaluate candidates' understanding of data science, artificial intelligence, mathematics, statistics, and computing concepts. The examination is conducted in English and has a total duration of 3 hours (180 minutes). The paper contains 65 questions carrying 100 marks.

GATE 2026 DA Exam Pattern

• Examination Mode: Computer Based Test (CBT)
• Total Questions: 65
• Total Marks: 100 Marks
• Examination Duration: 3 Hours
• General Aptitude: 10 Questions – 15 Marks
• DA Subject Section: 55 Questions – 85 Marks
• Question Marks: Each question carries either 1 or 2 marks.

Types of Questions

The GATE DA paper includes three types of questions:

• MCQ (Multiple Choice Questions): Four options are provided, and only one option is correct.
• MSQ (Multiple Select Questions): One or more options can be correct. All correct options must be selected to receive full marks.
• NAT (Numerical Answer Type): Candidates must calculate and enter the numerical answer without selecting any options.

Marking Scheme

There is negative marking only for incorrect MCQ answers. One-third mark is deducted for a wrong 1-mark MCQ, and two-thirds mark for a wrong 2-mark MCQ. There is no negative marking for MSQ and NAT questions. MSQ questions do not have partial marking.

Why Should You Practise This Question Paper?

Practising this GATE 2026 DA question paper helps you understand the actual examination pattern, question difficulty, and different types of problems asked in the exam. It allows you to test your conceptual knowledge, improve problem-solving skills, practise numerical calculations, and develop better time-management strategies. By attempting the questions and checking your solutions, you can identify your strong and weak topics and understand which concepts require more revision. Regular practice can also improve accuracy, confidence, and exam preparation. This question paper is a useful practice resource for students preparing for upcoming GATE examinations.

```

Question 1 1 Mark

Verbosity : Brevity :: Insolence : ______
Choose the word that best fills the blank.

Correct Answer: B. Respect
Verbosity means excessive use of words, while brevity means being concise. Insolence means rudeness or disrespect; respect is the contrasting word.

Question 2 1 Mark

The product of the digits of a three-digit number is 70. The sum of the digits of this three-digit number is ______.

Correct Answer: B. 14
Digits 2, 5 and 7 give a product of 2 × 5 × 7 = 70. Their sum is 2 + 5 + 7 = 14.

Question 3 1 Mark

The four pieces of a puzzle are shown in the figure. 

Which one of the figures labelled P, Q, R and S can be constructed by using each of the four pieces only once without overlaps?

The correct option cannot be reliably confirmed without checking the original puzzle diagram. The PDF text available here does not provide an answer key, so no answer has been assumed.

Question 4 1 Mark

Consider two distinct positive real numbers m and n, with m > n.

Let x = n<sup>log₁₀(m)</sup> and y = m<sup>log₁₀(n)</sup>.

The relation between x and y is ______.

Correct Answer: C. x = y
log₁₀(x) = log₁₀(n) × log₁₀(m).
log₁₀(y) = log₁₀(m) × log₁₀(n).
Since multiplication is commutative, the expressions are equal. Therefore x = y.

Question 5 1 Mark

“If his latest movie had been a commercial success, the actor would have made enough money to sponsor his next movie.”

Based only on the above sentence, which one of the following statements is true?

Answer note: The sentence uses counterfactual wording, which suggests that the movie was not commercially successful. However, the sentence alone does not strictly establish the actual outcome. Verify the intended option against an official answer key before marking it as correct.

Question 6 2 Marks

“My friend and I parted __ the door __ the cabin that I had rented __ the night.”

Choose the option with the correct sequence of words to fill the blanks.

Correct Answer: A. at; of; for
Correct sentence: “My friend and I parted at the door of the cabin that I had rented for the night.”

Question 7 2 Marks

Five integers are picked from 0 to 20, with possible repetitions, such that their mean is 12, median is 18, and they have a single mode of 20.

Ignoring permutations, the number of ways to pick these five integers is ______.

Correct Answer: B. 1
Mean = 12, so the total sum of five integers is 5 × 12 = 60.
In sorted order, the median (third number) is 18. Since 20 must be the unique mode, the only valid set is 0, 2, 18, 20, 20.
Its sum is 60, median is 18, and 20 is the unique mode. Ignoring permutations, there is 1 way.

Question 8 2 Marks

Rishi and Swathi are students of Class 5. Pavan and Tanvi are students of Class 4. Rishi and Pavan are boys. Swathi and Tanvi are girls. The four students played a total of three games of chess, one after another. A player who lost a game did not participate in any more games.

It was observed that:
(i) The first game was the only game where two students of the same class played against each other.
(ii) The students of Class 5 won more games than the students of Class 4.
(iii) The boys won two games and the girls won one game.

The student who did not lose any game is ______.

Correct Answer: D. Tanvi
A valid sequence is: Rishi beats Swathi, Rishi beats Pavan, and Tanvi beats Rishi. This gives two wins to the boys and one to the girls, with Class 5 winning two games and Class 4 winning one. Tanvi is the student who never loses a game in this sequence.

Question 9 2 Marks

P, Q, R, S, X and Y are distinct single-digit whole numbers from 0 to 9.

PQ is a two-digit number with Q in the units place and P in the tens place. Similarly, RS is a two-digit number.

PQ and RS are consecutive numbers and (PQ)<sup>2</sup> + (RS)<sup>2</sup> = XYP, where XYP is a three-digit number.

The value of Y is ______.

Correct Answer: C. 6
The consecutive two-digit numbers are 19 and 20.
19² + 20² = 361 + 400 = 761.
Therefore X = 7, Y = 6 and P = 1. All six digits P, Q, R, S, X and Y are distinct, as required. Hence Y = 6.

Question 10 2 Marks

In the given figure, P, Q and R are three points on a circle of radius 10 cm with O as its center, PQ = RQ, and angle PQR = 45°. The figure is representative.

The area of the shaded region PQRO is ______ cm².

Correct Answer: C. 50√2 cm²
The angle at the circumference is 45°, so the central angle POR = 2 × 45° = 90°.
Since PQ = RQ, the corresponding central angles POQ and QOR are each 135°.
Area of triangle PQR = 50(1 + √2) cm².
Area of triangle POR = ½ × 10 × 10 × sin 90° = 50 cm².
Shaded area = 50(1 + √2) − 50 = 50√2 cm².

Question 11 1 Mark

For a classification problem, Principal Component Analysis (PCA) has been used to reduce the dimensionality of a feature space from 100 to 10.

Which of the following options is true about the angle θ between the first and the tenth principal components?

Correct Answer: B. θ = 90°
PCA principal components are orthogonal to one another. Therefore, the first and tenth distinct principal-component directions are perpendicular, giving an angle of 90°.

Question 12 1 Mark

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 the 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 is the correct number of validation splits that will be generated?

Correct Answer: C. 900
Total samples = 1000. The first 100 are reserved for testing, leaving 1000 − 100 = 900 training samples.
LOOCV creates one validation split per training sample, so the number of validation splits is 900.

Question 13 1 Mark

Which of the following algorithms is NOT an example of uninformed search?

Correct Answer: C. A* Search
A* is an informed search algorithm because it uses a heuristic estimate along with the path cost to guide the search. BFS, DFS and depth-limited search are uninformed search methods.

Question 14 1 Mark

Which of the following statements is NOT true?
(The names of the predicates are intuitive.)

Correct Answer: C
“Each king is a person” should be represented as ∀x (IsKing(x) ⇒ IsPerson(x)).
The conjunction in option C says every object is both a king and a person, which is not the intended meaning. The implication is required.

Question 15 1 Mark

Consider that the quick sort algorithm is used to sort an array of n distinct randomly ordered elements. In every call, the pivot is chosen as the first element of the current subarray.

Let T(n) denote the expected time to sort the array. Assume that the time to partition is linear in the size of the current subarray.

Which recurrence relation correctly represents T(n) in this scenario?

Correct Answer: D
Since the array is randomly ordered and all elements are distinct, the pivot is equally likely to have any rank from 0 to n−1. For each possible rank k, the subproblem sizes are k and n−k−1.

Averaging over all pivot ranks gives:
T(n) = (1/n) Σ<sub>k=0</sub><sup>n−1</sup> [T(k) + T(n−k−1)] + O(n).

Question 16 1 Mark

Consider the given Python program:

def append_to_lst(val, lst=[]):
    lst.append(val)
    return lst

print(append_to_lst(1))
print(append_to_lst(2))
print(append_to_lst(3, []))

Which of the following is the correct output of this program?

Correct Answer: B
Python evaluates the default list only once when the function is defined. Therefore, the first two calls reuse the same list: [1], then [1, 2]. The third call explicitly passes a new empty list, so its output is [3].

Question 17 1 Mark

Let R(A, B, C, D, E) be a relational schema with functional dependency set F = {A → BC, CD → E, E → A}.

Which of the following statements is correct?

Correct Answer: A
D does not appear on the right-hand side of any dependency, so every candidate key must contain D.

A⁺ = {A, B, C}; adding D allows CD → E, so (AD)⁺ contains all attributes.
E → A → BC, so (ED)⁺ contains all attributes.
CD → E → A → BC, so (CD)⁺ contains all attributes.

Thus the candidate keys are AD, ED and CD.

Question 18 1 Mark

A 3-dimensional data cube visualizes Sales Quantity for each combination of Product Type, Month and Country.

To further visualize Sales Quantity for each combination of Product Type, Month and State, which OLAP operation should be performed?

Correct Answer: D. Drill-down
State represents a more detailed geographical level than Country. Drill-down moves from a higher-level summary to a more detailed level in a hierarchy, such as Country → State.

Question 19 1 Mark

Let M be a randomly chosen non-empty subset of S = {1, 2, 3, ..., 2026}.

Which of the following is the probability that the product of all the elements of M is even?

Correct Answer: C
There are 2<sup>2026</sup> − 1 non-empty subsets in total.
The product is odd only when every chosen element is odd. There are 1013 odd numbers from 1 to 2026, giving 2<sup>1013</sup> − 1 non-empty subsets with odd product.

Therefore,
P(even product) = 1 − (2<sup>1013</sup> − 1)/(2<sup>2026</sup> − 1)
= 2<sup>1013</sup>(2<sup>1013</sup> − 1)/(2<sup>2026</sup> − 1).

Question 20 1 Mark

Suppose that a computer program provides a non-negative and integer-valued random solution to the equation:

n₁ + n₂ + n₃ + n₄ = 20.

Which of the following is the probability that all of n₁, n₂, n₃ and n₄ in the provided solution are positive?

Correct Answer: A. C(19, 3) / C(23, 3)
The number of non-negative integer solutions to the equation is C(20 + 4 − 1, 4 − 1) = C(23, 3).

For all variables to be positive, let xᵢ = nᵢ − 1. Then x₁ + x₂ + x₃ + x₄ = 16, with each xᵢ non-negative.
The number of positive solutions is C(16 + 4 − 1, 3) = C(19, 3).

Probability = C(19, 3) / C(23, 3).

Question 21 1 Mark

Let

M = [ cos θ   −sin θ
sin θ   cos θ ]

be a 2 × 2 matrix, where θ = 2π/5, and

I₂ = [1   0;   0   1].

Which of the following options is equal to M<sup>2026</sup>?

Correct Answer: B. M

M is a rotation matrix. For a rotation matrix, raising it to a power multiplies the rotation angle by that power.

Since 2026 = 5 × 405 + 1,
2026 × (2π/5) = 810π + 2π/5.

The term 810π represents 405 complete rotations. Therefore, M<sup>2026</sup> = M.

Question 22 1 Mark

Consider a set

S₁ = {x = (x₁, x₂, x₃)<sup>T</sup> ∈ ℝ³ | x<sup>T</sup>x ≤ 16}.

Let S₂ be another set which is a subspace of ℝ³ with dimension two.

Which of the following gives the area of S₁ ∩ S₂?

Correct Answer: A. 16π

The condition x<sup>T</sup>x ≤ 16 means x₁² + x₂² + x₃² ≤ 16.

This describes a sphere of radius 4, including its interior. The intersection with a two-dimensional subspace through the origin is a circular disk of radius 4.

Area = πr² = π × 4² = 16π.

Question 23 1 Mark

In the following table, the Task column lists tasks related to machine learning. The Algorithm column lists algorithms.

Each task must be matched with an appropriate algorithm that can solve it. Denote a match as T:A.

Task                         Algorithm
T1 – Clustering              A1 – Markov Chain Monte Carlo
T2 – Classification          A2 – K-Medoid
T3 – Sampling                A3 – Linear Discriminant Analysis
T4 – Feature Extraction      A4 – Naive Bayes

Which of the following options gives the correct matching?

Correct Answer: B

• Clustering → K-Medoid (A2)
• Classification → Naive Bayes (A4)
• Sampling → Markov Chain Monte Carlo (A1)
• Feature Extraction → Linear Discriminant Analysis (A3)

Hence, the correct matching is T1:A2, T2:A4, T3:A1, T4:A3.

Question 24 1 Mark

Sentence X is said to entail Sentence Y if whenever X is TRUE, Y must also be TRUE.

Which of the following statements are correct if X entails Y?

Select any correct option.

Correct Answers: A, B and C

If X entails Y, whenever X is true, Y must be true.

A. X ⇒ Y expresses this implication, so it is correct.
B. X ∧ ¬Y cannot be true; therefore, it is false, so this is correct.
C. “If X then Y” is another way to express the same implication.
D. “If Y then X” is the converse and does not necessarily follow.

Hence, A, B and C are correct.

Question 25 1 Mark

You are given the following Pre-order and In-order traversals of a Binary Tree T with nodes E, F, G, P, Q, R and S.

Pre-order: P  Q  S  E  R  F  G
In-order:  S  Q  E  P  F  R  G

Which of the following statements are true about the Binary Tree T?

Select any correct option.

Correct Answers: A and B

In a pre-order traversal, the first node is the root. Therefore, P is the root of the tree.

From the in-order traversal, the left subtree of P contains S, Q and E; its pre-order sequence is Q, S, E. Thus Q has two children: S and E.

The right subtree of P has root R, with F in its left subtree and G in its right subtree.

Post-order visits the left subtree, then the right subtree, and finally the root. The resulting sequence is:
S, E, Q, F, G, R, P.

Hence A and B are correct; C and D are incorrect.

1 Mark

Question 26

Consider two relations r and s defined on the relational schemas R(A,B) and S(E,C), respectively. A is the primary key of R and E is a foreign key of S referencing A in R.

Which of the following operations will NEVER violate the foreign key constraint?

Correct Answer: B
Deleting records from relation S does not create a foreign key violation because it removes references rather than creating invalid references. Inserting into S or deleting referenced records from R may violate the constraint.

2 Marks

Question 27

Let f(x) = x³ − 3x² + 2 be a function defined on (−1, 3].

Which of the following statements is/are correct?

Correct Answers: B and D
f′(x) = 3x(x − 2), so the critical points are x = 0 and x = 2. At x = 2, the function has a minimum because the derivative changes from negative to positive. Also, f(1) = 1 − 3 + 2 = 0, so x = 1 is a root.
The function does not have exactly two roots in [−0.9, 0], and x = 0 is not its only maximum on the given domain.

2 Marks

Question 28

Suppose a random variable Z follows Normal(μ = 0, σ² = 1) distribution with probability density function g(z) and cumulative distribution function G(z). Another random variable Y follows t₁ distribution with probability density function h(y) and cumulative distribution function H(y). Let c be the positive real number for which g(c) = h(c).

Which of the following statements is/are correct?

Correct Answers: A and C
Both distributions are symmetric about zero, so G(0) = H(0) = 0.5. At zero, the standard normal density is 1/√(2π), while the t₁ (Cauchy) density is 1/π; these are not equal.
For the positive crossing point c, G(c) > H(c). By symmetry, G(−c) = 1 − G(c) and H(−c) = 1 − H(c), giving G(−c) < H(−c).

2 Marks

Question 29

Consider that for a supervised learning task, the objective function being minimized is f<sub>w</sub>(x) = wx, where x ∈ ℝ is the input and w ∈ ℝ is the parameter. Stochastic Gradient Descent with learning rate 0.10 is used for parameter updates.

Suppose that at the end of iteration i, the value of w becomes 10.00. Let x = 10.00 be the input for iteration (i + 1).

The value of w at the end of iteration (i + 1) is ______. (Rounded off to two decimal places)

Enter Answer:

Correct Answer: 9.00
The gradient is ∂f/∂w = x = 10.
SGD update: w<sub>new</sub> = w − η(∂f/∂w)
= 10 − 0.10 × 10 = 9.00.

2 Marks

Question 30

Consider a game tree for a two-player turn-taking minimax game. There are two players MAX and MIN. MAX prefers the state with the maximum value, while MIN prefers the state with the minimum value.

MAX starts at the root and has three strategies: 1, 2 and 3. Next, MIN also has three strategies: 1, 2 and 3. The game ends there. Both players always take optimal strategies throughout the game.

Terminal values from left to right under each root strategy are:
Strategy 1: 8, 6, −1
Strategy 2: 1, 5, 7
Strategy 3: −4, −3, −12

At the root, the best strategy for MAX is ______. (Answer in integer)

Enter Answer:

Correct Answer: 2
MIN chooses the minimum terminal value for each strategy:
Strategy 1: min(8, 6, −1) = −1
Strategy 2: min(1, 5, 7) = 1
Strategy 3: min(−4, −3, −12) = −12

MAX chooses the maximum among −1, 1 and −12. The maximum is 1, which corresponds to strategy 2.

1 Mark

Question 31

Let A be a sorted array containing 1000 distinct integers. You perform a recursive binary search on A to find an element y. Suppose each comparison checks whether the middle element computed during the current recursive step is equal to, less than, or greater than y.

The maximum number of comparisons that may have to be performed if y is not an element of A is _______. (Answer in integer)

Enter Answer:

Correct Answer: 10
Binary search approximately halves the remaining search interval after each comparison. For 1000 elements, the maximum number of comparisons for an unsuccessful search is ⌈log₂(1000 + 1)⌉ = ⌈log₂(1001)⌉ = 10.

1 Mark

Question 32

In a relational database, a B+ Tree Index is to be constructed for a relation on a key field. In a B+ Tree, a Node Pointer points to a sub-tree and a Data Record Pointer points to a block of database records.

Let:
Node size = 4096 bytes
Node Pointer size = 10 bytes
Search Key Field size = 11 bytes
Data Record Pointer size = 12 bytes

The maximum number of Node Pointers that can be present in a non-leaf node of the B+ Tree is _______. (Answer in integer)

Enter Answer:

Correct Answer: 195
Let p be the number of node pointers in a non-leaf node. There are p − 1 search keys, so:
10p + 11(p − 1) ≤ 4096
21p − 11 ≤ 4096
21p ≤ 4107
p ≤ 195.57...

Since p must be an integer, the maximum number of node pointers is 195.

1 Mark

Question 33

The number of bijections f(·) from the set S = {1, 2, 3, 4} to itself such that f(f(n)) = n, for all n ∈ S, is _______. (Answer in integer)

Enter Answer:

Correct Answer: 10
The condition f(f(n)) = n means that every element is either fixed or belongs to a pair that swaps its two elements.

Count the possibilities:
• No swaps: 1 way
• One swapped pair: C(4,2) = 6 ways
• Two swapped pairs: 3 ways

Total = 1 + 6 + 3 = 10.

1 Mark

Question 34

Let X be an exponentially distributed random variable with mean λ (> 0). If P(X > 5) = 0.35, then the conditional probability P(X > 10 | X > 5) is _______.

(Rounded off to two decimal places)

Enter Answer:

Correct Answer: 0.35
The exponential distribution has the memoryless property:
P(X > s + t | X > s) = P(X > t).

Therefore,
P(X > 10 | X > 5) = P(X > 5)
= 0.35.

1 Mark

Question 35

The value of the following infinite series is _______.

∑<sub>i=0</sub><sup>∞</sup> ∑<sub>j=1</sub><sup>∞</sup> 2<sup>−i</sup> 3<sup>−j</sup>

(Answer in integer)

Enter Answer:

Correct Answer: 1
Since the terms can be separated into two geometric series:

∑<sub>i=0</sub><sup>∞</sup> 2<sup>−i</sup> = 1/(1 − 1/2) = 2

∑<sub>j=1</sub><sup>∞</sup> 3<sup>−j</sup> = (1/3)/(1 − 1/3) = 1/2

Therefore, the double sum = 2 × 1/2 = 1.

2 Marks

Question 36

Let four points in three-dimensional space be:

P₁: [2, 3, −1], P₂: [3, 1, 1], P₃: [5, −2, 3] and P₄: [3, 3, 3].

Hierarchical Agglomerative Clustering is used to cluster the above points. If Manhattan Distance is used as the distance metric during clustering, which of the following options indicates the two points that will be merged first?

(A) P₁, P₂ (B) P₂, P₃ (C) P₃, P₄ (D) P₂, P₄

Solution:
Manhattan distance is calculated by adding the absolute differences of the coordinates.

d(P₁,P₂) = 1 + 2 + 2 = 5
d(P₁,P₄) = 1 + 0 + 4 = 5
d(P₂,P₄) = 0 + 2 + 2 = 4

The minimum distance is 4, between P₂ and P₄. Therefore, they will be merged first.
Correct Option: (D)

2 Marks

Question 37

Which of the following statements is true for Ridge Regression?

(A) The regularizer in the objective function is used to guard against scenarios where the model works well for the test data, but poorly for the training data. (B) The regularizer of Ridge Regression uses L₁ norm. (C) Ridge Regression aims to reduce the number of parameters that have negative values. (D) The regularizer of Ridge Regression may increase the bias of the model, but it helps in reducing the variance in predictions.

Solution:
Ridge Regression uses an L₂ regularization penalty. Regularization can increase model bias while reducing variance in predictions, helping to control overfitting.

Correct Option: (D)

2 Marks

Question 38

Assume that a Creative (C) person will Succeed (S) if the person is also Disciplined (D), but will not succeed otherwise. Consider the following statements:

(i) (C ∧ S) ⇔ D
(ii) C ⇒ (S ⇔ D)
(iii) C ⇔ ((D ⇒ S) ∨ ¬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

Solution:
The given condition is represented by C ⇒ (S ⇔ D): for a creative person, success is equivalent to being disciplined. Hence, statement (ii) is true.

Statement (i) is not equivalent to the given condition. In statement (iii), (D ⇒ S) ∨ ¬S simplifies to TRUE. Therefore, statement (iii) reduces to C, which is not always true.

Thus, only statement (ii) is true.
Correct Option: (B)

2 Marks

Question 39

A recursive function in Python is given below:

def mystery(n):
    if n <= 0:
        return 1
    else:
        return mystery(n-1) + mystery(n-2)

Now, consider the function call mystery(4). Assume that a typical runtime stack is used to manage function calls. Each function call is pushed onto the stack and removed only after it finishes execution.

Which option denotes the total number of function calls, including the initial call, to compute mystery(4)?

(A) 5 (B) 9 (C) 15 (D) 17

Solution:
Let T(n) denote the total number of function calls.

T(−1) = 1, T(0) = 1
T(1) = 1 + T(0) + T(−1) = 3
T(2) = 1 + T(1) + T(0) = 5
T(3) = 1 + T(2) + T(1) = 9
T(4) = 1 + T(3) + T(2) = 1 + 9 + 5 = 15.

Correct Option: (C) 15

2 Marks

Question 40

Consider a directed graph G = (V, E), where V is the finite set of vertices and E is the set of directed edges. G may contain cycles but there is no self-loop. Further, G may not be strongly connected.

Let Gᴿ be the graph obtained by reversing the directions of all the edges in G without changing the set of vertices.

Assume that BFS or DFS from any given vertex v visits only the reachable vertices from v in that graph.

Which of the following statements must always be true, regardless of the structure of G?

(A) If u is reachable in the BFS of Gᴿ from v, then u is also reachable in the DFS of G from v. (B) In Gᴿ, BFS from v visits exactly the same set of vertices as DFS from v in G. (C) The order of vertices in BFS of Gᴿ from v is the reverse of the order in DFS of G from v. (D) If u is reachable in the DFS of G from v, then v is also reachable in the BFS of Gᴿ from u.

Solution:
If u is reachable from v in G, there is a directed path from v to u. Reversing all edges creates a path from u to v in Gᴿ.

Therefore, v is reachable from u in Gᴿ, and BFS from u will visit v.
Correct Option: (D)

2 Marks

Question 41

Consider a B+ Tree where the maximum number of key values in each leaf node is 2 and the maximum number of pointers in each non-leaf node is 3. The content of the B+ Tree is shown below.

[ 5 | 8 ] / | \ / | \ [1 | 5] [7 | 8] [9 | 12] ↓ ↓ ↓ Leaf nodes linked from left to right

Which of the following options denotes the key value(s) stored in the root node after inserting a key value 3 in the given B+ Tree?

(A) 5 (B) 8 (C) 3 and 5 (D) 3, 5 and 8

Solution:
Initially, the root contains keys 5 and 8. Inserting 3 into the left leaf causes that leaf to overflow. The leaf is split, and the separator is inserted into the parent. The root then overflows because it can contain at most two keys.

Splitting the root promotes the middle separator, 5, to become the new root key.

Correct Option: (A) 5

2 Marks

Question 42

Consider the given relations X, Y and Z. Relation X has columns P, Q and R. Relation Y has columns P, Q and S. Relation Z has columns P and T.

Relation X
P	Q	R
P1	Q1	R1
P2	Q2	R2
P3	Q3	R2

Relation Y
P	Q	S
P1	Q1	2
P1	Q2	5
P2	Q1	6
P3	Q3	1

Relation Z
P	T
P1	T1
P3	T2
P4	T3
P4	NULL

Consider the relational algebra expression:

π<sub>P,R,S</sub> [ (σ<sub>Q=Q3 ∨ R=R2</sub>(X ⋈ Y)) ⋈ (σ<sub>S>1</sub>(Y ⋈ Z)) ]

Here, ⋈ denotes natural join. Which option is the correct output?

(A) Two rows (P1, R1, 2) and (P1, R1, 5) (B) Three rows (P1, R1, 2), (P1, R1, 5) and (P2, R2, 6) (C) One row (P1, R1, 2) (D) Zero rows

Solution:
First, X ⋈ Y matches rows on both P and Q. Applying the condition Q = Q3 OR R = R2 leaves the matching tuple (P3, Q3, R2, S=1).

In Y ⋈ Z, the condition S > 1 retains the matching tuples with P1 and S values 2 and 5. The P3 tuple has S = 1, so it is excluded.

The final natural join requires matching common attributes P, Q and S. The retained tuples have no match, so the result contains zero rows.

Correct Option: (D) Zero rows

2 Marks

Question 43

Consider the following concept hierarchies:

Time	Location	Product
Year	Country	Item Category
Month	State	Item Type
Day	City	Item

Which of the following options denotes the total number of possible data cuboids from these concept hierarchies?

(A) 4³ (B) 2³ (C) 2! (D) 4!

Solution:
There are three dimensions: Time, Location and Product. Each hierarchy has three levels, plus the ALL level representing aggregation over the entire dimension.

Thus, each dimension has 4 possible levels. The total number of cuboids is:

4 × 4 × 4 = 4³ = 64.

Correct Option: (A) 4³

2 Marks

Question 44

Let X and Y be two independent random variables. X follows Bernoulli(p = 0.3) distribution and Y follows Normal(μ = 0, σ² = 100) distribution.

Which of the following options is the variance of (2X − 1)Y?

(A) 100 (B) 90 (C) 49 (D) 21

Solution:
Since X is Bernoulli, X is either 0 or 1. Therefore, (2X − 1) is either −1 or +1, so its square is always 1.

Let W = (2X − 1)Y. Since X and Y are independent and E[Y] = 0:

Var(W) = E[W²] − (E[W])²
= E[(2X − 1)²Y²]
= E[(2X − 1)²] E[Y²]
= 1 × 100 = 100.

Correct Option: (A) 100

2 Marks

Question 45

Let

L = lim<sub>n→∞</sub> ∑<sub>k=0</sub><sup>n</sup> (e<sup>−n</sup> n<sup>k</sup> / k!)

Which of the following is the value of L?

(A) 0.5 (B) 1.0 (C) 0 (D) e⁻¹

Solution:
The expression is the cumulative probability P(N ≤ n), where N follows a Poisson distribution with parameter λ = n.

For a Poisson random variable with parameter n, the mean is n and the variance is n. As n tends to infinity, the standardized variable approaches a standard normal distribution.

The cutoff k = n is at the mean. Therefore, the limiting cumulative probability is 0.5.

Correct Option: (A) 0.5

Question 46 1 Mark

Single Correct MCQ — Select one option

Let γ₁, γ₂, γ₃ be the eigenvalues of the matrix

1	0	0
0	cos t	sin t
0	−sin t	cos t

where t ∈ [−π, π] is in radians.

Which one of the following options lists all the possible values of t satisfying

γ₁ + γ₂ + γ₃ = 1 + √2

(A) {π/3, −π/4} (B) {π/4, −π/3} (C) {π/4, −π/4} (D) {π/3, −π/3}

Solution:

The sum of eigenvalues of a matrix is equal to its trace.

γ₁ + γ₂ + γ₃ = 1 + cos t + cos t
1 + 2 cos t = 1 + √2
2 cos t = √2
cos t = √2/2

Since t ∈ [−π, π],
t = π/4 or t = −π/4.

Correct Option: (C)

Question 47 1 Mark

Multiple Select Questions — Select all correct options

Consider that 20 stories of Author X and 10 stories of Author Y were kept together without mentioning the names of the authors. A classifier was then asked to predict the author (X or Y) of each of these stories.

Let, out of X’s stories, 6 were classified as that of Y. On the other hand, out of Y’s stories, 2 were classified as that of X.

Considering X and Y as two classes, which of the following statements is/are true? (A) Accuracy of the classifier is 11/15. (B) Precision of Class X is higher than the Precision of Class Y. (C) Recall of Class X is higher than the Recall of Class Y. (D) Accuracy of the classifier is 14/15.

Solution:

Total stories = 20 + 10 = 30.

Correct predictions:
X = 20 − 6 = 14
Y = 10 − 2 = 8

Accuracy = (14 + 8)/30
= 22/30 = 11/15.

Precision of X = 14/(14 + 2)
= 14/16 = 0.875.

Precision of Y = 8/(8 + 6)
= 8/14 ≈ 0.571.

Therefore, Precision of X is higher than Precision of Y.

Recall of X = 14/20 = 0.7.
Recall of Y = 8/10 = 0.8.

Hence, (A) and (B) are correct.

Correct Options: (A), (B)

Question 48 1 Mark

Multiple Select Questions — Select all correct options

Let P(x) be a predicate.

Which of the following statements is/are NOT valid in first-order logic? (A) ∀x P(x) ⇒ ∃x P(x) (B) ∃x P(x) ⇒ ∀x P(x) (C) ∃x P(x) ⇔ ∀x P(x) (D) ∀x P(x) ⇒ ∃x ¬P(x)

Solution:

(A) ∀x P(x) ⇒ ∃x P(x)
If P(x) is true for every element, it is true for at least one element. This is valid for a non-empty domain.

(B) ∃x P(x) ⇒ ∀x P(x)
P(x) being true for some elements does not imply that it is true for every element. Hence, invalid.

(C) ∃x P(x) ⇔ ∀x P(x)
Existential and universal statements are not equivalent in general. Hence, invalid.

(D) ∀x P(x) ⇒ ∃x ¬P(x)
If P(x) is true for every element, then no element satisfies ¬P(x). Hence, invalid.

Correct Options: (B), (C), (D)

Question 49 1 Mark

Multiple Select Questions — Select all correct options

Consider the problem of sorting the given array in ascending order:

P = [1, 2, 3, 5, 4]

Consider two sorting algorithms Bubble Sort (BS) and Insertion Sort (IS).

Let N₁ be the total number of comparisons done by BS on the elements of P and N₂ be the total number of comparisons done by IS on the elements of P.

Which of the following options is/are correct? (A) N₁ = 10, N₂ = 4 (B) N₁ > N₂ (C) IS on P will perform only one swap (D) Both BS and IS on P will make at least one unnecessary comparison (i.e., comparing elements that are already in correct order)

Solution:

Given array:
P = [1, 2, 3, 5, 4]

Bubble Sort performs 10 comparisons under the standard fixed-pass implementation.

Insertion Sort comparisons:
Inserting 2: 1 comparison
Inserting 3: 1 comparison
Inserting 5: 1 comparison
Inserting 4: 2 comparisons

Therefore, N₂ = 5.
N₁ = 10 and N₂ = 5.

(A) is false because N₂ = 5, not 4.
(B) is true because N₁ > N₂.
(C) is true because only 4 and 5 need to exchange places.
(D) is true because both algorithms perform comparisons involving elements already in the correct relative order.

Correct Options: (B), (C), (D)

Question 50 1 Mark

Multiple Select Questions — Select all correct options

Consider the given Python program.

def outer():

    x = []

    def inner(val):

        x.append(val)

        return x

    return inner

f1 = outer()
f2 = outer()

print(f1(10))
# Line P

print(f1(20))
# Line Q

print(f2(30))
# Line R

print(f1(40))
# Line S

Which of the following options is/are correct? (A) f1 and f2 share the same list x (B) Output of Line Q is [10, 20] (C) Output of Line R is [10, 20, 30] (D) Output of Line S is [10, 20, 40]

Solution:

Each call to outer() creates a separate list x.

Therefore, f1 and f2 do not share the same list.

Line P:
[10]

Line Q:
[10, 20]

Line R:
[30]

Line S:
[10, 20, 40]

Hence, (B) and (D) are correct.

Correct Options: (B), (D)

Q.51

Consider a table Employee(EmpID, TeamID), where the column EmpID (ID of an employee) is the primary key. The column TeamID denotes the team ID of the team of which the employee is a member. TeamID is a NOT NULL column. We want to display the size of the team (denoted as TeamSize) in which each employee is a member by using SQL. As an example, the desired output for the given Employee table is also shown in tabular form. Which of the following is/are correct? Employee EmpID TeamID 1 8 2 8 3 8 4 7 5 7 6 9 Output EmpID TeamSize 1 3 2 3 3 3 4 2 5 2 6 1

Correct Answer: (A) Explanation: The subquery groups the Employee table by TeamID and calculates the number of employees in each team. The outer query joins this result with Employee using TeamID. Therefore, each employee receives the size of their team. Option (B) counts only the row matching the same EmpID because EmpID is a primary key. Option (C) groups by EmpID, so each employee gets a count of 1. Option (D) does not select TeamID in its subquery, so B.TeamID is unavailable for the join.

Q.52

Let M = (Iₙ − (1/n)11ᵀ) be a matrix, where 1 = (1,1,1, … ,1)ᵀ ∈ ℝⁿ and Iₙ is the identity matrix of order n. Which of the following options is/are correct?

Correct Answer: (A), (D) Explanation: M = Iₙ − (1/n)11ᵀ is symmetric because both Iₙ and 11ᵀ are symmetric. Hence, Mᵀ = M. Also, M² = M, so M is a projection matrix. The trace of M is n − 1, not n. Therefore, options (A) and (D) are correct.

Q.53

Let X₁, X₂, …, Xₙ be n independent random variables. Each of the random variables follows Normal(μ = 0, σ² = 1) distribution. Define X̄ = (1/n) ∑ᵢ₌₁ⁿ Xᵢ. Which of the following statements is/are correct?

Correct Answer: (A), (B), (C) Explanation: (A) The sum of squares of n independent standard normal random variables follows a Chi-square distribution with n degrees of freedom. (B) The sum of squared deviations from the sample mean follows a Chi-square distribution with n − 1 degrees of freedom. (C) X₁² + Xₙ² follows a Chi-square distribution with 2 degrees of freedom. A Chi-square distribution with 2 degrees of freedom is an exponential distribution with mean 2. (D) √n X̄ follows a standard normal distribution. Its square follows a Chi-square distribution with 1 degree of freedom, not 2.

Q.54

Let X be a discrete valued random variable with cumulative distribution function F(x). Which of the following statements is/are correct?

Correct Answer: (B), (C) Explanation: A cumulative distribution function is non-decreasing. For a discrete random variable, the CDF has jump discontinuities at values where the variable has positive probability. A CDF can equal zero, so it is not always strictly positive. A CDF is right-continuous, not left-continuous.

Q.55

Consider that Linear Ridge Regression is being used to learn a prediction function y_pred = wᵀx, where w, x ∈ ℝ² and Mean Absolute Error (MAE) is used to measure the prediction error. A weight of 0.20 is associated with the regularizer. At an intermediate step of the training process, assume that the parameter w = [−3.00, 4.00]ᵀ. In the next step, for the input x = [1.00, 2.00]ᵀ, the predicted value of y is noted. Let the relation between x = [x₁, x₂]ᵀ and the true value of y be y_true = x₁ + x₂. The value of the overall regularized loss function for this instance is _______. (Rounded off to two decimal places)

Answer: 7.00 Explanation: Predicted value: y_pred = wᵀx = (−3)(1) + (4)(2) = 5 True value: y_true = x₁ + x₂ = 1 + 2 = 3 MAE for this instance: |y_true − y_pred| = |3 − 5| = 2 Ridge regularization term: 0.20 × (w₁² + w₂²) = 0.20 × (9 + 16) = 5 Overall regularized loss: 2 + 5 = 7.00

Question 56

Numerical Answer

Consider a fully-connected feed-forward multi-layer perceptron. It has 30 neurons in the input layer, followed by two hidden layers and an output layer. The first hidden layer has 4 neurons and the second hidden layer has 3 neurons. The output layer has only one neuron. Assume that no bias parameters are used.

The number of learnable parameters in the multi-layer perceptron is __________ .
(Answer in integer)

Enter Answer:

Correct Answer: 135
Input to first hidden layer: \(30\times4=120\)
First to second hidden layer: \(4\times3=12\)
Second hidden layer to output: \(3\times1=3\)
Total learnable parameters \(=120+12+3=\boxed{135}\).

Question 57

Numerical Answer

A clinic specializes in testing for a disease D. The result of the test can be either positive or negative.

A study revealed that if a person suffers from the disease D, the test result in that clinic comes out positive 80% of the time, and negative 20% of the time. If a person is not suffering from the disease D, the test comes out positive 10% of the time and negative 90% of the time. It is also known that among the general population, the disease D occurs in 30% of the individuals.

If a person tests positive for D in that clinic, the probability that he/she actually suffers from the disease D is __________ . (Rounded off to two decimal places)

Enter Answer:

Correct Answer: 0.77
By Bayes' theorem,
\(P(D\mid +)=\frac{P(+\mid D)P(D)}{P(+\mid D)P(D)+P(+\mid \neg D)P(\neg D)}\)
\(=\frac{0.8\times0.3}{(0.8\times0.3)+(0.1\times0.7)}\)
\(=\frac{0.24}{0.31}\approx0.77\).

Question 58

Numerical Answer

Consider the given Python program.

def fun(L, i=0):
    if i >= len(L)-1:
        return 0
    if L[i] > L[i+1]:
        L[i+1], L[i] = L[i], L[i+1]
        return 1+fun(L, i+1)
    else:
        return fun(L, i+1)

data = [5, 3, 4, 1, 2]
count = 0
for _ in range(len(data)):
    count += fun(data)
print(count)

The output of the program is __________ . (Answer in integer)

Enter Answer:

Correct Answer: 7
Each call to fun(data) makes one left-to-right pass through adjacent elements and counts the swaps. Across the five passes, the swap counts are 4, 2, 1, 0 and 0.
Total \(=4+2+1+0+0=\boxed{7}\).

Question 59

Numerical Answer

Let there be two relations X and Y as shown. X has three columns P, Q and R. Y has two columns P and S.

X

P	Q	R
P1	Q1	R1
P2	Q2	R2
P3	Q3	R2

Y

P	S
P1	10
P1	15
P2	20
P3	1

Consider that the following tuple relational calculus expression is evaluated.

{t | t ∈ X ∧ ∃z ∈ X (t[P] = z[P]) ∧ ∃m ∈ Y (m[P] = t[P] ∧ m[S] > 1)}

The number of tuples that will be returned is __________ . (Answer in integer)

Enter Answer:

Correct Answer: 2
The condition requires a matching P value in relation Y with S greater than 1. P1 satisfies the condition because its S values include 10 and 15. P2 satisfies it because its S value is 20. P3 does not satisfy it because its only S value is 1. Therefore, two tuples from X are returned.

Question 60

Numerical Answer

Let Account be a relation as shown.

Account
AccNo	Balance
A1	5000
A2	5000
A3	10000
A4	15000
A5	18000

Consider the given SQL query.

SELECT AccNo FROM Account AS A
WHERE (SELECT COUNT(*) FROM Account AS B
       WHERE A.Balance < B.Balance) >= (SELECT COUNT(*)
       FROM Account AS C WHERE A.Balance > C.Balance)

The number of rows returned by the SQL query is __________ . (Answer in integer)

Enter Answer:

Correct Answer: 3
For A1 and A2, the number of accounts with greater balances is 3 and the number with lower balances is 0. For A3, both counts are 2. These three rows satisfy the condition. A4 and A5 do not satisfy it. Therefore, the query returns \(\boxed{3}\) rows.

```html

Question 61

Consider an ER model with the entities E1(A11,A12,A13) and E2(A21,A22,A23), where, A11, A12, A13 are the attributes of E1, and A21, A22, A23 are the attributes of E2. Let A22 be a multi-valued attribute. A11 and A21 are the primary keys of E1 and E2, respectively.

Let R12 be a many-to-many relationship between E1 and E2. Participation of both E1 and E2 in R12 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)

Solution:
1. Relation for entity E1.
2. Relation for entity E2.
3. A separate relation for the multi-valued attribute A22.
4. A separate relation for the many-to-many relationship R12.

Answer: 4

Question 62

For a given data set {x<sub>1</sub>, x<sub>2</sub>, ... , x<sub>n</sub>}, where n = 100, it is known that

1
2000   ∑<sup>n</sup><sub>i=1</sub> ∑<sup>n</sup><sub>j=1</sub> (x<sub>i</sub> − x<sub>j</sub>)<sup>2</sup> = 99

Let us denote

x̄ = 1
―
n ∑<sup>n</sup><sub>i=1</sub> x<sub>i</sub>

The value of

1
―
99 ∑<sup>n</sup><sub>i=1</sub>(x<sub>i</sub> − x̄)<sup>2</sup>

is __________ . (Answer in integer)

Solution:
Given:
(1/2000) ∑<sub>i=1</sub><sup>n</sup> ∑<sub>j=1</sub><sup>n</sup> (x<sub>i</sub> − x<sub>j</sub>)² = 99

Using the identity:
∑<sub>i=1</sub><sup>n</sup> ∑<sub>j=1</sub><sup>n</sup> (x<sub>i</sub> − x<sub>j</sub>)² = 2n ∑<sub>i=1</sub><sup>n</sup>(x<sub>i</sub> − x̄)²

Since n = 100:
(1/2000) × 200 × ∑(x<sub>i</sub> − x̄)² = 99
∑(x<sub>i</sub> − x̄)² = 990

Required value = 990/99 = 10

Answer: 10

Question 63

Let X be a random variable that follows Uniform(−1,1) distribution. The conditional distribution of the random variable Y given X = x is the Uniform(x<sup>2</sup> − 0.1, x<sup>2</sup> + 0.1) distribution.

The value of correlation(X, Y) is __________ . (Answer in integer)

Solution:
Given X follows Uniform(−1,1).
The conditional mean is:
E[Y | X=x] = x²

Therefore:
E[XY] = E[X × E(Y|X)] = E[X³] = 0
Also, E[X] = 0 and E[Y] = E[X²] = 1/3.

Thus Cov(X,Y) = E[XY] − E[X]E[Y] = 0.
Hence correlation(X,Y) = 0.

Answer: 0

Question 64

Let A<sub>5×5</sub> be a matrix such that each of its elements follows Bernoulli(p = 0.50) distribution independently.

The probability that the row-sum of the second row and the column-sum of the third column are both equal to 3 is ________ . (Rounded off to two decimal places)

Solution:
The second row and third column share one common element.

If the common element is 1:
P = 0.5 × (6/16) × (6/16)

If the common element is 0:
P = 0.5 × (4/16) × (4/16)

Total probability:
P = 0.5 × (6/16)² + 0.5 × (4/16)²
P = 26/256 = 0.1015625

Rounded off to two decimal places:
Answer: 0.10

Question 65

Let A = (I<sub>n</sub> − 1
n 11<sup>T</sup>) be a matrix, where 1 = (1,1,1, ... ,1)<sup>T</sup> ∈ ℝ<sup>n</sup> and I<sub>n</sub> is the identity matrix of order n.

The value of

max<sub>S</sub> x<sup>T</sup>Ax

where, S = {x ∈ ℝ<sup>n</sup> | x<sup>T</sup>x = 1}, is __________ . (Answer in integer)

Solution:
A = I<sub>n</sub> − (1/n)11<sup>T</sup> is a projection matrix onto the subspace orthogonal to the all-ones vector.

For n ≥ 2, its eigenvalues are 1 and 0.
The maximum value of x<sup>T</sup>Ax subject to x<sup>T</sup>x = 1 is the largest eigenvalue of A.

Answer: 1

```