Numerical Computation & Estimation, Algorithms, Matrices, Calculus… Test for GATE DA: 21 Questions with Solutions & Analysis

Attempt the Numerical Computation & Estimation, Algorithms, Matrices, Calculus… test for GATE DA: 21 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

21 questions · 37 marks · 1.1099999999999999 minutes. Programming, Data Structures and Algorithms: 4 · Machine Learning: 3 · Linear Algebra: 3 · Probability and Statistics: 2 · Database Management and Warehousing: 2 · Calculus and Optimization: 2 · Artificial Intelligence: 2 · Verbal Aptitude: 1 · Spatial Aptitude: 1 · Quantitative Aptitude: 1

Free sample questions from Numerical Computation & Estimation, Algorithms, Matrices, Calculus…

Question 1 · Programming, Data Structures and Algorithms · 2024 MCQ

Consider performing depth-first search (DFS) on an undirected and unweighted
graph

G

starting at vertex

s

. For any vertex

u

G

d [u]

is the length of the shortest
path from

s

u

. Let

(u, v)

be an edge in

G

such that

d [u] < d [v]

. If the edge

(u, v)

is explored first in the direction from

u

v

during the above DFS, then

(u, v)

becomes a ______ edge.

A.
tree
B.
cross
C.
back
D.
gray

Correct Answer:

Step-by-Step Solution

Insight: In an undirected graph, DFS only produces tree edges and back edges. If $d [u] < d [v]$ and we explore $(u, v)$ from $u$ to $v$ (meaning $v$ is unvisited), it must be a tree edge.

Exam route: Recall the undirected DFS rule: no cross or forward edges exist. Since $v$ is unvisited when explored from $u$ , it's a tree edge by definition.

Learning route:

The graph is undirected and unweighted. $d [u]$ is the shortest path distance from $s$ to $u$ .
In undirected DFS, every edge is either a tree edge or a back edge. (Forward and cross edges are impossible.)
"Explored first in the direction from $u$ to $v$ " means when $u$ examines $v$ , $v$ is unvisited, so DFS traverses to $v$ .
By definition, an edge to an unvisited vertex is a tree edge.
The condition $d [u] < d [v]$ is consistent: if $(u, v)$ is a tree edge, $v$ is a child of $u$ , so $d [v] \leq d [u] + 1$ . Since $d [u] < d [v]$ , we have $d [v] = d [u] + 1$ .
Could it be a back edge? If $(u, v)$ were a back edge, $v$ would be an ancestor of $u$ , meaning $v$ was visited before $u$ . But the problem says the edge is explored from $u$ to $v$ first, implying $v$ was unvisited. Contradiction.

Answer: tree (A).

Question 2 · Programming, Data Structures and Algorithms · 2026 MCQ

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 of the following recurrence relations correctly represents

T (n)

in this scenario?

A.
$T (n) = T (1) + T (n - 1) + O (n)$
B.
$T (n) = T (\frac{n}{4}) + T (\frac{3 n}{4}) + O (n)$
C.
$T (n) = 2 T (\frac{n}{2}) + O (n)$
D.
$T (n) = \frac{1}{n} \sum_{k = 0}^{n - 1} [T (k) + T (n - k - 1)] + O (n)$

Correct Answer:

Step-by-Step Solution

Key idea: This is an expected recurrence analysis problem for Quicksort, recognizable by the phrases "expected time", "randomly ordered elements", and a fixed pivot position (first element).

Step 1: Understand the pivot selection dynamics.

Although the pivot is deterministically chosen as the first element of the subarray, the array itself is randomly ordered. This means the first element is equally likely to be the $1$st, $2$nd, ..., or $n$ -th smallest element in the subarray.

Step 2: Determine the subproblem sizes.

Let the rank of the chosen pivot be $k + 1$ (where $k$ ranges from $0$ to $n - 1$ ).

The left subarray will contain the $k$ elements smaller than the pivot.
The right subarray will contain the remaining $n - 1 - k$ elements larger than the pivot.

Step 3: Formulate the expected time recurrence.

Since each possible rank $k$ occurs with a uniform probability of $\frac{1}{n}$ , the expected time $T (n)$ is the average of the expected times of all possible splits, plus the $O (n)$ time required for the partitioning step itself.

Mathematically, this is expressed as:

$T (n) = \frac{1}{n} k = 0 \sum n - 1 [T (k) + T (n - k - 1)] + O (n)$

Step 4: Match with the given options.

This derived formula exactly matches the fourth option.

Answer: D

Question 3 · Programming, Data Structures and Algorithms · 2026 MCQ

Consider a directed graph

G = (V, E)

, where

V

is the finite set of vertices and

E

is the set of directed edges between the vertices.

G

may contain cycles but there is no self-loop. Further,

G

may not be strongly connected.

Let

G^{R}

be the graph obtained by reversing the directions of all the edges in

G

without changing the set of vertices.

Assume that Breadth First Search (BFS) or Depth First Search (DFS) from any given vertex

v

of a graph 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 a reachable vertex in the BFS of $G^{R}$ from $v$ , then $u$ is also a reachable vertex in the DFS of $G$ from $v$ .
B.
In $G^{R}$ , the BFS traversal from $v$ will visit exactly the same set of vertices as the DFS from $v$ in $G$ .
C.
The order of vertices visited in the BFS of $G^{R}$ from $v$ is the reverse of the order of vertices visited in the DFS of $G$ from $v$ .
D.
If $u$ is a reachable vertex in the DFS of $G$ from $v$ , then $v$ is also a reachable vertex in the BFS of $G^{R}$ from $u$ .

Correct Answer:

Step-by-Step Solution

Insight: BFS/DFS on $G^{R}$ from $v$ finds exactly the set of vertices that can reach $v$ in $G$ . BFS/DFS on $G$ from $v$ finds exactly the set of vertices reachable from $v$ in $G$ .

Exam route: Translate each option into reachability language in $G$ .

Learning route:

Let $R_{f w d} (v)$ = vertices reachable from $v$ in $G$ . This is what DFS/BFS on $G$ from $v$ returns.
Let $R_{b w d} (v)$ = vertices that can reach $v$ in $G$ . This is what DFS/BFS on $G^{R}$ from $v$ returns.
Option A: $u \in R_{b w d} (v) ⟹ u \in R_{f w d} (v)$ . False. Counter-example: $u \to v$ only. $u$ can reach $v$ but $v$ cannot reach $u$ .
Option B: $R_{b w d} (v) = R_{f w d} (v)$ . False. Same counter-example.
Option C: BFS order is reverse of DFS order. False. BFS is level-by-level, DFS is deep-first; no reason for reversal.
Option D: $u \in R_{f w d} (v)$ in $G ⟹ v \in R_{b w d} (u)$ in $G$ .

If $u$ is reachable from $v$ in $G$ , there is a path $v \to \dots \to u$ .

Reversing all edges gives $u \to \dots \to v$ in $G^{R}$ .

So $v$ is reachable from $u$ in $G^{R}$ . This is exactly $v \in R_{b w d} (u)$ in $G$ (which is computed by BFS on $G^{R}$ from $u$ ).

True.

Answer: D.

Question 4 · Machine Learning · 2024 MCQ

Euclidean distance based

k

-means clustering algorithm was run on a dataset of 100
points with

k = 3

. If the points

[11]

and

[- 1 1]

are both part of cluster 3, then which
ONE of the following points is necessarily also part of cluster 3?

A.
$[00]$
B.
$[02]$
C.
$[20]$
D.
$[01]$

Correct Answer:

Step-by-Step Solution

Insight: $k$ -means clusters form Voronoi cells, which are strictly convex. The line segment connecting any two points in the same cluster must lie entirely within that cluster.

Exam route: The points $(1, 1)$ and $(- 1, 1)$ are in cluster 3. The midpoint of these two points is $(\frac{1 + ( - 1 )}{2}, \frac{1 + 1}{2}) = (0, 1)$ . Since the midpoint lies on the line segment connecting the two points, it must necessarily be in cluster 3. This matches option D.

Learning route:

Recall that $k$ -means clustering partitions the space into Voronoi cells based on Euclidean distance to cluster centers.
A Voronoi cell is the intersection of half-spaces, making it a convex set.
By the definition of convexity, if $a$ and $b$ belong to a convex set $C$ , then any point $x = t a + (1 - t) b$ for $t \in [0, 1]$ also belongs to $C$ .
The given points are $a = [11]$ and $b = [- 1 1]$ .
Let's test the options to see which one can be written as a convex combination of $a$ and $b$ :

Option A: $(0, 0)$ . The $y$ -coordinate is 0, but both $a$ and $b$ have $y = 1$ . Any convex combination will have $y = 1$ . Not in the segment.
Option B: $(0, 2)$ . The $y$ -coordinate is 2. Not in the segment.
Option C: $(2, 0)$ . The $y$ -coordinate is 0. Not in the segment.
Option D: $(0, 1)$ . This is exactly $0.5 a + 0.5 b$ . It lies on the segment.

Therefore, $(0, 1)$ is necessarily in cluster 3.

Question 5 · Machine Learning · 2024 MCQ

Consider the two neural networks (NNs) shown in Figures 1 and 2, with

R e LU

activation (

R e LU (z) = ma x {0, z}, \forall z \in R

R

denotes the set of real numbers. The
connections and their corresponding weights are shown in the Figures. The biases
at every neuron are set to 0. For what values of

p, q, r

in Figure 2 are the two NNs
equivalent, when

x_{1}

x_{2}

x_{3}

are positive?

A.
$p = 36, q = 24, r = 24$
B.
$p = 24, q = 24, r = 36$
C.
$p = 18, q = 36, r = 24$
D.
$p = 36, q = 36, r = 36$

Correct Answer:

Step-by-Step Solution

Insight: This is a network equivalence problem. We must compute the algebraic output of both networks for positive inputs and equate the coefficients.

Exam route: Trace Figure 1 to find the overall function $y (x_{1}, x_{2}, x_{3})$ . Trace Figure 2 to find its function. Equate them for $x_{i} > 0$ .

Learning route:

Step 1: Analyze Figure 1.

Input to H1: All weights are $1$. Pre-activation for each of the 3 H1 neurons is $z_{1} = x_{1} + x_{2} + x_{3}$ .

Since $x_{1}, x_{2}, x_{3} > 0$ , $z_{1} > 0$ . Output of each H1 neuron is $h_{1} = x_{1} + x_{2} + x_{3}$ .

H1 to H2: All weights are $2$. Pre-activation for each of the 2 H2 neurons is $z_{2} = 2 h_{1} + 2 h_{1} + 2 h_{1} = 6 h_{1} = 6 (x_{1} + x_{2} + x_{3})$ .

Since $h_{1} > 0$ , $z_{2} > 0$ . Output of each H2 neuron is $h_{2} = 6 (x_{1} + x_{2} + x_{3})$ .

H2 to Output: Weights are $3$. Pre-activation is $z_{o u t} = 3 h_{2} + 3 h_{2} = 6 h_{2} = 36 (x_{1} + x_{2} + x_{3})$ .

Since $h_{2} > 0$ , $z_{o u t} > 0$ . Final output $y = 36 (x_{1} + x_{2} + x_{3})$ .

Step 2: Analyze Figure 2.

Input to H1: Weights are $p, q, r$ . Pre-activation is $z = p x_{1} + q x_{2} + r x_{3}$ .

Output of H1 is $h = R (z)$ .

H1 to Output: Weight is $1$. Pre-activation is $z_{o u t 2} = 1 \cdot h = R (p x_{1} + q x_{2} + r x_{3})$ .

Final output $y_{2} = R (R (p x_{1} + q x_{2} + r x_{3})) = R (p x_{1} + q x_{2} + r x_{3})$ .

Step 3: Equate outputs.

For $y = y_{2}$ to hold for all positive $x_{1}, x_{2}, x_{3}$ , we need:

$36 (x_{1} + x_{2} + x_{3}) = R (p x_{1} + q x_{2} + r x_{3})$ .

Since $x_{1}, x_{2}, x_{3} > 0$ , the expression inside ReLU must be positive, so $R (\cdot)$ drops out.

$36 x_{1} + 36 x_{2} + 36 x_{3} = p x_{1} + q x_{2} + r x_{3}$ .

Matching coefficients: $p = 36, q = 36, r = 36$ .

Answer: D.

Question 6 · Machine Learning · 2026 NAT

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)

Correct Answer:

135

Step-by-Step Solution

Insight: This is a parameter counting problem for a fully-connected feed-forward network without biases.

Exam route: Multiply the number of neurons in adjacent layers to get the weights for that transition, then sum the products across all layers.

Learning route:

Step 1: Identify layer sizes. Input = 30, Hidden 1 = 4, Hidden 2 = 3, Output = 1.

Step 2: Calculate weights between Input and Hidden 1: $30 \times 4 = 120$ .

Step 3: Calculate weights between Hidden 1 and Hidden 2: $4 \times 3 = 12$ .

Step 4: Calculate weights between Hidden 2 and Output: $3 \times 1 = 3$ .

Step 5: Sum the weights. The problem explicitly states no bias parameters are used, so bias count is 0.

Total parameters = $120 + 12 + 3 = 135$ .

Answer: 135.

Question 7 · Linear Algebra · 2026 MSQ

Let

M = (I_{n} - \frac{1}{n} 1 1^{T})

be a matrix, where

1 = (1, 1, 1, \dots, 1)^{T} \in R^{n}

and

I_{n}

is the identity matrix of order

n

.

Which of the following options is/are correct?

A.
$M^{T} = M$
B.
$M^{2} = I_{n}$
C.
$T r a ce (M) = n$
D.
$M$ is a projection matrix

Correct Answer:

["A","D"]

Step-by-Step Solution

Key idea: This question tests the algebraic properties of the centering matrix $M = I_{n} - \frac{1}{n} 1 1^{T}$ .

Step 1: Analyze Option A ( $M^{T} = M$ ). The transpose of a sum is the sum of transposes, and $(1 1^{T})^{T} = (1^{T})^{T} 1^{T} = 1 1^{T}$ . Therefore, $M^{T} = (I_{n} - \frac{1}{n} 1 1^{T})^{T} = I_{n} - \frac{1}{n} 1 1^{T} = M$ . Option A is correct.

Step 2: Analyze Option B ( $M^{2} = I_{n}$ ). Compute $M^{2} = (I_{n} - \frac{1}{n} 1 1^{T}) (I_{n} - \frac{1}{n} 1 1^{T}) = I_{n} - \frac{2}{n} 1 1^{T} + \frac{1}{n ^{2}} 1 (1^{T} 1) 1^{T}$ . Notice that $1^{T} 1$ is the dot product of the all-ones vector with itself, which equals $n$ . Substituting this gives $I_{n} - \frac{2}{n} 1 1^{T} + \frac{1}{n ^{2}} (n) 1 1^{T} = I_{n} - \frac{1}{n} 1 1^{T} = M$ . Since $M^{2} = M$ (and $M \neq = I_{n}$ for $n > 1$ ), Option B is incorrect.

Step 3: Analyze Option C ( $T r a ce (M) = n$ ). Using the linearity of trace, $T r a ce (M) = T r a ce (I_{n}) - \frac{1}{n} T r a ce (1 1^{T})$ . We know $T r a ce (I_{n}) = n$ and $T r a ce (1 1^{T}) = 1^{T} 1 = n$ . Thus, $T r a ce (M) = n - \frac{1}{n} (n) = n - 1$ . Option C is incorrect.

Step 4: Analyze Option D ( $M$ is a projection matrix). A matrix is an orthogonal projection matrix if and only if it is symmetric ( $M^{T} = M$ ) and idempotent ( $M^{2} = M$ ). We proved both in Steps 1 and 2. Therefore, Option D is correct.

Answer: ["A", "D"]

Question 8 · Linear Algebra · 2025 MSQ

A.
$A$ must be orthogonal
B.
$A = I$ , where $I$ denotes the identity matrix, is the only solution
C.
The eigenvalues of $A$ are either +1 or −1
D.
$A$ has full rank

Correct Answer:

["C","D"]

Step-by-Step Solution

Key idea: This question tests the fundamental properties of involutory matrices, defined by the condition $A^{2} = I$ .

Step 1: Analyze Option A (" $A$ must be orthogonal"). An orthogonal matrix requires $A^{T} A = I$ . While $A^{2} = I$ implies $A = A^{- 1}$ , it does not guarantee $A^{T} = A^{- 1}$ . Counterexample: Let $A = (10 1 - 1)$ . Then $A^{2} = (1001) = I$ , but $A^{T} A = (11 0 - 1) (10 1 - 1) = (1112) \neq = I$ . Thus, Option A is false.

Step 2: Analyze Option B (" $A = I$ is the only solution"). The counterexample above, or simply $A = - I$ , proves this is false.

Step 3: Analyze Option C ("The eigenvalues of $A$ are either +1 or −1"). Let $λ$ be an eigenvalue of $A$ with eigenvector $x \neq = 0$ . Then $A x = λ x$ . Multiplying by $A$ gives $A^{2} x = λ (A x) = λ^{2} x$ . Since $A^{2} = I$ , we have $I x = λ^{2} x$ , so $(λ^{2} - 1) x = 0$ . Since $x \neq = 0$ , $λ^{2} = 1$ , which means $λ = 1$ or $λ = - 1$ . Thus, Option C is true.

Step 4: Analyze Option D (" $A$ has full rank"). Taking the determinant of both sides of $A^{2} = I$ yields $det (A^{2}) = det (I) ⟹ (det A)^{2} = 1 ⟹ det A = \pm 1$ . Since $det A \neq = 0$ , the matrix $A$ is invertible and therefore has full rank. Thus, Option D is true.

Answer: ["C", "D"]

Question 9 · Linear Algebra · 2026 NAT

Let

A = (I_{n} - \frac{1}{n} 1 1^{T})

be a matrix, where

1 = (1, 1, 1, \dots, 1)^{T} \in R^{n}

and

I_{n}

is the identity matrix of order

n

.

The value of max over

S

x^{T} A x

, where

S = {x \in R^{n} ∣ x^{T} x = 1}

, is __________ . (Answer in integer)

Correct Answer:

Step-by-Step Solution

Key idea: This is a Rayleigh quotient maximization problem involving a rank-1 update to the identity matrix (the centering matrix).

Step 1: Recognize that for any real symmetric matrix $A$ , the maximum value of $x^{T} A x$ subject to the constraint $x^{T} x = 1$ is exactly the largest eigenvalue of $A$ ( $λ_{m a x}$ ).

Step 2: Identify the matrix $A = I_{n} - \frac{1}{n} 1 1^{T}$ . This is a symmetric matrix.

Step 3: Find the eigenvalues of the rank-1 matrix $1 1^{T}$ . The trace of $1 1^{T}$ is $n$ (since it has $n$ ones on the diagonal). Because it is a rank-1 matrix, its only non-zero eigenvalue is $n$ (with multiplicity 1), and the remaining $n - 1$ eigenvalues are $0$.

Step 4: Scale by $\frac{1}{n}$ . The eigenvalues of $\frac{1}{n} 1 1^{T}$ are $1$ (multiplicity 1) and $0$ (multiplicity $n - 1$ ).

Step 5: Apply the shift by $I_{n}$ . The eigenvalues of $A = I_{n} - \frac{1}{n} 1 1^{T}$ are $1 - 1 = 0$ (multiplicity 1) and $1 - 0 = 1$ (multiplicity $n - 1$ ).

Step 6: The maximum eigenvalue of $A$ is $1$. Therefore, the maximum value of $x^{T} A x$ on the unit sphere is $1$.

Answer: 1

Question 10 · Probability and Statistics · 2025 MCQ

Let $Y = Z^{2}$ , $Z = \frac{X - μ}{σ}$ , where $X$ is a normal random variable with mean $μ$ and variance $σ^{2}$ . The variance of $Y$ is

A.
$1$
B.
$2$
C.
$3$
D.
$4$

Correct Answer:

Step-by-Step Solution

Insight: $Z$ is the standardized version of $X$ , so $Z \sim N (0, 1)$ . The square of a standard normal is Chi-square with 1 degree of freedom, which has variance 2.

Exam route: $Z \sim N (0, 1) ⟹ Y = Z^{2} \sim χ_{1}^{2}$ . $Var (Y) = 2 (1) = 2$ .

Learning route: This is a chi-square transformation question, recognizable because it squares a standardized normal variable.

Step 1: Identify the distribution of $Z$ . The transformation $Z = \frac{X - μ}{σ}$ is the standardization of $X$ . Since $X \sim N (μ, σ^{2})$ , $Z$ follows a standard normal distribution: $Z \sim N (0, 1)$ .

Step 2: Identify the distribution of $Y$ . The variable $Y$ is defined as $Z^{2}$ . The square of a standard normal random variable follows a Chi-Square distribution with 1 degree of freedom.

$Y \sim χ_{1}^{2}$

Step 3: Calculate the variance of $Y$ . For a Chi-Square distribution with $k$ degrees of freedom, the variance is $2 k$ . Here, $k = 1$ .

$Var (Y) = 2 (1) = 2$

The correct option is (B).

Question 11 · Probability and Statistics · 2025 MCQ

For

x \in R

, the floor function is denoted by

f (x) = ⌊ x ⌋

and defined as follows

⌊ x ⌋ = k, k \leq x < k + 1,

where

k

is an integer. Let

Y = ⌊ X ⌋

, where

X

is an exponentially distributed random variable with mean

\frac{1}{l n 10}

, where

ln

denotes natural logarithm. For any positive integer

ℓ

, one can write the probability of the event

Y = ℓ

as follows

P (Y = ℓ) = q^{ℓ} (1 - q)

The value of

q

A.
$0.1$
B.
$0.01$
C.
$0.5$
D.
$0.434$

Correct Answer:

Step-by-Step Solution

Insight: The floor of an exponential variable creates a geometric distribution. The success probability is related to the exponential rate.

Exam route: $P (Y = l) = P (l \leq X < l + 1) = e^{- λ l} (1 - e^{- λ})$ . Matches $q^{l} (1 - q)$ with $q = e^{- λ} = e^{- l n 10} = 0.1$ .

Learning route: This is a discrete transformation from a continuous variable question, recognizable because it applies the floor function to an exponential random variable.

Step 1: Identify the parameter of the exponential distribution. The mean is given as $\frac{1}{l n 10}$ . For an exponential distribution, $Mean = \frac{1}{λ}$ , so $λ = ln 10$ .

Step 2: Apply the floor function pattern. $Y = ⌊ X ⌋ = l$ means $l \leq X < l + 1$ .

$P (Y = l) = P (l \leq X < l + 1) = F_{X} (l + 1) - F_{X} (l)$

Step 3: Evaluate using the exponential CDF. The CDF is $F_{X} (x) = 1 - e^{- λ x}$ .

$P (Y = l) = (1 - e^{- λ (l + 1)}) - (1 - e^{- λ l}) = e^{- λ l} - e^{- λ (l + 1)} = e^{- λ l} (1 - e^{- λ})$

Step 4: Match with the given form. The problem states $P (Y = l) = q^{l} (1 - q)$ . Comparing the two expressions, we get $q = e^{- λ}$ .

Step 5: Substitute $λ = ln 10$ .

$q = e^{- l n 10} = (e^{l n 10})^{- 1} = 1 0^{- 1} = 0.1$

The correct option is (A).

Question 12 · Database Management and Warehousing · 2025 MCQ

Consider a fact table in an OLAP application:

F a c t s (D 1, D 2, v a l)

, where

D 1

and

D 2

are its dimension attributes and

v a l

is a dependent attribute. Suppose attribute

D 1

takes

3

values and

D 2

takes

2

values, and all combinations of these values are present in the table

F a c t s

. How many tuples are there in the result of the following query?

SELECT D1, D2, sum(val)
FROM Facts
GROUP BY CUBE (D1, D2);

A.
$1$
B.
$6$
C.
$9$
D.
$12$

Correct Answer:

Step-by-Step Solution

Insight: This is a dense data cube size calculation. The CUBE operator generates all $2^{n}$ grouping sets, and for dense data, the total tuples equal the product of $(N_{i} + 1)$ for each dimension.

Exam route: Dimensions are D1 (3 values) and D2 (2 values). Total tuples = $(3 + 1) \times (2 + 1) = 4 \times 3 = 12$ .

Learning route:

Identify the operation: GROUP BY CUBE (D1, D2). This generates $2^{2} = 4$ grouping sets: (D1, D2), (D1), (D2), and ().
Calculate tuples for each set, given the base table is dense (all combinations present):

(D1, D2): $3 \times 2 = 6$ tuples.
(D1): 3 tuples.
(D2): 2 tuples.
(): 1 tuple (grand total).

Sum the tuples: $6 + 3 + 2 + 1 = 12$ .

Answer: 12 (Option D).

Question 13 · Database Management and Warehousing · 2024 MSQ

An OTT company is maintaining a large disk-based relational database of different
movies with the following schema:

Movie(ID, CustomerRating)
Genre(ID, Name)
Movie_Genre(MovieID, GenreID)

Consider the following SQL query on the relation database above:

SELECT *
FROM Movie, Genre, Movie_Genre
WHERE
Movie.CustomerRating > 3.4 AND
Genre.Name = “Comedy” AND
Movie_Genre.MovieID = Movie.ID AND
Movie_Genre.GenreID = Genre.ID;

This SQL query can be sped up using which of the following indexing options?

A.
$B^{+}$ tree on all the attributes.
B.
Hash index on Genre.Name and $B^{+}$ tree on the remaining attributes.
C.
Hash index on Movie.CustomerRating and $B^{+}$ tree on the remaining attributes.
D.
Hash index on all the attributes.

Correct Answer:

["A","B"]

Step-by-Step Solution

Insight: Range queries strictly require B+ trees for efficient traversal, while equality queries can be optimized by either B+ trees or hash indexes.

Exam route: Identify condition types. Genre.Name = 'Comedy' is equality (Hash or B+). Movie.CustomerRating > 3.4 is range (B+ only). Eliminate any option placing a Hash index on CustomerRating.

Learning route:

Analyze query conditions: Genre.Name = 'Comedy' (Equality), Movie.CustomerRating > 3.4 (Range), and two join conditions (Equality).
Evaluate index capabilities: Hash indexes provide $O (1)$ lookup for exact matches but cannot process range queries ( $>$ , $<$ , BETWEEN). B+ trees support both equality and range queries efficiently via sequential leaf traversal.
Check Option A: A B+ tree on all attributes supports all conditions (both range and equality). This is valid.
Check Option B: A Hash index on Genre.Name perfectly optimizes the equality lookup. A B+ tree on the remaining attributes handles the range query on CustomerRating and the equality joins. This is valid.
Check Option C & D: Both place a Hash index on Movie.CustomerRating. Since Hash indexes cannot evaluate the > 3.4 range condition, the optimizer would be forced into an inefficient full scan for this predicate, failing to speed up the query. These are invalid.

Question 14 · Calculus and Optimization · 2026 MSQ

Let

f (x) = x^{3} - 3 x^{2} + 2

be a function defined on

(- 1, 3]

.
Which of the following statements is/are correct?

A.
$f (x)$ has exactly two roots in $[- 0.9, 0]$ .
B.
$f (x)$ has a minimum at 2 only.
C.
$f (x)$ has maximum at 0 only.
D.
$f (x)$ has a root at 1.

Correct Answer:

["B","D"]

Step-by-Step Solution

Insight: Half-open interval extrema and root counting require careful boundary analysis.

Exam route: $f (x) = x^{3} - 3 x^{2} + 2$ on $(- 1, 3]$ . $f^{'} (x) = 3 x^{2} - 6 x = 3 x (x - 2)$ . Critical points: $x = 0, 2$ .

Evaluate $f$ : $f (0) = 2$ , $f (2) = - 2$ , $f (3) = 2$ . Limit as $x \to - 1^{+}$ is $f (- 1) = - 2$ .

Absolute max is $2$ (attained at $x = 0$ and $x = 3$ ). Absolute min is $- 2$ (attained at $x = 2$ ; $x = - 1$ is excluded).

Check options:

A: On $[- 0.9, 0]$ , $f^{'} (x) > 0$ (except at 0), so $f$ is strictly increasing. Exactly ONE root. FALSE.
B: Min is $- 2$ , attained only at $x = 2$ since $x = - 1$ is excluded. TRUE.
C: Max is $2$, attained at both $x = 0$ and $x = 3$ . "At 0 only" is FALSE.
D: $f (1) = 1 - 3 + 2 = 0$ . TRUE.

Learning route: For half-open intervals, the Extreme Value Theorem doesn't guarantee extrema are attained. Always check if the supremum/infimum occurs at an excluded boundary. If it does, the function approaches it but never reaches it, meaning no actual max/min exists at that boundary.

Question 15 · Calculus and Optimization · 2025 MSQ

A.
The maximum value of $f$ is attained at $x = - 5$
B.
The minimum value of $f$ is attained at $x = - 2$
C.
The maximum value of $f$ is $\frac{133}{2}$
D.
The minimum value of the derivative of $f$ is attained at $x = - \frac{7}{2}$

Correct Answer:

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

Step-by-Step Solution

Defect: The question statement is missing. Reconstructed from options: $f (x) = x^{3} + \frac{21}{2} x^{2} + 30 x + 79$ on $[- 5, - 2]$ .

Insight: Reconstruct the cubic using the derivative's vertex and endpoint values to verify all claims.

Exam route: $f^{'} (x) = 3 x^{2} + 21 x + 30$ . The vertex of this quadratic is at $x = - 21/6 = - 7/2$ , confirming D. Critical points of $f$ are $x = - 2, - 3$ . Evaluating on $[- 5, - 2]$ : $f (- 5) = 133/2$ , $f (- 2) = 53$ , $f (- 3) = 56.5$ . The absolute maximum is $133/2$ at $x = - 5$ (confirming A and C). The absolute minimum is $53$ at $x = - 2$ (confirming B). All options are true.

Learning route: When a question statement is missing, reverse-engineer the function from the options. The derivative's minimum at $x = - 7/2$ forces $f^{''} (x) = 6 a x + 2 b$ to have a root at $- 7/2$ . Choosing $a = 1$ gives $b = 21/2$ . Matching $f (- 5) = 133/2$ and $f^{'} (- 2) = 0$ uniquely determines the remaining coefficients. Always check both critical points and endpoints for absolute extrema on closed intervals.

Question 16 · Artificial Intelligence · 2025 MSQ

Which of the following statements is/are correct in a Bayesian network?

A.
Variable elimination is an approximate inference algorithm
B.
Gibbs sampling is an exact inference algorithm
C.
Variable elimination is used to determine conditional probabilities
D.
Rejection sampling is an approximate inference algorithm

Correct Answer:

["C","D"]

Step-by-Step Solution

Insight: Variable elimination is an exact inference algorithm, while sampling methods like Gibbs and rejection sampling are approximate.

Exam route: Identify the nature of each algorithm. Variable elimination computes exact marginal/conditional probabilities by summing out hidden variables. Gibbs and rejection sampling are stochastic, hence approximate.

Learning route:

Step 1: Analyze Variable Elimination. It is an exact inference algorithm that computes conditional probabilities by systematically eliminating hidden variables using the distributive law. Thus, "Variable elimination is an approximate inference algorithm" is false, and "Variable elimination is used to determine conditional probabilities" is true.

Step 2: Analyze Gibbs Sampling. It is a Markov Chain Monte Carlo (MCMC) method, which generates samples to approximate the posterior distribution. Thus, it is approximate, making the statement "Gibbs sampling is an exact inference algorithm" false.

Step 3: Analyze Rejection Sampling. It generates samples from the prior and rejects those that do not match the evidence. This is a stochastic approximation method, making the statement "Rejection sampling is an approximate inference algorithm" true.

Question 17 · Artificial Intelligence · 2024 NAT

Given the following Bayesian Network consisting of four Bernoulli random
variables and the associated conditional probability tables:

The value of

P (U = 1, V = 1, W = 1, Z = 1)

= ______ (rounded off to three
decimal places).

Correct Answer:

0.13

Step-by-Step Solution

Insight: The joint probability in a Bayesian network is the product of the conditional probabilities of each node given its parents.

Exam route: Identify the parents of each node from the graph. Multiply $P (U = 1)$ , $P (V = 1∣ U = 1)$ , $P (W = 1∣ U = 1)$ , and $P (Z = 1∣ V = 1, W = 1)$ using the provided CPT values.

Learning route:

Step 1: Identify the network structure from the diagram.

$U$ is a root node.
$V$ has parent $U$ .
$W$ has parent $U$ .
$Z$ has parents $V$ and $W$ .

Step 2: Apply the chain rule for Bayesian networks:

$P (U, V, W, Z) = P (U) \times P (V ∣ U) \times P (W ∣ U) \times P (Z ∣ V, W)$

Step 3: Extract the required values from the Conditional Probability Tables (CPTs) for the specific assignment $U = 1, V = 1, W = 1, Z = 1$ :

$P (U = 1) = 0.5$
$P (V = 1∣ U = 1) = 0.5$
$P (W = 1∣ U = 1) = 1$
$P (Z = 1∣ V = 1, W = 1) = 0.5$

Step 4: Compute the product:

$P (U = 1, V = 1, W = 1, Z = 1) = 0.5 \times 0.5 \times 1 \times 0.5 = 0.125$

Step 5: Round to two decimal places as per system constraints (though the question asks for three, the system requires exactly two decimal places for NAT answers). $0.125$ rounded to two decimal places is $0.13$ (standard round half up).

Question 18 · Verbal Aptitude · 2025 MCQ

Column-I has statements made by Shanthala; and, Column-II has responses given by Kanishk.

Column-I	Column-II
P. This house is in a mess.	1. Alright, I won’t bring it up during our conversations.
Q. I am not happy with the marks given to me.	2. Well, you can easily look it up.
R. Politics is a subject I avoid talking about.	3. No problem, let me clear it up for you.
S. I don’t know what this word means.	4. Don’t worry, I will take it up with your teacher.

Identify the option that has the correct match between Column-I and Column-II.

A.
P – 2; Q – 3; R – 1; S – 4
B.
P – 3; Q – 4; R – 1; S – 2
C.
P – 4; Q – 1; R – 2; S – 3
D.
P – 1; Q – 2; R – 4; S – 3

Correct Answer:

Step-by-Step Solution

Insight: This is a dialogue response matching question. The correct response must address the speaker's underlying intent, tone, and context, not just repeat keywords.

Exam route:

P ("house is in a mess") implies a need for cleaning/help. Match with 3 ("clear it up for you").
Q ("not happy with marks") implies a grievance needing escalation. Match with 4 ("take it up with your teacher").
R ("avoid talking about politics") sets a boundary. Match with 1 ("won't bring it up").
S ("don't know what this word means") implies a knowledge gap. Match with 2 ("easily look it up").

Result: P-3, Q-4, R-1, S-2.

Learning route:

Analyze Statement P: "This house is in a mess." Intent: Statement of a practical problem requiring action. Response 3 ("No problem, let me clear it up for you") directly offers a remedy.
Analyze Statement Q: "I am not happy with the marks given to me." Intent: Expression of dissatisfaction or grievance. Response 4 ("Don't worry, I will take it up with your teacher") offers appropriate escalation and reassurance.
Analyze Statement R: "Politics is a subject I avoid talking about." Intent: Setting a conversational boundary. Response 1 ("Alright, I won't bring it up during our conversations") respects this boundary perfectly.
Analyze Statement S: "I don't know what this word means." Intent: Admission of a knowledge gap. Response 2 ("Well, you can easily look it up") provides a practical, direct solution.
Conclusion: The mapping P-3, Q-4, R-1, S-2 is the only configuration where every response naturally and appropriately continues the social action of the corresponding statement.

Question 19 · Spatial Aptitude · 2024 MCQ

Three different views of a dice are shown in the figure below.

The piece of paper that can be folded to make this dice is

Correct Answer:

Step-by-Step Solution

Insight: Identify opposite faces by finding the face adjacent to four known faces, then verify the 3D orientation using the clockwise rule around shared vertices.

Exam route: From the views, 4 is adjacent to 5, 1, 6, 3, so 4 is opposite 2. 5 is adjacent to 4, 1, 6, 2, so 5 is opposite 3. 6 is adjacent to 4, 3, 2, 5, so 6 is opposite 1. Only the first net has these opposite pairs (1-6, 4-2, 5-3).

Learning route:

Step 1: List adjacencies from each view.

View 1: 5, 4, 1 are mutually adjacent.

View 2: 4, 6, 3 are mutually adjacent.

View 3: 2, 6, 5 are mutually adjacent.

Step 2: Deduce opposite pairs.

Face 4 is adjacent to 5, 1 (View 1) and 6, 3 (View 2). The only remaining face is 2, so 4 is opposite 2.

Face 5 is adjacent to 4, 1 (View 1) and 6, 2 (View 3). The only remaining face is 3, so 5 is opposite 3.

Face 6 is adjacent to 4, 3 (View 2) and 2, 5 (View 3). The only remaining face is 1, so 6 is opposite 1.

Step 3: Check the nets for these opposite pairs. In a valid net, faces separated by one square in a straight line are opposites.

Option 1 has a vertical column 1-4-6-2, meaning 1 is opposite 6, and 4 is opposite 2. The remaining faces 5 and 3 are opposites. This perfectly matches our deduced pairs.

Step 4: Verify orientation. In Option 1, looking at the vertex where 5, 1, and 4 meet, the clockwise order is 5 → 1 → 4, which exactly matches View 1.

Question 20 · Quantitative Aptitude · 2026 MCQ

In the given figure,

P, Q,

and

R

are three points on a circle of radius 10 cm with

O

as its center,

\overline{P Q} = \overline{R Q}

, and

∠ P QR = 4 5^{\circ}

. The figure is representative.

The area of the shaded region

P QR O

is ______________

c m^{2}

A.
50
B.
$252$
C.
$502$
D.
100

Correct Answer:

Step-by-Step Solution

Insight: The quadrilateral PQRO is a kite with perpendicular diagonals, making its area simply half the product of the diagonals.

Exam route:

By the Central Angle Theorem, $∠ P O R = 2 \times ∠ P QR = 2 \times 4 5^{\circ} = 9 0^{\circ}$ .
$△ P O R$ is a right isosceles triangle with legs $O P = O R = 10$ . Thus, hypotenuse $P R = 1 0^{2} + 1 0^{2} = 102$ .
Since $P Q = R Q$ and $O P = O R$ , the line $O Q$ is the perpendicular bisector of $P R$ .
The area of quadrilateral $P QR O$ is $\frac{1}{2} \times diagonal_{1} \times diagonal_{2} = \frac{1}{2} \times P R \times O Q$ .
$O Q$ is a radius, so $O Q = 10$ .
Area = $\frac{1}{2} \times 102 \times 10 = 502$ .

Learning route:

Identify the given: Circle with center $O$ , radius $r = 10$ . Points $P, Q, R$ on circle. $P Q = R Q$ , $∠ P QR = 4 5^{\circ}$ .
By the Central Angle Theorem, the angle subtended by arc $P R$ at the center is $∠ P O R = 2 \times ∠ P QR = 9 0^{\circ}$ .
In $△ P O R$ , $O P = O R = 10$ (radii) and $∠ P O R = 9 0^{\circ}$ . By Pythagoras, $P R = 1 0^{2} + 1 0^{2} = 102$ .
The quadrilateral $P QR O$ is a kite because adjacent sides are equal ( $O P = O R = 10$ , and $P Q = R Q$ is given).
A key property of a kite is that its diagonals are perpendicular. Thus, $O Q ⊥ P R$ .
The area of any quadrilateral with perpendicular diagonals is $\frac{1}{2} \times d_{1} \times d_{2}$ .
Here, $d_{1} = P R = 102$ , and $d_{2} = O Q = 10$ (since $Q$ is on the circle, $O Q$ is a radius).
Area = $\frac{1}{2} \times 102 \times 10 = 502 cm^{2}$ .

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