GATE DA 2025 Question Paper with Solutions: 65 Questions, Answer Key & Section-wise Analysis

GATE DA 2025 previous year paper: 65 questions with answer key and detailed solutions, section-wise breakdown and free sample questions.

Paper breakdown

65 questions · 100 marks · 3 minutes. Probability and Statistics: 13 · Programming, Data Structures and Algorithms: 9 · Linear Algebra: 9 · Machine Learning: 7 · Database Management and Warehousing: 7 · Calculus and Optimization: 6 · Quantitative Aptitude: 4 · Verbal Aptitude: 3 · Artificial Intelligence: 3 · Spatial Aptitude: 2 · Analytical Aptitude: 2

Free sample questions from GATE DA 2025 Question Paper

Question 1 · Probability and Statistics · 2025 MCQ

Suppose $X$ and $Y$ are random variables. The conditional expectation of $X$ given $Y$ is denoted by $E [X ∣ Y]$ . Then $E [E [X ∣ Y]]$ equals

A.
$E [X ∣ Y]$
B.
$\frac{E [ X ]}{E [ Y ]}$
C.
$E [X]$
D.
$E [Y]$

Question 2 · Probability and Statistics · 2025 MCQ

Let

X

be a continuous random variable whose cumulative distribution function (CDF)

F_{X} (x)

, for some

t

, is given as follows:

F_{X} (x) = ⎩ ⎨ ⎧ 0 \frac{x - t}{4 - t} 1 x \leq t t \leq x \leq 4 x \geq 4

If the median of

X

3

, then what is the value of

t

A.
$2$
B.
$1$
C.
$- 1$
D.
$0$

Question 3 · Probability and Statistics · 2025 MCQ

Let

X = a Z + b

, where

Z

is a standard normal random variable, and

a, b

are two unknown constants. It is given that

E [X] = 1, E [(X - E [X]) Z] = - 2, E [(X - E [X])^{2}] = 4,

where

E [X]

denotes the expectation of random variable

X

. The values of

a, b

are:

A.
$a = - 2, b = 1$
B.
$a = 2, b = - 1$
C.
$a = - 2, b = - 1$
D.
$a = 1, b = 1$

Question 4 · Programming, Data Structures and Algorithms · 2025 MCQ

Consider a hash table of size

10

with indices

{0, 1, ..., 9}

, with the hash function

h (x) = 3 x (mod 10),

where linear probing is used to handle collisions. The hash table is initially empty and then the following sequence of keys is inserted into the hash table:

1, 4, 5, 6, 14, 15

. The indices where the keys

14

and

15

are stored are, respectively

A.
$2$ and $5$
B.
$2$ and $6$
C.
$4$ and $5$
D.
$4$ and $6$

Correct Answer:

Step-by-Step Solution

Key idea: This is a hash table tracing question, recognizable because it provides a specific hash function, collision resolution strategy, and a sequence of keys to insert.

Step 1: Understand the hash function and table size. The table has size 10 (indices 0 to 9). The hash function is $h (x) = 3 x (mod 10)$ .

Step 2: Trace the insertion of each key sequentially using linear probing.

Insert 1: $h (1) = 3 (1) (mod 10) = 3$ . Index 3 is empty. Place 1 at index 3.
Insert 4: $h (4) = 3 (4) (mod 10) = 12 (mod 10) = 2$ . Index 2 is empty. Place 4 at index 2.
Insert 5: $h (5) = 3 (5) (mod 10) = 15 (mod 10) = 5$ . Index 5 is empty. Place 5 at index 5.
Insert 6: $h (6) = 3 (6) (mod 10) = 18 (mod 10) = 8$ . Index 8 is empty. Place 6 at index 8.
Insert 14: $h (14) = 3 (14) (mod 10) = 42 (mod 10) = 2$ . Index 2 is occupied (by 4).
Probe 1: $(2 + 1) (mod 10) = 3$ . Index 3 is occupied (by 1).
Probe 2: $(3 + 1) (mod 10) = 4$ . Index 4 is empty. Place 14 at index 4.
Insert 15: $h (15) = 3 (15) (mod 10) = 45 (mod 10) = 5$ . Index 5 is occupied (by 5).
Probe 1: $(5 + 1) (mod 10) = 6$ . Index 6 is empty. Place 15 at index 6.

Step 3: Identify the final indices. Key 14 is at index 4, and key 15 is at index 6.

Answer: D

Question 5 · Programming, Data Structures and Algorithms · 2025 MCQ

Consider the following Python declarations of two lists.

A=[1,2,3]
B=[4,5,6]

Which one of the following statements results in

A = [1, 2, 3, 4, 5, 6]

A.
A.extend(B)
B.
A.append(B)
C.
A.update(B)
D.
A.insert(B)

Correct Answer:

Step-by-Step Solution

Insight: The question tests the difference between adding an iterable as a single nested object versus flattening its elements into the existing list.

Exam route: Eliminate options that use non-existent list methods or wrong argument counts. Between the remaining two, recall that extend flattens while append nests.

Learning route:

A.extend(B): The extend method iterates over the argument B and adds each element (4, 5, 6) one by one to A. The final list is [1, 2, 3, 4, 5, 6]. This matches the requirement.
A.append(B): The append method adds the entire object B as a single element at the end of A. The final list becomes [1, 2, 3, [4, 5, 6]]. This does not match.
A.update(B): The update method is not defined for Python lists (it is used for sets and dictionaries). Calling this raises an AttributeError.
A.insert(B): The insert method requires exactly two arguments: an index and an element. Providing only one argument raises a TypeError.

Therefore, A.extend(B) is the only statement that produces the desired flat list.

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

For which of the following inputs does binary search take time $O (lo g n)$ in the worst case?

A.
An array of $n$ integers in any order
B.
A linked list of $n$ integers in any order
C.
An array of $n$ integers in increasing order
D.
A linked list of $n$ integers in increasing order

Correct Answer:

Step-by-Step Solution

Insight: Binary search requires both sorted data (for elimination) and random access (for $O (1)$ mid calculation).

Exam route: Check options for "sorted" and "array". Only "array in increasing order" satisfies both.

Learning route:

Binary search achieves $O (lo g n)$ time complexity by halving the search space at each step.

This requires two strict preconditions:

The data must be sorted, so we can mathematically guarantee which half contains the target.
The data structure must support $O (1)$ random access, so we can find the middle element in constant time.

Arrays provide contiguous memory allocation, allowing $O (1)$ random access via index math.

Linked lists require $O (n)$ traversal to find the middle node, collapsing the time complexity to $O (n lo g n)$ or worse.

Therefore, only a sorted array (increasing order) guarantees $O (lo g n)$ worst-case time.

Question 7 · Linear Algebra · 2025 MCQ

The number of additions and multiplications involved in performing Gaussian elimination on any $n \times n$ upper triangular matrix is of the order

A.
$O (n)$
B.
$O (n^{2})$
C.
$O (n^{3})$
D.
$O (n^{4})$

Correct Answer:

Step-by-Step Solution

Insight: An upper triangular matrix already has all the zeros that forward elimination would create, so the cubic phase collapses to zero work. The only cost left is the quadratic back-substitution phase.

Exam route:

Recognise the matrix structure: $a_{ij} = 0$ for all $i > j$ .
Forward elimination has nothing to eliminate below any pivot $\Rightarrow$ cost $= 0$ .
The task "perform Gaussian elimination" means the complete solve (elimination + back substitution).
Back substitution on an $n \times n$ upper triangular system uses $\sum_{i = 1}^{n} (n - i) = \frac{n ( n - 1 )}{2}$ multiplications and the same number of additions.
Total additions + multiplications $= n (n - 1) = O (n^{2})$ .

Learning route:

Gaussian elimination has two phases. Phase 1 (forward elimination) converts $A \to U$ by zeroing entries below each pivot. For a general dense $n \times n$ matrix this costs $\frac{2 ( n ^{3} - n )}{3}$ additions and multiplications, i.e. $O (n^{3})$ . Phase 2 (back substitution) solves $U x = c$ from the bottom row upward, costing $n (n - 1)$ additions and multiplications, i.e. $O (n^{2})$ .

When the input is already upper triangular, every entry below the diagonal is already zero. At pivot step $k$ , there are no entries in column $k$ below row $k$ to remove, so no row updates are performed. Forward elimination contributes exactly $0$ operations.

The system still must be solved, so back substitution runs in full. Variable $x_{i}$ requires $n - i$ multiplications and $n - i$ additions. Summing over $i = 1, \dots, n$ gives $\frac{n ( n - 1 )}{2}$ of each, for a combined total of $n (n - 1)$ , which is $O (n^{2})$ .

The common wrong path is to memorise "Gaussian elimination is $O (n^{3})$ " and answer $O (n^{3})$ without inspecting the matrix structure. That ignores the fact that the cubic cost comes entirely from the row updates that this matrix does not need.

Verification: For $n = 3$ , back substitution does $2 + 1 + 0 = 3$ multiplications and $3$ additions, total $6 = 3 (2)$ , matching $n (n - 1)$ . Growth is clearly quadratic, not cubic.

Question 8 · Linear Algebra · 2025 MCQ

The sum of the elements in each row of $A \in R^{n \times n}$ is $1$ . If $B = A^{3} - 2 A^{2} + A$ , which one of the following statements is correct (for $x \in R^{n}$ )?

A.
The equation $B x = 0$ has no solution
B.
The equation $B x = 0$ has exactly two solutions
C.
The equation $B x = 0$ has infinitely many solutions
D.
The equation $B x = 0$ has a unique solution

Correct Answer:

Step-by-Step Solution

Key idea: The row sum condition directly gives an eigenvalue and its eigenvector, which can be evaluated in the matrix polynomial.

Step 1: The statement "sum of the elements in each row of $A$ is $1$ " means that if we multiply $A$ by the all-ones column vector $1 = [1, 1, \dots, 1]^{T}$ , the result is $1$ . Thus, $A 1 = 1 \cdot 1$ .

Step 2: This implies $λ = 1$ is an eigenvalue of $A$ , and $1$ is a corresponding non-zero eigenvector.

Step 3: We are given $B = A^{3} - 2 A^{2} + A$ . We can factor this polynomial as $B = A (A^{2} - 2 A + I) = A (A - I)^{2}$ .

Step 4: Evaluate $B 1$ . Since $(A - I) 1 = A 1 - 1 = 1 - 1 = 0$ , we have $(A - I)^{2} 1 = (A - I) 0 = 0$ .

Step 5: Therefore, $B 1 = A (A - I)^{2} 1 = A 0 = 0$ .

Step 6: Since $1$ is a non-zero vector and $B 1 = 0$ , the homogeneous system $B x = 0$ has a non-trivial solution. Any homogeneous system with a non-trivial solution has infinitely many solutions.

Answer: The equation $B x = 0$ has infinitely many solutions.

Question 9 · Linear Algebra · 2025 MSQ

Which of the following statements is/are correct?

A.
$R^{n}$ has a unique set of orthonormal basis vectors
B.
$R^{n}$ does not have a unique set of orthonormal basis vectors
C.
Linearly independent vectors in $R^{n}$ are orthonormal
D.
Orthonormal vectors $R^{n}$ are linearly independent

Correct Answer:

["B","D"]

Step-by-Step Solution

Insight: Orthonormal bases are not unique (any rotation works), but orthonormality strictly guarantees linear independence.

Exam route: Evaluate each option. A is false because we can rotate the standard basis. B is true. C is false because independent vectors need not be orthogonal. D is true because orthonormal vectors are always independent.

Learning route:

An orthonormal basis for $R^{n}$ requires vectors to be mutually orthogonal and unit length. The standard basis is one, but any rotation of it (e.g., in $R^{2}$ , using $(cos θ, sin θ)$ and $(- sin θ, cos θ)$ ) is another. Thus, it is not unique (A is false, B is true).
For C, consider vectors $(1, 0)$ and $(1, 1)$ in $R^{2}$ . They are linearly independent, but their dot product is $1 \neq = 0$ , so they are not orthogonal. Thus, independent vectors are not necessarily orthonormal.
For D, let $v_{1}, \dots, v_{k}$ be orthonormal. Suppose $\sum c_{i} v_{i} = 0$ . Taking the dot product with $v_{j}$ gives $c_{j} (v_{j} \cdot v_{j}) = 0 ⟹ c_{j} = 0$ . Thus, they are linearly independent.

Question 10 · Machine Learning · 2025 MCQ

Consider designing a linear classifier

y = s i g n (f (x; w, b)), f (x; w, b) = w^{T} x + b

on a dataset

D = {(x_{1}, y_{1}), (x_{2}, y_{2}), ..., (x_{N}, y_{N})}

x_{i} \in R^{d}

y_{i} \in {+ 1, - 1}

i = 1, 2, ..., N

. Recall that the sign function outputs

+ 1

if the argument is positive, and

- 1

if the argument is non-positive. The parameters

w

and

b

are updated as per the following training algorithm:

w_{n e w} = w_{o l d} + y_{n} x_{n}, b_{n e w} = b_{o l d} + y_{n}

whenever

s i g n (f (x_{n}; w_{o l d}, b_{o l d})) \neq = y_{n}

. In other words, whenever the classifier wrongly predicts a sample

(x_{n}, y_{n})

from the dataset,

w_{o l d}

gets updated to

w_{n e w}

, and likewise

b_{o l d}

gets updated to

b_{n e w}

. Consider the case

(x_{n}, + 1)

f (x_{n}; w_{o l d}, b_{o l d}) < 0

. Then

A.
$f (x_{n}; w_{n e w}, b_{n e w}) > f (x_{n}; w_{o l d}, b_{o l d})$
B.
$f (x_{n}; w_{n e w}, b_{n e w}) < f (x_{n}; w_{o l d}, b_{o l d})$
C.
$f (x_{n}; w_{n e w}, b_{n e w}) = f (x_{n}; w_{o l d}, b_{o l d})$
D.
$y_{n} f (x_{n}; w_{o l d}, b_{o l d}) > 1$

Question 11 · Machine Learning · 2025 MSQ

Let

C_{1}

and

C_{2}

be two sets of objects. Let

D (x, y)

be a measure of dissimilarity between two objects

x

and

y

. Consider the following definitions of dissimilarity between

C_{1}

and

C_{2}

DIS - 1 (C_{1}, C_{2}) = x \in C_{1}, y \in C_{2} max D (x, y)

DIS - 2 (C_{1}, C_{2}) = x \in C_{1}, y \in C_{2} min D (x, y)

Which of the following statements is/are correct?

A.
Single Linkage Clustering uses DIS-1
B.
Single Linkage Clustering uses DIS-2
C.
Complete Linkage Clustering uses DIS-2
D.
Complete Linkage Clustering uses DIS-1

Question 12 · Machine Learning · 2025 NAT

Given data

{(- 1, 1), (2, - 5), (3, 5)}

of the form

(x, y)

, we fit a model

y = w x

using linear least-squares regression. The optimal value of

w

is
(Round off to three decimal places)

Question 13 · Database Management and Warehousing · 2025 MCQ

If a relational decomposition is not dependency-preserving, which one of the following relational operators will be executed more frequently in order to maintain the dependencies?

A.
Selection
B.
Projection
C.
Join
D.
Set union

Question 14 · Database Management and Warehousing · 2025 MCQ

Consider the following three relations:

Car (model, year, serial, color)
Make (maker, model)
Own (owner, serial)

A tuple in Car represents a specific car of a given model, made in a given year, with a serial number and a color. A tuple in Make specifies that a maker company makes cars of a certain model. A tuple in Own specifies that an owner owns the car with a given serial number. Keys are underlined; (owner, serial) together form key for Own. (

⋈

denotes natural join)

π_{o w n er} (O w n ⋈ (σ_{co l or = ‘‘ r e d^{''}} (C a r ⋈ (σ_{mak er = ‘‘ A B C^{''}} M ak e))))

Which one of the following options describes what the above expression computes?

A.
All owners of a red car, a car made by ABC, or a red car made by ABC
B.
All owners of more than one car, where at least one car is red and made by ABC
C.
All owners of a red car made by ABC
D.
All red cars made by ABC

Question 15 · Database Management and Warehousing · 2025 NAT

On a relation named Loan of a bank:

Loan
loan number	branch name	amount
L11	Banjara Hills	90000
L14	Kondapur	50000
L15	SR Nagar	40000
L22	SR Nagar	25000
L23	Balanagar	80000
L25	Kondapur	70000
L19	SR Nagar	65000

the following SQL query is executed.

SELECT L1.loan number
FROM Loan L1
WHERE L1.amount > (SELECT MAX (L2.amount)
                    FROM Loan L2
                    WHERE L2.branch name = ’SR Nagar’);

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

Question 16 · Calculus and Optimization · 2025 MCQ

Let $f (x) = \frac{e ^{x} - e ^{- x}}{2}$ , $x \in R$ . Let $f^{(k)} (a)$ denote the $k^{t h}$ derivative of $f$ evaluated at $a$ . What is the value of $f^{(10)} (0)$ ? (Note: ! denotes factorial)

A.
$0$
B.
$1$
C.
$\frac{1}{10 !}$
D.
$\frac{2}{10 !}$

Correct Answer:

Step-by-Step Solution

Insight: The function is the hyperbolic sine, $sinh (x)$ , whose derivatives cycle every two steps.

Exam route: Recognize $f (x) = sinh (x)$ . The $n$ -th derivative of $sinh (x)$ is $sinh (x)$ for even $n$ and $cosh (x)$ for odd $n$ . Since 10 is even, $f^{(10)} (x) = sinh (x)$ . Evaluate at $x = 0$ to get $sinh (0) = 0$ .

Learning route:

Write $f (x) = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x}$ .

Differentiate 10 times using the rule $\frac{d ^{n}}{d x ^{n}} e^{a x} = a^{n} e^{a x}$ .

$f^{(10)} (x) = \frac{1}{2} (1)^{10} e^{x} - \frac{1}{2} (- 1)^{10} e^{- x} = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x} = f (x)$ .

Substitute $x = 0$ : $f^{(10)} (0) = f (0) = \frac{e ^{0} - e ^{0}}{2} = 0$ .

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 · Calculus and Optimization · 2025 NAT

$t \to + \infty lim t^{2} + t - t =$

(Round off to one decimal place)

Correct Answer:

0.50

Step-by-Step Solution

Insight: This is an $\infty - \infty$ indeterminate form. Rationalise by multiplying and dividing by the conjugate $t^{2} + t + t$ , or equivalently factor out $t$ and use the binomial approximation.

Exam route: Rationalise → simplify → divide numerator and denominator by $t$ → substitute $t \to \infty$ .

Learning route:

This is an infinity-minus-infinity limit question, recognisable because both $t^{2} + t \to \infty$ and $t \to \infty$ , giving the indeterminate form $\infty - \infty$ .

Method 1 — Conjugate rationalisation (recommended for exams):

Step 1 — Multiply and divide by the conjugate:

$t^{2} + t - t = \frac{( t ^{2} + t - t ) ( t ^{2} + t + t )}{t ^{2} + t + t}$

Step 2 — Simplify the numerator using $(a - b) (a + b) = a^{2} - b^{2}$ :

$= \frac{( t ^{2} + t ) - t ^{2}}{t ^{2} + t + t} = \frac{t}{t ^{2} + t + t}$

Step 3 — Divide numerator and denominator by $t$ (valid since $t > 0$ ):

$= \frac{1}{1 + \frac{1}{t} + 1}$

Step 4 — Take the limit as $t \to \infty$ (so $1/ t \to 0$ ):

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

Method 2 — Binomial expansion:

Factor $t$ from the square root: $t^{2} + t = t 1 + 1/ t$ .

Using $(1 + u)^{1/2} \approx 1 + u /2$ with $u = 1/ t$ :

$t (1 + \frac{1}{2 t}) - t = t + \frac{1}{2} - t = \frac{1}{2}$

Trap check: A common error is to conclude the limit is $0$ by arguing "both terms go to infinity, so they cancel." This ignores the sub-leading $1/2$ term that survives.

Verification: At $t = 1000$ : $1000000 + 1000 - 1000 = 1001000 - 1000 \approx 1000.4999 - 1000 = 0.4999 \approx 0.5$ . Confirmed.

The answer is $0.5$ (rounded to one decimal place: $0.5$ ).

Question 19 · Quantitative Aptitude · 2025 MCQ

4 \times 4

digital image has pixel intensities

(U)

as shown in the figure. The number of pixels with

U \leq 4

is:

A.
3
B.
8
C.
11
D.
9

Correct Answer:

Step-by-Step Solution

Key idea: This is a systematic counting question on a matrix, recognisable because it provides a grid of numerical values and asks for the count of cells satisfying a specific inequality condition.

Step 1: Identify the condition. We need to count cells where the pixel intensity U \leq 4.

Step 2: Examine each row systematically.

Row 1: 0, 1, 0, 2 (All 4 values are \leq 4) -> Count = 4
Row 2: 4, 7, 3, 3 (4, 3, 3 are \leq 4) -> Count = 3
Row 3: 5, 5, 4, 4 (4, 4 are \leq 4) -> Count = 2
Row 4: 6, 7, 3, 2 (3, 2 are \leq 4) -> Count = 2

Step 3: Sum the counts. Total = 4 + 3 + 2 + 2 = 11.

Answer: C

Question 20 · 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 21 · Quantitative Aptitude · 2025 MCQ

If a real variable $x$ satisfies $3^{x^{2}} = 27 \times 9^{x}$ , then the value of $\frac{2 ^{x^{2}}}{( 2 ^{x} ) ^{2}}$ is:

A.
$2^{- 1}$
B.
$2^{0}$
C.
$2^{3}$
D.
$2^{15}$

Correct Answer:

Step-by-Step Solution

Insight: Unify the bases to 3 to find a relation between $x^{2}$ and $x$ , then substitute this relation directly into the target expression without solving for $x$ .

Exam route: Convert $27$ and $9^{x}$ to base 3. Equate exponents to get $x^{2} - 2 x = 3$ . Simplify the target expression to $2^{x^{2} - 2 x}$ and substitute 3.

Learning route:

Given $3^{x^{2}} = 27 \times 9^{x}$ .

Express all terms with base 3: $27 = 3^{3}$ and $9^{x} = (3^{2})^{x} = 3^{2 x}$ .

So, $3^{x^{2}} = 3^{3} \times 3^{2 x} = 3^{3 + 2 x}$ .

Since the bases are equal, equate the exponents:

$x^{2} = 3 + 2 x ⟹ x^{2} - 2 x = 3$ .

Now look at the target expression: $\frac{2 ^{x^{2}}}{( 2 ^{x} ) ^{2}}$ .

Using exponent laws, $(2^{x})^{2} = 2^{2 x}$ .

So the expression becomes $\frac{2 ^{x^{2}}}{2 ^{2 x}} = 2^{x^{2} - 2 x}$ .

Substitute $x^{2} - 2 x = 3$ into the exponent:

$2^{3} = 8$ .

The value is $2^{3}$ , which matches option C.

Question 22 · Verbal Aptitude · 2025 MCQ

Courage : Bravery :: Yearning : __________

Select the most appropriate option to complete the analogy.

A.
Longing
B.
Yelling
C.
Yawning
D.
Glaring

Correct Answer:

Step-by-Step Solution

Insight: This is a synonymic analogy. The two words in the first pair share a nearly identical meaning, and the second pair must replicate this synonymic relationship.

Exam route: "Courage" and "Bravery" are synonyms denoting strength in facing danger. "Yearning" means a strong desire or longing. The synonym for "Yearning" among the options is "Longing".

Learning route:

Analyze Pair A: Courage : Bravery. Both are nouns representing the quality of being brave and facing fear. The relationship is direct synonymy.
Construct Bridge: Word A is a direct synonym for Word B.
Apply to Pair B: Yearning is a noun meaning an intense feeling of loss or lack, and a deep longing for something.
Evaluate options:

Longing: A yearning desire. This is a direct synonym.
Yelling: Shouting loudly. This is an unrelated physical action.
Yawning: Opening the mouth wide due to tiredness. This is an unrelated physical action.
Glaring: Staring fiercely. This is an unrelated physical action.

Conclusion: "Longing" is the only word that maintains the synonymic relationship, while the others are phonetic or random distractors.

Question 23 · Verbal Aptitude · 2025 MCQ

We __________ tennis in the lawn when it suddenly started to rain.

Select the most appropriate option to complete the above sentence.

A.
have been playing
B.
had been playing
C.
would have been playing
D.
could be playing

Correct Answer:

Step-by-Step Solution

Insight: The sentence describes a past action that was ongoing up until another past event interrupted it.

Exam route: The interrupting event "started to rain" is in the simple past. The main action must be in a past tense that shows ongoing duration up to that point. Among the options, only "had been playing" (past perfect continuous) fits this past duration requirement.

Learning route:

Step 1: Anchor the time. The phrase "started to rain" places the entire sentence in the past.

Step 2: Identify the aspect. The action of playing tennis was ongoing for a duration before the rain started. This requires a continuous aspect in the past.

Step 3: Evaluate options. "have been playing" is present perfect continuous (wrong time). "would have been playing" is conditional (no conditional context). "could be playing" is present modal (wrong time). "had been playing" is past perfect continuous, correctly showing an action ongoing up to a past reference point.

Question 24 · 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 25 · 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

Question 26 · 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 $y \in (- \infty, 5]$

Question 27 · Artificial Intelligence · 2025 MCQ

The state graph shows the action cost along the edges and the heuristic function

h

associated with each state.

Suppose

A^{*}

algorithm is applied on this state graph using priority queue to store the frontier. In what sequence are the nodes expanded?

A.
S,A,E,C,B,D,G
B.
S,E,A,C,B,D,G
C.
S,A,E,B,C,D,G
D.
S,A,B,E,C,D,G

Question 28 · Spatial Aptitude · 2025 MCQ

In the given figure, the numbers associated with the rectangle, triangle, and ellipse are

1, 2,

and

3

, respectively. Which one among the given options is the most appropriate combination of

P, Q,

and

R

A.
$P = 6; Q = 5; R = 3$
B.
$P = 5; Q = 6; R = 3$
C.
$P = 3; Q = 6; R = 6$
D.
$P = 5; Q = 3; R = 6$

Question 29 · Spatial Aptitude · 2025 MCQ

A regular dodecagon (12-sided regular polygon) is inscribed in a circle of radius

r

cm as shown in the figure. The side of the dodecagon is

d

cm. All the triangles (numbered 1 to 12) in the figure are used to form squares of side

r

cm and each numbered triangle is used only once to form a square.
The number of squares that can be formed and the number of triangles required to form each square, respectively, are:

Note: The figure shown is representative.

A.
3; 4
B.
4; 3
C.
3; 3
D.
3; 2

Question 30 · Analytical Aptitude · 2025 MCQ

Weight of a person can be expressed as a function of their age. The function usually varies from person to person. Suppose this function is identical for two brothers, and it monotonically increases till the age of

50

years and then it monotonically decreases. Let

a_{1}

and

a_{2}

(in years) denote the ages of the brothers and

a_{1} < a_{2}

.
Which one of the following statements is correct about their age on the day when they attain the same weight?

A.
$a_{1} < a_{2} < 50$
B.
$a_{1} < 50 < a_{2}$
C.
$50 < a_{1} < a_{2}$
D.
Either $a_{1} = 50$ or $a_{2} = 50$

Correct Answer:

Step-by-Step Solution

Key idea: This is a monotonicity-based logical deduction question, recognizable by the function's trend (increasing then decreasing) and the equality of outputs for two different inputs.

Step 1: Understand the function's behavior. The weight function $W (a)$ monotonically increases for $a \leq 50$ and monotonically decreases for $a > 50$ . This creates a single peak at $a = 50$ .

Step 2: Analyze the condition $W (a_{1}) = W (a_{2})$ with $a_{1} < a_{2}$ .

Step 3: Evaluate the positions of $a_{1}$ and $a_{2}$ relative to the peak.

If both $a_{1} < a_{2} \leq 50$ , they lie on the monotonically increasing segment. For a strictly monotonic function, $W (a_{1}) < W (a_{2})$ , so they cannot be equal.

If both $50 \leq a_{1} < a_{2}$ , they lie on the monotonically decreasing segment. Similarly, $W (a_{1}) > W (a_{2})$ , so they cannot be equal.

Step 4: The only way for $W (a_{1}) = W (a_{2})$ with $a_{1} \neq = a_{2}$ is if one age is on the increasing slope and the other is on the decreasing slope.

Step 5: Since $a_{1} < a_{2}$ , it must be that $a_{1} < 50$ and $a_{2} > 50$ .

Step 6: This matches the condition $a_{1} < 50 < a_{2}$ .

Answer: B

Question 31 · 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 ( $\land$ ).

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