GATE DA 2026 Question Paper with Solutions: 65 Questions, Answer Key & Section-wise Analysis

GATE DA 2026 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: 12 · Database Management and Warehousing: 11 · Programming, Data Structures and Algorithms: 9 · Machine Learning: 9 · Quantitative Aptitude: 6 · Linear Algebra: 5 · Artificial Intelligence: 5 · Verbal Aptitude: 3 · Calculus and Optimization: 2 · Analytical Aptitude: 2 · Spatial Aptitude: 1

Free sample questions from GATE DA 2026 Question Paper

Question 1 · Probability and Statistics · 2026 MCQ

Let

M

be a randomly chosen non-empty subset of

S = {1, 2, 3, \dots, 2026}

.
Which of the following is the probability that the product of all the elements of

M

is even?

A.
$2^{1013} (2^{1013} - 1) / 2^{2026}$
B.
$2^{1013} / 2^{2026}$
C.
$2^{1013} (2^{1013} - 1) / (2^{2026} - 1)$
D.
$1/ (2^{2026} - 1)$

Correct Answer:

Step-by-Step Solution

Insight: The product of a subset is even if at least one element is even. It is much easier to calculate the complement: the product is odd if and only if ALL elements in the subset are odd.

Exam route: Total non-empty subsets of $S$ is $2^{2026} - 1$ . There are 1013 odd numbers in $S$ . Non-empty subsets containing only odd numbers is $2^{1013} - 1$ . Probability of odd product is $\frac{2 ^{1013} - 1}{2 ^{2026} - 1}$ . Probability of even product is $1 - \frac{2 ^{1013} - 1}{2 ^{2026} - 1} = \frac{2 ^{2026} - 2 ^{1013}}{2 ^{2026} - 1} = \frac{2 ^{1013} ( 2 ^{1013} - 1 )}{2 ^{2026} - 1}$ .

Learning route: The set $S = {1, 2, \dots, 2026}$ contains 1013 odd numbers and 1013 even numbers. A subset product is odd if and only if every element in the subset is odd. The number of non-empty subsets formed entirely from the 1013 odd numbers is $2^{1013} - 1$ . The total number of non-empty subsets of $S$ is $2^{2026} - 1$ . The probability that a randomly chosen non-empty subset has an odd product is $\frac{2 ^{1013} - 1}{2 ^{2026} - 1}$ . Using the complement rule, the probability that the product is even is $1 - \frac{2 ^{1013} - 1}{2 ^{2026} - 1} = \frac{2 ^{2026} - 1 - ( 2 ^{1013} - 1 )}{2 ^{2026} - 1} = \frac{2 ^{2026} - 2 ^{1013}}{2 ^{2026} - 1}$ . Factoring out $2^{1013}$ gives $\frac{2 ^{1013} ( 2 ^{1013} - 1 )}{2 ^{2026} - 1}$ .

Question 2 · Probability and Statistics · 2026 MCQ

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

n_{1} + n_{2} + n_{3} + n_{4} = 20

.
Which of the following is the probability that all of

n_{1}, n_{2}, n_{3}, n_{4}

in the provided solution are positive?

A.
$(3 19) / (3 23)$
B.
$(4 20) / (4 24)$
C.
$(3 20) / (3 23)$
D.
$(4 19) / (4 24)$

Correct Answer:

Step-by-Step Solution

Insight: This is a classic Stars and Bars problem. We need the ratio of positive integer solutions to non-negative integer solutions for the same equation.

Exam route: Total non-negative solutions to $n_{1} + n_{2} + n_{3} + n_{4} = 20$ is $(4 - 1 20 + 4 - 1) = (3 23)$ . For positive solutions ( $n_{i} \geq 1$ ), give 1 to each variable, leaving a sum of 16. The number of non-negative solutions for the remaining 16 is $(4 - 1 16 + 4 - 1) = (3 19)$ . The probability is the ratio $(3 19) / (3 23)$ .

Learning route: The equation is $n_{1} + n_{2} + n_{3} + n_{4} = 20$ . For non-negative integers ( $n_{i} \geq 0$ ), we use the Stars and Bars formula: $(k - 1 n + k - 1)$ . Here $n = 20, k = 4$ , so total solutions = $(3 23)$ . For positive integers ( $n_{i} \geq 1$ ), we can substitute $m_{i} = n_{i} - 1 \geq 0$ . The equation becomes $m_{1} + m_{2} + m_{3} + m_{4} = 16$ . The number of solutions is $(4 - 1 16 + 4 - 1) = (3 19)$ . Since the program provides a random non-negative solution, each of the $(3 23)$ solutions is equally likely. The probability that all are positive is the ratio of favorable to total solutions: $(3 19) / (3 23)$ .

Question 3 · Probability and Statistics · 2026 MSQ

Suppose a random variable

Z

follows

N or ma l (μ = 0, σ^{2} = 1)

distribution with probability density function

g (z)

and cumulative distribution function

G (z)

.
Another random variable

Y

follows

t_{1}

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?

A.
$G (0) = H (0)$
B.
$G (c) < H (c)$
C.
$G (- c) < H (- c)$
D.
$g (0) = h (0)$

Question 4 · Database Management and Warehousing · 2026 MCQ

Let R(A,B,C,D,E) be a relational schema with functional dependency set

F = {A \to B C, C D \to E, E \to A}

.
Which of the following statements is correct?

A.
AD, ED and CD are the only candidate keys of R.
B.
AD and ED are the only candidate keys of R.
C.
A, E and CD are the only candidate keys of R.
D.
A and CD are the only candidate keys of R.

Question 5 · Database Management and Warehousing · 2026 MCQ

Consider that the visualization of a 3-dimensional data cube is showing Sales Quantity for each combination of the attributes Product Type, Month and Country.
From this, if we want to further visualize the Sales Quantity for each combination of Product Type, Month and State, which of the following OLAP operations should be performed?

A.
Slicing
B.
Dicing
C.
Roll-up
D.
Drill-down

Question 6 · Database Management and Warehousing · 2026 MSQ

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?

A.
Inserting records into relation r
B.
Deleting records from relation s
C.
Deleting records from relation r
D.
Inserting records into relation s

Question 7 · 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 8 · Programming, Data Structures and Algorithms · 2026 MCQ

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?

A. [1]
[2]
[3]
B. [1]
[1, 2]
[3]
C. [1]
[2]
[1, 2, 3]
D. [1]
[1, 2]
[1, 3]

Correct Answer:

Step-by-Step Solution

Insight: Default arguments are evaluated exactly once at definition time; mutating a mutable default object persists its state across all subsequent calls that use the default.

Exam route: Call 1 uses the default empty list, appends 1, and returns [1]. The default list is now [1]. Call 2 uses that same default list, appends 2, and returns [1, 2]. Call 3 explicitly passes a new empty list, bypassing the default, appends 3, and returns [3].

Learning route: This is a classic mutable default argument question, recognizable by the lst=[] parameter. When Python executes the def statement, it creates a single list object and stores it as the default.

Step 1: In the first call append_to_lst(1), no list is provided, so lst binds to the default []. The append method mutates this default object to [1]. The function returns [1].

Step 2: In the second call append_to_lst(2), no list is provided, so lst binds to the exact same default object, which is now [1]. The append method mutates it to [1, 2]. The function returns [1, 2].

Step 3: In the third call append_to_lst(3, []), a new list is explicitly provided. The parameter lst binds to this new object, completely ignoring the default. The append method mutates the new list to [3]. The function returns [3].

The final output is [1], then [1, 2], then [3].

Answer: Option B.

Question 9 · Programming, Data Structures and Algorithms · 2026 MSQ

You are given the following Pre-order and In-order traversals of a Binary Tree T with nodes E, F, G, P, Q, R, S.
Pre-order: P Q S E R F G
In-order: S Q E P F R G
Which of the following statements is/are true about the Binary Tree T?

A.
Node P is the root of T
B. The Post-order traversal of T is:
S E Q F G R P
C.
Node Q has only one child
D.
The left subtree of node R contains the node G

Correct Answer:

["A","B"]

Step-by-Step Solution

Insight: Reconstruct the tree using the standard Pre-order and In-order split method.

Exam route: Root is P. Left In-order has 3 elements, Right has 3. Split Pre-order accordingly. Build left and right subtrees. Check options.

Learning route:

Step 1: Root is the first element in Pre-order: P.

Step 2: Find P in In-order. Left of P is S, Q, E (size 3). Right of P is F, R, G (size 3).

Step 3: Split remaining Pre-order (Q, S, E, R, F, G) into Left Pre-order (Q, S, E) and Right Pre-order (R, F, G).

Step 4: Left subtree: Pre (Q, S, E), In (S, Q, E). Root is Q. Left In is S, Right In is E. So Q has left child S, right child E.

Step 5: Right subtree: Pre (R, F, G), In (F, R, G). Root is R. Left In is F, Right In is G. So R has left child F, right child G.

Step 6: Evaluate options.

A) P is root. (True)

B) Post-order is Left Post + Right Post + Root. Left Post: S, E, Q. Right Post: F, G, R. Total: S, E, Q, F, G, R, P. (True)

C) Q has two children (S and E). (False)

D) G is in the right subtree of R. (False)

Question 10 · Machine Learning · 2026 MCQ

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?

A.
$θ = 0^{\circ}$
B.
$θ = 9 0^{\circ}$
C.
$9 0^{\circ} < θ \leq 18 0^{\circ}$
D.
$0 < θ < 9 0^{\circ}$

Question 11 · Machine Learning · 2026 MCQ

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

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

Question 12 · Machine Learning · 2026 MSQ

In the following table, the Task column lists a few tasks related to machine learning. The Algorithm column lists a few algorithms.
Each entry “t” from the Task column is to be matched with an appropriate entry “a” from the Algorithm column such that the task “t” can be solved using the algorithm “a”. Denote such 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 is/are the correct matching(s)?

A.
T1:A4, T2:A3, T3:A1, T4:A2
B.
T1:A2, T2:A4, T3:A1, T4:A3
C.
T1:A3, T2:A4, T3:A1, T4:A2
D.
T1:A4, T2:A2, T3:A1, T4:A3

Question 13 · Quantitative Aptitude · 2026 MCQ

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

A.
12
B.
14
C.
16
D.
18

Correct Answer:

Step-by-Step Solution

Insight: The product of three single digits is 70, so factorise 70 into three single-digit factors first; the sum follows immediately.

Exam route: $70 = 2 \times 5 \times 7$ . These are the only three single-digit factors (any other factorisation like $1 \times 7 \times 10$ uses a non-digit). Sum $= 2 + 5 + 7 = 14$ .

Learning route:

Step 1: Prime factorise $70 = 2 \times 5 \times 7$ .

Step 2: We need three single digits $a, b, c \in {1, 2, \dots, 9}$ such that $a \times b \times c = 70$ .

Step 3: Since $70 = 2 \times 5 \times 7$ , and all three are single digits, the only valid triple is ${2, 5, 7}$ . Any attempt to introduce a 1 (e.g. $1 \times 5 \times 14$ ) forces a factor $\geq 10$ , which is not a digit.

Step 4: Sum $= 2 + 5 + 7 = 14$ .

Trap: Using $1 \times 7 \times 10$ gives sum 18, but 10 is not a single digit.

Verification: $2 \times 5 \times 7 = 70$ ✓, and $2 + 5 + 7 = 14$ ✓.

Question 14 · Quantitative Aptitude · 2026 MCQ

Consider two distinct positive real numbers

m, n

, with

m > n

.
Let

x = n^{l o g_{10} (m)}

and

y = m^{l o g_{10} (n)}

. The relation between

x

and

y

is _______.

A.
$x > y$
B.
$x < y$
C.
$x = y$
D.
$x = lo g_{10} (y)$

Correct Answer:

Step-by-Step Solution

Insight: Taking the logarithm of both expressions reveals that their log values are identical due to the commutative property of multiplication.

Exam route: Apply $lo g_{10}$ to both $x$ and $y$ . Use the power rule to bring the exponents down. Compare the results.

Learning route:

Given $x = n^{l o g_{10} m}$ and $y = m^{l o g_{10} n}$ .

Take $lo g_{10}$ of both sides for $x$ :

$lo g_{10} x = lo g_{10} (n^{l o g_{10} m}) = (lo g_{10} m) \times (lo g_{10} n)$ .

Take $lo g_{10}$ of both sides for $y$ :

$lo g_{10} y = lo g_{10} (m^{l o g_{10} n}) = (lo g_{10} n) \times (lo g_{10} m)$ .

Since multiplication is commutative, $(lo g_{10} m) \times (lo g_{10} n) = (lo g_{10} n) \times (lo g_{10} m)$ .

Therefore, $lo g_{10} x = lo g_{10} y$ .

Since the logarithm function is one-to-one, this implies $x = y$ .

The relation is $x = y$ , matching option C.

Question 15 · Quantitative Aptitude · 2026 MCQ

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 _____

A.
0
B.
1
C.
2
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: This is a statistical constraint satisfaction question, recognisable because it asks to reconstruct a specific data set given its mean, median, and mode.

Step 1: Use the mean to find the sum. For 5 integers with a mean of 12, the sum is $5 \times 12 = 60$ .

Step 2: Use the median to fix the middle value. Sorting the 5 integers as $x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \leq x_{5}$ , the median is $x_{3}$ . Thus, $x_{3} = 18$ .

Step 3: Use the mode to fix the highest values. The mode is 20, meaning 20 must appear more frequently than any other number. Since $x_{3} = 18$ , the only way to include 20 is if $x_{4} = 20$ and $x_{5} = 20$ .

Step 4: Find the remaining sum. We have $x_{1} + x_{2} + 18 + 20 + 20 = 60$ , which simplifies to $x_{1} + x_{2} = 2$ .

Step 5: Find integer pairs for $x_{1}$ and $x_{2}$ . Since the integers are from 0 to 20 and $x_{1} \leq x_{2} \leq 18$ , the possible pairs summing to 2 are $(0, 2)$ and $(1, 1)$ .

Step 6: Apply the "single mode" constraint. If the pair is $(1, 1)$ , the set is ${1, 1, 18, 20, 20}$ , which has two modes (1 and 20). This violates the single mode condition. If the pair is $(0, 2)$ , the set is ${0, 2, 18, 20, 20}$ , which has a single mode of 20.

Answer: There is exactly 1 valid way to pick these integers.

Question 16 · Linear Algebra · 2026 MCQ

Let

M = (cos θ s in θ - s in θ cos θ)

be a

2 \times 2

matrix, where

θ = \frac{2 π}{5}

, and

I_{2} = (1001)

.
Which of the following options is equal to

M^{2026}

A.
$M^{2}$
B.
$M$
C.
$M^{- 1}$
D.
$I_{2}$

Correct Answer:

Step-by-Step Solution

Insight: $M = R (2 π /5)$ is a rotation by $2 π /5$ , so $M^{5} = I$ . Reduce the exponent $2026$ modulo $5$ .

Exam route: $2026 = 5 \times 405 + 1$ , so $M^{2026} = (M^{5})^{405} \cdot M^{1} = I^{405} \cdot M = M$ .

Learning route:

Identify $M$ as the standard counter-clockwise rotation matrix $R (θ)$ with $θ = \frac{2 π}{5}$ .
A rotation by $θ$ raised to the $n$ -th power is a rotation by $n θ$ : $M^{n} = R (n θ)$ .
Compute $n θ = 2026 \times \frac{2 π}{5} = \frac{4052 π}{5}$ .
Reduce modulo $2 π$ : $\frac{4052}{5} = 810 + \frac{2}{5}$ , so $\frac{4052 π}{5} = 810 π + \frac{2 π}{5} = 405 (2 π) + \frac{2 π}{5}$ .
The effective angle is $\frac{2 π}{5} = θ$ , so $M^{2026} = R (θ) = M$ .
Alternatively, $M^{5} = R (5 \cdot 2 π /5) = R (2 π) = I$ . Then $2026 mod 5 = 1$ , so $M^{2026} = M^{1} = M$ .

Question 17 · Linear Algebra · 2026 MCQ

Consider a set

S_{1} = {x = (x_{1}, x_{2}, x_{3})^{T} \in R^{3} ∣ x^{T} x \leq 16}

. Let

S_{2}

be another set which is a subspace of

R^{3}

with dimension two.
Which of the following gives the area of

S_{1} \cap S_{2}

A.
$16 π$
B.
$4 π$
C.
$4 π^{2}$
D.
$16 π^{2}$

Correct Answer:

Step-by-Step Solution

Insight: The intersection of a solid n-dimensional ball with a k-dimensional subspace passing through the origin is a solid k-dimensional ball of the exact same radius.

Exam route: $x^{T} x \leq 16$ means $∥ x ∥_{2} \leq 4$ , so it's a solid 3D ball of radius $R = 4$ . $S_{2}$ is a 2D subspace (a plane through the origin). The intersection is a 2D solid disk of radius 4. Area = $π R^{2} = 16 π$ .

Learning route:

Identify $S_{1}$ : The condition $x^{T} x \leq 16$ is equivalent to $∥ x ∥_{2}^{2} \leq 16$ , which means $∥ x ∥_{2} \leq 4$ . This defines a solid 3-dimensional $L_{2}$ ball centered at the origin with radius $R = 4$ .
Identify $S_{2}$ : A subspace of $R^{3}$ with dimension 2 is a flat plane that passes exactly through the origin.
Find the intersection $S_{1} \cap S_{2}$ : Slicing a solid 3D sphere with a plane that passes through its exact center (the origin) yields a solid 2D disk (a "great disk") with the same radius $R = 4$ .
Calculate the area: The area of a 2D disk of radius $R$ is $π R^{2}$ . Substituting $R = 4$ , we get Area = $π (4)^{2} = 16 π$ .

Question 18 · Linear Algebra · 2026 MCQ

Let

γ_{1}, γ_{2}, γ_{3}

be the eigenvalues of the matrix

100 0 cos t - s in t 0 s in t cos t

, where

t \in [- π, π]

is in radians.

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

t

satisfying

γ_{1} + γ_{2} + γ_{3} = 1 + 2

A.
${\frac{π}{3}, - \frac{π}{4}}$
B.
${\frac{π}{4}, - \frac{π}{3}}$
C.
${\frac{π}{4}, - \frac{π}{4}}$
D.
${\frac{π}{3}, - \frac{π}{3}}$

Correct Answer:

Step-by-Step Solution

Insight: The matrix is block-diagonal — a $1 \times 1$ block $[1]$ and a $2 \times 2$ rotation block — so the eigenvalues are the union of the blocks' eigenvalues.

Exam route: Eigenvalues of the $1 \times 1$ block: ${1}$ . Eigenvalues of the $2 \times 2$ block $[cos t - sin t sin t cos t]$ : solve $(cos t - λ)^{2} + sin^{2} t = 0 \Rightarrow λ = cos t \pm i sin t = e^{\pm i t}$ . Sum of all three eigenvalues (the trace) is $1 + 2 cos t$ . Set $1 + 2 cos t = 1 + 2 \Rightarrow cos t = \frac{1}{2}$ . For $t \in [- π, π]$ , this gives $t = \pm \frac{π}{4}$ .

Learning route:

Identify the block structure: $A = diag (1, R (- t))$ where $R (- t)$ is a clockwise rotation by $t$ .
The spectrum of a block-diagonal matrix is the union of the spectra of its blocks.
Block 1 contributes $γ_{1} = 1$ .
Block 2 has characteristic polynomial $λ^{2} - 2 cos t \cdot λ + 1 = 0$ , giving $γ_{2} = e^{i t}, γ_{3} = e^{- i t}$ .
Sum: $γ_{1} + γ_{2} + γ_{3} = 1 + e^{i t} + e^{- i t} = 1 + 2 cos t$ .
Equation: $1 + 2 cos t = 1 + 2 \Rightarrow cos t = \frac{2}{2}$ .
In $[- π, π]$ , $cos t = \frac{2}{2}$ has exactly two solutions: $t = \frac{π}{4}$ and $t = - \frac{π}{4}$ .
The set of all such $t$ is ${\frac{π}{4}, - \frac{π}{4}}$ , matching option C.

Question 19 · Artificial Intelligence · 2026 MCQ

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

A.
Breadth First Search
B.
Depth First Search
C.
A* Search
D.
Depth-limited Search

Question 20 · Artificial Intelligence · 2026 MCQ

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

A.
$\forall x \forall y C l a ss ma t e (x, y) \Rightarrow C l a ss ma t e (y, x)$
B.
$\forall x L ik es (x, I cecr e am) \Rightarrow \neg\exists x \neg L ik es (x, I cecr e am)$
C.
“Each king is a person” is equivalent to $\forall x I sK in g (x) \land I s P er so n (x)$
D.
“All humans are mortal” is equivalent to $\forall x I sH u man (x) \Rightarrow I s M or t a l (x)$

Question 21 · Artificial Intelligence · 2026 MSQ

Sentence

X

is said to entail Sentence

Y

if whenever

X

T R U E

Y

also must hold

T R U E

.
Which of the following statements is/are correct if

X

entails

Y

A.
$X \Rightarrow Y$
B.
$X \land \neg Y$ is $F A L S E$
C.
if $X$ then $Y$
D.
if $Y$ then $X$

Question 22 · Verbal Aptitude · 2026 MCQ

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

A.
Innocence
B.
Respect
C.
Solace
D.
Wealth

Correct Answer:

Step-by-Step Solution

Insight: This is an antonymic analogy. The relationship in the first pair is direct opposition, which must be mirrored precisely in the second pair.

Exam route: "Verbosity" (excessive wordiness) is the direct opposite of "Brevity" (conciseness). Therefore, we need the direct opposite of "Insolence" (rude disrespect). "Respect" is the precise antonym.

Learning route:

Analyze Pair A: Verbosity : Brevity. Verbosity means using too many words; Brevity means using few. They are polar opposites regarding communication style.
Construct Bridge: Word A is the excessive form of a trait, and Word B is the deficient or opposite form of that same trait.
Apply to Pair B: Insolence means rude arrogance or disrespectful behavior. We need the word representing polite modesty or regard for others.
Evaluate options:

Innocence: Lack of guilt or wrongdoing. Not the direct opposite of rudeness.
Respect: Regard for the feelings, wishes, or rights of others. This is the direct opposite of insolence.
Solace: Comfort in a time of distress. Unrelated to social conduct.
Wealth: Abundance of valuable resources. Unrelated domain.

Conclusion: "Respect" perfectly mirrors the antonymic relationship established in the first pair.

Question 23 · Verbal Aptitude · 2026 MCQ

‘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?

A.
The actor will certainly sponsor his next movie.
B.
His latest movie was a commercial success.
C.
The actor made enough money from his latest movie.
D.
His latest movie was not commercially successful.

Correct Answer:

Step-by-Step Solution

Insight: The sentence uses a counterfactual conditional ("If ... had been ..., ... would have ..."), which logically implies that the condition did not occur and the result did not happen.

Exam route: Identify the grammatical markers "had been" and "would have". This immediately signals a counterfactual. The reality is the opposite of the condition: the movie was NOT a commercial success. Match this directly to Option D.

Learning route:

Recognize the structure: "If P had been true, Q would have happened."
Understand counterfactuals: This specific grammatical construction is used to describe hypothetical situations in the past that did not actually occur. Therefore, P is false, and Q is false.
Apply to the text: P = "latest movie was a commercial success". Since the sentence says "had been", the reality is that the latest movie was NOT commercially successful. Consequently, Q (making enough money) also did not happen.
Evaluate options: Option D directly states this unstated reality. Options A, B, and C all mistakenly treat the hypothetical condition or result as factual.

Question 24 · Verbal Aptitude · 2026 MCQ

‘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.

A.
at; of; for
B.
for; at; of
C.
of; for; in
D.
in; of; for

Correct Answer:

Step-by-Step Solution

Insight: This is a multi-blank preposition question requiring independent analysis of spatial points, possession, and duration.

Exam route: Blank 1: "parted at the door" (specific point). Blank 2: "door of the cabin" (possession/association). Blank 3: "rented for the night" (duration). Sequence: at; of; for.

Learning route:

Step 1: Analyze the first blank. "Parted" happens at a specific physical location (the door). The preposition "at" is used for exact points.

Step 2: Analyze the second blank. The relationship between "door" and "cabin" is one of possession or association. The preposition "of" correctly denotes this relationship.

Step 3: Analyze the third blank. "The night" here refers to a duration of time for which the cabin was rented. The preposition "for" is used to denote a period or duration.

Step 4: Verify the sequence "at; of; for" against the options. Option A matches perfectly.

Question 25 · 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 26 · Calculus and Optimization · 2026 NAT

The value of $\sum_{i = 0}^{\infty} \sum_{j = 1}^{\infty} 2^{- i} 3^{- j}$ is ______________ . (Answer in integer)

Correct Answer:

1.00

Step-by-Step Solution

Insight: The summand $2^{- i} 3^{- j}$ is a product of a function of $i$ alone and a function of $j$ alone, so the double sum separates into the product of two independent geometric series.

Exam route: Split $\sum_{i} \sum_{j} 2^{- i} 3^{- j} = (\sum_{i = 0}^{\infty} 2^{- i}) (\sum_{j = 1}^{\infty} 3^{- j})$ . Evaluate each geometric series using $S = \frac{a}{1 - r}$ , then multiply.

Learning route:

This is a separable double summation question, recognisable because the general term $a_{i, j} = 2^{- i} \cdot 3^{- j}$ factors cleanly into $f (i) \cdot g (j)$ .

Step 1 — Apply the separation rule (Fubini for positive terms):

$i = 0 \sum \infty j = 1 \sum \infty 2^{- i} 3^{- j} = (i = 0 \sum \infty (\frac{1}{2})^{i}) \times (j = 1 \sum \infty (\frac{1}{3})^{j})$

Step 2 — First geometric series ( $i$ starts at $0$ , ratio $r = \frac{1}{2}$ ):

$S_{1} = \frac{1}{1 - \frac{1}{2}} = 2$

Step 3 — Second geometric series ( $j$ starts at $1$ , ratio $r = \frac{1}{3}$ , first term $a = \frac{1}{3}$ ):

$S_{2} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$

Step 4 — Multiply:

$S = 2 \times \frac{1}{2} = 1$

Trap check: A common mistake is to start the $j$ -sum at $0$ instead of $1$ , which would give $S_{2} = \frac{1}{1 - 1/3} = \frac{3}{2}$ and a wrong total of $3$ . Always read the lower limit carefully.

Verification: Evaluate the inner sum first: $\sum_{j = 1}^{\infty} 2^{- i} 3^{- j} = 2^{- i} \cdot \frac{1}{2}$ . Then $\sum_{i = 0}^{\infty} 2^{- i} \cdot \frac{1}{2} = \frac{1}{2} \cdot 2 = 1$ . Confirmed.

The answer is $1$ .

Question 27 · Analytical Aptitude · 2026 MCQ

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. The games were played 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, and
(iii) the boys won two games and the girls won one game.
The student who did not lose any game is __________.

A.
Pavan
B.
Rishi
C.
Swathi
D.
Tanvi

Correct Answer:

Step-by-Step Solution

Key idea: This is a constraint-based sequencing and elimination question, recognizable by the sequential games, elimination rules, and conditions on wins/classes.

Step 1: Understand the game format. There are 4 students and 3 games. A loser is eliminated. For all 4 students to play, the format must be: Game 1 (A vs B), Game 2 (Winner of G1 vs C), Game 3 (Winner of G2 vs D).

Step 2: Use condition (i). Game 1 is the ONLY game where two students of the same class played. The same-class pairs are (Rishi, Swathi) in Class 5 and (Pavan, Tanvi) in Class 4. Thus, Game 1 must be either (R, S) or (P, T).

Step 3: Test if Game 1 is (P, T). If Class 4 plays each other in G1, Class 4 wins G1. Condition (ii) states Class 5 won more games than Class 4. Since there are 3 games total, Class 5 must win 2 and Class 4 must win exactly 1. This means Class 5 must win G2 and G3. If Class 5 wins G2, the winner is R or S. Then Game 3 would be (R or S) vs the remaining (S or R), which is a same-class game. This violates condition (i). Thus, Game 1 cannot be (P, T).

Step 4: Test if Game 1 is (R, S). Game 1 is R vs S. Class 5 wins G1. To satisfy the quotas (Class 5 wins 2, Class 4 wins 1; Boys win 2, Girls win 1), Rishi (Boy) must win G1. If Swathi won, we would need 2 Boy wins from the remaining games, forcing Class 4 to win 2 games, violating (ii). So Rishi wins G1.

Step 5: Determine Game 2 and 3. Rishi plays Game 2 against Pavan or Tanvi. If Rishi loses G2, the Class 4 student wins, and Class 4 would end up winning 2 games total (violating ii). Thus, Rishi must win G2.

Step 6: Now Rishi has won G1 and G2 (2 Boy wins, 2 Class 5 wins). The quotas for Boys and Class 5 are fully met. Therefore, Game 3 must be won by a Girl from Class 4 to satisfy the remaining quotas (1 Girl win, 1 Class 4 win).

Step 7: Game 3 is Rishi vs Tanvi. Tanvi must win. Since Tanvi won her only game and didn't play before, she never lost any game.

Answer: D

Question 28 · Analytical Aptitude · 2026 MCQ

Assume that a Creative

(C)

person will Succeed

(S)

if the person is also Disciplined

(D)

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

(i)

C \land S \Leftrightarrow D

(ii)

C \Rightarrow (S ⟺ D)

(iii)

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

Which of the following options is correct?

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

Correct Answer:

Step-by-Step Solution

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

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

Learning route:

Step 1: Parse the sentence structure.

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

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

Step 2: Translate the "if" part.

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

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

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

Step 4: Combine 2 and 3.

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

Step 5: Apply the context.

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

Step 6: Evaluate the given statements.

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

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

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

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

Question 29 · Spatial Aptitude · 2026 MCQ

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

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

A.
P
B.
Q
C.
R
D.
S

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