Numerical Computation & Estimation, Data Structures, Algorithms, Vector… Test for GATE DA: 23 Questions with Solutions & Analysis

Attempt the Numerical Computation & Estimation, Data Structures, Algorithms, Vector… test for GATE DA: 23 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

23 questions · 31 marks · 0.9299999999999999 minutes. Programming, Data Structures and Algorithms: 4 · Probability and Statistics: 4 · Verbal Aptitude: 3 · Machine Learning: 2 · Linear Algebra: 2 · Database Management and Warehousing: 2 · Calculus and Optimization: 2 · Artificial Intelligence: 2 · Spatial Aptitude: 1 · Quantitative Aptitude: 1

Free sample questions from Numerical Computation & Estimation, Data Structures, Algorithms, Vector…

Question 1 · 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 2 · Programming, Data Structures and Algorithms · 2026 NAT

Let A be a sorted array containing 1000 distinct integers. You perform a recursive binary search on A to find an element y. Suppose each comparison checks whether the middle element computed during the current recursive step is equal to, less than, or greater than y.
The maximum number of comparisons that may have to be performed if y is not an element of A is _______ . (Answer in integer)

Correct Answer:

10.00

Step-by-Step Solution

Insight: A 3-way comparison counts as exactly 1 comparison per step. The max comparisons is simply the depth of the recursion tree.

Exam route: Use the formula $⌊ lo g_{2} n ⌋ + 1$ . For $n = 1000$ , $⌊ 9.96 ⌋ + 1 = 10$ .

Learning route:

The problem specifies a 3-way comparison (equal, less, greater). This counts as 1 comparison per recursive step.

The maximum number of comparisons is the maximum depth of the recursion tree, which occurs when the element is not found or is at the deepest leaf.

The exact formula for the maximum depth (number of comparisons) is $⌊ lo g_{2} n ⌋ + 1$ .

Given $n = 1000$ :

$lo g_{2} 1000 \approx 9.965$ .

$⌊ 9.965 ⌋ = 9$ .

Total comparisons = $9 + 1 = 10$ .

If the problem had specified two separate checks (e.g., if A[mid] == y then else if A[mid] < y), the worst case would be 2 comparisons per step, yielding 20. But the 3-way comparison explicitly avoids this trap.

Question 3 · Programming, Data Structures and Algorithms · 2024 MCQ

Let

F (n)

denote the maximum number of comparisons made while searching for
an entry in a sorted array of size

n

using binary search.
Which ONE of the following options is TRUE?

A.
$F (n) = F (⌊ n /2 ⌋) + 1$
B.
$F (n) = F (⌊ n /2 ⌋) + F (⌈ n /2 ⌉)$
C.
$F (n) = F (⌊ n /2 ⌋)$
D.
$F (n) = F (n - 1) + 1$

Correct Answer:

Step-by-Step Solution

Insight: Binary search divides the problem into one half, not both. The recurrence must reflect a single recursive call plus the current step.

Exam route: Identify the worst-case subproblem size as $⌊ n /2 ⌋$ . Add 1 for the current comparison. Match with $F (n) = F (⌊ n /2 ⌋) + 1$ .

Learning route:

In binary search, we compare the target with the middle element. This takes 1 comparison.

Then, we discard half the array and recurse on the other half.

The worst-case scenario is when the remaining half is as large as possible.

For an array of size $n$ , the two halves have sizes $⌊(n - 1) /2 ⌋$ and $⌈(n - 1) /2 ⌉$ . The maximum of these is $⌊ n /2 ⌋$ .

Thus, the maximum number of comparisons $F (n)$ satisfies the recurrence:

$F (n) = F (⌊ n /2 ⌋) + 1$ .

Option A matches this. Option B implies searching both halves. Option C misses the current comparison. Option D is for linear search.

Question 4 · Probability and Statistics · 2024 MCQ

Consider the following statements:
(i) The mean and variance of a Poisson random variable are equal.
(ii) For a standard normal random variable, the mean is zero and the
variance is one.
Which ONE of the following options is correct?

A.
Both (i) and (ii) are true
B.
(i) is true and (ii) is false
C.
(ii) is true and (i) is false
D.
Both (i) and (ii) are false

Correct Answer:

Step-by-Step Solution

Insight: This is a direct recall question testing the fundamental properties of two standard distributions: Poisson and Standard Normal.

Exam route: Poisson mean = variance = $λ$ . Standard Normal mean = 0, variance = 1. Both statements are true.

Learning route: This is a property verification question, recognizable because it lists basic statistical properties of standard distributions and asks to identify the true ones.

Step 1: Evaluate statement (i). The Poisson distribution models the number of events in a fixed interval. Its probability mass function is $P (X = k) = \frac{e ^{- λ} λ ^{k}}{k !}$ . The mean is $E (X) = λ$ and the variance is $V a r (X) = λ$ . Therefore, the mean and variance are equal. Statement (i) is true.

Step 2: Evaluate statement (ii). The standard normal distribution is a special case of the normal distribution where the mean $μ = 0$ and the variance $σ^{2} = 1$ . Its PDF is $f (z) = \frac{1}{2 π} e^{- z^{2} /2}$ . Therefore, the mean is zero and the variance is one. Statement (ii) is true.

Step 3: Combine the evaluations. Since both statements are true, the correct option is (A).

Question 5 · Probability and Statistics · 2025 MCQ

It is given that

P (X \geq 2) = 0.25

for an exponentially distributed random variable

X

with

E [X] = \frac{1}{λ}

, where

E [X]

denotes the expectation of

X

. What is the value of

λ

?
(ln denotes natural logarithm)

A.
$ln 2$
B.
$ln 4$
C.
$ln 3$
D.
$ln 0.25$

Correct Answer:

Step-by-Step Solution

Insight: Use the survival function $P (X \geq x) = e^{- λ x}$ . Setting $x = 2$ and equating to 0.25 allows direct extraction of $λ$ using natural logarithms.

Exam route: $e^{- 2 λ} = 0.25 = 1/4 ⟹ - 2 λ = ln (1/4) = - ln 4 = - 2 ln 2 ⟹ λ = ln 2$ .

Learning route: This is a tail probability parameter extraction question, recognizable because it gives a specific probability $P (X \geq x)$ and asks for the rate parameter $λ$ .

Step 1: Recall the survival function for an exponential distribution. The probability that $X$ is greater than or equal to $x$ is $P (X \geq x) = e^{- λ x}$ .

Step 2: Substitute the given values into the survival function. We are given $P (X \geq 2) = 0.25$ .

$e^{- 2 λ} = 0.25$

Step 3: Solve for $λ$ using natural logarithms. Note that $0.25 = \frac{1}{4}$ .

$- 2 λ = ln (\frac{1}{4})$

$- 2 λ = ln (1) - ln (4) = - ln (4)$

$- 2 λ = - 2 ln (2)$

$λ = ln (2)$

The correct option is (A).

Question 6 · Probability and Statistics · 2026 NAT

Let $X$ be an exponentially distributed random variable with mean $λ (> 0)$ . If $P (X > 5) = 0.35$ , then the conditional probability $P (X > 10∣ X > 5)$ is ___________ . (Rounded off to two decimal places)

Correct Answer:

0.35

Step-by-Step Solution

Insight: The exponential distribution is memoryless. The conditional probability $P (X > 10∣ X > 5)$ simply equals the unconditional probability $P (X > 10 - 5) = P (X > 5)$ , which is already given.

Exam route: By the memoryless property, $P (X > 10∣ X > 5) = P (X > 5) = 0.35$ .

Learning route: This is a conditional probability question for the exponential distribution, recognizable because it asks for $P (X > a ∣ X > b)$ . The memoryless property provides an instant shortcut.

Step 1: Identify the distribution and the given information. $X$ is exponentially distributed. We are given $P (X > 5) = 0.35$ . Note that the mean is given as $λ$ , which means the rate parameter is $1/ λ$ , but we don't actually need to find the rate parameter.

Step 2: Recognize the memoryless property. For any exponential random variable, the probability of waiting an additional time $t$ , given that you have already waited time $s$ , is the same as the probability of waiting time $t$ from the start.

Mathematically: $P (X > s + t ∣ X > s) = P (X > t)$ .

Step 3: Apply the property to the specific values. We need $P (X > 10∣ X > 5)$ . Here, $s = 5$ and $s + t = 10$ , which means $t = 5$ .

$P (X > 10∣ X > 5) = P (X > 5)$

Step 4: Substitute the given value. Since $P (X > 5) = 0.35$ , the conditional probability is also exactly 0.35.

Question 7 · 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 8 · 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 9 · Verbal Aptitude · 2024 MCQ

If ‘→’ denotes increasing order of intensity, then the meaning of the words
[sick → infirm → moribund] is analogous to [silly → _______ → daft].
Which one of the given options is appropriate to fill the blank?

A.
frown
B.
fawn
C.
vein
D.
vain

Correct Answer:

Step-by-Step Solution

Insight: This question tests semantic gradients, requiring a word that fits the increasing intensity of cognitive foolishness between "silly" and "daft", while strictly maintaining part-of-speech parallelism.

Exam route: Identify the progression in the first pair: sick (mild) → infirm (moderate) → moribund (extreme). Apply this to the second pair: silly (mild) → ? → daft (extreme). Evaluate options by part of speech. "Vain" is the only adjective among the choices, making it the only structurally viable bridge for a personal quality.

Learning route:

Analyze the source vector: "sick" (general unwellness) → "infirm" (significant weakness) → "moribund" (near death). This is a strict increasing intensity gradient of physical decline.
Analyze the target vector: "silly" (mild lack of sense) → [blank] → "daft" (extreme foolishness). We need an intermediate stage of cognitive inadequacy.
Evaluate options by grammatical category:

"frown": Noun/verb (facial expression). Fails part-of-speech match.
"fawn": Noun/verb (young deer / show affection). Fails part-of-speech match.
"vein": Noun (blood vessel / style). Fails part-of-speech match.
"vain": Adjective (having excessive pride or producing no result). In the context of this specific exam question, it is the only adjective provided, serving as the structural bridge for a personal quality.

Conclusion: "vain" is the only grammatically and structurally appropriate option to complete the analogy.

Question 10 · Machine Learning · 2024 MCQ

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

Column 1	Column 2
(p) Principal Component Analysis	(i) Discriminative Model
(q) Naïve Bayes Classification	(ii) Dimensionality Reduction
(r) Logistic Regression	(iii) Generative Model

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

Correct Answer:

Step-by-Step Solution

Insight: This is a model family classification question.

Exam route: PCA reduces dimensions (ii). Naive Bayes models joint probability (iii). Logistic Regression models conditional probability (i). Match them directly.

Learning route:

Step 1: Identify PCA. It projects data to a lower-dimensional space, preserving variance. It is a Dimensionality Reduction technique. So (p) -> (ii).

Step 2: Identify Naive Bayes. It calculates class priors and feature likelihoods, modeling $P (X ∣ Y) P (Y)$ . It is a Generative Model. So (q) -> (iii).

Step 3: Identify Logistic Regression. It directly models the decision boundary $P (Y ∣ X)$ without modeling the feature distribution. It is a Discriminative Model. So (r) -> (i).

Step 4: Combine the matches: (p)-(ii), (q)-(iii), (r)-(i). This corresponds to option C.

Question 11 · Machine Learning · 2024 NAT

Given the two-dimensional dataset consisting of 5 data points from two classes
(circles and squares) and assume that the Euclidean distance is used to measure the
distance between two points. The minimum odd value of

k

k

-nearest neighbor
algorithm for which the diamond (⋄) shaped data point is assigned the label square
is ______.

Correct Answer:

5.00

Step-by-Step Solution

Insight: The minimum odd $k$ is found by sorting the squared Euclidean distances from the query point to all training points and finding the smallest odd $k$ where the majority class among the $k$ nearest neighbors is the target class.

Exam route: Identify coordinates from the SVG. Query is at (2, 1). Calculate squared distances to all 5 points. Sort them. Count classes cumulatively for odd $k$ until "Square" wins.

Learning route:

Step 1: Extract coordinates from the SVG grid.

Circles are at (1, 1) and (1, 2).

Squares are at (2, 3), (3, 4), and (4, 3).

The query Diamond is at (2, 1).

Step 2: Calculate squared Euclidean distances from the query (2, 1) to each point:

To Circle (1, 1): $(2 - 1)^{2} + (1 - 1)^{2} = 1$
To Circle (1, 2): $(2 - 1)^{2} + (1 - 2)^{2} = 2$
To Square (2, 3): $(2 - 2)^{2} + (1 - 3)^{2} = 4$
To Square (4, 3): $(2 - 4)^{2} + (1 - 3)^{2} = 8$
To Square (3, 4): $(2 - 3)^{2} + (1 - 4)^{2} = 10$

Step 3: Sort the neighbors by ascending distance:

1st: Circle (dist $^{2}$ = 1)

2nd: Circle (dist $^{2}$ = 2)

3rd: Square (dist $^{2}$ = 4)

4th: Square (dist $^{2}$ = 8)

5th: Square (dist $^{2}$ = 10)

Step 4: Evaluate the majority class for odd values of $k$ :

For $k = 1$ : 1 Circle, 0 Squares $\to$ Label is Circle.
For $k = 3$ : 2 Circles, 1 Square $\to$ Label is Circle.
For $k = 5$ : 2 Circles, 3 Squares $\to$ Label is Square.

Step 5: The minimum odd value of $k$ for which the label is Square is 5.

Answer: 5.00

Question 12 · Linear Algebra · 2025 MSQ

Let $A = I_{n} + x x^{T}$ , where $I_{n}$ is the $n \times n$ identity matrix and $x \in R^{n}$ , $x^{T} x = 1$ . Which of the following options is/are correct?

A.
Rank of $A$ is $n$
B.
$A$ is invertible
C.
$0$ is an eigenvalue of $A$
D.
$A^{- 1}$ has a negative eigenvalue

Correct Answer:

["A","B"]

Step-by-Step Solution

Key idea: The eigenvalues of a rank-one update $I + x x^{T}$ can be found directly from the eigenvalues of the rank-one matrix $x x^{T}$ .

Step 1: Analyze the rank-one matrix $M = x x^{T}$ . For any vector $v$ orthogonal to $x$ (i.e., $x^{T} v = 0$ ), we have $M v = x (x^{T} v) = x (0) = 0$ . Since the orthogonal complement of $x$ in $R^{n}$ has dimension $n - 1$ , $λ = 0$ is an eigenvalue of $M$ with multiplicity at least $n - 1$ .

Step 2: For the vector $x$ itself, $M x = (x x^{T}) x = x (x^{T} x)$ . We are given $x^{T} x = 1$ , so $M x = 1 \cdot x$ . Thus, $λ = 1$ is an eigenvalue of $M$ with multiplicity 1.

Step 3: The eigenvalues of $A = I_{n} + M$ are simply the eigenvalues of $M$ shifted by 1 (since $I_{n} v = 1 \cdot v$ for all $v$ ).

Step 4: Therefore, the eigenvalues of $A$ are $1 + 0 = 1$ (with multiplicity $n - 1$ ) and $1 + 1 = 2$ (with multiplicity 1).

Step 5: Evaluate the options:

Since all eigenvalues of $A$ are non-zero (they are 1 and 2), $A$ is invertible. (Option B is correct)
An invertible $n \times n$ matrix has full rank, so the rank of $A$ is $n$ . (Option A is correct)
$0$ is not an eigenvalue of $A$ . (Option C is incorrect)
The eigenvalues of $A^{- 1}$ are the reciprocals of the eigenvalues of $A$ , which are $1/1 = 1$ and $1/2$ . Neither is negative. (Option D is incorrect)

Answer: Rank of $A$ is $n$ , and $A$ is invertible.

Question 13 · Linear Algebra · 2024 MSQ

Which of the following statements is/are TRUE?
Note:

R

denotes the set of real numbers.

A. There exist $M \in R^{3 \times 3}$ , $p \in R^{3}$ , and $q \in R^{3}$ such that $M x = p$ has a unique
solution and $M x = q$ has infinite solutions.
B. There exist $M \in R^{3 \times 3}$ , $p \in R^{3}$ , and $q \in R^{3}$ such that $M x = p$ has no solutions
and $M x = q$ has infinite solutions.
C. There exist $M \in R^{2 \times 3}$ , $p \in R^{2}$ , and $q \in R^{2}$ such that $M x = p$ has a unique
solution and $M x = q$ has infinite solutions.
D. There exist $M \in R^{3 \times 2}$ , $p \in R^{3}$ , and $q \in R^{3}$ such that $M x = p$ has a unique
solution and $M x = q$ has no solutions.

Correct Answer:

["B","D"]

Step-by-Step Solution

Key idea: This question tests the consistency conditions of linear systems $A x = b$ based on the shape of matrix $A$ (square, wide, or tall) and its rank.

Step 1: Analyze Option A ( $M \in R^{3 \times 3}$ ).

For a square matrix $M$ , if $M x = p$ has a unique solution, then $det (M) \neq = 0$ ( $M$ is invertible).

If $M$ is invertible, $M x = q$ must also have a unique solution ( $x = M^{- 1} q$ ) for ANY $q$ .

It is impossible for an invertible matrix to have infinite solutions for any RHS.

Thus, Option A is FALSE.

Step 2: Analyze Option B ( $M \in R^{3 \times 3}$ ).

Can $M x = p$ have no solution and $M x = q$ have infinite solutions?

Yes. Let $M$ be singular (Rank < 3).

Example: $M = 100010000$ .

Let $p = 001$ . System is inconsistent (no solution) because $0 \neq = 1$ .
Let $q = 000$ . System is $x = 0, y = 0, 0 = 0$ . $z$ is free. Infinite solutions.

Thus, Option B is TRUE.

Step 3: Analyze Option C ( $M \in R^{2 \times 3}$ ).

Can $M x = p$ have a unique solution?

$M$ has 3 columns (variables) and 2 rows (equations).

Max Rank is 2.

Number of free variables = $3 - Rank (M) \geq 3 - 2 = 1$ .

If a solution exists, there is at least 1 free variable, implying infinite solutions.

A unique solution is impossible for a wide matrix ( $n > m$ ).

Thus, Option C is FALSE.

Step 4: Analyze Option D ( $M \in R^{3 \times 2}$ ).

Can $M x = p$ have a unique solution?

$M$ has 2 columns. Max Rank is 2.

If Rank( $M$ )=2, the columns are linearly independent.

If $p$ is in the column space, the solution is unique (0 free variables).

Can $M x = q$ have no solution?

Yes, if $q$ is not in the column space (which is a 2D plane in $R^{3}$ ).

So, Unique Solution AND No Solution are both possible for different RHS vectors.

Thus, Option D is TRUE.

Answer: Options B and D are TRUE.

Question 14 · 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

Correct Answer:

Step-by-Step Solution

Insight: A rule that no single fragment can see must be re-checked by bringing the fragments back together — that repeated recombination is the price of losing dependency preservation. Exam route: Not preserved means some FD has its attributes split across fragments, so it cannot be validated inside one table. The DBMS must reconstruct the original tuples on every risky update, which is the join. Option C. Learning route: Step 1 — what dependency preservation means. After decomposing

R

into

R_{1}, R_{2}, \dots, R_{n}

, with

F_{i}

the dependencies that hold in

R_{i}

, the decomposition preserves dependencies when

(F_{1} \cup F_{2} \cup \dots \cup F_{n})^{+} = F^{+}

. In plain words: every original business rule can be checked by looking at the small tables individually. Step 2 — what actually breaks when it fails. Some original rule

X \to Y

has attributes spread over two or more fragments, so no single fragment contains all of

X \cup Y

. A constraint checker that scans one table at a time simply cannot see the rule. Step 3 — what the system must then do. To test the rule it has to line up rows of the fragments on their common attributes and evaluate the rule on the combined result. That reconstruction has to happen whenever an update could violate the rule, so it runs repeatedly and is computationally expensive — precisely the cost decomposition was meant to remove. Step 4 — why the other operators cannot do the job. Selection filters rows within one table; it cannot reach attributes living in another table. Projection removes columns, and in fact it is the operator that creates the fragments — it cannot reassemble anything. Set union stacks rows of two relations that must already share a schema, and fragments of a decomposition generally do not. None of the three can bring attributes from different tables into one working relation. Second pass from the trade-off table: a 3NF decomposition is always lossless and always dependency-preserving, whereas a BCNF decomposition is always lossless but may sacrifice preservation. The textbook cost quoted for that sacrifice is exactly this repeated recombination of fragments at update time. Same conclusion: option C. The tempting wrong path: picking Projection, because projection is the operator associated with decomposition in every worked example. Following that thought gives option B, and it breaks at one line — projection produces the fragments, it never re-tests a rule whose attributes sit in two different fragments. Generalisation: "preservation" is a statement about where a rule can be checked; if checking needs attributes from two tables, those tables must first be brought back together. Verification: if the decomposition had been dependency-preserving, every rule would be checkable inside a single fragment and no recombination would ever be needed, so the frequency of the recombining operator would be zero. The question is built entirely on that contrast, and only one operator performs recombination. Answer: C.

Question 15 · Database Management and Warehousing · 2024 MSQ

Given the relational schema

R = (U, V, W, X, Y, Z)

and the set of functional
dependencies:

{U \to V, U \to W, W X \to Y, W X \to Z, V \to X}

Which of the following functional dependencies can be derived from the above
set?

A.
$V W \to Y Z$
B.
$W X \to Y Z$
C.
$V W \to U$
D.
$V W \to Y$

Correct Answer:

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

Step-by-Step Solution

Insight: "Which FD can be derived?" is a closure question — $X \to Y$ follows from $F$ exactly when $Y \subseteq X^{+}$ .

Exam route: Close the left side of each option. $V W^{+} = {V, W, X, Y, Z}$ and $W X^{+} = {W, X, Y, Z}$ . $U$ never appears on any right-hand side, so no closure that lacks $U$ can ever contain $U$ . Options A, B and D follow; C does not.

Learning route:

Step 0 — the trigger. The phrase "can be derived from the above set" with no data instance means logical implication. Because Armstrong's axioms are sound and complete, the attribute closure is a decision procedure: it never misses a true rule and never invents a false one. So we close, we do not guess.

Step 1 — compute $W X^{+}$ . Start with ${W, X}$ . The rule $W X \to Y$ has its left side inside the set, so add $Y$ . The rule $W X \to Z$ also applies, so add $Z$ . Nothing else fires: $U \to V$ and $U \to W$ need $U$ , and $V \to X$ needs $V$ . Hence $W X^{+} = {W, X, Y, Z}$ . Since ${Y, Z} \subseteq W X^{+}$ , option B is derivable.

Step 2 — compute $V W^{+}$ . Start with ${V, W}$ . The rule $V \to X$ applies, giving ${V, W, X}$ . Now both $W X \to Y$ and $W X \to Z$ fire, giving ${V, W, X, Y, Z}$ . No rule produces $U$ , so the closure stops there: $V W^{+} = {V, W, X, Y, Z}$ . It contains $Y$ and it contains $Z$ , so option A ( $V W \to Y Z$ ) and option D ( $V W \to Y$ ) are both derivable.

Step 3 — option C. The attribute $U$ appears only on left-hand sides in $F$ . An attribute can enter a closure only as the right side of some rule, so $U$ can never be added unless it was in the starting set. $V W$ does not contain $U$ , therefore $U \in / V W^{+}$ and $V W \to U$ is not derivable.

Second pass by pure axioms (back-substitution): from $V \to X$ , augmentation with $W$ gives $V W \to X W$ ; transitivity with $W X \to Y$ gives $V W \to Y$ , and with $W X \to Z$ gives $V W \to Z$ ; union of those two gives $V W \to Y Z$ (option A), and decomposition of it gives $V W \to Y$ (option D). Option B is nothing but the union of the two given rules $W X \to Y$ and $W X \to Z$ . Option C would require running $U \to V$ and $U \to W$ backwards, and functional dependency has no converse.

The tempting wrong path: a student sees $U \to V$ and $U \to W$ , reasons "so $V$ and $W$ together must pin down $U$ ", and marks C as well. The break is at the very first line of that chain — implication is not reversible. Carrying it through produces the answer set A, B, C, D, which collapses the moment you notice that $U$ is on the right-hand side of nothing in $F$ .

Generalisation: if an option's right side contains an attribute that never appears on any right-hand side of $F$ , and the left side does not already contain it, the option is dead on sight.

Verification: $V W^{+} = {V, W, X, Y, Z}$ contains $Y$ and $Z$ , so A and D survive; $W X^{+} = {W, X, Y, Z}$ contains $Y$ and $Z$ , so B survives; neither closure contains $U$ , so C dies. Final answer: A, B, D.

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 · 2024 NAT

Let

f : R \to R

be the function

f (x) = \frac{1}{1 + e ^{- x}}

.
The value of the derivative of

f

x

where

f (x) = 0.4

is ______
(rounded off to two decimal places).
Note:

R

denotes the set of real numbers.

Correct Answer:

0.24

Step-by-Step Solution

Insight: The derivative of the sigmoid function $f (x)$ can be expressed entirely in terms of $f (x)$ itself: $f^{'} (x) = f (x) (1 - f (x))$ .

Exam route: Use the identity $f^{'} (x) = f (x) (1 - f (x))$ . Substitute $f (x) = 0.4$ to get $0.4 \times 0.6 = 0.24$ .

Learning route:

Given $f (x) = (1 + e^{- x})^{- 1}$ .

Using the chain rule, $f^{'} (x) = - 1 (1 + e^{- x})^{- 2} \cdot (- e^{- x}) = \frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}$ .

Rewrite this as $\frac{1}{1 + e ^{- x}} \cdot \frac{e ^{- x}}{1 + e ^{- x}}$ .

Notice that $\frac{e ^{- x}}{1 + e ^{- x}} = \frac{1 + e ^{- x} - 1}{1 + e ^{- x}} = 1 - \frac{1}{1 + e ^{- x}} = 1 - f (x)$ .

Thus, $f^{'} (x) = f (x) (1 - f (x))$ .

Given $f (x) = 0.4$ , we have $f^{'} (x) = 0.4 (1 - 0.4) = 0.4 \times 0.6 = 0.24$ .

Question 18 · 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 $\forall x I sH u man (x) \Rightarrow I s M or t a l (x)$

Correct Answer:

Step-by-Step Solution

Insight: "Each king is a person" requires a universal quantifier paired with implication, not conjunction.

Exam route: Check each option for logical validity. Option C uses $\forall x (I sK in g (x) \land I s P er so n (x))$ , which falsely claims everything in the universe is a king and a person. This is the only invalid statement.

Learning route:

Step 1: Analyze Option A. The symmetry of the classmate relation is correctly formalized as $\forall x \forall y (C l a ss ma t e (x, y) ⟹ C l a ss ma t e (y, x))$ . This is a valid representation.

Step 2: Analyze Option B. $\neg\exists x \neg L ik es (x, I cecr e am)$ is logically equivalent to $\forall xL ik es (x, I cecr e am)$ . Thus, the statement is $P ⟹ P$ , which is a tautology and always true.

Step 3: Analyze Option C. "Each king is a person" restricts the property of being a person to the subset of kings. The correct translation is $\forall x (I sK in g (x) ⟹ I s P er so n (x))$ . The given form $\forall x (I sK in g (x) \land I s P er so n (x))$ means "Everything in the universe is a king and a person", which is factually and logically incorrect for this translation.

Step 4: Analyze Option D. "All humans are mortal" is the classic example correctly translated as $\forall x (I sH u man (x) ⟹ I s M or t a l (x))$ .

Question 19 · Artificial Intelligence · 2026 MSQ

Let

P (x)

be a predicate.

Which of the following statements is/are NOT valid in first-order logic?

A.
$\forall x P (x) \Rightarrow \exists x P (x)$
B.
$\exists x P (x) \Rightarrow \forall x P (x)$
C.
$\exists x P (x) \Leftrightarrow \forall x P (x)$
D.
$\forall x P (x) \Rightarrow \exists x \neg P (x)$

Correct Answer:

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

Step-by-Step Solution

Insight: Validity requires a statement to be true in all non-empty domains. Existential does not imply universal, and universal contradicts existential negation.

Exam route: A is valid (non-empty domain assumption). B is invalid (counterexample: some but not all). C is invalid (same reason). D is invalid (contradiction if all are P).

Learning route:

Step 1: Recall that standard First-Order Logic assumes domains are non-empty.

Step 2: Evaluate Option A: $\forall x P (x) ⟹ \exists x P (x)$ . If everything in the domain has property $P$ , and the domain has at least one element, then there exists something with property $P$ . This is valid.

Step 3: Evaluate Option B: $\exists x P (x) ⟹ \forall x P (x)$ . Construct a counterexample. Let the domain be ${a, b}$ , with $P (a) = True$ and $P (b) = False$ . The antecedent $\exists x P (x)$ is True, but the consequent $\forall x P (x)$ is False. True $⟹$ False is False. Thus, not valid.

Step 4: Evaluate Option C: $\exists x P (x) \Leftrightarrow \forall x P (x)$ . Using the same counterexample, True $\Leftrightarrow$ False is False. Thus, not valid.

Step 5: Evaluate Option D: $\forall x P (x) ⟹ \exists x \neg P (x)$ . If $\forall x P (x)$ is True, then every element has property $P$ . This means $\exists x \neg P (x)$ must be False. True $⟹$ False is False. Thus, not valid.

Question 20 · Spatial Aptitude · 2024 MCQ

Visualize two identical right circular cones such that one is inverted over the other
and they share a common circular base. If a cutting plane passes through the vertices
of the assembled cones, what shape does the outer boundary of the
resulting cross-section make?

A.
A rhombus
B.
A triangle
C.
An ellipse
D.
A hexagon

Correct Answer:

Step-by-Step Solution

Insight: A vertical cut through the apex of a single right circular cone yields an isosceles triangle. Two identical cones base-to-base cut through both apexes yields two such triangles sharing a base.

Exam route: The cutting plane is vertical and passes through both vertices. For the top cone, this creates an isosceles triangle. For the bottom inverted cone, it creates an identical isosceles triangle. They share the common circular base's diameter. A quadrilateral with four equal sides (the slant heights of the identical cones) is a rhombus.

Learning route:

Step 1: Deconstruct the solid. The object is a double cone (two identical right circular cones sharing a common circular base).

Step 2: Define the cut. The plane is vertical and passes through both the top apex and the bottom apex.

Step 3: Analyze the top half. A vertical plane passing through the apex of a right circular cone intersects the base along a diameter, forming an isosceles triangle.

Step 4: Analyze the bottom half. By identical reasoning, the bottom inverted cone also yields an isosceles triangle.

Step 5: Combine. The two triangles share the common base (the diameter of the middle circle). Because the cones are identical, their slant heights are equal. Therefore, all four sides of the resulting quadrilateral are equal in length.

Step 6: Conclusion. A quadrilateral with four equal sides is a rhombus.

Question 21 · Quantitative Aptitude · 2024 MCQ

Let the minimum, maximum, mean and standard deviation values for the attribute
income of data scientists be ₹46000, ₹170000, ₹96000, and ₹21000, respectively.
The z-score normalized income value of ₹106000 is closest to which ONE of the
following options?

A.
0.217
B.
0.476
C.
0.623
D.
2.304

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct Z-score calculation question, recognisable because it provides the mean, standard deviation, and a specific data point, asking for its normalized value.

Step 1: Identify the given parameters. The raw value $x = 106000$ , the mean $μ = 96000$ , and the standard deviation $σ = 21000$ . (Note: The minimum and maximum values are distractors for Z-score normalization).

Step 2: Recall the Z-score formula, which measures how many standard deviations a value is from the mean: $z = \frac{x - μ}{σ}$ .

Step 3: Substitute the values into the formula: $z = \frac{106000 - 96000}{21000}$ .

Step 4: Simplify the numerator: $106000 - 96000 = 10000$ .

Step 5: Divide by the standard deviation: $z = \frac{10000}{21000} = \frac{10}{21} \approx 0.47619$ .

Answer: The closest option is 0.476.

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