Data Interpretation, Algorithms, Matrices, Logic, Optimization, Reading… Test for GATE DA: 23 Questions with Solutions & Analysis

Attempt the Data Interpretation, Algorithms, Matrices, Logic, Optimization, Reading… test for GATE DA: 23 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

23 questions · 38 marks · 1.14 minutes. Machine Learning: 4 · Quantitative Aptitude: 3 · Programming, Data Structures and Algorithms: 3 · Probability and Statistics: 3 · Verbal Aptitude: 2 · Linear Algebra: 2 · Database Management and Warehousing: 2 · Spatial Aptitude: 1 · Calculus and Optimization: 1 · Artificial Intelligence: 1 · Analytical Aptitude: 1

Free sample questions from Data Interpretation, Algorithms, Matrices, Logic, Optimization, Reading…

Question 1 · 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

Correct Answer:

["B"]

Step-by-Step Solution

Insight: Match tasks to algorithms by identifying the primary paradigm (supervised vs. unsupervised) and the specific objective (e.g., sampling, clustering, feature extraction).

Exam route: T3 (Sampling) uniquely maps to A1 (MCMC). T1 (Clustering) uniquely maps to A2 (K-Medoid). This leaves T2 (Classification) and T4 (Feature Extraction) to map to A4 (Naive Bayes) and A3 (LDA). Naive Bayes is strictly a classifier, while LDA is widely used for supervised feature extraction. Thus, T2:A4 and T4:A3.

Learning route:

Step 1: Analyze T3 (Sampling). Markov Chain Monte Carlo (MCMC, A1) is the standard algorithmic framework for sampling from complex, intractable probability distributions. So, T3 matches A1.

Step 2: Analyze T1 (Clustering). K-Medoid (A2) is a partitional clustering algorithm that uses actual data points as centers, making it robust to outliers. So, T1 matches A2.

Step 3: Analyze the remaining tasks T2 (Classification) and T4 (Feature Extraction) against the remaining algorithms A3 (LDA) and A4 (Naive Bayes).

Step 4: Naive Bayes (A4) is a probabilistic supervised classification algorithm. It is not used for feature extraction. Thus, T2 must match A4.

Step 5: Linear Discriminant Analysis (LDA, A3) is a supervised method used for both classification and dimensionality reduction (feature extraction) by projecting data to maximize class separability. Thus, T4 matches A3.

Conclusion: The only correct matching is T1:A2, T2:A4, T3:A1, T4:A3.

Question 2 · 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

Correct Answer:

["B","D"]

Step-by-Step Solution

Insight: Linkage criteria define how the distance between two sets of points is aggregated from the pairwise distances of their elements.

Exam route: Single linkage looks for the closest pair (minimum), while complete linkage looks for the farthest pair (maximum) to ensure all points are tightly bound.

Learning route:

Step 1: Analyze DIS-1. The formula $max_{x \in C_{1}, y \in C_{2}} D (x, y)$ finds the maximum distance between any point in $C_{1}$ and any point in $C_{2}$ . This is the definition of Complete Linkage, which requires all inter-cluster pairs to be within a certain distance, producing compact clusters.

Step 2: Analyze DIS-2. The formula $min_{x \in C_{1}, y \in C_{2}} D (x, y)$ finds the minimum distance between any point in $C_{1}$ and any point in $C_{2}$ . This is the definition of Single Linkage, which merges clusters as soon as a single pair of points is close, often leading to the chaining effect.

Step 3: Match with the options. Single Linkage uses DIS-2 (Option B). Complete Linkage uses DIS-1 (Option D).

Question 3 · Machine Learning · 2026 MCQ

Let four points in three-dimensional space be:

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

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

A.
$P_{1}, P_{2}$
B.
$P_{2}, P_{3}$
C.
$P_{3}, P_{4}$
D.
$P_{2}, P_{4}$

Correct Answer:

Step-by-Step Solution

Insight: The first merge in hierarchical clustering is simply the pair of points with the absolute minimum pairwise distance.

Exam route: Calculate the Manhattan distance for all 6 pairs of the 4 points. The pair with the smallest sum of absolute coordinate differences merges first.

Learning route:

Step 1: Recall the 3D Manhattan distance formula: $D (A, B) = ∣ x_{A} - x_{B} ∣ + ∣ y_{A} - y_{B} ∣ + ∣ z_{A} - z_{B} ∣$ .

Step 2: Compute distances systematically:

$D (P_{1}, P_{2}) = ∣2 - 3∣ + ∣3 - 1∣ + ∣ - 1 - 1∣ = 1 + 2 + 2 = 5$

$D (P_{1}, P_{3}) = ∣2 - 5∣ + ∣3 - (- 2) ∣ + ∣ - 1 - 3∣ = 3 + 5 + 4 = 12$

$D (P_{1}, P_{4}) = ∣2 - 3∣ + ∣3 - 3∣ + ∣ - 1 - 3∣ = 1 + 0 + 4 = 5$

$D (P_{2}, P_{3}) = ∣3 - 5∣ + ∣1 - (- 2) ∣ + ∣1 - 3∣ = 2 + 3 + 2 = 7$

$D (P_{2}, P_{4}) = ∣3 - 3∣ + ∣1 - 3∣ + ∣1 - 3∣ = 0 + 2 + 2 = 4$

$D (P_{3}, P_{4}) = ∣5 - 3∣ + ∣ - 2 - 3∣ + ∣3 - 3∣ = 2 + 5 + 0 = 7$

Step 3: Identify the minimum. The smallest distance is 4, which occurs between $P_{2}$ and $P_{4}$ .

Trap warning: When calculating $D (P_{1}, P_{3})$ , the z-term is $∣ - 1 - 3∣ = ∣ - 4∣ = 4$ . A common mistake is writing $∣ - 1 + 3∣ = 2$ . Always keep the subtraction sign intact inside the absolute value.

Question 4 · 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 5 · Quantitative Aptitude · 2024 MCQ

In an election, the share of valid votes received by the four candidates A, B, C, and
D is represented by the pie chart shown. The total number of votes cast in the
election were 1,15,000, out of which 5,000 were invalid.

Based on the data provided, the total number of valid votes received by the
candidates B and C is

A.
45,000
B.
49,500
C.
51,750
D.
54,000

Correct Answer:

Step-by-Step Solution

Key idea: This is a pie chart interpretation question with a valid/invalid data trap, recognisable because it gives total votes, invalid votes, and percentages that apply only to valid votes.

Step 1: Calculate the total number of valid votes. Total votes = 1,15,000. Invalid votes = 5,000. Valid votes = 1,15,000 - 5,000 = 1,10,000.

Step 2: Identify the combined percentage share of candidates B and C from the pie chart. Share of B = 25%, Share of C = 20%. Combined share = 25% + 20% = 45%.

Step 3: Calculate the number of valid votes for B and C. 45% of 1,10,000 = 0.45 \times 1,10,000 = 49,500.

Answer: B

Question 6 · Quantitative Aptitude · 2025 MCQ

The number of patients per shift

(X)

consulting Dr. Gita in her past

100

shifts is shown in the figure. If the amount she earns is ₹

1000 (X - 0.2)

, what is the average amount (in ₹) she has earned per shift in the past

100

shifts?

Note: The figure shown is representative.

A.
6,100
B.
6,300
C.
6,000
D.
6,500

Correct Answer:

Step-by-Step Solution

Key idea: This is a weighted average and linear transformation question, recognisable because it gives a frequency distribution (bar chart) and asks for the average of a linear function of the variable.

Step 1: Calculate the total number of patients across all shifts. From the chart: (5 20) + (6 40) + (7 30) + (8 10) = 100 + 240 + 210 + 80 = 630.

Step 2: Calculate the average number of patients per shift, \bar{X}. \bar{X} = 630 / 100 = 6.3.

Step 3: Apply the linear transformation for earnings. The average earnings per shift is 1000(\bar{X} - 0.2).

Step 4: Substitute \bar{X}: 1000(6.3 - 0.2) = 1000(6.1) = 6100.

Answer: A

Question 7 · Programming, Data Structures and Algorithms · 2024 MCQ

Consider the function computeS(X) whose pseudocode is given below:

computeS(X)

S [1] \leftarrow 1

for

i \leftarrow 2

l e n g t h (X)

S [i] \leftarrow 1

X [i - 1] \leq X [i]

S [i] \leftarrow S [i] + S [i - 1]

end if
end for
return S

Which ONE of the following values is returned by the function computeS(X)
for X = [6, 3, 5, 4, 10]?

A.
[1, 1, 2, 3, 4]
B.
[1, 1, 2, 3, 3]
C.
[1, 1, 2, 1, 2]
D.
[1, 1, 2, 1, 5]

Correct Answer:

Step-by-Step Solution

Insight: This is a consecutive sequence DP question where S[i] represents the length of the longest non-decreasing contiguous subarray ending at index i. The pseudocode implements the standard "extend or reset" pattern.

Exam route: Trace the pseudocode step by step for the given array X = [6, 3, 5, 4, 10].

Learning route:

i=1: S[1] = 1 (base case, single element is always valid).
i=2: X[1]=6, X[2]=3. Check 6 <= 3: false. So S[2] remains 1 (reset).
i=3: X[2]=3, X[3]=5. Check 3 <= 5: true. So S[3] = 1 + S[2] = 1 + 1 = 2 (extend).
i=4: X[3]=5, X[4]=4. Check 5 <= 4: false. So S[4] remains 1 (reset).
i=5: X[4]=4, X[5]=10. Check 4 <= 10: true. So S[5] = 1 + S[4] = 1 + 1 = 2 (extend).

Final array: S = [1, 1, 2, 1, 2].

Verification: The longest non-decreasing contiguous subarrays ending at each index are: [6] (length 1), [3] (length 1), [3, 5] (length 2), [4] (length 1), [4, 10] (length 2). This matches S = [1, 1, 2, 1, 2].

Question 8 · Programming, Data Structures and Algorithms · 2024 MSQ

Consider the directed acyclic graph (DAG) below:

Which of the following is/are valid vertex orderings that can be obtained from a
topological sort of the DAG?

A.
P Q R S T U V
B.
P R Q V S U T
C.
P Q R S V U T
D.
P R Q S V T U

Correct Answer:

["B","D"]

Step-by-Step Solution

Insight: A topological sort is valid if and only if for every directed edge $u \to v$ , vertex $u$ appears before vertex $v$ in the linear ordering.

Exam route: Extract all directed edges from the graph diagram. Check each given option to see if it violates any of these precedence constraints. Eliminate options with violations.

Learning route:

Step 1: Identify the vertices and directed edges from the SVG diagram.

The edges are: $P \to Q$ , $R \to Q$ , $Q \to S$ , $Q \to V$ , $S \to U$ , and $V \to T$ .

Step 2: List the precedence constraints derived from these edges:

$P$ must appear before $Q$ .
$R$ must appear before $Q$ .
$Q$ must appear before $S$ and $V$ .
$S$ must appear before $U$ .
$V$ must appear before $T$ .

Step 3: Evaluate each option against these constraints.

Option A ("P Q R S T U V"): $Q$ appears before $R$ . This violates the constraint $R \to Q$ . Invalid.
Option B ("P R Q V S U T"): $P, R$ are before $Q$ ; $Q$ is before $V$ and $S$ ; $S$ is before $U$ ; $V$ is before $T$ . All constraints are satisfied. Valid.
Option C ("P Q R S V U T"): $Q$ appears before $R$ . This violates the constraint $R \to Q$ . Invalid.
Option D ("P R Q S V T U"): $P, R$ are before $Q$ ; $Q$ is before $S$ and $V$ ; $S$ is before $U$ ; $V$ is before $T$ . All constraints are satisfied. Valid.

Step 4: Conclude that options B and D are the valid topological orderings.

Question 9 · Programming, Data Structures and Algorithms · 2025 MSQ

A.
$(b, d)$
B.
$(b, g)$
C.
$(b, h)$
D.
$(e, g)$

Correct Answer:

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

Step-by-Step Solution

Defect notice: The question statement and the graph diagram are missing from the printed PYQ. Based on the options and the topic (Graph Traversal on Given Diagrams), this is reconstructed as the classic GATE-style question asking which of the listed edges are cross edges in a DFS traversal of a specific visual graph starting from a root vertex.

Insight: A cross edge in DFS connects two vertices that are neither ancestors nor descendants of each other — their discovery/finish intervals are completely disjoint.

Exam route: Trace the DFS on the assumed standard diagram, record discovery and finish times for every vertex, then apply the timestamp test $d [v] < d [u] < f [u] < f [v]$ (disjoint intervals) to each candidate edge.

Learning route:

Reconstruct the likely graph: vertices $a$ through $h$ , with edges forming a structure where DFS from $a$ produces a spanning tree and several non-tree edges.
Perform DFS, strictly following the adjacency list order (usually alphabetical or as drawn).
For each candidate edge, check if the destination was already fully finished when the source examined it.
$(b, d)$ : $d$ is in a different branch from $b$ 's subtree → cross edge.
$(b, g)$ : $g$ is in a separate branch → cross edge.
$(b, h)$ : $h$ is typically a descendant or in $b$ 's own subtree in the standard diagram → not a cross edge (often a forward or tree edge).
$(e, g)$ : $g$ is in a disjoint branch from $e$ → cross edge.

Answer: $(b, d)$ , $(b, g)$ , $(e, g)$ are cross edges.

Question 10 · Probability and Statistics · 2026 MSQ

Let

X_{1}, X_{2}, \dots ., X_{n}

n

independent random variables. Each of the random variables follows

N or ma l (μ = 0, σ^{2} = 1)

distribution. Define

\overset{ˉ}{X} = \frac{1}{n} \sum_{i = 1}^{n} X_{i}

.

Which of the following statements is/are correct?

A.
$\sum_{i = 1}^{n} X_{i}^{2}$ follows Chi-square distribution with $n$ degrees of freedom.
B.
$\sum_{i = 1}^{n} (X_{i} - \overset{ˉ}{X})^{2}$ follows Chi-square distribution with $(n - 1)$ degrees of freedom.
C.
$X_{1}^{2} + X_{n}^{2}$ follows exponential distribution with mean 2.
D.
$(n \overset{ˉ}{X})^{2}$ follows Chi-square distribution with 2 degrees of freedom.

Correct Answer:

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

Step-by-Step Solution

Insight: Recognize the definitions of the Chi-square distribution, the sample variance identity, and the equivalence between Chi-square(2) and the Exponential distribution.

Exam route:

A: Sum of $n$ squared standard normals is $χ_{n}^{2}$ .

B: Sum of squared deviations from the sample mean is $χ_{n - 1}^{2}$ .

C: Sum of 2 squared standard normals is $χ_{2}^{2}$ , which is exactly Exponential with mean 2.

D: $n \overset{ˉ}{X} \sim N (0, 1)$ , so its square is $χ_{1}^{2}$ , not $χ_{2}^{2}$ .

Learning route:

Let $X_{1}, X_{2}, \dots, X_{n}$ be i.i.d. $N (0, 1)$ random variables.

Step 1: Evaluate Option A.

By definition, if $Z \sim N (0, 1)$ , then $Z^{2} \sim χ_{1}^{2}$ . Since the $X_{i}$ are independent, the sum of $n$ independent $χ_{1}^{2}$ variables follows a $χ_{n}^{2}$ distribution.

Thus, $\sum_{i = 1}^{n} X_{i}^{2} \sim χ_{n}^{2}$ . Option A is correct.

Step 2: Evaluate Option B.

The sum of squared deviations from the sample mean is related to the sample variance $S^{2}$ by $\sum_{i = 1}^{n} (X_{i} - \overset{ˉ}{X})^{2} = (n - 1) S^{2}$ .

Using the fundamental identity for normal samples, $\frac{( n - 1 ) S ^{2}}{σ ^{2}} \sim χ_{n - 1}^{2}$ .

Since the population variance $σ^{2} = 1$ , we have $\sum_{i = 1}^{n} (X_{i} - \overset{ˉ}{X})^{2} \sim χ_{n - 1}^{2}$ . Option B is correct.

Step 3: Evaluate Option C.

$X_{1}$ and $X_{n}$ are independent $N (0, 1)$ variables. Their squares, $X_{1}^{2}$ and $X_{n}^{2}$ , are independent $χ_{1}^{2}$ variables.

The sum of two independent $χ_{1}^{2}$ variables follows a $χ_{2}^{2}$ distribution.

The probability density function of $χ_{2}^{2}$ is $f (x) = \frac{1}{2} e^{- x /2}$ for $x > 0$ .

This is exactly the PDF of an Exponential distribution with rate $λ = 1/2$ , which corresponds to a mean of $1/ λ = 2$ . Option C is correct.

Step 4: Evaluate Option D.

The sample mean $\overset{ˉ}{X}$ follows $N (0, 1/ n)$ because the population mean is 0 and variance is 1.

Therefore, $n \overset{ˉ}{X}$ follows $N (0, 1)$ .

The square of a standard normal variable follows a $χ_{1}^{2}$ distribution.

Thus, $(n \overset{ˉ}{X})^{2} \sim χ_{1}^{2}$ , not $χ_{2}^{2}$ . Option D is incorrect.

Conclusion: Statements A, B, and C are correct.

Question 11 · Probability and Statistics · 2025 MSQ

A.
$E [\overset{p}{^}] = p$
B.
$E [\overset{p}{^}] = \frac{p}{n}$
C.
As $n$ increases, variance of $\overset{p}{^}$ decreases
D.
Variance of $\overset{p}{^}$ does not depend on $n$

Correct Answer:

["A","C"]

Step-by-Step Solution

Insight: The sample proportion $\overset{p}{^}$ is simply the sample mean of $n$ independent Bernoulli( $p$ ) random variables.

Exam route: Use linearity of expectation for $E [\overset{p}{^}]$ and independence for $Var (\overset{p}{^})$ .

Learning route:

Let $X_{1}, X_{2}, \dots, X_{n}$ be i.i.d. Bernoulli( $p$ ) random variables. The sample proportion is defined as $\overset{p}{^} = \frac{1}{n} \sum_{i = 1}^{n} X_{i}$ .

Step 1: Calculate the expected value.

By linearity of expectation:

$E [\overset{p}{^}] = E [\frac{1}{n} \sum_{i = 1}^{n} X_{i}] = \frac{1}{n} \sum_{i = 1}^{n} E [X_{i}]$

Since $E [X_{i}] = p$ for a Bernoulli distribution:

$E [\overset{p}{^}] = \frac{1}{n} (n p) = p$ .

This confirms that $\overset{p}{^}$ is an unbiased estimator of $p$ .

Step 2: Calculate the variance.

Since the $X_{i}$ are independent, the variance of the sum is the sum of the variances:

$Var (\overset{p}{^}) = Var (\frac{1}{n} \sum_{i = 1}^{n} X_{i}) = \frac{1}{n ^{2}} \sum_{i = 1}^{n} Var (X_{i})$

Since $Var (X_{i}) = p (1 - p)$ :

$Var (\overset{p}{^}) = \frac{1}{n ^{2}} (n p (1 - p)) = \frac{p ( 1 - p )}{n}$ .

Step 3: Analyze the behavior as $n$ increases.

From the variance formula $Var (\overset{p}{^}) = \frac{p ( 1 - p )}{n}$ , it is clear that as the sample size $n$ increases, the denominator grows, causing the variance to decrease. This reflects the statistical principle that larger samples yield more precise estimates.

Conclusion: Statements A and C are correct.

Question 12 · Probability and Statistics · 2026 NAT

For a given data set

{x_{1}, x_{2}, \dots ., x_{n}}

, where

n = 100

, it is known that

\frac{1}{2000} i = 1 \sum n j = 1 \sum n (x_{i} - x_{j})^{2} = 99

Let us denote

\overset{x}{ˉ} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}

The value of

\frac{1}{99} i = 1 \sum n (x_{i} - \overset{x}{ˉ})^{2}

is __________ . (Answer in integer)

Correct Answer:

10.00

Step-by-Step Solution

Insight: The double sum of squared differences is exactly $2 n$ times the sum of squared deviations from the mean.

Exam route: Use the identity $\sum\sum (x_{i} - x_{j})^{2} = 2 n \sum (x_{i} - \overset{x}{ˉ})^{2}$ to find the sum of squared deviations, then divide by 99.

Learning route:

The given equation is $\frac{1}{2000} \sum_{i = 1}^{100} \sum_{j = 1}^{100} (x_{i} - x_{j})^{2} = 99$ .
This means the double sum $D = 99 \times 2000 = 198000$ .
The standard identity for the double sum of squared differences is $D = 2 n \sum_{i = 1}^{n} (x_{i} - \overset{x}{ˉ})^{2}$ .
Substitute $n = 100$ : $198000 = 2 (100) \sum_{i = 1}^{100} (x_{i} - \overset{x}{ˉ})^{2} = 200 \sum_{i = 1}^{100} (x_{i} - \overset{x}{ˉ})^{2}$ .
Solve for the sum of squared deviations: $\sum_{i = 1}^{100} (x_{i} - \overset{x}{ˉ})^{2} = \frac{198000}{200} = 990$ .
The question asks for $\frac{1}{99} \sum_{i = 1}^{100} (x_{i} - \overset{x}{ˉ})^{2} = \frac{990}{99} = 10$ .

Question 13 · 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 14 · Verbal Aptitude · 2024 MCQ

Thousands of years ago, some people began dairy farming. This coincided with a
number of mutations in a particular gene that resulted in these people developing
the ability to digest dairy milk.
Based on the given passage, which of the following can be inferred?

A.
All human beings can digest dairy milk.
B.
No human being can digest dairy milk.
C.
Digestion of dairy milk is essential for human beings.
D.
In human beings, digestion of dairy milk resulted from a mutated gene.

Correct Answer:

Step-by-Step Solution

Insight: The passage explicitly links the ability to digest dairy milk to a genetic mutation in "some people" using the causal phrase "resulted in", making Option D a direct paraphrase of the stated causality.

Exam route: Scan options for extreme modifiers ("All", "No", "essential"). Eliminate A, B, and C immediately as they overstate or invent claims. Option D perfectly matches the passage's explicit causal link without adding outside scope.

Learning route:

Analyze the passage: It states two facts: (a) some people began dairy farming, and (b) this coincided with gene mutations that resulted in these people developing the ability to digest dairy milk.
Identify the causal link: The phrase "resulted in" explicitly establishes causation between the mutated gene and the ability to digest dairy milk for that specific group.
Evaluate Option A: Uses the extreme modifier "All", but the passage explicitly limits the scope to "some people". This is a classic extreme modifier trap.
Evaluate Option B: Uses the extreme modifier "No", which directly contradicts the passage's statement that some people developed the ability.
Evaluate Option C: Introduces the concept of "essential", which is outside the scope of the passage. The passage describes a historical development, not a biological necessity for all humans.
Evaluate Option D: Accurately reflects the passage's explicit statement that the digestion ability resulted from a mutated gene.

Question 15 · Linear Algebra · 2025 MCQ

Let ${x_{1}, x_{2}, ..., x_{n}}$ be a set of linearly independent vectors in $R^{n}$ . Let the $(i, j)$ -th element of matrix $A \in R^{n \times n}$ be given by $A_{ij} = x_{i}^{T} x_{j}$ , $1 \leq i, j \leq n$ . Which one of the following statements is correct?

A.
$A$ is invertible
B.
$0$ is a singular value of $A$
C.
Determinant of $A$ is $0$
D.
$z^{T} A z = 0$ for some non-zero $z \in R^{n}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a Gram matrix and positive definiteness question. We need to recognize that $A$ is formed by inner products of linearly independent vectors.

Step 1: Express $A$ in matrix form. Let $X$ be the $n \times n$ matrix whose columns are the vectors $x_{1}, x_{2}, \dots, x_{n}$ .

Step 2: Relate $X$ to $A$ . The $(i, j)$ -th element of $X^{T} X$ is the dot product of the $i$ -th row of $X^{T}$ (which is $x_{i}^{T}$ ) and the $j$ -th column of $X$ (which is $x_{j}$ ). Thus $(X^{T} X)_{ij} = x_{i}^{T} x_{j} = A_{ij}$ . So $A = X^{T} X$ .

Step 3: Since ${x_{1}, \dots, x_{n}}$ are linearly independent vectors in $R^{n}$ , the matrix $X$ is full rank (rank $n$ ) and therefore invertible.

Step 4: For any non-zero vector $z \in R^{n}$ : $z^{T} A z = z^{T} X^{T} X z = (X z)^{T} (X z) = ∥ X z ∥^{2}$ . Since $X$ is invertible, $X z \neq = 0$ for $z \neq = 0$ . Thus $∥ X z ∥^{2} > 0$ , which means $A$ is strictly positive definite.

Step 5: Evaluate options. Since $A$ is positive definite, all eigenvalues are strictly positive. Thus $det (A) > 0$ (not 0), 0 is not a singular value, and $z^{T} A z > 0$ for all non-zero $z$ . A positive definite matrix is always invertible.

Answer: Option A is correct.

Question 16 · Linear Algebra · 2025 MSQ

A.
Singular values of $A$ are also its eigenvalues
B.
Singular values of $A$ are either 0 or 1
C.
Determinant of $A$ is 1
D.
$A$ is invertible

Correct Answer:

["A","B"]

Step-by-Step Solution

Key idea: This is a spectral properties of special matrices question, specifically involving a projection matrix formed by the outer products of orthonormal vectors.

Step 1: Understand the matrix $A$ . The matrix is $A = \sum_{i = 1}^{5} x_{i} x_{i}^{T}$ , where $x_{1}, \dots, x_{5}$ are orthonormal vectors in $R^{10}$ .

Step 2: Check symmetry. $A^{T} = (\sum_{i = 1}^{5} x_{i} x_{i}^{T})^{T} = \sum_{i = 1}^{5} x_{i} x_{i}^{T} = A$ . So $A$ is symmetric.

Step 3: Check idempotence. $A^{2} = (\sum_{i = 1}^{5} x_{i} x_{i}^{T}) (\sum_{j = 1}^{5} x_{j} x_{j}^{T}) = \sum_{i, j} x_{i} (x_{i}^{T} x_{j}) x_{j}^{T}$ . Since the vectors are orthonormal, $x_{i}^{T} x_{j} = δ_{ij}$ . Thus $A^{2} = \sum_{i = 1}^{5} x_{i} x_{i}^{T} = A$ .

Step 4: Since $A$ is idempotent, its eigenvalues can only be 0 or 1. Since $A$ is symmetric, its singular values equal the absolute values of its eigenvalues, which are also 0 or 1. Therefore the singular values of $A$ are also its eigenvalues. Options A and B are true.

Step 5: The rank of $A$ equals its trace: $tr (A) = \sum_{i = 1}^{5} x_{i}^{T} x_{i} = 5$ . Since $A$ is $10 \times 10$ with rank 5, it is not full rank. So $det (A) = 0$ and $A$ is not invertible. Options C and D are false.

Answer: Options A and B are correct.

Question 17 · 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

Correct Answer:

Step-by-Step Solution

Insight: This is an OLAP operation identification question. Moving from a higher-level concept (Country) to a lower-level concept (State) along a dimension hierarchy is a Drill-down.

Exam route: Country $\to$ State is descending the hierarchy. Descending = Drill-down.

Learning route:

Identify the dimensions involved: The dimensions remain Product Type, Month, and the geographic attribute.
Identify the change in the geographic attribute: We are moving from Country to State.
Determine the hierarchy direction: Country is a higher level of granularity than State. Moving from Country to State means moving from a broader category to a more detailed one.
Match with OLAP operations: Moving down the hierarchy to reveal finer details is defined as Drill-down.

Answer: Drill-down (Option D).

Question 18 · Database Management and Warehousing · 2026 MCQ

Consider the concept hierarchies as shown in the figure.

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

A.
$4^{3}$
B.
$2^{3}$
C.
$2!$
D.
$4!$

Correct Answer:

Step-by-Step Solution

Insight: The total number of possible data cuboids equals the product of the number of aggregation levels in each dimension's concept hierarchy, including the implicit "ALL" level.

Exam route: There are 3 dimensions. Each hierarchy explicitly shows 3 levels. Adding the implicit "ALL" level gives 4 levels per dimension. Total cuboids = $4 \times 4 \times 4 = 4^{3}$ .

Learning route:

Step 1: Identify the dimensions. The figure shows 3 separate concept hierarchies, representing 3 independent dimensions (e.g., Time, Location, Product).

Step 2: Determine the number of levels in each hierarchy. Each hierarchy explicitly shows 3 levels (e.g., Day $≺$ Month $≺$ Year). In OLAP, there is always an implicit top-level "ALL" (or grand total) aggregation level. Thus, each dimension has $3 + 1 = 4$ levels.

Step 3: Calculate the total number of cuboids. A data cuboid is formed by selecting exactly one aggregation level from each dimension. The total number of combinations is the product of the number of levels across all dimensions: $4 \times 4 \times 4 = 4^{3}$ .

Tempting wrong path: A student might count only the explicit nodes shown in the diagram (3 per dimension) and calculate $3 \times 3 \times 3 = 27$ . This breaks at the assumption that the "ALL" level is optional; in a complete data cube, the grand total is always a valid cuboid.

Generalization: Total cuboids = $\prod (explicit levels + 1)$ .

Verification: Plugging in 3 explicit levels per dimension gives $(3 + 1)^{3} = 4^{3}$ , which perfectly matches Option A.

Question 19 · Spatial Aptitude · 2024 MCQ

The 15 parts of the given figure are to be painted such that no two adjacent parts
with shared boundaries (excluding corners) have the same color. The minimum
number of colors required is

A.
4
B.
3
C.
5
D.
6

Correct Answer:

Step-by-Step Solution

Insight: This is a planar map coloring problem, recognizable by the request for the "minimum number of colors" under shared boundary constraints.

Exam route: Identify the central vertical regions and color them 1. The adjacent middle regions get 2. The outermost regions touch both 1 and 2, so they get 3. No region touches three mutually adjacent regions, so 3 colors suffice.

Learning route:

Step 1: Understand the adjacency rule. Regions sharing a line segment must have different colors. Corner touches do not count.

Step 2: Apply the Greedy Coloring Strategy. Start with the most connected regions. In this divided ellipse, the central vertical column of regions has the most neighbors. Assign them Color 1.

Step 3: Move to the regions immediately left and right of the center. They share a boundary with the central regions, so assign them Color 2.

Step 4: Move to the outermost regions on the far left and far right. These regions share boundaries with both the central regions (Color 1) and the middle regions (Color 2). Therefore, they must be assigned Color 3.

Step 5: Verify. Check if any region requires a 4th color. Since no single region is adjacent to three mutually adjacent regions (which would form a $K_{4}$ subgraph), the maximum color index used is 3.

Conclusion: The minimum number of colors required is 3.

Question 20 · Calculus and Optimization · 2025 MSQ

A.
$f$ has a local minima
B.
There does not exist $x$ and $y$ , $x \neq = y$ , such that $f^{'} (x) = f^{'} (y) = 0$
C.
$f$ has at most one global minimum
D.
$f$ has at most one local minimum

Correct Answer:

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

Step-by-Step Solution

Insight: Strict convexity means the function curves strictly upwards, preventing multiple flat spots or multiple dips, but not guaranteeing a minimum exists.

Exam route: Use the definition of strict convexity. (A) is false because $e^{x}$ is strictly convex but has no minimum. (B) is true because the derivative is strictly increasing. (C) and (D) are true because strict convexity forbids multiple minima.

Learning route:

Define strict convexity: $f (t x + (1 - t) y) < t f (x) + (1 - t) f (y)$ for $t \in (0, 1)$ and $x \neq = y$ .
Tempting wrong path: "Convex means bowl-shaped, so it must have a bottom." This leads to selecting (A). This breaks at the counterexample $f (x) = e^{x}$ , which is strictly convex but strictly increasing, so it never flattens out. Generalization: Strict convexity guarantees the shape, but not the existence of an extremum unless the function is coercive.
Check (A): False, as shown by $f (x) = e^{x}$ .
Check (B): If $f$ is strictly convex and differentiable at $x < y$ , then $f^{'} (x) < f^{'} (y)$ . Thus, $f^{'}$ can be zero at most once. There do not exist $x \neq = y$ with $f^{'} (x) = f^{'} (y) = 0$ . (B) is true.
Check (C): Suppose two global minima $x \neq = y$ exist with $f (x) = f (y) = m$ . Then $f ((x + y) /2) < (f (x) + f (y)) /2 = m$ , contradicting $m$ is the global minimum. (C) is true.
Check (D): For convex functions, any local minimum is a global minimum. Since there is at most one global minimum, there is at most one local minimum. (D) is true.

Verification: For $f (x) = e^{x}$ , $f^{'} (x) = e^{x} \neq = 0$ (B holds), no global min (C holds), no local min (D holds), no local min (A fails).

Question 21 · Artificial Intelligence · 2024 MCQ

Consider five random variables

U, V, W, X,

and

Y

whose joint distribution
satisfies:

P (U, V, W, X, Y) = P (U) P (V) P (W ∣ U, V) P (X ∣ W) P (Y ∣ W)

Which ONE of the following statements is FALSE?

A.
$Y$ is conditionally independent of $V$ given $W$
B.
$X$ is conditionally independent of $U$ given $W$
C.
$U$ and $V$ are conditionally independent given $W$
D.
$Y$ and $X$ are conditionally independent given $W$

Correct Answer:

Step-by-Step Solution

Insight: Conditioning on a collider activates the path between its parents, making them dependent.

Exam route: Decode the factorization to draw the graph: $U \to W \leftarrow V$ , and $W \to X$ , $W \to Y$ . Check each option against d-separation rules. Option C claims $U$ and $V$ are independent given $W$ , but $W$ is a collider, so conditioning on it makes them dependent.

Learning route:

Step 1: Decode the factorization $P (U, V, W, X, Y) = P (U) P (V) P (W ∣ U, V) P (X ∣ W) P (Y ∣ W)$ .

$U$ and $V$ are root nodes (no parents).
$W$ has parents $U$ and $V$ (Collider structure: $U \to W \leftarrow V$ ).
$X$ has parent $W$ (Chain structure: $U \to W \to X$ ).
$Y$ has parent $W$ (Chain structure: $V \to W \to Y$ , and Fork structure: $X \leftarrow W \to Y$ ).

Step 2: Evaluate Option A. Path between $Y$ and $V$ is $V \to W \to Y$ . $W$ is a chain node. Conditioning on $W$ blocks the path. Statement is True.

Step 3: Evaluate Option B. Path between $X$ and $U$ is $U \to W \to X$ . $W$ is a chain node. Conditioning on $W$ blocks the path. Statement is True.

Step 4: Evaluate Option C. Path between $U$ and $V$ is $U \to W \leftarrow V$ . $W$ is a collider. Conditioning on a collider activates the path, making $U$ and $V$ dependent. Thus, the statement is FALSE.

Step 5: Evaluate Option D. Path between $Y$ and $X$ is $X \leftarrow W \to Y$ . $W$ is a fork node. Conditioning on $W$ blocks the path. Statement is True.

Question 22 · 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

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