GATE DA 2025 — Solved Walkthrough
A curated set of fully-worked GATE DA 2025 problems across every subject — see exactly how each concept turns into an exam question and its verified answer.
One paper is a single sample, and the last lesson warned that a single sample can mislead. Here is the second data point. GATE DA 2025 — the stream’s second-ever paper — is where the exam’s habits start to show through the noise: read it beside 2024 and you begin to see which corners of the syllabus the setters return to every year, which traps they reuse, and where the real difficulty consistently sits. Probability rewards Bayes and total-expectation again; linear algebra again hides its answer in an eigenvalue observation; and again, almost every problem turns on recognising the shortcut rather than grinding. Watch the pattern firm up.
Probability & Statistics
Bayes with three unequal-prior boxes. Three boxes hold white and black balls with priors (1/2, 1/6, 1/3). A white ball is drawn. Find P(Box 2 | white).
Apply Bayes with total probability in the denominator — three terms, one per box:
P(Box 2 | white) = P(white | Box 2)·P(Box 2)
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Σ over all three boxes of P(white | Boxᵢ)·P(Boxᵢ)Plugging the box compositions and the priors (1/2, 1/6, 1/3) and simplifying gives 0.25. The recipe never changes: write priors and likelihoods, build the evidence denominator by total probability, divide.
Answer: P(Box 2 | white) = 0.25.
→ Taught in Bayes’ Theorem
Law of total expectation. A joint setup asks for E[ E[X | Y] ]. Evaluate it.
This is a one-line recognition problem. Averaging the conditional expectation E[X | Y] over Y recovers the plain expectation of X:
E[ E[X | Y] ] = E[X] (law of total expectation)Spot the nested expectation and you skip all the table algebra — the answer is just E[X], no computation needed.
Answer: E[ E[X | Y] ] = E[X].
→ Taught in Joint, Marginal & Conditional Distributions
CLT on a sum of 300 Bernoulli(0.25). Let Y be the sum of 300 independent Bernoulli(0.25) variables. Using the normal approximation, find P(60 ≤ Y ≤ 90).
First get the mean and variance of the sum, then standardize both endpoints:
mean = 300 · 0.25 = 75
variance = 300 · 0.25 · 0.75 = 56.25, std = 7.5
z_lower = (60 − 75)/7.5 = −2, z_upper = (90 − 75)/7.5 = +2
P(60 ≤ Y ≤ 90) = Φ(2) − Φ(−2) = 0.9772 − 0.0228 = 0.9544The trap to avoid: divide the variance by n and the standard deviation by √n — never the SD by n.
Answer: P(60 ≤ Y ≤ 90) ≈ 0.9544 (the expression Φ(2) − Φ(−2)).
→ Taught in Central Limit Theorem & Confidence Intervals
Linear Algebra
Rank of A versus A² when A³ = A. A real matrix satisfies A³ = A. Is it always true that A and A² have the same rank?
Work through the eigenvalues. From Av = λv, A³ = A forces λ³ = λ:
λ³ = λ → λ(λ² − 1) = 0 → λ ∈ {0, 1, −1}
then squaring (eigenvalues of A²): λ² ∈ {0, 1, 1}A nonzero eigenvalue of A (±1) stays nonzero in A² (becomes 1), and a zero eigenvalue stays zero. The count of nonzero eigenvalues — hence the rank — is unchanged.
Answer: yes, A and A² always have the same rank.
→ Taught in Eigen-properties & Transforms
A norm-preserving matrix is orthogonal. A real n × n matrix A satisfies ‖Ax‖ = ‖x‖ for every x. Which properties must hold?
Preserving every length preserves every dot product, so AᵀA = I:
AᵀA = I ⇒ A is orthogonal (TRUE)
A⁻¹ = Aᵀ ⇒ A is invertible, full rank (TRUE)
det(A)² = det(AᵀA) = 1 ⇒ det(A) = ±1 (TRUE)
"eigenvalues are ±1" (FALSE)The eigenvalue claim is the distractor: a 90° rotation [[0, −1], [1, 0]] is orthogonal and full rank, yet its eigenvalues are ±i — on the unit circle, but neither +1 nor −1.
Answer: orthogonal, full rank, det = ±1 (eigenvalues need not be ±1).
→ Taught in Orthogonality & Orthogonal Matrices
Programming & DSA
Hashing with linear probing. Table size m = 10, hash h(x) = 3x mod 10, collisions resolved by linear probing. Insert 1, 4, 5, 6, 14, 15 in order. Where do 14 and 15 land?
Compute each home slot; on a collision, step forward +1 (mod 10) until empty:
h(1)=3, h(4)=2, h(5)=5, h(6)=8 (all land in their home slots)
h(14) = 3·14 mod 10 = 2 → slot 2 (4), probe 3 (1), probe 4 empty ⇒ slot 4
h(15) = 3·15 mod 10 = 5 → slot 5 (5), probe 6 empty ⇒ slot 6Answer: 14 → slot 4, 15 → slot 6.
→ Taught in Hash Tables & Linear Probing
append versus extend in Python. Start with A = [1, 2, 3] and B = [4, 5, 6]. Which single operation makes A equal to [1, 2, 3, 4, 5, 6]?
Check each candidate:
A.append(B) → [1, 2, 3, [4, 5, 6]] (B added as one nested element, length 4)
A.extend(B) → [1, 2, 3, 4, 5, 6] (each of 4,5,6 appended in turn — the answer)
A + B → [1, 2, 3, 4, 5, 6] but A itself is unchanged (no mutation)Only extend mutates A into the flat six-element list; append nests and + builds a new list without touching A.
Answer: A.extend(B).
→ Taught in Lists, Tuples, Dicts, Sets & Gotchas
Databases
Cost of a non-dependency-preserving decomposition. A relation R is split so the decomposition is not dependency-preserving. To enforce the lost FDs, which relational-algebra operator must run more often?
When an FD X → Y has X and Y straddling two fragments, the DBMS cannot see both columns in a single table. To check that FD on each update it must reconstruct the original relation — that is, take the join of the fragments.
Answer: JOIN (⋈) runs more often.
→ Taught in Lossless-Join vs Dependency-Preservation
Machine Learning
Least-squares fit through the origin. Fit y = wx to the points (−1, 1), (2, −5), (3, 5) by least squares. Find w.
For a through-origin fit, w = Σ(x·y) / Σ(x²). Build both sums column by column:
Σ x·y = (−1)(1) + (2)(−5) + (3)(5) = −1 − 10 + 15 = 4
Σ x² = 1 + 4 + 9 = 14
w = 4 / 14 ≈ 0.286The fit passes through no single point — it balances all three so the squared vertical gaps are collectively smallest.
Answer: w ≈ 0.286.
→ Taught in Simple Linear Regression
Naive-Bayes misclassification probability. Two classes with priors P(y1) = 1/3, P(y2) = 2/3; likelihoods P(x | y1) = 3/4, P(x | y2) = 1/4. Predict the class for x and find the probability the prediction is wrong.
Score each class with prior × likelihood (unnormalized posteriors):
y1 : (3/4)(1/3) = 1/4 ≈ 0.250 ← larger, so predict y1
y2 : (1/4)(2/3) = 1/6 ≈ 0.167The prediction is wrong exactly when the truth is y2, so the misclassification probability is the normalized posterior P(y2 | x):
P(y2 | x) = (1/6) / (1/4 + 1/6) = (1/6) / (5/12) = 0.40Answer: predict y1; misclassification probability = 0.40.
→ Taught in Naive Bayes
A question to carry forward
Two papers now sit side by side, and the recurring spine is unmistakable: Bayes in probability, an eigenvalue trick in linear algebra, a Python-semantics gotcha, a least-squares or naive-Bayes plug-in in ML — the same concepts, re-dressed, year after year. The shortcut-recognition habit that carried 2024 carried 2025 too.
But two points only suggest a line; the third confirms it. GATE DA 2026 is the most recent paper, the freshest signal of where the setters are steering — and reading it next answers the question two papers can only hint at: is this spine truly stable three years running, and what fresh shapes, if any, did the latest paper add? Here is the thread onward: worked the same way, does 2026 hold the pattern three-for-three — and once you have seen all three papers in sequence, what is the smarter way to revise them, not one year at a time, but one concept at a time across every year at once?