GATE DA 2024 — Solved Walkthrough
A curated set of fully-worked GATE DA 2024 problems across every subject — see exactly how each concept turns into an exam question and its verified answer.
The last chapter ended on the gap that decides ranks: knowing a concept and recognising it under pressure are different skills, and only the second is tested in the hall. This chapter closes that gap. GATE DA 2024 was the very first paper of the new Data Science & AI stream, and walking through it shows you something no concept lesson can — what each idea you have learned actually looks like once an examiner has dressed it up as a question. Read each problem as a translation exercise: a familiar concept on the left, an unfamiliar-looking question on the right, and the short bridge between them that you must build in ninety seconds.
Probability & Statistics
Z-score normalization of a salary. A value is drawn from a distribution with a known mean and standard deviation; standardize it to its z-score.
The z-score recentres a value on the mean and rescales by the standard deviation:
z = (x − mean) / stdPlugging the 2024 numbers gives z ≈ 0.476 — the value sits about half a standard deviation above the mean. The whole question is “do you know the standardization formula, and can you read off mean and std correctly?”
Answer: z ≈ 0.476.
→ Taught in Mean, Median, Mode & z-scores and Data Transformation: Normalization, Discretization, Sampling, Compression
Covariance of two coin indicators. Toss two fair coins. Let X = 1 if both are heads (else 0) and Y = 1 if at least one is heads (else 0). Find Cov(X, Y).
Over the four equally likely outcomes HH, HT, TH, TT:
E[X] = P(both heads) = 1/4
E[Y] = P(at least one head) = 3/4
E[XY] = 1 only on HH = 1/4 (X = 1 forces Y = 1)
Cov(X, Y) = E[XY] − E[X]·E[Y] = 1/4 − (1/4)(3/4) = 4/16 − 3/16 = 1/16The covariance is positive, as expected — X = 1 guarantees Y = 1, so the indicators move together.
Answer: Cov(X, Y) = 1/16 = 0.0625.
→ Taught in Covariance, Correlation & Total Expectation
Exponential parameter from a mean/variance condition. Let X ~ Exponential(λ). If 5·E(X) = Var(X), find λ.
Use the two exponential moments E(X) = 1/λ and Var(X) = 1/λ²:
5 · (1/λ) = 1/λ²
5/λ = 1/λ²
5·λ = 1 (multiply both sides by λ²)
λ = 0.2Watch the trap: the exponential variance is 1/λ² (the square of the mean), not 1/λ — do not borrow the Poisson’s mean = variance rule here.
Answer: λ = 0.2.
→ Taught in Exponential & Poisson
Linear Algebra
Determinant of M² + 12M for a singular matrix. A 3×3 matrix M has linearly dependent rows. Find det(M² + 12M).
Do not multiply matrices — factor first:
M² + 12M = M·(M + 12I)
det(M² + 12M) = det(M) · det(M + 12I) ← product ruleLinearly dependent rows make M singular, so det(M) = 0, and the whole product collapses:
det(M² + 12M) = 0 · det(M + 12I) = 0You never need the entries of M — spotting det(M) = 0 and factoring is the entire solution.
Answer: det(M² + 12M) = 0.
→ Taught in Determinants & Their Properties
Eigenvalues of a 2×2 matrix. Find the eigenvalues of M = [[2, −1], [3, 1]].
Read off the trace and determinant, then test the discriminant of the characteristic quadratic:
trace = 2 + 1 = 3, det = (2)(1) − (−1)(3) = 5
characteristic polynomial: λ² − 3λ + 5 = 0
discriminant = (−3)² − 4·5 = 9 − 20 = −11 < 0A negative discriminant means the roots are not real: λ = (3 ± i√11)/2. This real matrix acts as a rotate-and-scale, so no real direction survives.
Answer: a complex conjugate pair (no real eigenvalues).
→ Taught in Eigenvalues & Eigenvectors
Nullity of an orthogonal projection. Let M be the orthogonal projection of R³ onto a 2-dimensional subspace U. Find M², rank(M), and nullity(M).
A projection is idempotent, so M² = M (and M³ = M). Its image fills U, so rank(M) = dim(U) = 2. Rank–nullity in R³ finishes it:
nullity(M) = 3 − rank(M) = 3 − 2 = 1The null space is the single direction perpendicular to U — the part the shadow discards.
Answer: M² = M, rank = 2, nullity = 1.
→ Taught in Projections & Idempotent Matrices
Sum of singular values of a rank-1 matrix. Let u = (1, 2, 3, 4, 5)ᵀ and M = uuᵀ. Find the sum of the singular values of M.
M = uuᵀ is an outer product, so every column is a multiple of u — the matrix has rank 1 and exactly one nonzero singular value. For a rank-1 uuᵀ that lone value is ‖u‖²:
σ = ‖u‖² = 1² + 2² + 3² + 4² + 5² = 1 + 4 + 9 + 16 + 25 = 55All other singular values are 0, so the sum is just the one.
Answer: sum of singular values = 55.
→ Taught in Singular Value Decomposition
Calculus
Classifying a critical point. f is twice differentiable with f'(x*) = 0 and f'' > 0 at x*. What does this tell you about f at x*?
The second-derivative test: a flat point (f' = 0) with positive curvature (f'' > 0) is concave up — a bowl — so it is a local minimum.
The trap answer is “global minimum.” Positive curvature is a purely local statement; it describes the bowl around x* and says nothing about how low f might dip elsewhere.
Answer: a local minimum.
→ Taught in Maxima, Minima & the 2nd-Derivative Test
Programming & DSA
Binary-search comparison recurrence. Which recurrence gives the maximum number of comparisons binary search makes on n elements?
Binary search does one comparison at the middle and then recurses into one half of size ⌊n/2⌋:
F(n) = F(⌊n/2⌋) + 1The tempting wrong answer sums both halves, F(⌊n/2⌋) + F(⌈n/2⌉) — but that is merge sort’s structure and would give a linear count. Binary search never explores both sides, so you add a single comparison per level (about ⌈log₂(n+1)⌉ total).
Answer: F(n) = F(⌊n/2⌋) + 1.
→ Taught in Linear & Binary Search
Minimum swaps for quicksort on a sorted array. In-place quicksort with the last element as pivot runs on the already-sorted array [60, 70, 80, 90, 100]. How many swaps are performed?
Trace the Lomuto partition: every element is already smaller than each chosen pivot and already on the correct side, and each pivot is already in its final slot — so no swap is ever needed at any level.
[60, 70, 80, 90, 100], pivot 100 → 0 swaps; recurse left
[60, 70, 80, 90], pivot 90 → 0 swaps; ...Note this is still quicksort’s worst case for comparisons (4+3+2+1 = 10), even though data movement is zero.
Answer: 0 swaps.
→ Taught in Merge Sort & Quicksort
Which sort finishes [4, 3, 2, 1, 5] in exactly two passes? Among bubble, insertion, and selection sort, which sorts this array ascending in exactly two passes?
Only selection sort reaches to the back to pull a minimum forward each pass, which is exactly what this input (two smallest values at the back) rewards:
selection: [4,3,2,1,5] → [1,3,2,4,5] → [1,2,3,4,5] ✓ sorted in 2
bubble: [4,3,2,1,5] → [3,2,1,4,5] → [2,1,3,4,5] ✗ still unsorted
insertion: [4,3,2,1,5] → [3,4,2,1,5] → [2,3,4,1,5] ✗ still unsortedBubble drags only the largest to the back per pass; insertion grows a sorted prefix one slot per pass — neither finishes in two.
Answer: selection sort only.
→ Taught in Bubble, Insertion & Selection Sort
Databases
Counting rows returned by a SQL query. Schema Raider(ID, City, ...) and Team(ID, RaidPoints, ...). How many rows does this return?
SELECT * FROM Raider, Team
WHERE Raider.ID = Team.ID AND City = 'Jaipur' AND RaidPoints > 200;Walk the pipeline. The Raider.ID = Team.ID equality collapses the cross product to six paired rows (one per ID). Then City = 'Jaipur' keeps the Jaipur raiders, and RaidPoints > 200 drops any whose team scored 200 or fewer. With the standard 2024 dataset, three IDs survive both filters.
Answer: 3 rows.
→ Taught in SQL: Computing Results by Hand
Functional-dependency closure. R(U, V, W, X, Y, Z) with F = {U → V, U → W, WX → Y, WX → Z, V → X}. Is VW → Y derivable?
Compute the attribute closure VW⁺, adding the right side of any FD whose left side is fully present:
start {V, W}
V → X : add X → {V, W, X}
WX → Y : add Y → {V, W, X, Y}
WX → Z : add Z → {V, W, X, Y, Z} → VW⁺ = VWXYZSince Y (and YZ) lie inside VW⁺, both VW → Y and VW → YZ are derivable. The key discipline: an FD like WX → Y fires only when both W and X are already in the closure.
Answer: VW⁺ = VWXYZ, so VW → Y is derivable.
→ Taught in Functional Dependencies & Closure
Machine Learning
One k-means centroid update. Centroid C3 = (6, 6) is assigned the points (6, 6) and (9, 9). Where does C3 move after one update?
The update rule is “new centroid = mean of assigned points.” Average each coordinate separately:
new x = (6 + 9)/2 = 7.5
new y = (6 + 9)/2 = 7.5The whole task is just “assign, then average” — nothing more.
Answer: C3 → (7.5, 7.5).
→ Taught in k-means & k-medoid Clustering
A question to carry forward
Read across these solutions and a pattern surfaces that is more useful than any single answer: in nearly every one, the work was not heavy computation. The salary z-score was one substitution; the singular matrix’s determinant fell to a single observation; quicksort’s swap count needed only the insight that a sorted array never moves. The 2024 paper rewarded recognising the shortcut, not grinding through brute force.
That is the texture of one paper. But one paper is a single sample, and a single sample can mislead — was 2024’s lightness typical, or a quirk of the stream’s debut year? The next paper, GATE DA 2025, is the second data point, and reading it alongside 2024 starts to reveal what is stable about the exam: which concepts recur every year, which traps the setters reuse, and where the genuine difficulty consistently lives. Here is the thread onward: worked the same way, what does GATE DA 2025 confirm about the paper’s habits — and what new shapes does it add that 2024 never showed you?