GATE DA 2024 — Solved Walkthrough

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) / std

Plugging 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 = 1/16

The 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         = 0.2

Watch 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 rule

Linearly dependent rows make M singular, so det(M) = 0, and the whole product collapses:

det(M² + 12M) = 0 · det(M + 12I) = 0

You 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 < 0

A 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 = 1

The 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 = 55

All 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⌋) + 1

The 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 unsorted

Bubble 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⁺ = VWXYZ

Since 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 = 7.5

The whole task is just “assign, then average” — nothing more.

Answer: C3 → (7.5, 7.5).

→ Taught in k-means & k-medoid Clustering