GATE 2026 DA (Data Science & AI) question paper with solutions

Verbosity (using too many words) is the opposite of brevity (using few words). So the pair is a pair of antonyms. The opposite of insolence (rude, disrespectful behaviour) is respect. Innocence, solace and wealth are not opposites of insolence. Answer: (B) Respect.

Concept: Analogy by antonymy. The trap is picking a word that merely sounds related (innocence) instead of checking the exact relation in the first pair.

70 = 2 x 5 x 7. Each digit is between 1 and 9 (a zero would make the product 0). Any other split needs a factor of 10, 14, 35 or 70, which is not a single digit, and using 1 forces two digits multiplying to 70, which is impossible with single digits. So the digits are 2, 5 and 7 in some order, and their sum is 2 + 5 + 7 = 14. Answer: (B) 14.

Concept: Prime factorisation of the product fixes the digits. The trap is forgetting to rule out splits that use the digit 1.

Take log base 10 of both. log x = log₁₀ m · log₁₀ n. log y = log₁₀ n · log₁₀ m. The two logarithms are equal, and log is one-to-one, so x = y for every pair of positive m, n (the condition m > n does not matter). Answer: (C) x = y.

Concept: Identity a^(log b) = b^(log a). The trap is assuming the larger base gives the larger value and picking x < y or x > y.

The sentence is a third conditional: 'If ... had been ..., ... would have ...'. This form describes an unreal past situation, so it implies the condition did not happen: the latest movie was not a commercial success. Hence the actor did not make that money, and nothing says he will sponsor the next movie. Answer: (D) His latest movie was not commercially successful.

Concept: Reading counterfactual conditionals. The trap is treating 'had been' as a statement of fact and choosing (B) or (C).

We part AT a place (at the door). The door belongs to the cabin: the door OF the cabin. A cabin is rented FOR a period: for the night. Sequence: at; of; for. Answer: (A) at; of; for.

Concept: Prepositions of place, possession and duration. The trap is 'in the night', which describes when something happens, not the length of a rental.

Sort them a ≤ b ≤ c ≤ d ≤ e. Median c = 18, sum = 5 x 12 = 60. A single mode of 20 needs 20 to occur at least twice, and only d and e can exceed 18, so d = e = 20. Then a + b = 60 - 18 - 40 = 2 with a ≤ b ≤ 18. Options: (0, 2) or (1, 1). (1, 1) makes 1 appear twice, tying with 20, so the mode is no longer single. Only {0, 2, 18, 20, 20} works. Answer: (B) 1.

Concept: Using mean, median and mode together as constraints. The trap is counting (1, 1, 18, 20, 20), which has two modes.

If game 1 were Pavan vs Tanvi, its Class 4 winner must lose game 2 to a Class 5 student (Class 5 needs 2 wins), leaving Rishi vs Swathi for game 3 - a second same-class game. So game 1 is Rishi vs Swathi. Its winner must play both remaining games (otherwise game 3 is Pavan vs Tanvi) and must win game 2 so Class 5 has 2 wins. If Swathi won game 1 and game 2, girls already have 2 wins - not allowed. So Rishi wins game 1 (boy) and game 2 (boy), and game 3 must be won by a girl: Rishi beats Pavan in game 2 and loses to Tanvi in game 3. Tanvi never lost. Answer: (D) Tanvi.

Concept: Case analysis under elimination rules. The trap is assuming the strongest player (Rishi) stays unbeaten, which would give the boys three wins.

n² + (n+1)² < 1000 forces n ≤ 21. P and R must differ, so the two numbers cannot share a tens digit: that leaves 19 & 20 or 20 & 21 (or 10 & 11 type pairs, which share tens digit 1). 20 & 21 share tens digit 2. So the pair is 19 and 20: 361 + 400 = 761 = XYP, giving P = 1, so PQ = 19, RS = 20, X = 7, Y = 6. Digits 1, 9, 2, 0, 7, 6 are all distinct. Answer: (C) 6.

Concept: Bounding then using the distinct-digit condition. The trap is missing that consecutive numbers in the same decade share the tens digit, which violates distinctness.

Split PQRO along OQ into triangles OPQ and ORQ. Since PQ = RQ, OQ bisects angle PQR, so each triangle has angle 22.5° at Q. In isosceles triangle OPQ (OP = OQ = 10), the angle at P is also 22.5°, so angle POQ = 180 - 45 = 135°. Area of OPQ = ½ · 10 · 10 · sin 135° = 50 · (√2/2) = 25√2. Both triangles are congruent, so area = 2 · 25√2 = 50√2 cm². Answer: (C) 50√2.

Concept: Inscribed angle and isosceles triangles at the centre. The trap is using the central angle POR = 90° and computing a sector or triangle POR instead of the two triangles through Q.

Principal components are eigenvectors of the (symmetric) covariance matrix, chosen to be mutually orthonormal. Any two distinct principal components, including the first and the tenth, are orthogonal, so the angle between them is 90°. Answer: (B) θ = 90°.

Concept: PCA directions are orthogonal. The trap is thinking the angle depends on how much variance each component explains.

The 100 test samples are held out and not used in model selection, leaving 900 samples for cross-validation. LOOCV leaves out one sample at a time as the validation set, so it produces one split per sample: 900 splits. The number of classes (10) and the dimension (512) are irrelevant. Answer: (C) 900.

Concept: LOOCV creates as many folds as training samples. The trap is counting all 1000 samples including the test set.

Uninformed (blind) search uses only the problem definition: BFS, DFS and depth-limited search expand nodes without any estimate of distance to the goal. A* uses f(n) = g(n) + h(n), where h is a heuristic estimate of cost to the goal, so it is an informed search. Answer: (C) A* Search.

Concept: Informed vs uninformed search is about using a heuristic. The trap is thinking depth-limited search is informed because it uses a limit.

(A) Being a classmate is symmetric, so it is true. (B) ∀x P(x) is equivalent to ¬∃x ¬P(x), so the implication holds. (D) is the standard translation of a universal statement. (C) ∀x IsKing(x) ∧ IsPerson(x) says everything is a king and a person; the correct form is ∀x IsKing(x) ⇒ IsPerson(x). Answer: (C).

Concept: With ∀ the connective is ⇒, with ∃ it is ∧. The classic trap is writing ∀ with ∧.

With randomly ordered input, the first element is equally likely to be any rank. If it is the (k+1)-th smallest, the partitions have sizes k and n - k - 1, each with probability 1/n for k = 0..n-1. Averaging gives T(n) = (1/n) Σ [T(k) + T(n - k - 1)] + O(n), which solves to O(n log n). (A) is the worst case, (B) and (C) are particular fixed splits. Answer: (D).

Concept: Expected-case recurrence averages over all pivot ranks. The trap is choosing the balanced-split recurrence, which describes one ideal case, not the expectation.

A default argument is evaluated once, when the function is defined, so every call that omits lst shares the same list. Call 1 appends 1 to it: [1]. Call 2 appends 2 to the same list: [1, 2]. Call 3 passes a fresh empty list, so it returns [3]. Answer: (B) [1] [1, 2] [3].

Concept: Mutable default arguments persist across calls. The trap is expecting a fresh list each time ([2] on the second call).

D appears on no right-hand side, so every key contains D. AD+ = {A,D} → BC → CD gives E: all attributes. ED+: E → A, A → BC: all. CD+: CD → E, E → A, A → B: all. Single D alone gives only D. BD+ = {B,D}, not a key. Each of AD, ED, CD is minimal (A, E, C alone do not give D). Any other key containing D would be a superset of these. Answer: (A) AD, ED and CD.

Concept: Attributes never on a right side must be in every key. The trap is (C), forgetting that D can never be derived.

In the location concept hierarchy, State lies below Country (finer granularity). Moving from Country to State goes down the hierarchy to more detailed data, which is drill-down. Roll-up would go the other way (State to Country); slice and dice select sub-cubes without changing granularity. Answer: (D) Drill-down.

Concept: Drill-down = finer level, roll-up = coarser level. The trap is confusing them because the number of dimensions stays the same.

There are 2²⁰²⁶ - 1 non-empty subsets, each equally likely. The product is odd exactly when every element is odd. S has 1013 odd numbers, giving 2¹⁰¹³ - 1 non-empty all-odd subsets. Even-product subsets: (2²⁰²⁶ - 1) - (2¹⁰¹³ - 1) = 2²⁰²⁶ - 2¹⁰¹³ = 2¹⁰¹³(2¹⁰¹³ - 1). Probability = 2¹⁰¹³(2¹⁰¹³ - 1)/(2²⁰²⁶ - 1). Answer: (C).

Concept: Complement counting. The trap is dividing by 2²⁰²⁶, forgetting that the empty set is excluded from the sample space.

Taking each solution as equally likely: by stars and bars, non-negative solutions of n₁ + n₂ + n₃ + n₄ = 20 number C(20 + 3, 3) = C(23,3). Positive solutions: substitute mᵢ = nᵢ - 1 ≥ 0, so m₁ + ... + m₄ = 16, giving C(16 + 3, 3) = C(19,3). Probability = C(19,3)/C(23,3). Answer: (A).

Concept: Stars and bars for non-negative vs positive solutions. The trap is using C(n,k) with k = 4 instead of k - 1 = 3 bars.

M is rotation by θ, so Mᵏ is rotation by kθ. M⁵ is rotation by 2π, which is I₂. 2026 = 5 · 405 + 1, so M²⁰²⁶ = (M⁵)⁴⁰⁵ · M = M. Answer: (B) M.

Concept: Powers of a rotation matrix are periodic. The trap is computing 2026 mod 5 wrongly or expecting I₂ because 2026 is even.

S₁ is the solid ball of radius 4 centred at the origin. A two-dimensional subspace of ℝ³ is a plane through the origin, so it cuts the ball through its centre. The intersection is a disc of radius 4, with area π · 4² = 16π. Answer: (A) 16π.

Concept: A subspace always contains the origin, so the cut is a great disc. The trap is taking the radius as 16 or treating the plane as arbitrary.

K-Medoid is a clustering algorithm (T1:A2). Naive Bayes is a classifier (T2:A4). MCMC draws samples from a distribution (T3:A1). LDA projects data onto directions that best separate classes, so it is used for supervised feature extraction / dimensionality reduction (T4:A3). Only option B has all four. Options A and C send K-Medoid to feature extraction, D sends it to classification. Answer: (B).

Concept: Knowing what each algorithm does. The trap is that LDA can also classify, so one may pair it with classification and leave Naive Bayes stranded.

X entails Y means no model makes X true and Y false. That is exactly: X ⇒ Y is valid (A), X ∧ ¬Y is unsatisfiable, i.e. always FALSE (B), and 'if X then Y' (C). The converse 'if Y then X' (D) need not hold: Y can be true while X is false. Answer: (A), (B), (C).

Concept: Deduction theorem: entailment equals validity of the implication. The trap is accepting the converse.

Pre-order starts with P, so P is the root. In-order splits into left {S, Q, E} and right {F, R, G}. Left pre-order Q S E: Q is its root with S left and E right. Right pre-order R F G: R is its root with F left and G right. Post-order: S E Q F G R P. So (A) true, (B) true, (C) false (Q has two children), (D) false (G is R's right child). Answer: (A), (B).

Concept: Rebuilding a tree from pre-order and in-order. The trap is misreading which side of R holds G.

s is the referencing relation, r the referenced one. Inserting into r only adds possible targets, so no s tuple loses its match. Deleting from s removes referencing tuples, which can never leave a dangling reference. Deleting from r can remove an A value still used by some E in s, and inserting into s can add an E value absent from r - both can violate. Answer: (A), (B).

Concept: Referential integrity: only deletes/updates on the referenced table and inserts/updates on the referencing table can break it. The trap is mixing up the two directions.

f(1) = 1 - 3 + 2 = 0, so (D) holds; f(x) = (x - 1)(x² - 2x - 2), other roots 1 ± √3 ≈ -0.732 and 2.732. Only -0.732 lies in [-0.9, 0], so (A) is false. f'(x) = 3x² - 6x = 0 at x = 0, 2. f(0) = 2, f(2) = -2, f(3) = 2, and f(x) → -2 as x → -1 but x = -1 is excluded. So the minimum value -2 is attained only at x = 2 (B true), while the maximum value 2 is attained at both x = 0 and x = 3 (C false). Answer: (B), (D).

Concept: Extrema on a half-open interval: check endpoints and excluded points. The trap is forgetting the endpoint x = 3 ties with x = 0, and that x = -1 is not in the domain.

Both densities are symmetric about 0, so G(0) = H(0) = 0.5 (A true). g(0) = 1/√(2π) ≈ 0.399 and h(0) = 1/π ≈ 0.318 (t₁ is Cauchy), so (D) false. t₁ has heavier tails: h > g for |y| > c and h < g near 0. Hence P(Y > c) > P(Z > c), i.e. H(c) < G(c), so (B) false. By symmetry P(Y < -c) > P(Z < -c), i.e. G(-c) < H(-c) (C true). Answer: (A), (C).

Concept: Heavy tails of the t distribution versus the normal. The trap is assuming a heavier-tailed CDF is larger everywhere; it is smaller on the right tail.

The gradient of f_w(x) = wx with respect to w is x = 10. SGD update: w ← w - η · ∂f/∂w = 10.00 - 0.10 · 10.00 = 9.00. Answer: 9.00 (accepted range 8.9 to 9.1).

Concept: Gradient is taken with respect to the parameter, not the input. The trap is differentiating with respect to x and getting w.

MIN picks the minimum leaf under each MAX move: strategy 1 gives min(8, 6, -1) = -1, strategy 2 gives min(1, 5, 7) = 1, strategy 3 gives min(-4, -3, -12) = -12. MAX picks the largest of -1, 1, -12, which is 1, reached by strategy 2. Answer: 2.

Concept: Minimax backs up min values at MIN nodes first. The trap is MAX greedily chasing the largest leaf (8) under strategy 1.

Each step makes one (three-way) comparison and recurses on at most the larger half. Worst-case sizes: 1000 → 500 → 250 → 125 → 62 → 31 → 15 → 7 → 3 → 1, then the range becomes empty. That is 10 comparisons, i.e. ⌊log₂ 1000⌋ + 1 = 9 + 1 = 10. Answer: 10.

Concept: Unsuccessful binary search takes ⌊log₂ n⌋ + 1 probes in the worst case. The trap is answering ⌈log₂ 1000⌉ = 10 for the wrong reason or 9 by forgetting the last probe.

A non-leaf node with p node pointers holds p - 1 keys and no record pointers. Need 10p + 11(p - 1) ≤ 4096, so 21p ≤ 4107 and p ≤ 195.57. Maximum p = 195. Answer: 195.

Concept: Internal B+ tree nodes carry only child pointers and keys. The trap is including the data record pointer size, which belongs to leaf nodes.

f(f(n)) = n means f is an involution: a permutation whose cycles all have length 1 or 2. Count by number of 2-cycles: none: 1 (identity); one 2-cycle: C(4,2) = 6; two 2-cycles: 3 ways to pair four elements. Total 1 + 6 + 3 = 10. Answer: 10.

Concept: Involutions consist of fixed points and transpositions. The trap is forgetting the identity or the double-transposition cases.

P(X > 10 | X > 5) = P(X > 10)/P(X > 5) = e^(-10/λ)/e^(-5/λ) = e^(-5/λ) = P(X > 5) = 0.35, by the memoryless property. Answer: 0.35 (accepted range 0.34 to 0.36).

Concept: Memorylessness of the exponential distribution. The trap is computing λ first or squaring 0.35.

The double sum factorises: (Σ_{i≥0} (1/2)^i)(Σ_{j≥1} (1/3)^j) = [1/(1 - 1/2)] · [(1/3)/(1 - 1/3)] = 2 · (1/2) = 1. Answer: 1.

Concept: Separable double geometric series. The trap is the starting index: j starts at 1, not 0.

Manhattan distances: P1-P2 = 1 + 2 + 2 = 5; P1-P3 = 3 + 5 + 4 = 12; P1-P4 = 1 + 0 + 4 = 5; P2-P3 = 2 + 3 + 2 = 7; P2-P4 = 0 + 2 + 2 = 4; P3-P4 = 2 + 5 + 0 = 7. Agglomerative clustering merges the closest pair first: P2 and P4 at distance 4. Answer: (D) P2, P4.

Concept: First merge = smallest pairwise distance. The trap is using Euclidean distance or missing a sign when taking absolute differences.

Ridge adds λ‖w‖₂² (squared L₂ norm, not L₁) to shrink weights. It guards against overfitting - good on training, poor on test - which is the reverse of (A). It does not target negative weights (C). Shrinking weights adds bias but lowers variance, which is (D). Answer: (D).

Concept: Regularisation trades bias for variance. The trap is (B): L₁ is Lasso, not Ridge.

The rule: for a creative person, success holds exactly when discipline holds, i.e. C ⇒ (S ⇔ D) - statement (ii). (i) fails: a non-creative disciplined person makes D true and C ∧ S false, yet the rule says nothing about non-creative people. (iii): (D ⇒ S) ∨ ¬S = ¬D ∨ S ∨ ¬S, which is always true, so (iii) reduces to C ⇔ TRUE, i.e. just C - not the rule. Answer: (B) Only (ii).

Concept: Translating a conditional rule scoped to a condition. The trap is (i), which wrongly forces D to imply creativity and success.

Let C(n) be the number of calls made by mystery(n). C(n) = 1 for n ≤ 0, else C(n) = 1 + C(n-1) + C(n-2). C(0) = C(-1) = 1; C(1) = 1 + 1 + 1 = 3; C(2) = 1 + 3 + 1 = 5; C(3) = 1 + 5 + 3 = 9; C(4) = 1 + 9 + 5 = 15. Answer: (C) 15.

Concept: Counting nodes of a recursion tree. The trap is forgetting that mystery(1) still calls mystery(0) and mystery(-1), since the base case is n ≤ 0.

BFS and DFS both visit exactly the vertices reachable from the start. If u is reachable from v in G, there is a path v → ... → u; reversing every edge gives a path u → ... → v in Gᴿ, so v is reachable from u in Gᴿ (D true). (A)/(B) fail for G with the single edge v → u: in Gᴿ nothing is reachable from v, but u is reachable in G. (C) fails because traversal orders are unrelated. Answer: (D).

Concept: Reachability reverses under edge reversal. The trap is assuming reversed graphs have the same reachable sets from the same source.

3 goes to leaf [1, 5], which overflows as [1, 3, 5] and splits into two leaves, pushing a new separator (3, as the largest key of the left leaf [1, 3] | [5]) into the root. The root would then hold 3, 5, 8 with four children, exceeding 3 pointers, so it splits: [3] and [8] become two internal nodes and the middle key 5 moves up into a new root. The same result follows if the leaf splits as [1] | [3, 5]. New root: 5. Answer: (A) 5.

Concept: Cascading splits in a B+ tree: leaf keys are copied up, internal keys are moved up. The trap is stopping after the leaf split and answering 3, 5 and 8.

X ⋈ Y joins on (P, Q): (P1, Q1, R1, 2) and (P3, Q3, R2, 1); (P2, Q2) has no match in Y. Selecting Q = Q3 or R = R2 keeps only (P3, Q3, R2, 1). Y ⋈ Z joins on P: (P1, Q1, 2, T1), (P1, Q2, 5, T1), (P3, Q3, 1, T2). Selecting S > 1 keeps the two P1 rows. Joining the left result (P3 only) with the right (P1 only) on P, Q, S gives no common tuple. Answer: (D) Zero rows.

Concept: Natural join matches on all common attributes. The trap is joining on P alone or forgetting that the outer join also matches on Q and S.

For an n-dimensional cube where dimension i has Lᵢ hierarchy levels, the number of cuboids is Π(Lᵢ + 1); the +1 is the virtual top level 'all' (dimension aggregated away). Here each of the 3 dimensions has 3 levels, so the count is (3 + 1)³ = 4³ = 64. Answer: (A) 4³.

Concept: Cuboid count with hierarchies. The trap is 2³, which counts cuboids only when each dimension has a single level.

Let W = 2X - 1, which is ±1, so W² = 1. By independence, E[WY] = E[W]E[Y] = E[W] · 0 = 0. Var(WY) = E[W²Y²] - (E[WY])² = E[W²]E[Y²] - 0 = 1 · 100 = 100. Answer: (A) 100.

Concept: Variance of a product of independent variables with a zero-mean factor. The trap is computing Var(2X - 1) = 0.84 and multiplying by 100 to get 84 or similar.

The sum is P(Nₙ ≤ n) where Nₙ ~ Poisson(n). Nₙ is a sum of n independent Poisson(1) variables, with mean n and variance n. By the central limit theorem, (Nₙ - n)/√n → N(0, 1), so P(Nₙ ≤ n) = P((Nₙ - n)/√n ≤ 0) → Φ(0) = 0.5. Answer: (A) 0.5.

Concept: Recognising a Poisson CDF and applying the CLT. The trap is answering 1 because the full series Σ n^k/k! equals e^n; here the sum stops at k = n.

The sum of eigenvalues equals the trace: 1 + cos t + cos t = 1 + 2cos t. Setting 1 + 2cos t = 1 + √2 gives cos t = 1/√2. In [-π, π], t = π/4 or t = -π/4. Answer: (C) {π/4, -π/4}.

Concept: Trace equals sum of eigenvalues, so no eigenvalue computation is needed. The trap is forgetting the negative solution of cos t = 1/√2.

Confusion counts: X predicted X = 14, X predicted Y = 6, Y predicted Y = 8, Y predicted X = 2. Accuracy = (14 + 8)/30 = 22/30 = 11/15 (A true, D false). Precision X = 14/(14 + 2) = 0.875; precision Y = 8/(8 + 6) ≈ 0.571, so (B) true. Recall X = 14/20 = 0.70; recall Y = 8/10 = 0.80, so (C) false. Answer: (A), (B).

Concept: Precision divides by predicted counts, recall by actual counts. The trap is swapping the two denominators.

In standard first-order semantics the domain is non-empty. (A) If P holds for every element, it holds for some element: valid. (B) Some element satisfying P does not make all satisfy it: not valid. (C) contains (B) as one direction: not valid. (D) If P holds for all x, then ¬P(x) holds for none, so the implication fails whenever ∀x P(x) is true: not valid. Answer: (B), (C), (D).

Concept: Universal implies existential (non-empty domain), not the reverse. The trap is thinking (D) is valid by misreading it as ∀x P(x) ⇒ ¬∃x ¬P(x).

Insertion sort: insert 2 (1 comparison), 3 (1), 5 (1), 4 (compare with 5 and shift, then compare with 3 and stop: 2). N2 = 5, with one swap/shift (C true), so (A) false. Bubble sort: the plain version makes 4 + 3 + 2 + 1 = 10 comparisons; even with an early-exit flag it makes 4 + 3 = 7. Either way N1 > N2 (B true). Both compare already-ordered pairs such as (1, 2) (D true). Answer: (B), (C), (D).

Concept: Counting comparisons on a nearly sorted array. The trap is counting only 4 comparisons for insertion sort, missing the final comparison that stops the insertion of 4.

Each call to outer() creates a new list x, and the returned inner closure keeps its own x. So f1 and f2 have separate lists (A false). Line P: [10]. Line Q: [10, 20] (B true). Line R uses f2's own list: [30] (C false). Line S: f1's list becomes [10, 20, 40] (D true). Answer: (B), (D).

Concept: Closures capture the enclosing scope of the call that created them. The trap is thinking all closures from the same function share one list.

(A) The derived table gives (TeamID, size) per team; joining each employee to its team's row yields the required output. (B) The extra condition A.EmpID = B.EmpID pairs each employee only with itself, so every count is 1. (C) Grouping by the primary key EmpID gives 1 for everyone. (D) The derived table has no TeamID column, so B.TeamID is invalid and the query errors. Answer: (A).

Concept: Aggregate per group, then join back. The trap is (B): the self-join condition on EmpID destroys the grouping by team.

M is the centering matrix. (A) Iₙ and 𝟏𝟏ᵀ are symmetric, so M is symmetric. M² = Iₙ - (2/n)𝟏𝟏ᵀ + (1/n²)𝟏(𝟏ᵀ𝟏)𝟏ᵀ = Iₙ - (2/n)𝟏𝟏ᵀ + (1/n)𝟏𝟏ᵀ = M, so M² = M, not Iₙ (B false). Trace = n - (1/n) · n = n - 1 (C false). Symmetric and idempotent means M is an (orthogonal) projection (D true). Answer: (A), (D).

Concept: Properties of the centering matrix. The trap is confusing idempotent (M² = M) with involutory (M² = I).

(A) A sum of n squared independent standard normals is χ²ₙ: true. (B) The centred sum of squares loses one degree of freedom (Cochran's theorem): χ²ₙ₋₁, true. (C) X₁² + Xₙ² is χ²₂, whose density is (1/2)e^(-x/2), i.e. exponential with mean 2: true. (D) X̄ ~ N(0, 1/n), so √n X̄ ~ N(0, 1) and its square is χ²₁, not χ²₂: false. Answer: (A), (B), (C).

Concept: Chi-square facts and χ²₂ = Exp(mean 2). The trap is (D), giving the wrong degrees of freedom for a single squared normal.

F(x) = P(X ≤ x). Below the smallest support point F(x) = 0, so it is non-negative but not always positive (A false). Any CDF is non-decreasing (B true). For a discrete variable F is a step function that jumps by P(X = a) at each support point a (C true). A CDF defined with ≤ is right continuous; at a jump it is not left continuous (D false). Answer: (B), (C).

Concept: CDF properties. The trap is (D): with F(x) = P(X ≤ x), continuity is from the right, not the left.

y_pred = (-3)(1) + (4)(2) = 5. y_true = 1 + 2 = 3. Absolute error = |5 - 3| = 2. Ridge penalty = 0.20 · ‖w‖₂² = 0.20 · (9 + 16) = 5. Total loss = 2 + 5 = 7.00. Answer: 7.00 (accepted range 6.9 to 7.1).

Concept: Ridge uses the squared L₂ norm of the weights. The trap is using ‖w‖ = 5 (not squared), giving 3.00.

Without biases, each pair of consecutive fully connected layers contributes (size of one) x (size of next) weights. 30 x 4 = 120, 4 x 3 = 12, 3 x 1 = 3. Total = 120 + 12 + 3 = 135. Answer: 135.

Concept: Counting weights between layers. The trap is adding bias terms (which would give 143) or counting neurons instead of connections.

P(+) = P(+|D)P(D) + P(+|not D)P(not D) = 0.8 · 0.3 + 0.1 · 0.7 = 0.24 + 0.07 = 0.31. By Bayes, P(D|+) = 0.24/0.31 ≈ 0.774. Answer: 0.77 (accepted range 0.76 to 0.78).

Concept: Bayes' theorem with a base rate. The trap is answering the sensitivity 0.80 and ignoring false positives.

fun is one bubble-sort pass that swaps adjacent out-of-order pairs and returns the number of swaps; the list is modified in place. The loop runs 5 passes, enough to sort a 5-element list, so count equals the total number of adjacent swaps, which equals the number of inversions. Inversions in [5, 3, 4, 1, 2]: 5 with 3, 4, 1, 2 (4); 3 with 1, 2 (2); 4 with 1, 2 (2); total 8. Pass by pass: 4, 2, 2, 0, 0. Answer: 8.

Concept: Bubble sort swaps equal the inversion count. The trap is counting comparisons instead of swaps, or thinking the list is copied between calls.

The condition ∃z ∈ X (t[P] = z[P]) is always true (take z = t). So the result is the tuples of X whose P value has some Y tuple with S > 1. P1: S = 10, 15, yes. P2: S = 20, yes. P3: only S = 1, no. TRC returns a set, so the answer is 2 tuples (P1, Q1, R1) and (P2, Q2, R2). The key also accepts 3, which matches counting P1 once per matching Y tuple (bag semantics). Answer: 2 (accepted range 2 to 3).

Concept: Existential quantifiers filter, they do not multiply tuples. The trap is counting P1 twice because two Y tuples match it.

For each account, compare (number of strictly larger balances) with (number of strictly smaller balances). A1: 3 larger, 0 smaller, kept. A2: same as A1, kept. A3 (10000): 2 larger, 2 smaller, kept (2 ≥ 2). A4: 1 larger, 3 smaller, rejected. A5: 0 larger, 4 smaller, rejected. Rows returned: A1, A2, A3. Answer: 3.

Concept: Correlated subqueries evaluated per outer row. The trap is treating the equal balances 5000 and 5000 as smaller/larger than each other; strict comparisons ignore ties.

E1 → E1(A11, A12, A13). E2 without the multi-valued attribute → E2(A21, A23). The multi-valued attribute needs its own relation → E2_A22(A21, A22) with key (A21, A22); keeping it inside E2 would repeat A23 and break normal form. The many-to-many relationship needs its own relation → R12(A11, A21). Total participation on an M:N relationship does not allow merging R12 into either entity. Total: 4. Answer: 4.

Concept: Mapping M:N relationships and multi-valued attributes. The trap is merging R12 into an entity because of total participation, which only works for 1:1 or 1:N.

Write xᵢ - xⱼ = (xᵢ - x̄) - (xⱼ - x̄). Expanding, the cross term sums to zero, so Σᵢ Σⱼ (xᵢ - xⱼ)² = 2n Σᵢ (xᵢ - x̄)². With n = 100: 200 · SS / 2000 = 99, so SS/10 = 99 and SS = 990. Then (1/99) · 990 = 10. Answer: 10.

Concept: Pairwise squared differences equal 2n times the sum of squared deviations. The trap is forgetting the factor 2 (pairs are counted in both orders).

E[Y | X] = x², so E[XY] = E[X · E[Y | X]] = E[X³] = 0, since X is symmetric about 0. Also E[X] = 0. Cov(X, Y) = E[XY] - E[X]E[Y] = 0, and both variances are positive, so the correlation is 0, even though Y clearly depends on X. Answer: 0.

Concept: Zero correlation does not mean independence; correlation only captures linear dependence. The trap is expecting a non-zero value because Y is a function of X plus noise.

Row 2 and column 3 share the entry a₂₃; the other 4 entries in each are independent. If a₂₃ = 1 (prob 1/2), each needs 2 ones among 4: C(4,2)/16 = 6/16, so this case gives (1/2)(6/16)² = 36/512. If a₂₃ = 0, each needs 3 ones among 4: 4/16, giving (1/2)(4/16)² = 16/512. Total = 52/512 ≈ 0.1016. Answer: 0.10 (accepted range 0.08 to 0.12).

Concept: Conditioning on the shared cell. The trap is treating the row and column sums as independent and squaring C(5,3)/32.

A is symmetric and idempotent (the centering projection), so its eigenvalues are 0 (eigenvector 𝟏) and 1 (any vector orthogonal to 𝟏, multiplicity n - 1). The maximum of xᵀAx over unit vectors is the largest eigenvalue (Rayleigh quotient), which is 1, attained e.g. at x = (1, -1, 0, ..., 0)/√2. Answer: 1.

Concept: Rayleigh quotient maximum equals the largest eigenvalue; projections have eigenvalues 0 and 1. The trap is answering n - 1 (the trace) instead of the largest eigenvalue.