GATE DA 2026 question paper with solutions

GATE DA 2026 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: B
2. Q.2MCQ · 1 markOfficial answer: B
3. Q.3MCQ · 1 markOfficial answer: A
4. Q.4MCQ · 1 markOfficial answer: C
5. Q.5MCQ · 1 markOfficial answer: D
6. Q.6MCQ · 2 marksOfficial answer: A
7. Q.7MCQ · 2 marksOfficial answer: B
8. Q.8MCQ · 2 marksOfficial answer: D
9. Q.9MCQ · 2 marksOfficial answer: C
10. Q.10MCQ · 2 marksOfficial answer: C

Data Science and Artificial Intelligence Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: B

   Machine Learning › PCA

   Principal components are eigenvectors of the symmetric covariance matrix, so they form an orthonormal set. Any two distinct components are perpendicular, whichever two you pick and however many you keep.

   WATCH OUT FOR

   Thinking that the first and the last retained components are somehow "far apart" or correlated, or that dimensionality reduction tilts the retained axes. PCA axes are always mutually orthogonal, and discarding components does not change that.

   Full solution →
2. Q.12MCQ · 1 markOfficial answer: C

   Machine Learning › Bias–variance & CV

   LOOCV makes one split per sample in the pool used for model selection. The held-out test set is excluded from that pool, so the number of splits is $1000-100=900$.

   WATCH OUT FOR

   Answering 1000 by counting every sample, which pulls the test data into model selection. Another slip is to link the split count to the number of classes (10) or the feature dimension (512), which have no bearing on it.

   Full solution →
3. Q.13MCQ · 1 markOfficial answer: C

   Artificial Intelligence › Search

   Uninformed search uses no goal-distance estimate. Only a heuristic $h(n)$, as in $f(n)=g(n)+h(n)$ for A*, makes a search informed.

   WATCH OUT FOR

   Misreading the question: it asks which algorithm is NOT uninformed, so the answer is the informed one. Choosing a blind search such as DFS answers the opposite question.

   Full solution →
4. Q.14MCQ · 1 markOfficial answer: C

   Artificial Intelligence › Logic

   A universal quantifier over a restricted class ("each king", "all humans") pairs with an implication $\Rightarrow$, never a conjunction $\wedge$. An existential statement ("some king is a person") pairs with $\wedge$.

   WATCH OUT FOR

   Writing $\forall x(P(x)\wedge Q(x))$ for "every $P$ is a $Q$". That claims every object in the domain is a $P$ and a $Q$, which is far stronger than the English sentence.

   Full solution →
5. Q.15MCQ · 1 markOfficial answer: D

   Data Structures and Algorithms › Design techniques

   An expected-time recurrence for quick sort averages over all $n$ equally likely pivot ranks. Each split $(k,n-k-1)$ has probability $\frac{1}{n}$, and the linear partition cost is added on top.

   WATCH OUT FOR

   Choosing a fixed-split recurrence. Option (A) is the worst case, while (B) and (C) assume a constant-ratio split, such as the balanced $\frac{1}{2}:\frac{1}{2}$ split. Expected time needs the average over every split, not one assumed split.

   Full solution →
6. Q.16MCQ · 1 markOfficial answer: B

   Data Structures and Algorithms › Python

   A default argument such as lst=[] is evaluated once, when the function is defined, so every call that omits it shares one mutable list. A list passed explicitly is a different object and does not touch the shared default.

   WATCH OUT FOR

   Assuming lst=[] creates a fresh empty list on every call, which gives [1], [2], [3] (option A). The second call reuses the list that already holds 1, so it prints [1, 2].

   Full solution →
7. Q.17MCQ · 1 markOfficial answer: A

   Database Management and Warehousing › Normal forms

   First classify attributes: one that never appears on a right-hand side must be in every key, and one that never appears on a left-hand side is in no key. Then take attribute closures of the remaining combinations and keep the minimal ones.

   WATCH OUT FOR

   Treating $A$, $E$ or $CD$ as keys because $A\to BC$ and $E\to A$ look like they reach everything. These never derive $D$, which appears on no right-hand side, so $A$ and $E$ alone cannot be keys. Another slip is stopping after finding $AD$ and $CD$ and missing $ED$.

   Full solution →
8. Q.18MCQ · 1 markOfficial answer: D

   Database Management and Warehousing › Data warehousing

   Changing the level of a dimension's hierarchy is drill-down (to a finer level) or roll-up (to a coarser level). Slicing and dicing only select part of the cube and keep the level of detail.

   WATCH OUT FOR

   Picking dicing or slicing because the question says 'for each combination of' attributes. These operations only restrict values. They do not change the level of detail. Another slip is reversing the direction and choosing roll-up, but State is finer than Country, so the move is down the hierarchy.

   Full solution →
9. Q.19MCQ · 1 markOfficial answer: C

   Probability and Statistics › Counting & axioms

   Use the complement. A product is odd only if every element is odd. Count non-empty subsets of the $1013$ odd numbers, and divide by the number of non-empty subsets of $S$.

   WATCH OUT FOR

   Dividing by $2^{2026}$ (counting the empty set as an outcome) gives option (A). The question says $M$ is non-empty, so the denominator is $2^{2026}-1$. A second slip is to forget to remove the empty set from the all-odd count, which should be $2^{1013}-1$.

   Full solution →
10. Q.20MCQ · 1 markOfficial answer: A

    Probability and Statistics › Counting & axioms

    Count with stars and bars. All non-negative solutions: $\left(\genfrac{}{}{0ex}{}{n+k-1}{k-1}\right)$. For positive solutions, subtract 1 from each variable first, so the sum falls from $n$ to $n-k$.

    WATCH OUT FOR

    Using $\left(\genfrac{}{}{0ex}{}{20}{3}\right)$ or $\left(\genfrac{}{}{0ex}{}{19}{4}\right)$ for the count, which comes from mixing up "$n+k-1$ choose $k-1$" with "$n$ choose $k$". Another slip is forgetting to subtract 4 from 20 for the positive case.

    Full solution →
11. Q.21MCQ · 1 markOfficial answer: B

    Linear Algebra › Special matrices

    A rotation matrix by $\theta =\frac{2\pi }{5}$ has order 5, since $M^5=I_2$. Reduce the exponent modulo 5: $2026\equiv 1$.

    WATCH OUT FOR

    Reducing the exponent incorrectly, for example reading $2026\equiv 2$ or $2026\equiv 0(\mathrm{mod}5)$. That gives $M^2$ or $I_2$. The remainder of $2026$ on division by 5 is 1.

    Full solution →
12. Q.22MCQ · 1 markOfficial answer: A

    Linear Algebra › Vector spaces

    A two-dimensional subspace of ${\mathbb{R}}^3$ is a plane through the origin. The origin is the centre of the ball $x^{T}x\le 16$, so every such plane cuts the ball in a disc of the full radius, $4$.

    WATCH OUT FOR

    Treating $S_2$ as a general plane that might miss the centre, and concluding the area depends on $S_2$. A subspace must contain the origin, so the cross-section is always the largest disc. Another slip is taking the radius as $16$ instead of $\sqrt{16}=4$, which gives $256\pi$. A third is confusing the disc area $\pi r^2$ with the surface area or volume of the ball.

    Full solution →
13. Q.23MSQ · 1 markOfficial answer: B

    Machine Learning › LDA, SVM & trees

    Match each task to the algorithm built for it. Clustering goes with K-Medoid, classification with Naive Bayes, sampling with Markov Chain Monte Carlo, and feature extraction with Linear Discriminant Analysis, which projects onto class-separating directions. Clustering is the quickest way to eliminate the wrong options.

    WATCH OUT FOR

    Pairing clustering with a supervised method such as Naive Bayes or Linear Discriminant Analysis, or pairing Linear Discriminant Analysis with classification only and forgetting its use as a feature-extraction (dimensionality-reduction) step. It can also be used to classify, but that does not make T1:A3 or T1:A4 valid.

    Full solution →
14. Q.24MSQ · 1 markOfficial answer: A, B, C

    Artificial Intelligence › Logic

    $X$ entails $Y$ exactly when $X\Rightarrow Y$ is true in every model. That is the same as saying $X\wedge \neg Y$ is never true. Options A, B and C are three ways of writing this one condition.

    WATCH OUT FOR

    Treating 'if $Y$ then $X$' as also following. That is the converse, and entailment goes in one direction only.

    Full solution →
15. Q.25MSQ · 1 markOfficial answer: A, B

    Data Structures and Algorithms › Trees & heaps

    The first node of the pre-order is the root; its position in the in-order splits the remaining nodes into the left and right subtrees. Repeat on each part, using the subtree sizes to cut the pre-order.

    WATCH OUT FOR

    Splitting the pre-order at the wrong place, or treating a node as having only one child without checking both sides of it in the in-order. Here $Q$ has $S$ on its left and $E$ on its right, so (C) is false. Another slip is thinking $G$ is in $R$'s left subtree, when it appears after $R$ in the in-order.

    Full solution →
16. Q.26MSQ · 1 markOfficial answer: A, B

    Database Management and Warehousing › Normal forms

    A foreign key only constrains the child's values. Adding a parent row or removing a child row cannot create a dangling reference. Adding a child row or removing a parent row can.

    WATCH OUT FOR

    Treating deletion from the parent as safe because the key is "still unique", or forgetting that inserting into the child needs a matching parent value. Both of these can violate the constraint.

    Full solution →
17. Q.27MSQ · 1 markOfficial answer: B, D

    Calculus and Optimization › Maxima & minima

    On a restricted domain, find the global extrema by comparing the critical-point values with the values at the endpoints. An endpoint can tie with a critical value, and an excluded endpoint is never attained.

    WATCH OUT FOR

    Taking $x=0$ as the only maximum and ignoring that $f(3)=2$ equals $f(0)$ because $3$ is included. A second slip is to treat $f(-1)=-2$ as a second minimum, although $-1$ is excluded from the domain. A third is to count both roots $1\pm \sqrt{3}$ as lying in $[-0.9,0]$, when only $1-\sqrt{3}\approx -0.732$ does.

    Full solution →
18. Q.28MSQ · 1 markOfficial answer: A, C

    Probability and Statistics › Continuous distributions

    A $t_1$ distribution is the standard Cauchy distribution, whose tails are heavier than the normal's. The two densities cross once for $c>0$, at $c\approx 1.85$. Beyond that point the Cauchy has more tail mass, so $G(-c)<H(-c)$ and $G(c)>H(c)$. At $0$, both CDFs equal $\frac{1}{2}$ by symmetry, but the peak densities differ.

    WATCH OUT FOR

    Assuming that because $g(c)=h(c)$ the CDFs must also be ordered the same way at $c$ and $-c$. Both curves are symmetric, so the lower-tail comparison is the opposite of the comparison at $+c$. Another slip is to think $g(0)=h(0)$ because both bells are centred at $0$.

    Full solution →
19. Q.29NAT · 1 markOfficial answer: 8.90 to 9.10

    Machine Learning › Regression

    For a linear model $f_w(x)=wx$ the gradient with respect to $w$ is just the input $x$. One SGD step is therefore $w\leftarrow w-\eta x$.

    WATCH OUT FOR

    Differentiating with respect to $x$ instead of $w$ gives $w$, i.e. $10$, and a wrong update of $10-0.1\times 10$ for the wrong reason. Another slip is to multiply the learning rate by $wx=100$ and get $0$.

    Full solution →
20. Q.30NAT · 1 markOfficial answer: 2

    Artificial Intelligence › Search

    In minimax, back up values from the leaves. MIN nodes take the minimum of their children and MAX nodes take the maximum. The question asks for the edge label (the strategy), not the game value.

    WATCH OUT FOR

    Giving the game value (1) instead of the strategy number (2). The other slip is taking the maximum at the MIN level, which would pick strategy 1 (leaf 8 beats 7 and -3).

    Full solution →
21. Q.31NAT · 1 markOfficial answer: 10

    Data Structures and Algorithms › Sorting & hashing

    Binary search on $n$ elements halves the range at each comparison. An unsuccessful search runs until the range is empty, so the worst case is $\lfloor {\log }_{2}n\rfloor +1$ comparisons when each step uses one three-way comparison.

    WATCH OUT FOR

    Answering $\lceil {\log }_{2}1000\rceil =10$ by coincidence, or counting the final empty-range check as an extra comparison and answering 11. Another slip is using ${\log }_{2}1000\approx 9.97$ and writing 9 or 10 without checking the halving sizes.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: 195

    Database Management and Warehousing › Indexing

    An internal node with $n$ node pointers has $n-1$ search keys and no data record pointers. Solve $n\cdot p+(n-1)\cdot k\le \text{block size}$ and take the floor.

    WATCH OUT FOR

    Counting $n$ keys instead of $n-1$ gives $21n\le 4096$, so $n=195$ by coincidence here but a wrong method. Adding the 12-byte data record pointer to a non-leaf node is also wrong, because those pointers are held only in leaf nodes. Forgetting to take the floor of $195.57$ is a third slip.

    Full solution →
23. Q.33NAT · 1 markOfficial answer: 10

    Probability and Statistics › Counting & axioms

    $f(f(n))=n$ means $f$ is an involution, so its cycles have length 1 or 2. Count permutations by the number of 2-cycles: $1+6+3=10$.

    WATCH OUT FOR

    Counting only the permutations made of swaps and forgetting the identity and the single-swap cases. Another slip is counting the two-pair case as $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ instead of $3$, which counts each unordered pair of pairs twice.

    Full solution →
24. Q.34NAT · 1 markOfficial answer: 0.34 to 0.36

    Probability and Statistics › Continuous distributions

    The exponential distribution is memoryless, so $P(X>10\mid X>5)=P(X>5)$. The given $0.35$ is the answer directly, and $\lambda$ never has to be found.

    WATCH OUT FOR

    Solving for $\lambda$ first and then computing $P(X>10)$ alone, which gives $0.35^2\approx 0.12$. That forgets to divide by $P(X>5)$, so it is not a conditional probability.

    Full solution →
25. Q.35NAT · 1 markOfficial answer: 1

    Calculus and Optimization › Taylor series

    A double sum of positive terms whose summand factors into a function of $i$ times a function of $j$ equals the product of the two single sums. Each factor here is a geometric series, ${\sum }_{k\ge k_0}{r}^k=r^{k_0}/(1-r)$ for $|r|<1$.

    WATCH OUT FOR

    Using the formula $1/(1-r)$ for the $j$ series, which starts at $j=1$, not $j=0$. That gives $3/2$ in place of $1/2$ and a final answer of $3$. The first term of the $j$ series is $\frac{1}{3}$, not $1$.

    Full solution →
26. Q.36MCQ · 2 marksOfficial answer: D

    Machine Learning › Clustering

    The first merge in agglomerative clustering takes the closest pair of individual points. The linkage rule only matters from the second merge onward, so just compute all pairwise distances in the stated metric and take the minimum.

    WATCH OUT FOR

    Using Euclidean distance instead of Manhattan, or dropping a minus sign when subtracting coordinates such as $-1$ and $-2$. Another slip is to stop at the first small value, such as $d(P1,P2)=5$, without checking all six pairs. $P1,P4$ also has distance 5, and the true minimum is 4.

    Full solution →
27. Q.37MCQ · 2 marksOfficial answer: D

    Machine Learning › Regression

    Ridge adds a squared $L_2$ penalty $\lambda \Vert \mathbf{w}{\Vert }_2^2$. It shrinks the weights, which adds a little bias and lowers variance (the bias-variance trade-off). It is a guard against overfitting.

    WATCH OUT FOR

    Reading (A) carelessly. Overfitting means good on training data and poor on test data, and (A) says the opposite. Another slip is mixing up Ridge ($L_2$) with Lasso ($L_1$).

    Full solution →
28. Q.38MCQ · 2 marksOfficial answer: B

    Artificial Intelligence › Logic

    A statement of the form "if X then (S iff D)" is a conditional. It says nothing when the condition $C$ is false, so any statement that must hold for non-creative people as well is not guaranteed.

    WATCH OUT FOR

    Reading the assumption as a full biconditional between $D$ and success for everyone, and so accepting (i). The assumption applies only to creative people, so $C=\text{F}$, $D=\text{T}$ breaks (i). Another slip is missing that the right side of (iii) is a tautology, which turns (iii) into plain $C$.

    Full solution →
29. Q.39MCQ · 2 marksOfficial answer: C

    Data Structures and Algorithms › Python

    Count activations, not the returned value. The recurrence is $C(n)=1+C(n-1)+C(n-2)$ with $C(n)=1$ for $n\le 0$. Note that n <= 0 makes both $n=0$ and negative $n$ base cases.

    WATCH OUT FOR

    Treating only n == 0 as a leaf, or forgetting that mystery(n-2) can reach $n=-1$, which is also a leaf. Another slip is answering 5 (the depth plus one) or 9 (the count for mystery(3)), or counting only the leaves (8) or only the internal calls (7).

    Full solution →
30. Q.40MCQ · 2 marksOfficial answer: D

    Data Structures and Algorithms › Graph algorithms

    Reversing all edges reverses every path, so $v⇝u$ in $G$ if and only if $u⇝v$ in $G^R$. BFS and DFS both visit exactly the reachable set, so only statements phrased through this path reversal always hold.

    WATCH OUT FOR

    Treating reachability from $v$ in $G^R$ as the same as reachability from $v$ in $G$ (options A and B), or assuming BFS and DFS orders are related by reversal (option C). Only the start and end of the path swap, not the starting vertex.

    Full solution →
31. Q.41MCQ · 2 marksOfficial answer: A

    Database Management and Warehousing › Indexing

    A leaf split copies the separator up and keeps it in the leaf. A non-leaf split moves the middle key up. Here the inserted 3 overflows the leaf, which makes the root overflow, and the middle key 5 becomes the new root.

    WATCH OUT FOR

    Stopping after the leaf split and concluding the root is $3,5,8$ or $3,5$ (options D or C). With at most 3 pointers per non-leaf node, 3 keys and 4 pointers cannot stay in one node, so the root must split too. Another mistake is to copy the middle key up from the internal node instead of moving it, which would leave 5 duplicated in the internal level.

    Full solution →
32. Q.42MCQ · 2 marksOfficial answer: D

    Database Management and Warehousing › Relational algebra

    A natural join keeps only tuples that agree on every common attribute. Evaluate each side of the final join separately and compare them on the shared attributes $P,Q,S$.

    WATCH OUT FOR

    Reading the final join as being on $P$ alone, or forgetting that the selection with $Q=Q3\vee R=R2$ removes the $P1$ row from the left side. Another slip is to treat $Y\bowtie Z$ as the only source of rows and read off $(P1,R1,2)$ and $(P1,R1,5)$, ignoring that the left side contains only $P3$.

    Full solution →
33. Q.43MCQ · 2 marksOfficial answer: A

    Database Management and Warehousing › Data warehousing

    With concept hierarchies, a cuboid picks one level per dimension, including the top level 'all'. The count is $\prod (L_i+1)$, where $L_i$ is the number of levels of dimension $i$ excluding 'all'.

    WATCH OUT FOR

    Using $2^n$, the count for dimensions without hierarchies, and choosing option (B). Another slip is counting only the $3$ named levels per dimension and forgetting the virtual level 'all', which gives $3^3$ and is not an option.

    Full solution →
34. Q.44MCQ · 2 marksOfficial answer: A

    Probability and Statistics › Random variables

    For independent $W$ and $Y$ with $E[Y]=0$, $\mathrm{Var}(WY)=E[W^2]E[Y^2]$. Here $2X-1$ takes only the values $\pm 1$, so $E[W^2]=1$ and the variance equals $\mathrm{Var}(Y)$, whatever $p$ is.

    WATCH OUT FOR

    Using $\mathrm{Var}(2X-1)\mathrm{Var}(Y)=0.84\times 100=84$, which wrongly treats the variance of a product as the product of the variances. Another slip is using $4p(1-p)\cdot 100$ and ignoring that $E[(2X-1{)}^2]=1$, not $0.84$.

    Full solution →
35. Q.45MCQ · 2 marksOfficial answer: A

    Probability and Statistics › CLT & tests

    The sum is the Poisson($n$) cdf evaluated at its own mean $n$. A Poisson($n$) variable is a sum of $n$ independent Poisson(1) variables, so the central limit theorem makes it approximately normal, and the normal cdf at its mean is $\frac{1}{2}$.

    WATCH OUT FOR

    Reading the sum as the whole Poisson series and answering $1$ (option B). That would need the sum to run to infinity. It stops at $k=n$, which is the mean, so only about half the probability mass is included. Another trap is answering $e^{-1}$ by looking only at the $e^{-n}$ factor.

    Full solution →
36. Q.46MCQ · 2 marksOfficial answer: C

    Linear Algebra › Eigenvalues

    The sum of the eigenvalues is the trace, so there is no need to find the eigenvalues. Here the trace is $1+2\cos t$. Solve $\cos t=\frac{\sqrt{2}}{2}$ on $[-\pi ,\pi ]$, where both $t$ and $-t$ qualify because cosine is even.

    WATCH OUT FOR

    Computing the characteristic polynomial and the complex eigenvalues $\cos t\pm i\sin t$ takes longer and invites sign errors. Another slip is to give only the positive angle, or to pair $\frac{\pi }{4}$ with $-\frac{\pi }{3}$ because $\sin t$ also appears in the matrix. The trace involves only $\cos t$, so $t$ and $-t$ always give the same sum.

    Full solution →
37. Q.47MSQ · 2 marksOfficial answer: A, B

    Machine Learning › Bias–variance & CV

    Build the 2x2 confusion matrix from the counts. Precision divides by the column total (everything predicted as that class); recall divides by the row total (everything that truly is that class); accuracy divides the diagonal by the grand total.

    WATCH OUT FOR

    Mixing up precision and recall by dividing by the wrong total. Another slip is computing accuracy as 14/15 or from one class only. Here the class with the higher recall (Y) has the lower precision, so swapping the two definitions gives the wrong verdict on (B) and (C).

    Full solution →
38. Q.48MSQ · 2 marksOfficial answer: B, C, D

    Artificial Intelligence › Logic

    A formula is valid only if no interpretation falsifies it. Because first-order domains are non-empty, $\forall \to \exists$ is valid, but $\exists \to \forall$ and anything claiming $\forall xP(x)$ forces $\exists x\neg P(x)$ each have a simple counter-model.

    WATCH OUT FOR

    Reading "NOT valid" as "never true", or confusing satisfiable with valid. A statement can be true in some models (as (B) and (D) are) and still not be valid. Also, forgetting that the question asks for the NOT valid ones and picking (A).

    Full solution →
39. Q.49MSQ · 2 marksOfficial answer: B, C, D

    Data Structures and Algorithms › Sorting & hashing

    Count comparisons directly. Insertion sort on an almost-sorted array costs one comparison per already-placed element plus the extra comparisons for the one displaced element. Bubble sort needs at least two passes here, because the early-exit pass that confirms sortedness is still a full pass of comparisons.

    WATCH OUT FOR

    Counting only the comparisons that lead to a swap, which gives N2 = 1 or 4. Also assuming IS needs only n - 1 = 4 comparisons. Inserting 4 needs two comparisons, so N2 = 5. Another slip is taking N1 = 4 by forgetting the second pass that confirms the array is sorted.

    Full solution →
40. Q.50MSQ · 2 marksOfficial answer: B, D

    Data Structures and Algorithms › Python

    A closure captures the variable of the specific call to outer() that created it. Every call to outer() makes a fresh list, so each returned function keeps its own list. append changes that list in place, so the earlier values stay in it.

    WATCH OUT FOR

    Assuming f1 and f2 share one list because they come from the same outer definition, which leads to choosing (A) and (C). The list is created inside outer, so it is new on every call and is not shared.

    Full solution →
41. Q.51MSQ · 2 marksOfficial answer: A

    Database Management and Warehousing › SQL

    To attach a group size to every row, aggregate with GROUP BY on the grouping column (TeamID) in a derived table, keep that column in the derived table's SELECT list, and join it back to the base table on that column.

    WATCH OUT FOR

    Grouping by the primary key (EmpID), as in (B) and (C), gives one row per group and a count of 1. Dropping TeamID from the derived table, as in (D), leaves nothing to join on, so the query is invalid.

    Full solution →
42. Q.52MSQ · 2 marksOfficial answer: A, D

    Linear Algebra › Special matrices

    $\frac{1}{n}11^T$ is the orthogonal projection onto $1$, so $I_n$ minus it is the orthogonal projection onto the complement: symmetric, idempotent, with trace equal to its rank $n-1$.

    WATCH OUT FOR

    Forgetting that a projection matrix satisfies $M^2=M$ rather than $M^2=I_n$, or taking $\mathrm{tr}(M)=n$ by ignoring the subtracted rank-one part, whose trace is 1.

    Full solution →
43. Q.53MSQ · 2 marksOfficial answer: A, B, C

    Probability and Statistics › Continuous distributions

    A sum of squares of independent standard normals is chi-square with one degree of freedom per independent term. Replacing the true mean with $\bar{X}$ costs one degree of freedom. ${\chi }_2^2$ is exponential with mean 2.

    WATCH OUT FOR

    Judging (D) from the factor $\sqrt{n}$ or from the sum, and giving it 2 or $n$ degrees of freedom. $\sqrt{n}\bar{X}$ is a single standard normal, so its square is ${\chi }_1^2$. Another slip is to treat (B) as ${\chi }_n^2$ and forget the lost degree of freedom.

    Full solution →
44. Q.54MSQ · 2 marksOfficial answer: B, C

    Probability and Statistics › Random variables

    Any CDF $F(x)=P(X\le x)$ is non-decreasing and right continuous, and its jump at a point equals the point mass there. A discrete variable has such jumps, so $F$ has jump discontinuities and is not left continuous.

    WATCH OUT FOR

    Mixing up right and left continuity. Because $F$ uses $\le$, it is right continuous, not left continuous. Another slip is calling $F$ "positive" when it is only non-negative, since it equals $0$ to the left of the smallest value of $X$.

    Full solution →
45. Q.55NAT · 2 marksOfficial answer: 6.90 to 7.10

    Machine Learning › Regression

    Regularized loss = data-fit term (here absolute error, because MAE is specified) + $\lambda \Vert w{\Vert }_2^2$. Compute the prediction first, then the error, then the penalty on the given weight vector.

    WATCH OUT FOR

    Using squared error instead of absolute error (giving $4+5=9$), or penalising with $\lambda \Vert w{\Vert }_2$ (not squared) or $\frac{\lambda }{2}\Vert w{\Vert }^2$, or forgetting the penalty term altogether (giving 2).

    Full solution →
46. Q.56NAT · 2 marksOfficial answer: 135

    Machine Learning › Neural networks

    With no biases, count one weight per connection between adjacent layers: the sum of the products of consecutive layer sizes, $30\cdot 4+4\cdot 3+3\cdot 1$.

    WATCH OUT FOR

    Adding a bias for each neuron out of habit (which would give $135+4+3+1=143$), or adding the layer sizes instead of multiplying consecutive pairs.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 0.76 to 0.78

    Probability and Statistics › Conditional & Bayes

    The question asks for $P(D\mid +)$ while the data give $P(+\mid D)$, so reverse the conditioning with Bayes' theorem. Find the denominator $P(+)$ first, using total probability over $D$ and $\bar{D}$.

    WATCH OUT FOR

    Reporting $P(+\mid D)=0.80$ as the answer, which confuses the two conditional probabilities. Another slip is dividing $0.24$ by $0.8+0.1$ and ignoring the prior weights $0.3$ and $0.7$ in the denominator.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 8

    Data Structures and Algorithms › Python

    fun is one bubble-sort pass that returns its swap count, and the loop makes 5 passes on the same mutated list. The total is the number of inversions of the original list: $4+2+2=8$.

    WATCH OUT FOR

    Treating each call as acting on a fresh copy of [5, 3, 4, 1, 2], which would give $5\times 4=20$. The list is mutated in place and the loop passes the same object every time. Another slip is stopping after the first pass (4) and forgetting that later passes still swap.

    Full solution →
49. Q.59NAT · 2 marksOfficial answer: 2 to 3

    Database Management and Warehousing › Relational algebra

    The first existential is trivially true because $t$ can serve as $z$. The second is a semi-join-like test on $X$ against $Y$ with the filter $S>1$. The result is a set of tuples from $X$, so each qualifying tuple of $X$ appears once.

    WATCH OUT FOR

    Counting 3 tuples by treating $(P3,1)$ as a match even though $1>1$ is false. Counting 3 by treating the expression as a join, so that $P1$ appears twice. A join would repeat P1, but this expression returns tuples $t$ of $X$ only.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 3

    Database Management and Warehousing › SQL

    For each outer row, a correlated subquery counts the rows with a strictly larger balance and the rows with a strictly smaller balance. The row qualifies when the larger-count is at least the smaller-count. Rows with equal balance fall in neither count.

    WATCH OUT FOR

    Counting tied balances (the two 5000 rows) as greater or smaller. Both comparisons are strict, so a tied row counts on neither side. Another slip is reversing the two counts, which would keep A3, A4 and A5 and still give 3 rows by coincidence of a different set. Evaluate each row explicitly.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 4

    Database Management and Warehousing › ER model

    In ER-to-relational mapping, each entity gets a relation, each multi-valued attribute gets its own relation, and each many-to-many relationship gets its own relation. Total participation only affects foreign-key nullability when a relationship is merged, so it does not change the count here.

    WATCH OUT FOR

    Counting only 3 relations by folding the multi-valued attribute $A_{22}$ into $E2$, or by merging $R12$ into an entity relation because participation is total. Both break first normal form. Another slip is to count only the two entities plus the relationship and miss the separate relation for $A_{22}$.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 10

    Probability and Statistics › Descriptive statistics

    The double sum of squared pairwise differences equals $2n$ times the sum of squared deviations from the mean, ${\sum }_i{\sum }_j(x_i-x_j{)}^2=2n{\sum }_i(x_i-\bar{x}{)}^2$. This turns the given condition into the sample variance (divisor $n-1=99$).

    WATCH OUT FOR

    Forgetting the factor 2 (writing $n{S}_{xx}$ instead of $2n{S}_{xx}$), which gives 20. Another slip is to divide the double sum by $n^2$ and treat it as a variance with divisor $n$ instead of $n-1$.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 0

    Probability and Statistics › Descriptive statistics

    Find $E[Y\mid X]=X^2$, then use the tower property to get $E[XY]=E[X^3]=0$. $X$ is symmetric about 0, so the covariance and hence the correlation are zero, although $Y$ depends on $X$.

    WATCH OUT FOR

    Assuming that a strong dependence of $Y$ on $X$ (here through $X^2$) means a non-zero correlation. Correlation detects only linear association. Zero correlation does not imply independence.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 0.08 to 0.12

    Probability and Statistics › Discrete distributions

    The row and the column overlap in one entry, $a_{23}$. Condition on that entry. Given it, the other 8 entries are independent, so the row event and the column event are independent and each is a Binomial$(4,\frac{1}{2})$ probability.

    WATCH OUT FOR

    Multiplying $P(\text{row sum}=3)\cdot P(\text{column sum}=3)=(10/32{)}^2$ as if the events were independent. They are not, because they share $a_{23}$. The result happens to round to the same $0.10$ here, but the method is wrong.

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 1

    Linear Algebra › Special matrices

    $I_n-\frac{1}{n}11^T$ is the centering matrix, a symmetric projection with eigenvalues $0$ (once) and $1$ ($n-1$ times). The Rayleigh quotient maximum is the largest eigenvalue, $1$.

    WATCH OUT FOR

    Taking the single eigenvalue $0$ (from the direction $1$) as the answer, or forgetting that the maximum is the largest eigenvalue rather than the smallest. Another slip is to think the value depends on $n$.

    Full solution →