GATE DA 2024 question paper with solutions

GATE DA 2024 answer key

General Aptitude Q.1 to Q.10

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

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

1. Q.11MCQ · 1 markOfficial answer: A

   Probability and Statistics › Continuous distributions

   A Poisson($\lambda$) variable has mean $\lambda$ and variance $\lambda$. The standard normal $N(0,1)$ is the normal with mean 0 and variance 1. Both statements hold.

   WATCH OUT FOR

   Confusing the second parameter of $N(\mu ,{\sigma }^2)$ with the standard deviation, or doubting that a Poisson variable's variance equals its mean. Both statements are standard facts, so there is nothing to trap you.

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

   Probability and Statistics › Counting & axioms

   With 3 tosses, heads plus tails equals 3. Two events that each need at least 2 of the 3 tosses cannot both occur, so they are mutually exclusive and the intersection is empty.

   WATCH OUT FOR

   Multiplying $P(T)P(S)=0.25$ as if $T$ and $S$ were independent, and picking (C). Independent coins do not make $T$ and $S$ independent. In fact $S$ is determined by the same tosses and cannot occur when $T$ does.

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

   Linear Algebra › Eigenvalues

   For a $2\times 2$ real matrix, find the sign of the discriminant of ${\lambda }^2-\mathrm{tr}(\mathbf{M})\lambda +\det (\mathbf{M})$. A negative discriminant means a complex conjugate pair, with no real eigenvalue.

   WATCH OUT FOR

   Computing the determinant as $2-3=-1$ by dropping the sign of the $-1$ entry. That gives ${\lambda }^2-3\lambda -1=0$, which has real roots of opposite sign and wrongly points to (D). The correct determinant is $(2)(1)-(-1)(3)=5$.

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

   Data Structures and Algorithms › Graph algorithms

   In undirected DFS, an edge first explored from $u$ to $v$ is a tree edge if $v$ is white and a back edge if $v$ is gray. It can never be a cross edge, and 'gray' is a vertex colour, not an edge type.

   WATCH OUT FOR

   Choosing 'cross' or 'gray': in a DFS of an undirected graph every edge is either a tree edge or a back edge, never a cross edge, and 'gray' is a vertex colour, not an edge type.

   The question and its official answer →
5. Q.15MCQ · 1 markOfficial answer: A

   Calculus and Optimization › Maxima & minima

   At a stationary point, $f^{\prime \prime }(x^{\ast })>0$ (second-derivative test) guarantees a local minimum only. It says nothing about global behaviour.

   WATCH OUT FOR

   Choosing "global minimum" because the second derivative is positive. $f^{\prime \prime }(x^{\ast })>0$ is a local condition. Global minimality needs convexity everywhere, i.e. $f^{\prime \prime }\ge 0$ on all of $\mathbb{R}$.

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

   Data Structures and Algorithms › Linear structures

   Match each item to the structure that defines it: FIFO is the queue discipline, LIFO is the stack discipline, and key-based lookup is what a hash table is built for.

   WATCH OUT FOR

   Swapping FIFO and LIFO, so that stacks are paired with first-in-first-out (option D). Remember that a queue serves the oldest item and a stack serves the newest.

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

   Machine Learning › LDA, SVM & trees

   Support vectors are the points exactly on the margin. Find the maximum-margin hyperplane (the bisector of the closest pair of points in the two convex hulls), then pick the points with $y_i({\mathbf{w}}^{\top }{\mathbf{x}}_i+b)=1$.

   WATCH OUT FOR

   Assuming the far points $(2,2)$ and $(-2,-2)$ must be support vectors because they look extreme. Support vectors are the points closest to the boundary, not the farthest ones; points well outside the margin have zero multiplier.

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

   Machine Learning › Logistic, k-NN, naive Bayes

   A generative model learns the joint distribution $P(\mathbf{x},y)$, usually as $P(y)P(\mathbf{x}\mid y)$ (Naive Bayes). A discriminative model learns $P(y\mid \mathbf{x})$ directly (logistic regression). PCA is unsupervised and has no class model, so it is dimensionality reduction.

   WATCH OUT FOR

   Swapping Naive Bayes and logistic regression because both are called classifiers that output class probabilities. What separates them is whether $P(\mathbf{x}\mid y)$ is modelled (generative) or only $P(y\mid \mathbf{x})$ (discriminative). Another slip is labelling PCA a generative model.

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

   Machine Learning › Clustering

   Each $k$-means cluster is a Voronoi cell, which is convex. Any point on the segment between two members of a cluster is therefore also in that cluster.

   WATCH OUT FOR

   Picking $(0,0)$ because it is the origin or the "centre" of the data. The origin is not on the segment between the two points, so a centroid can sit near it and claim it for another cluster.

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

    Machine Learning › Logistic, k-NN, naive Bayes

    Conditional independence given the class turns the joint over $K$ attributes into $K$ separate one-parameter Bernoulli distributions per class. Count the free parameters: 1 for the prior, plus $K\times 2$ for the likelihoods.

    WATCH OUT FOR

    Counting the full joint table $2^K$ per class, which ignores the independence assumption. Another slip is counting both $P(X_i=0\mid y)$ and $P(X_i=1\mid y)$, or both class priors, instead of only the free parameters.

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

    Data Structures and Algorithms › Sorting & hashing

    In open addressing, an insertion probes exactly as an unsuccessful search does, until the first empty slot, so it has the same bound $\frac{1}{1-\alpha }$.

    WATCH OUT FOR

    Picking $\ln \frac{1}{1-\alpha }$ (option A) by mixing up the insertion cost with the successful-search cost, which is about $\frac{1}{\alpha }\ln \frac{1}{1-\alpha }$. An insertion always needs at least one probe and its cost blows up as $\alpha \to 1$.

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

    Machine Learning › LDA, SVM & trees

    Maximising the ratio $\frac{u^T{S}_{B}u}{u^T{S}_{W}u}$ gives the generalised eigenproblem $S_{B}u=\lambda S_{W}u$. With $S_W$ invertible this becomes the ordinary eigenproblem for $S_W^{-1}{S}_B$, and the maximum $J$ is its largest eigenvalue.

    WATCH OUT FOR

    Swapping the matrices and writing $S_{W}u=\lambda S_{B}u$, or multiplying in the wrong order ($S_B{S}_W$). The inverse must multiply $S_B$ from the left, so that it comes from $S_{B}u=\lambda S_{W}u$.

    Full solution →
13. Q.23MCQ · 1 markOfficial answer: D

    Artificial Intelligence › Search

    Admissible means $0\le h\le h^{\ast }$ at every node. The absolute difference of two values in $[0,h^{\ast }]$ stays in $[0,h^{\ast }]$, but a sum, product or ratio can exceed $h^{\ast }$.

    WATCH OUT FOR

    Choosing $h_1+h_2$ because combining two good heuristics seems to give a stronger one. Adding two admissible estimates can double-count the remaining cost and overestimate. Only the maximum (or minimum) of admissible heuristics is guaranteed admissible.

    Full solution →
14. Q.24MCQ · 1 markOfficial answer: C

    Artificial Intelligence › Uncertainty

    Read the factorisation as a Bayesian network and apply d-separation. Chains and forks are blocked by conditioning on the middle node, but a collider such as $U\to W\leftarrow V$ is opened by conditioning on $W$ (or on a descendant of it). Marginally independent parents therefore become dependent given the child.

    WATCH OUT FOR

    Assuming that because $U$ and $V$ are independent a priori (the factorisation has $P(U)P(V)$), they stay independent once $W$ is observed. This ignores the explaining-away effect at a collider.

    Full solution →
15. Q.25MCQ · 1 markOfficial answer: C

    Artificial Intelligence › Search

    $\alpha$ is the highest value MAX can already guarantee along the path, and $\beta$ is the lowest value MIN can already force. A branch is pruned as soon as $\alpha \ge \beta$.

    WATCH OUT FOR

    Reading "best choice for MIN" as the highest value. Best for MIN is the lowest value, so $\beta$ tracks the lowest value and only decreases. Mixing up the two bounds also flips which comparison ($\le \alpha$ at MIN nodes, $\ge \beta$ at MAX nodes) triggers pruning.

    Full solution →
16. Q.26MCQ · 1 markOfficial answer: C

    Database Management and Warehousing › Relational algebra

    $R\subseteq S$ is checked as $R\setminus S=\varphi$, with the relation that must be contained on the left. "Either ... or" means the union.

    WATCH OUT FOR

    Using the intersection for "either or" (option A), which demands membership in both relations. Or reversing the difference (options B and D), which tests whether Defender/Forward names are in Team.

    Full solution →
17. Q.27MCQ · 1 markOfficial answer: B

    Database Management and Warehousing › Data transformation

    Z-score normalization uses only the mean and standard deviation: z = (x − mean) / standard deviation. Minimum and maximum are used by min-max normalization, not here.

    WATCH OUT FOR

    Using the minimum and maximum (min-max scaling) gives about 0.484, which is near 0.476 and can tempt you to pick the wrong method. Check that the question asks for the z-score.

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

    Data Structures and Algorithms › Trees & heaps

    Inorder together with either preorder or postorder always rebuilds a binary tree. Preorder with postorder fails only because of nodes with a single child, and a full binary tree has none, so it works there too.

    WATCH OUT FOR

    Remembering the rule that preorder plus postorder cannot rebuild a tree and so rejecting (C). That rule applies to general binary trees, and the question specifies a full binary tree.

    Full solution →
19. Q.29MSQ · 1 markOfficial answer: B, C, D

    Artificial Intelligence › Logic

    An implication fails only when its antecedent is true and its consequent false. So make the antecedent true, which here fixes both $x$ and $y$, and test the consequent at that single assignment.

    WATCH OUT FOR

    Writing $y\Rightarrow x$ when the converse $x\Rightarrow y$ is meant, or the reverse. In (A) the antecedent forces $y=T$ and $x=F$, which makes $y\Rightarrow x$ false. In (D) the antecedent forces $x=T$, which makes $y\Rightarrow x$ true.

    Full solution →
20. Q.30NAT · 1 markOfficial answer: 0

    Data Structures and Algorithms › Design techniques

    With the last element as pivot, an already sorted array makes every pivot the maximum of its subarray, so every element is already on the correct side of it. Nothing needs to be exchanged, so the number of real swaps is zero (the sorted input is the best case for swaps, though the worst case for the recursion depth and comparisons).

    WATCH OUT FOR

    Counting the self-swaps of a blind Lomuto loop (4+3+2+1 element-with-itself exchanges, plus the final pivot placement) and answering 10 or 14. Those do not exchange two distinct elements, and the question asks for the minimum. Another slip is to treat sorted input as the worst case and expect many swaps; it is the worst case for comparisons and depth, not for real swaps.

    Full solution →
21. Q.31NAT · 1 markOfficial answer: 3

    Database Management and Warehousing › SQL

    A comma join with an equality condition on the key is an equi-join. Apply each filter to the joined rows, and count the rows that satisfy all three conditions together.

    WATCH OUT FOR

    Counting the three Jaipur rows without checking RaidPoints, or treating RaidPoints > 200 as >= 200 and then wrongly including Sunil. Sunil belongs to Patna, so he is excluded either way.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: 17

    Data Structures and Algorithms › Linear structures

    Trace the deque operation by operation. a is overwritten at every assignment, so only the last removal (removeLast()) decides the final value.

    WATCH OUT FOR

    Reporting the value from the first or second removal (10 or 32), or treating insertFirst as an append at the back. Another slip is to forget that a is reassigned, so only the last removal counts.

    Full solution →
23. Q.33NAT · 1 markOfficial answer: 0.24

    Calculus and Optimization › Limits & continuity

    For the logistic sigmoid, $f^{\prime }(x)=f(x)(1-f(x))$, so the derivative can be read straight from the function value without finding $x$.

    WATCH OUT FOR

    Solving $\frac{1}{1+e^{-x}}=0.4$ for $x$ and then substituting into the derivative is slower and invites rounding errors. Another slip is dropping the minus sign from the chain rule on $e^{-x}$, or forgetting the square in the denominator.

    Full solution →
24. Q.34NAT · 1 markOfficial answer: 42

    Probability and Statistics › Descriptive statistics

    A mean is the total divided by the count, so recover the total ($n\bar{x}$), add the new value, and divide by $n+1$. Equivalently, the mean moves by $\frac{x_{\text{new}}-\bar{x}}{n+1}$.

    WATCH OUT FOR

    Dividing the new sum by 50 instead of 51, or averaging the old mean with the new value as $(40+142)/2=91$. Both ignore that the old mean carries the weight of 50 points.

    Full solution →
25. Q.35NAT · 1 markOfficial answer: 0

    Linear Algebra › Eigenvalues

    Factor ${\mathbf{M}}^2+12\mathbf{M}=\mathbf{M}(\mathbf{M}+12\mathbf{I})$. If $\det (\mathbf{M})=0$, the whole product has determinant $0$, so the second factor never has to be computed.

    WATCH OUT FOR

    Computing ${\mathbf{M}}^2+12\mathbf{M}$ entry by entry and then taking its determinant. This is long and error-prone. Another slip is to check only whether the trace or an individual entry is zero, instead of checking $\det (\mathbf{M})$.

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

    Probability and Statistics › Random variables

    Model the wait as a two-state Markov chain (last throw even or not) and solve the two first-step equations for the expected remaining throws.

    WATCH OUT FOR

    Answering $\frac{1}{p^2}=4$ by treating pairs of throws as independent trials. An odd throw after one even throw wipes out the progress, and overlapping runs are allowed, so the true expectation is larger, at 6.

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

    Calculus and Optimization › Limits & continuity

    Differentiable implies continuous. At each junction, match both the function value and the derivative of the middle quadratic with those of the outer line. This gives four linear equations in $a,b,c$ (two of them consistent).

    WATCH OUT FOR

    Imposing only continuity. Continuity alone gives $b=0$ and $4a+c=2$, which leaves a whole family of solutions, and options (B) and (C) are tempting for that reason. The derivative conditions fix $a=\frac{1}{4}$.

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

    Data Structures and Algorithms › Python

    The recursion counts the nodes of a tree stored as a dictionary of child lists. Each call returns 1 for itself plus the counts of its children, and a key missing from the dictionary is a leaf that returns 1. The output is 1 + (1 + 3) + (1 + 3) = 9.

    WATCH OUT FOR

    Counting only the keys of the dictionary (3) or only the leaves (6), or forgetting that a missing key still returns 1 and so counts as a node. Another slip is returning 0 for leaves, which would give 3 instead of 9.

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

    Data Structures and Algorithms › Python

    S[i] is the length of the non-decreasing run ending at position i. The comparison is X[i-1] <= X[i], and S[i] is reset to 1 whenever it fails. The value added is S[i-1], which is the previous position's already-computed run length.

    WATCH OUT FOR

    Using S[i-1] + 1 unconditionally, or counting a run only when the sequence strictly increases, gives the wrong array. Another slip is forgetting that S[i] restarts at 1 after every descent (6 to 3, and 5 to 4). That slip leads to the cumulative-looking distractors in options (A), (B) and (D).

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

    Data Structures and Algorithms › Sorting & hashing

    Binary search makes one comparison at the middle, then continues in only one part, and the larger part has $\lfloor n/2\rfloor$ elements. The worst case therefore satisfies $F(n)=F(\lfloor n/2\rfloor )+1$, which gives $\lfloor {\log }_{2}n\rfloor +1$ comparisons.

    WATCH OUT FOR

    Choosing (B) by copying the merge sort recurrence, which adds the costs of both halves, or choosing (C) by forgetting that the middle comparison costs one step each time.

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

    Data Structures and Algorithms › Python

    A recursive two-pointer swap: swap the ends of the window, then recurse on the window shrunk by one at each end, until the pointers meet or cross. That is in-place reversal of the segment.

    WATCH OUT FOR

    Reading the single swap D[s1], D[s2] = D[s2], D[s1] and picking (D) without noticing the recursive call repeats the swap on the inner pairs. Another error is thinking swaps imply sorting. No comparison between elements ever happens, so the result depends only on positions, never on values.

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

    Machine Learning › Clustering

    In single linkage, the distance between two clusters is the smallest pairwise distance across them. Merge the closest pair repeatedly, recomputing cluster distances with a minimum.

    WATCH OUT FOR

    Merging $x_3$ with $x_4$ as a pair, because $d(x_3,x_4)=2$ looks small. That ignores that $x_4$ and $x_5$ already merged at distance 1, so $x_3$ joins the cluster $\{x_4,x_5\}$ rather than $x_4$ alone.

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

    Machine Learning › Neural networks

    With positive inputs, zero biases and positive weights, every pre-activation is positive. ReLU then acts as the identity, so Figure 1 collapses to a linear map. The coefficient of each input equals the sum, over all paths from that input to the output, of the product of the weights along the path. $x_1$ reaches 3 first-layer neurons and each first-layer neuron gives $2\cdot 2\cdot 3=12$ downstream, so $p=3\times 12=36$. $x_2$ and $x_3$ each reach only 2 first-layer neurons, so $q=r=2\times 12=24$.

    WATCH OUT FOR

    Assuming the first layer is fully connected (9 edges) because the crossing lines look dense. That gives $36(x_1+x_2+x_3)$ and option (D). Count the edges leaving each input: $x_1$ has 3, but $x_2$ and $x_3$ have only 2 each.

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

    Artificial Intelligence › Search

    When the question dictates the expansion order (smallest number first), that rule overrides the usual queue or stack behaviour of BFS and DFS. With the same order, both searches expand the same states and reach the goal after the same number of expansions.

    WATCH OUT FOR

    Applying the textbook behaviour of BFS (level by level) and DFS (deepest first) and concluding they differ, without noticing that the question forces the ascending-number order on both. Another slip is to count the goal itself as an expansion for one search and not the other.

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

    Data Structures and Algorithms › Sorting & hashing

    Simulate each algorithm for exactly two passes and compare with the sorted array. Selection sort fixes one element per pass in its final place. Here the two smallest elements, $1$ and $2$, are fixed in two passes, and the rest, $[3,4,5]$, already happens to be in order.

    WATCH OUT FOR

    Assuming that bubble sort or insertion sort also finishes in two passes. Bubble sort carries only the largest element to the end each pass, so the small elements $1$ and $2$ near the right move left by just one place per pass. Insertion sort after two passes has only sorted the first three elements and has not yet looked at $1$.

    Full solution →
36. Q.46MSQ · 2 marksOfficial answer: A, B, D

    Database Management and Warehousing › Normal forms

    $P\to Q$ is derivable exactly when $Q\subseteq P^{+}$. Compute the closure of the left side by repeatedly firing every dependency whose left side is already inside the set. Here $V\to X$ is what lets $VW$ reach $WX$ and so pull in $Y$ and $Z$.

    WATCH OUT FOR

    Stopping the closure of $VW$ at $\{V,W,X\}$ and forgetting to fire $WX\to Y$ and $WX\to Z$ afterwards. Another slip is to think $VW\to U$ holds because $U$ determines $V$ and $W$. The converse does not follow, since nothing determines $U$.

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

    Linear Algebra › Vector spaces

    A subspace must contain the zero vector and be closed under addition and all real scalar multiples. A span of vectors and the solution set of a homogeneous linear system $A\mathbf{x}=0$ are always subspaces.

    WATCH OUT FOR

    Treating (B) as a span because it is built from two vectors. The squared coefficients ${\alpha }^2,{\beta }^2$ can only be non-negative, so negative multiples are missing. For (D), the mistake is forgetting that a plane $5x_1+2x_3=-4$ misses the origin.

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

    Linear Algebra › Linear systems

    Uniqueness of a solution depends only on the nullity of $\mathbf{M}$ (independent columns), while existence depends on whether the right-hand side lies in the column space. So the same $\mathbf{M}$ can give different outcomes for different right-hand sides only when the nullity is positive (infinite versus none) or when the system is tall (unique versus none).

    WATCH OUT FOR

    Thinking that one matrix cannot give different outcomes for different right-hand sides. For a square invertible matrix every right-hand side behaves alike, but a singular square matrix (B) or a tall matrix with independent columns (D) can have a solution for one vector and none for another. In (C), forgetting that a wide matrix always has a non-trivial null space.

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

    Linear Algebra › Special matrices

    A projection matrix is idempotent (${\mathbf{M}}^2=\mathbf{M}$), its rank equals $\dim U$, and rank-nullity gives $\dim N(\mathbf{M})=3-\dim U$.

    WATCH OUT FOR

    Confusing the dimension of the subspace being projected onto with the dimension of the null space. For a 1-dimensional $U$ the null space has dimension 2, not 1.

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

    Calculus and Optimization › Maxima & minima

    Find the stationary points from $f^{\prime }(x)=0$, then classify each one with the second-derivative test (or the sign change of $f^{\prime }$): $f^{\prime \prime }>0$ gives a local minimum and $f^{\prime \prime }<0$ gives a local maximum.

    WATCH OUT FOR

    Mixing up the sign of $f^{\prime \prime }$. Here $f^{\prime \prime }(-1)=4>0$, so $x=-1$ is a local minimum, and option (C) is a trap. Another slip is to treat $x=0$ as a minimum, when $f^{\prime \prime }(0)=-3<0$ makes it a maximum.

    Full solution →
41. Q.51MSQ · 2 marksOfficial answer: B, D

    Data Structures and Algorithms › Graph algorithms

    An ordering is a valid topological sort exactly when, for every directed edge $u\to v$, $u$ appears before $v$. Check each edge against each option's positions. The only freedom is the relative order of independent branches, here $P$ versus $R$, and the $S\to U$ chain versus the $V\to T$ chain.

    WATCH OUT FOR

    Checking only that $P$ comes first and $Q$ comes early, and missing that $R$ must also precede $Q$. Options (A) and (C) put $Q$ right after $P$, before $R$. Another slip is forgetting that $S\to U$ and $V\to T$ are separate chains, so their order relative to each other is free.

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

    Data Structures and Algorithms › Trees & heaps

    In a binary tree, $n_0=n_2+1$ (leaves = degree-2 nodes + 1). Bound the node counts by the longest path, which gives the minimum, and by full levels, which gives the maximum.

    WATCH OUT FOR

    Counting height in nodes instead of edges, which breaks the lower bounds in B and C. Another slip is assuming a tree of height $H$ has at least $H$ leaves. A chain has only one leaf.

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

    Machine Learning › LDA, SVM & trees

    Two classes in the plane are linearly separable exactly when their convex hulls do not intersect. A point inside the other class's hull (panel ii) or crossing diagonals (the XOR layout, panel iii) rule out separation, and one explicit separating line proves it.

    WATCH OUT FOR

    Judging panel (iv) by eye from its tilted layout and calling it inseparable. Because the points are not on a grid, a slanted line is needed, and one such line separates the classes. The other slip is treating the XOR layout as separable because each class is symmetric.

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

    Artificial Intelligence › Logic

    \"All X are Y except Z\" means two implications: not-Z implies Y, and Z implies not-Y. Each implication is equivalent to its contrapositive, so $s1\equiv s2$ and $s3\equiv s4$. A pair is correct exactly when it contains one sentence from $\{s1,s2\}$ and one from $\{s3,s4\}$.

    WATCH OUT FOR

    Treating $s1\wedge s2$ or $s3\wedge s4$ as the full assertion. Each of these pairs holds two equivalent sentences, so it states only one half of the assertion. Another slip is to miss that $s2$ is the contrapositive of $s1$ and so drop option (C).

    Full solution →
45. Q.55MSQ · 2 marksOfficial answer: A, B

    Database Management and Warehousing › Indexing

    Match index type to predicate type. Hash indexes serve only equality, whereas B${}^{+}$ trees serve equality and range. A range predicate such as CustomerRating > 3.4 needs a B${}^{+}$ tree on that attribute.

    WATCH OUT FOR

    Assuming a hash index is always faster than a B${}^{+}$ tree and choosing a hash index on the attribute with the range condition (option C), or hashing everything (option D). Hashing cannot help with a $>$ comparison.

    Full solution →
46. Q.56NAT · 2 marksOfficial answer: 0.125

    Probability and Statistics › Continuous distributions

    For independent uniform variables the point $(X,Y)$ is uniform on a rectangle, so the probability is the area of the region $x\ge y$ divided by the rectangle's area.

    WATCH OUT FOR

    Treating the two intervals as if they overlapped on $[2,3]$ only and taking $P=\frac{1}{2}$ there, or integrating over $x\in [1,3]$ and forgetting that $Y\ge 2$ makes $P(Y\le x)=0$ for $x<2$. The area must be divided by the full rectangle area $4$, not by $1$.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 0.2

    Probability and Statistics › Continuous distributions

    For an exponential distribution with rate $\lambda$, the mean is $1/\lambda$ and the variance is $1/{\lambda }^2$, so the condition $5E(X)=\mathrm{Var}(X)$ reduces to $5/\lambda =1/{\lambda }^2$.

    WATCH OUT FOR

    Using the mean and variance of an exponential written in terms of the scale (mean) parameter, or taking the variance as $1/\lambda$ instead of $1/{\lambda }^2$. That gives $5/\lambda =1/\lambda$, which has no solution.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 0.25

    Probability and Statistics › Conditional & Bayes

    The question gives $P(\bar{T})$ and the likelihoods $P(S\mid T)$, $P(S\mid \bar{T})$. Get the prior $P(T)$ from the complement rule, get $P(S)$ by total probability over $T$ and $\bar{T}$, then reverse the conditioning with Bayes' theorem.

    WATCH OUT FOR

    Using $P(\bar{T})=0.6$ as if it were $P(T)$. That gives $P(S)=0.3\times 0.6+0.6\times 0.4=0.42$ and $P(T\mid S)=0.18/0.42\approx 0.43$. Another slip is dividing by $P(S\mid T)$ or by $P(T)$ instead of $P(S)$.

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

    Probability and Statistics › Random variables

    Given $X=x$, the conditional density is proportional to the joint density with $x$ held fixed. On $0<y<x$ that is proportional to $y$, so $f_{Y\mid X}(y\mid x)=2y/x^2$ and $E[Y\mid X=x]=2x/3$. Any constant in front of the joint density cancels in this ratio.

    WATCH OUT FOR

    One mistake is taking $Y$ from $0$ to $2$ instead of $0$ to $x$. Another is using the marginal mean $E[Y]$ instead of the conditional one. A third is to assume, without checking, that $2xy$ is a valid density: it integrates to $4$ on this triangle, so $x^3$ is not the true marginal of $X$ (the valid one is $x^3/4$). A fourth is to read the conditional as uniform on $(0,1.5)$, which would give $0.75$.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 0.5

    Calculus and Optimization › Limits & continuity

    Split the logarithm of the product into $\ln (1+x^2)+\ln \cos x$, then use the standard expansions $\ln (1+u)\approx u$ and $\cos x\approx 1-x^2/2$ so each piece divided by $x^2$ has a finite limit.

    WATCH OUT FOR

    Forgetting the $\ln \cos x$ contribution and answering $1$ (treating $\cos x\to 1$ as if $\ln \cos x$ were negligible next to $x^2$), or getting its sign wrong. $\ln \cos x$ is of order $x^2$, so it changes the limit by $-\frac{1}{2}$.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 55

    Linear Algebra › LU & SVD

    $\mathbf{u}{\mathbf{u}}^T$ has rank 1, so it has a single nonzero singular value, equal to ${\mathbf{u}}^T\mathbf{u}=\Vert \mathbf{u}{\Vert }^2$. The sum of the singular values is therefore the trace, $\sum u_i^2=55$.

    WATCH OUT FOR

    Taking the single singular value to be the norm $\Vert \mathbf{u}\Vert =\sqrt{55}\approx 7.42$ instead of $\Vert \mathbf{u}{\Vert }^2=55$. The matrix is $\mathbf{u}{\mathbf{u}}^T$, so the squared norm appears. Another slip is adding the entries of $\mathbf{u}$ (15).

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 0.12 to 0.13

    Machine Learning › LDA, SVM & trees

    Information gain is the parent's entropy minus the size-weighted average entropy of the children. Count the class split in the whole table and in each Pitch branch, then compute the binary entropies.

    WATCH OUT FOR

    Using the wrong class counts per branch, for example miscounting the Green wins on pace-friendly pitches. Another slip is forgetting to weight each branch by its share of the matches, or using natural log instead of base 2.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 5

    Machine Learning › Logistic, k-NN, naive Bayes

    In $k$-NN, rank all training points by Euclidean distance from the query and take a majority vote among the $k$ nearest. Compute the distances once, then test each odd $k$ in turn.

    WATCH OUT FOR

    Counting only the two squares nearest to the query, or assuming that the square at $(3,2)$ is farther than the circle at $(1,2)$. These two are tied at $\sqrt{2}$, so at $k=3$ both are included, and the circles still win 2 to 1. With 5 points in total, the only way for squares to win a majority is to use all of them.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 0.125

    Artificial Intelligence › Uncertainty

    A Bayesian network's joint probability is the product of each node's conditional probability given its parents. Read the one row of each table that matches the assignment asked for, then multiply.

    WATCH OUT FOR

    Reading the wrong row of a table, for example using the $V=0,W=0$ row of the $Z$ table. Another slip is to treat $V$ and $W$ as independent of $U$ and drop the $U$ conditioning. Here $W$ is fully determined by $U$, so $P(W=1\mid U=1)=1$, not $0.5$.

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 0.062 to 0.063

    Probability and Statistics › Descriptive statistics

    Write each indicator's expectation as a probability, then use $\mathrm{Cov}(X,Y)=E[XY]-E[X]E[Y]$. Because $X=1$ forces $Y=1$, the product $XY$ equals $X$.

    WATCH OUT FOR

    Treating $X$ and $Y$ as independent and answering $0$. They are dependent, since "both heads" is a special case of "at least one head". Another slip is computing $E[XY]$ as $E[X]E[Y]$.

    Full solution →