Numerical Computation & Estimation, Programming Fundamentals, Matrices… Test for GATE DA: 23 Questions with Solutions & Analysis

Attempt the Numerical Computation & Estimation, Programming Fundamentals, Matrices… test for GATE DA: 23 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

23 questions · 35 marks · 1.05 minutes. Programming, Data Structures and Algorithms: 5 · Probability and Statistics: 4 · Database Management and Warehousing: 4 · Machine Learning: 3 · Quantitative Aptitude: 2 · Linear Algebra: 2 · Artificial Intelligence: 2 · Calculus and Optimization: 1

Free sample questions from Numerical Computation & Estimation, Programming Fundamentals, Matrices…

Question 1 · Programming, Data Structures and Algorithms · 2025 MCQ

Consider the following Python declarations of two lists.

A=[1,2,3]
B=[4,5,6]

Which one of the following statements results in

A = [1, 2, 3, 4, 5, 6]

A.
A.extend(B)
B.
A.append(B)
C.
A.update(B)
D.
A.insert(B)

Correct Answer:

Step-by-Step Solution

Insight: The question tests the difference between adding an iterable as a single nested object versus flattening its elements into the existing list.

Exam route: Eliminate options that use non-existent list methods or wrong argument counts. Between the remaining two, recall that extend flattens while append nests.

Learning route:

A.extend(B): The extend method iterates over the argument B and adds each element (4, 5, 6) one by one to A. The final list is [1, 2, 3, 4, 5, 6]. This matches the requirement.
A.append(B): The append method adds the entire object B as a single element at the end of A. The final list becomes [1, 2, 3, [4, 5, 6]]. This does not match.
A.update(B): The update method is not defined for Python lists (it is used for sets and dictionaries). Calling this raises an AttributeError.
A.insert(B): The insert method requires exactly two arguments: an index and an element. Providing only one argument raises a TypeError.

Therefore, A.extend(B) is the only statement that produces the desired flat list.

Question 2 · Programming, Data Structures and Algorithms · 2024 MCQ

Consider the following Python code:
def count(child_dict, i):
if i not in child_dict.keys():
return 1
ans = 1
for j in child_dict[i]:
ans += count(child_dict, j)
return ans
child_dict = dict()
child_dict[0] = [1,2]
child_dict[1] = [3,4,5]
child_dict[2] = [6,7,8]
print(count(child_dict,0))
Which ONE of the following is the output of this code?

A.
6
B.
1
C.
8
D.
9

Correct Answer:

Step-by-Step Solution

Insight: The function computes the total number of nodes in a tree represented by an adjacency list, where each node counts itself (ans = 1) and adds the counts of its children via return values.

Exam route: Draw the tree. Root 0 has children 1, 2. Node 1 has 3, 4, 5. Node 2 has 6, 7, 8. Total nodes = 1 (root) + 1 (node 1) + 3 (leaves of 1) + 1 (node 2) + 3 (leaves of 2) = 9.

Learning route: Trace using stack frames. Frame 0 calls Frame 1 and Frame 2. Frame 1 calls 3, 4, 5 which return 1 each. Frame 1 returns $1 + 1 + 1 + 1 = 4$ . Frame 2 similarly returns 4. Frame 0 returns $1 + 4 + 4 = 9$ . The key is that ans = 1 resets in every frame, so it is return-value aggregation, not a shared global counter.

Question 3 · Programming, Data Structures and Algorithms · 2026 MCQ

A recursive function in Python is given.

def mystery(n):
    if n <= 0:
        return 1
    else:
        return mystery(n-1) + mystery(n-2)

Now, consider the following function call:

mystery(4)

Assume that a typical runtime stack is used to manage function calls. Each function call is pushed onto the stack and removed only after it finishes execution.

Which of the following options denotes the total number of function calls (i.e., the total number of stack activations), including the initial call, to compute mystery(4)?

A.
5
B.
9
C.
15
D.
17

Correct Answer:

Step-by-Step Solution

Insight: The total number of function calls follows a recurrence relation $C (n) = 1 + C (n - 1) + C (n - 2)$ , where the $+ 1$ accounts for the current function call itself.

Exam route: Compute $C (n)$ iteratively from base cases. $C (0) = 1, C (- 1) = 1$ . Then $C (1) = 1 + 1 + 1 = 3$ , $C (2) = 1 + 3 + 1 = 5$ , $C (3) = 1 + 5 + 3 = 9$ , $C (4) = 1 + 9 + 5 = 15$ .

Learning route: Draw the recursion tree for mystery(4). The root is 1 call. It branches to mystery(3) and mystery(2). Count all nodes in this call tree. Total nodes = 15. Note that mystery(0) and mystery(-1) are base cases that make 1 call each and return immediately without further branching.

Question 4 · Probability and Statistics · 2024 MCQ

The probability of a boy or a girl being born is 1/2. For a family having only
three children, what is the probability of having two girls and one boy?

A.
$\frac{3}{8}$
B.
$\frac{1}{8}$
C.
$\frac{1}{4}$
D.
$\frac{1}{2}$

Correct Answer:

Step-by-Step Solution

Insight: This is a classic binomial probability problem where the probability of success (girl) is 0.5, and we want exactly 2 successes in 3 trials.

Exam route: 3 children, exactly 2 girls. Number of arrangements = 3C2 = 3. Total outcomes = 2^3 = 8. Probability = 3/8.

Learning route: The sample space for a family of 3 children has $2^{3} = 8$ equally likely outcomes (since prob of boy/girl is 1/2). We want exactly 2 girls and 1 boy. The favorable outcomes are GGB, GBG, and BGG. There are 3 such outcomes. The probability is the ratio of favorable to total outcomes: $3/8$ . Alternatively, using the binomial formula: $(2 3) (1/2)^{2} (1/2)^{1} = 3 \times 1/8 = 3/8$ .

Question 5 · Probability and Statistics · 2024 MCQ

Three fair coins are tossed independently.

T

is the event that two or more tosses
result in heads.

S

is the event that two or more tosses result in tails.
What is the probability of the event

T \cap S

A.
0
B.
0.5
C.
0.25
D.
1

Correct Answer:

Step-by-Step Solution

Insight: In 3 tosses, you cannot have 2 or more heads AND 2 or more tails simultaneously because that requires at least 4 tosses.

Exam route: T requires >= 2 heads, S requires >= 2 tails. Total coins = 3. Max heads + Max tails = 3. The intersection is empty. Probability = 0.

Learning route: The sample space for 3 fair coins has 8 equally likely outcomes. Event T (>= 2 heads) consists of {HHH, HHT, HTH, THH}. Event S (>= 2 tails) consists of {TTT, TTH, THT, HTT}. The intersection T ∩ S requires an outcome to have at least 2 heads and at least 2 tails, which means at least 4 coins. Since we only have 3 coins, T ∩ S = ∅. Therefore, P(T ∩ S) = 0.

Question 6 · Probability and Statistics · 2026 MCQ

Let

M

be a randomly chosen non-empty subset of

S = {1, 2, 3, \dots, 2026}

.
Which of the following is the probability that the product of all the elements of

M

is even?

A.
$2^{1013} (2^{1013} - 1) / 2^{2026}$
B.
$2^{1013} / 2^{2026}$
C.
$2^{1013} (2^{1013} - 1) / (2^{2026} - 1)$
D.
$1/ (2^{2026} - 1)$

Correct Answer:

Step-by-Step Solution

Insight: The product of a subset is even if at least one element is even. It is much easier to calculate the complement: the product is odd if and only if ALL elements in the subset are odd.

Exam route: Total non-empty subsets of $S$ is $2^{2026} - 1$ . There are 1013 odd numbers in $S$ . Non-empty subsets containing only odd numbers is $2^{1013} - 1$ . Probability of odd product is $\frac{2 ^{1013} - 1}{2 ^{2026} - 1}$ . Probability of even product is $1 - \frac{2 ^{1013} - 1}{2 ^{2026} - 1} = \frac{2 ^{2026} - 2 ^{1013}}{2 ^{2026} - 1} = \frac{2 ^{1013} ( 2 ^{1013} - 1 )}{2 ^{2026} - 1}$ .

Learning route: The set $S = {1, 2, \dots, 2026}$ contains 1013 odd numbers and 1013 even numbers. A subset product is odd if and only if every element in the subset is odd. The number of non-empty subsets formed entirely from the 1013 odd numbers is $2^{1013} - 1$ . The total number of non-empty subsets of $S$ is $2^{2026} - 1$ . The probability that a randomly chosen non-empty subset has an odd product is $\frac{2 ^{1013} - 1}{2 ^{2026} - 1}$ . Using the complement rule, the probability that the product is even is $1 - \frac{2 ^{1013} - 1}{2 ^{2026} - 1} = \frac{2 ^{2026} - 1 - ( 2 ^{1013} - 1 )}{2 ^{2026} - 1} = \frac{2 ^{2026} - 2 ^{1013}}{2 ^{2026} - 1}$ . Factoring out $2^{1013}$ gives $\frac{2 ^{1013} ( 2 ^{1013} - 1 )}{2 ^{2026} - 1}$ .

Question 7 · Database Management and Warehousing · 2025 MCQ

Consider the following three relations:

Car (model, year, serial, color)
Make (maker, model)
Own (owner, serial)

A tuple in Car represents a specific car of a given model, made in a given year, with a serial number and a color. A tuple in Make specifies that a maker company makes cars of a certain model. A tuple in Own specifies that an owner owns the car with a given serial number. Keys are underlined; (owner, serial) together form key for Own. (

⋈

denotes natural join)

π_{o w n er} (O w n ⋈ (σ_{co l or = ‘‘ r e d^{''}} (C a r ⋈ (σ_{mak er = ‘‘ A B C^{''}} M ak e))))

Which one of the following options describes what the above expression computes?

A.
All owners of a red car, a car made by ABC, or a red car made by ABC
B.
All owners of more than one car, where at least one car is red and made by ABC
C.
All owners of a red car made by ABC
D.
All red cars made by ABC

Correct Answer:

Step-by-Step Solution

Insight: The expression nests joins and selections from inside out. Each operation restricts the data further, culminating in owners of cars that are both red and made by ABC.

Exam route: Trace from the innermost selection outward, identifying what each step filters or joins.

Learning route:

Step 1: $σ_{mak er = " A B C "} M ak e$ → models made by ABC.

Step 2: $C a r ⋈ (result)$ → cars of those models (joined on model).

Step 3: $σ_{co l or = " r e d "} (result)$ → red cars among those (further filtered).

Step 4: $O w n ⋈ (result)$ → ownership records for those cars (joined on serial).

Step 5: $π_{o w n er} (result)$ → owner names.

Interpretation: The expression finds all owners who own a car that is both red and made by ABC.

Why others fail:

Option A suggests a union (red OR ABC OR both), but the joins act as intersection (AND).
Option B adds a condition about "more than one car", which is not present.
Option D omits the Own join and owner projection, so it would return cars, not owners.

Verification: If a car is red and made by ABC, and an owner owns it, that owner appears in the result. This matches option C.

Question 8 · Database Management and Warehousing · 2024 MCQ

Consider a database that includes the following relations:
Defender(name, rating, side, goals)
Forward(name, rating, assists, goals)
Team(name, club, price)
Which ONE of the following relational algebra expressions checks that every name
occurring in Team appears in either Defender or Forward, where

ϕ

denotes the
empty set?

A.
$Π_{nam e} (Team) ∖ (Π_{nam e} (Defender) \cap Π_{nam e} (Forward)) = ϕ$
B.
$(Π_{nam e} (Defender) \cap Π_{nam e} (Forward)) ∖ Π_{nam e} (Team) = ϕ$
C.
$Π_{nam e} (Team) ∖ (Π_{nam e} (Defender) \cup Π_{nam e} (Forward)) = ϕ$
D.
$(Π_{nam e} (Defender) \cup Π_{nam e} (Forward)) ∖ Π_{nam e} (Team) = ϕ$

Correct Answer:

Step-by-Step Solution

Insight: "Every name in Team appears in Defender or Forward" is a subset statement: $π_{nam e} (T e am) \subseteq π_{nam e} (D e f e n d er) \cup π_{nam e} (F or w a r d)$ .

Exam route: Rewrite subset as set difference equals empty set: $A \subseteq B ⟺ A ∖ B = ϕ$ .

Learning route:

Step 1: Let $T = π_{nam e} (T e am)$ , $D = π_{nam e} (D e f e n d er)$ , $F = π_{nam e} (F or w a r d)$ .

Step 2: The English condition is: every element of $T$ is in $D$ or in $F$ . That is $T \subseteq (D \cup F)$ .

Step 3: The standard relational algebra test for $A \subseteq B$ is $A ∖ B = ϕ$ .

Step 4: Substitute: $π_{nam e} (T e am) ∖ (π_{nam e} (D e f e n d er) \cup π_{nam e} (F or w a r d)) = ϕ$ .

Step 5: This matches option C exactly.

Why others fail:

Option A uses $\cap$ instead of $\cup$ , which would mean every team name is in both D and F.
Option B tests $(D \cap F) \subseteq T$ , the reverse direction with intersection.
Option D tests $(D \cup F) \subseteq T$ , the reverse direction.

Verification: If T = {a, b}, D = {a, b, c}, F = {b, d}, then D ∪ F = {a, b, c, d}, T \ (D ∪ F) = ∅. Correct.

Question 9 · Database Management and Warehousing · 2026 MCQ

Consider the given relations

X, Y

and

Z

. The relation

X

has three columns

P, Q

and

R

. The relation

Y

has three columns

P, Q

and

S

. The relation

Z

has two columns

P

and

T

X

P

Q

R

P1 Q1 R1
P2 Q2 R2
P3 Q3 R2

Y

P

Q

S

P1 Q1 2
P1 Q2 5
P2 Q1 6
P3 Q3 1

Z

P

T

P1 T1
P3 T2
P4 T3
P4 NULL

Consider the relational algebra expression

P R S \prod [(σ_{(Q = Q 3 \lor R = R 2)} [X ⋈ Y]) ⋈ (σ_{(S > 1)} [Y ⋈ Z])]

where

⋈

denotes natural join operation.

Which of the following options is the correct output for the given expression?

A.
Two rows (P1, R1, 2) and (P1, R1, 5)
B.
Three rows (P1, R1, 2), (P1, R1, 5) and (P2, R2, 6)
C.
One row (P1, R1, 2)
D.
Zero rows

Correct Answer:

Step-by-Step Solution

Insight: Evaluate each sub-expression independently, then join the two results on their common attributes (P, Q, S). The join yields zero rows because the surviving tuples have no matching keys.

Exam route: Compute left side and right side of the outer join separately, then check for matching (P, Q, S) values.

Learning route:

Step 1: Compute $X ⋈ Y$ (natural join on P, Q).

X: (P1,Q1,R1), (P2,Q2,R2), (P3,Q3,R2)

Y: (P1,Q1,2), (P1,Q2,5), (P2,Q1,6), (P3,Q3,1)

Matches: (P1,Q1) → (P1,Q1,R1,2); (P3,Q3) → (P3,Q3,R2,1).

Result: {(P1,Q1,R1,2), (P3,Q3,R2,1)}.

Step 2: Apply $σ_{Q = Q 3 \lor R = R 2}$ .

(P1,Q1,R1,2): Q=Q1 (F), R=R1 (F) → F.

(P3,Q3,R2,1): Q=Q3 (T) → T.

Left result: {(P3,Q3,R2,1)}.

Step 3: Compute $Y ⋈ Z$ (natural join on P).

Y: (P1,Q1,2), (P1,Q2,5), (P2,Q1,6), (P3,Q3,1)

Z: (P1,T1), (P3,T2), (P4,T3), (P4,NULL)

Matches: P1 → (P1,Q1,2,T1), (P1,Q2,5,T1); P3 → (P3,Q3,1,T2).

Result: {(P1,Q1,2,T1), (P1,Q2,5,T1), (P3,Q3,1,T2)}.

Step 4: Apply $σ_{S > 1}$ .

S=2 > 1 → T; S=5 > 1 → T; S=1 > 1 → F.

Right result: {(P1,Q1,2,T1), (P1,Q2,5,T1)}.

Step 5: Join left and right on common attributes (P, Q, S).

Left: (P3, Q3, R2, S=1).

Right: (P1, Q1, S=2, T1), (P1, Q2, S=5, T1).

No match on P (P3 vs P1), Q (Q3 vs Q1/Q2), or S (1 vs 2/5).

Final result: Zero rows.

Verification: The left side has P3, Q3, S=1. The right side has P1, Q1/Q2, S=2/5. No overlap on any common attribute.

Question 10 · Machine Learning · 2026 NAT

Consider that for a supervised learning task, the objective function being minimized is

f_{w} (x) = w x

, where

x \in R

is the input and

w \in R

is the parameter. Stochastic Gradient Descent with learning rate of 0.10 is used for parameter updates.

Suppose that at the end of iteration

i

, the value of

w

becomes 10.00.

Let

x = 10.00

be the input for iteration

(i + 1)

.

The value of

w

at the end of iteration

(i + 1)

is __________ . (Rounded off to two decimal places)

Correct Answer:

9.00

Step-by-Step Solution

Insight: In SGD, the parameter update is $w_{n e w} = w_{o l d} - α \times gradient$ . The gradient of $f_{w} (x) = w x$ with respect to $w$ is simply $x$ .

Exam route: Gradient = $x = 10.00$ . Update = $10.00 - 0.10 \times 10.00 = 9.00$ .

Learning route:

Identify the objective function to minimize: $f_{w} (x) = w x$ .
Compute the gradient with respect to the parameter $w$ : $\frac{\partial f _{w} ( x )}{\partial w} = x$ .
Substitute the current input for iteration $(i + 1)$ : $x = 10.00$ . So, the gradient is $10.00$.
Apply the SGD update rule: $w_{i + 1} = w_{i} - α \times gradient$ .
Calculate: $w_{i + 1} = 10.00 - 0.10 \times 10.00 = 10.00 - 1.00 = 9.00$ .

Question 11 · Machine Learning · 2026 MCQ

Which of the following statements is true for Ridge Regression?

A.
The regularizer in the objective function of Ridge Regression is used to guard against scenarios where the model works well for the test data, but poorly for the training data.
B.
The regularizer of Ridge Regression uses $L_{1}$ norm.
C.
Ridge Regression aims to reduce the number of parameters that have negative values.
D.
The regularizer of Ridge Regression may increase the bias of the model, but it helps in reducing the variance in predictions.

Correct Answer:

Step-by-Step Solution

Insight: Ridge regression uses L2 regularization, which shrinks weights to reduce model variance at the cost of a slight increase in bias, thereby preventing overfitting.

Exam route: Evaluate each option. A is backwards (overfitting is good on train, bad on test). B is Lasso. C is nonsensical. D correctly describes the bias-variance tradeoff.

Learning route:

Option A is false. Regularization guards against overfitting, which is when a model works well on training data but poorly on test data, not the other way around.
Option B is false. Ridge Regression uses the $L_{2}$ norm ( $\sum w_{j}^{2}$ ). The $L_{1}$ norm is used by Lasso Regression.
Option C is false. The L2 penalty shrinks all weights symmetrically towards zero, regardless of whether they are positive or negative.
Option D is true. By adding a penalty term, we restrict the model's flexibility. This intentionally introduces a small amount of bias to achieve a larger reduction in variance, improving generalization to unseen data.

Question 12 · Machine Learning · 2026 NAT

Consider that Linear Ridge Regression is being used to learn a prediction function

y_{p r e d} = w^{T} x

, where

w, x \in R^{2}

and Mean Absolute Error (MAE) is used to measure the prediction error. A weight of 0.20 is associated with the regularizer.

At an intermediate step of the training process, assume that the parameter

w = [- 3.00, 4.00]^{T}

. In the next step, for the input

x = [1.00, 2.00]^{T}

, the predicted value of

y

is noted. Let the relation between

x = [x_{1}, x_{2}]^{T}

and the true value of

y

y_{t r u e} = x_{1} + x_{2}

.

The value of the overall regularized loss function for this instance is _______ . (Rounded off to two decimal places)

Correct Answer:

7.00

Step-by-Step Solution

Insight: The regularized loss for a single instance is the sum of the instance's prediction error (MAE) and the L2 regularization penalty on the weights.

Exam route: Calculate MAE = |y_true - y_pred|. Calculate L2 penalty = \lambda ||w||_2^2. Add them together.

Learning route:

Compute y_true: The relation is y_true = x_1 + x_2. For x = [1.00, 2.00]^T, y_true = 1.00 + 2.00 = 3.00.
Compute y_pred: y_pred = w^T x = (-3.00)(1.00) + (4.00)(2.00) = -3.00 + 8.00 = 5.00.
Compute MAE: |y_true - y_pred| = |3.00 - 5.00| = |-2.00| = 2.00.
Compute L2 penalty: \lambda ||w||_2^2 = 0.20 \times ((-3.00)^2 + (4.00)^2) = 0.20 \times (9.00 + 16.00) = 0.20 \times 25.00 = 5.00.
Total regularized loss: MAE + L2 penalty = 2.00 + 5.00 = 7.00.

Question 13 · Quantitative Aptitude · 2026 MCQ

The product of the digits of a three-digit number is 70. The sum of the digits of this three-digit number is _____

A.
12
B.
14
C.
16
D.
18

Correct Answer:

Step-by-Step Solution

Insight: The product of three single digits is 70, so factorise 70 into three single-digit factors first; the sum follows immediately.

Exam route: $70 = 2 \times 5 \times 7$ . These are the only three single-digit factors (any other factorisation like $1 \times 7 \times 10$ uses a non-digit). Sum $= 2 + 5 + 7 = 14$ .

Learning route:

Step 1: Prime factorise $70 = 2 \times 5 \times 7$ .

Step 2: We need three single digits $a, b, c \in {1, 2, \dots, 9}$ such that $a \times b \times c = 70$ .

Step 3: Since $70 = 2 \times 5 \times 7$ , and all three are single digits, the only valid triple is ${2, 5, 7}$ . Any attempt to introduce a 1 (e.g. $1 \times 5 \times 14$ ) forces a factor $\geq 10$ , which is not a digit.

Step 4: Sum $= 2 + 5 + 7 = 14$ .

Trap: Using $1 \times 7 \times 10$ gives sum 18, but 10 is not a single digit.

Verification: $2 \times 5 \times 7 = 70$ ✓, and $2 + 5 + 7 = 14$ ✓.

Question 14 · Quantitative Aptitude · 2026 MCQ

P, Q, R, S, X,

and

Y

are distinct single-digit whole numbers taking values from 0 to 9.

P Q

is a two-digit number with

Q

being in the units place and

P

in the tens place. Similarly,

R S

is a two-digit number.
It is known that

P Q

and

R S

are consecutive numbers and

(P Q)^{2} + (R S)^{2} = X Y P,

with

X Y P

being a three-digit number.
The value of

Y

is __________

A.
4
B.
5
C.
6
D.
7

Correct Answer:

Step-by-Step Solution

Insight: Let $P Q = n$ and $R S = n + 1$ (consecutive). Then $n^{2} + (n + 1)^{2} = 2 n^{2} + 2 n + 1 = X Y P$ . The units digit of this expression must equal $P$ , and $P$ is the tens digit of $n$ . This double role of $P$ is the key constraint.

Exam route:

Step 1: $2 n^{2} + 2 n + 1 \equiv P (mod 10)$ , where $n = 10 P + Q$ .

Step 2: Try $P = 1$ (since $n \leq 21$ for $X Y P$ to be 3-digit). $2 n^{2} + 2 n \equiv 0 (mod 10) ⟹ n (n + 1) \equiv 0 (mod 5) ⟹ n \equiv 0$ or $4 (mod 5)$ .

Step 3: $n \in {10, 14, 15, 19}$ . Check each:

$n = 10$ : $221 ⟹ X = 2, Y = 2$ (not distinct).
$n = 14$ : $421 ⟹ X = 4, Y = 2, P = 1, Q = 4$ . But $X = Q = 4$ (not distinct).
$n = 15$ : $481 ⟹ X = 4, Y = 8, P = 1, Q = 5$ . $R S = 16 ⟹ R = 1 = P$ (not distinct).
$n = 19$ : $761 ⟹ X = 7, Y = 6, P = 1, Q = 9$ . $R S = 20 ⟹ R = 2, S = 0$ . Digits ${1, 9, 2, 0, 7, 6}$ all distinct. ✓

Step 4: $Y = 6$ .

Learning route:

The constraint that $P$ is both the tens digit of $P Q$ and the units digit of $X Y P$ is the pivot. We express $X Y P = 100 X + 10 Y + P$ and $P Q = 10 P + Q$ . The equation $n^{2} + (n + 1)^{2} = 100 X + 10 Y + P$ forces $2 n^{2} + 2 n + 1 \equiv P (mod 10)$ . Since $n = 10 P + Q$ , we test small $P$ (as $n \leq 21$ for 3-digit result). Only $n = 19$ yields all distinct digits.

Trap: Ignoring that $P$ must be the tens digit of $P Q$ leads to $n = 20 ⟹ 841 ⟹ Y = 4$ , but $P = 2 \neq = 1$ .

Verification: $1 9^{2} + 2 0^{2} = 361 + 400 = 761$ . $X = 7, Y = 6, P = 1$ . $P Q = 19, R S = 20$ . All digits ${1, 9, 2, 0, 7, 6}$ distinct. ✓

Question 15 · Linear Algebra · 2025 MCQ

The number of additions and multiplications involved in performing Gaussian elimination on any $n \times n$ upper triangular matrix is of the order

A.
$O (n)$
B.
$O (n^{2})$
C.
$O (n^{3})$
D.
$O (n^{4})$

Correct Answer:

Step-by-Step Solution

Insight: An upper triangular matrix already has all the zeros that forward elimination would create, so the cubic phase collapses to zero work. The only cost left is the quadratic back-substitution phase.

Exam route:

Recognise the matrix structure: $a_{ij} = 0$ for all $i > j$ .
Forward elimination has nothing to eliminate below any pivot $\Rightarrow$ cost $= 0$ .
The task "perform Gaussian elimination" means the complete solve (elimination + back substitution).
Back substitution on an $n \times n$ upper triangular system uses $\sum_{i = 1}^{n} (n - i) = \frac{n ( n - 1 )}{2}$ multiplications and the same number of additions.
Total additions + multiplications $= n (n - 1) = O (n^{2})$ .

Learning route:

Gaussian elimination has two phases. Phase 1 (forward elimination) converts $A \to U$ by zeroing entries below each pivot. For a general dense $n \times n$ matrix this costs $\frac{2 ( n ^{3} - n )}{3}$ additions and multiplications, i.e. $O (n^{3})$ . Phase 2 (back substitution) solves $U x = c$ from the bottom row upward, costing $n (n - 1)$ additions and multiplications, i.e. $O (n^{2})$ .

When the input is already upper triangular, every entry below the diagonal is already zero. At pivot step $k$ , there are no entries in column $k$ below row $k$ to remove, so no row updates are performed. Forward elimination contributes exactly $0$ operations.

The system still must be solved, so back substitution runs in full. Variable $x_{i}$ requires $n - i$ multiplications and $n - i$ additions. Summing over $i = 1, \dots, n$ gives $\frac{n ( n - 1 )}{2}$ of each, for a combined total of $n (n - 1)$ , which is $O (n^{2})$ .

The common wrong path is to memorise "Gaussian elimination is $O (n^{3})$ " and answer $O (n^{3})$ without inspecting the matrix structure. That ignores the fact that the cubic cost comes entirely from the row updates that this matrix does not need.

Verification: For $n = 3$ , back substitution does $2 + 1 + 0 = 3$ multiplications and $3$ additions, total $6 = 3 (2)$ , matching $n (n - 1)$ . Growth is clearly quadratic, not cubic.

Question 16 · Linear Algebra · 2025 NAT

Let

D = {x^{(1)}, ..., x^{(n)}}

be a dataset of

n

observations where each

x^{(i)} \in R^{100}

. It is given that

\sum_{i = 1}^{n} x^{(i)} = 0

. The covariance matrix computed from

D

has eigenvalues

λ_{i} = 100 2^{- i}

1 \leq i \leq 100

. Let

u \in R^{100}

be the direction of maximum variance with

u^{T} u = 1

.

The value of

\frac{1}{n} i = 1 \sum n (u^{T} x^{(i)})^{2} =

(Answer in integer)

Correct Answer:

Step-by-Step Solution

Key idea: This is a PCA maximum variance direction problem. The expression given is the definition of variance along a direction, which is maximized by the largest eigenvalue of the covariance matrix.

Step 1: Recognize that the dataset has mean zero ( $\sum_{i = 1}^{n} x^{(i)} = 0$ ). Thus, the covariance matrix $C$ is given by $C = \frac{1}{n} \sum_{i = 1}^{n} x^{(i)} (x^{(i)})^{T}$ .

Step 2: The expression to evaluate is $\frac{1}{n} \sum_{i = 1}^{n} (u^{T} x^{(i)})^{2}$ . Notice that $(u^{T} x^{(i)})^{2} = (u^{T} x^{(i)}) (x^{(i) T} u) = u^{T} (x^{(i)} (x^{(i)})^{T}) u$ .

Step 3: Substitute this into the sum: $\frac{1}{n} \sum_{i = 1}^{n} u^{T} (x^{(i)} (x^{(i)})^{T}) u = u^{T} (\frac{1}{n} \sum_{i = 1}^{n} x^{(i)} (x^{(i)})^{T}) u = u^{T} C u$ .

Step 4: The problem states that $u$ is the direction of maximum variance with $u^{T} u = 1$ . By the properties of PCA, the maximum variance along any unit vector is the largest eigenvalue of the covariance matrix $C$ .

Step 5: The eigenvalues are given as $λ_{i} = 100 \cdot 2^{- i}$ for $1 \leq i \leq 100$ . The largest eigenvalue occurs at $i = 1$ , which is $λ_{1} = 100 \cdot 2^{- 1} = 50$ .

Step 6: Therefore, $u^{T} C u = λ_{1} = 50$ .

Answer: 50

Question 17 · Artificial Intelligence · 2026 MCQ

Which of the following algorithms is NOT an example of uninformed search?

A.
Breadth First Search
B.
Depth First Search
C.
A* Search
D.
Depth-limited Search

Correct Answer:

Step-by-Step Solution

Insight: Uninformed search algorithms rely solely on the problem definition (states, actions, costs) without any domain-specific knowledge to guide the search toward the goal.

Exam route: Identify which algorithm uses a heuristic function $h (n)$ . BFS, DFS, and Depth-limited Search explore blindly. A* Search explicitly uses $f (n) = g (n) + h (n)$ , making it an informed (heuristic) search.

Learning route:

Step 1: Define Uninformed Search: Algorithms that have no additional information about states beyond the problem definition. They do not know if a move brings them closer to the goal. Examples: BFS, DFS, Uniform Cost Search, Depth-limited Search, Iterative Deepening DFS.

Step 2: Define Informed Search: Algorithms that use problem-specific knowledge (a heuristic function $h (n)$ ) to estimate the cost to the goal and prioritize promising paths. Examples: A* Search, Greedy Best-First Search.

Step 3: Evaluate the options: Breadth First Search, Depth First Search, and Depth-limited Search are all classic uninformed strategies. A* Search is the quintessential informed search algorithm.

Answer: A* Search.

Question 18 · Artificial Intelligence · 2024 MCQ

Consider a state space where the start state is number 1. The successor function for
the state numbered n returns two states numbered n+1 and n+2. Assume that the
states in the unexpanded state list are expanded in the ascending order of numbers
and the previously expanded states are not added to the unexpanded state list.
Which ONE of the following statements about breadth-first search (BFS) and
depth-first search (DFS) is true, when reaching the goal state number 6?

A.
BFS expands more states than DFS.
B.
DFS expands more states than BFS.
C.
Both BFS and DFS expand equal number of states.
D.
Both BFS and DFS do not reach the goal state number 6.

Correct Answer:

Step-by-Step Solution

Insight: The problem explicitly overrides standard BFS/DFS selection rules by stating "states in the unexpanded state list are expanded in the ascending order of numbers".

Exam route: Since both algorithms are forced to pick the smallest numbered state from the unexpanded list at every step, their expansion sequence will be identical. Thus, they expand the same number of states.

Learning route:

Step 1: Initialize unexpanded list with start state: {1}.

Step 2: Pop 1 (min). Successors: 2, 3. Unexpanded: {2, 3}.

Step 3: Pop 2 (min). Successors: 3, 4. Unexpanded: {3, 4}.

Step 4: Pop 3 (min). Successors: 4, 5. Unexpanded: {4, 5}.

Step 5: Pop 4 (min). Successors: 5, 6. Unexpanded: {5, 6}.

Step 6: Pop 5 (min). Successors: 6, 7. Unexpanded: {6, 7}.

Step 7: Pop 6 (min). Goal state 6 reached.

Expanded states for both BFS and DFS: 1, 2, 3, 4, 5, 6. Total = 6 states.

Answer: Both BFS and DFS expand equal number of states.

Question 19 · Calculus and Optimization · 2026 NAT

The value of $\sum_{i = 0}^{\infty} \sum_{j = 1}^{\infty} 2^{- i} 3^{- j}$ is ______________ . (Answer in integer)

Correct Answer:

1.00

Step-by-Step Solution

Insight: The summand $2^{- i} 3^{- j}$ is a product of a function of $i$ alone and a function of $j$ alone, so the double sum separates into the product of two independent geometric series.

Exam route: Split $\sum_{i} \sum_{j} 2^{- i} 3^{- j} = (\sum_{i = 0}^{\infty} 2^{- i}) (\sum_{j = 1}^{\infty} 3^{- j})$ . Evaluate each geometric series using $S = \frac{a}{1 - r}$ , then multiply.

Learning route:

This is a separable double summation question, recognisable because the general term $a_{i, j} = 2^{- i} \cdot 3^{- j}$ factors cleanly into $f (i) \cdot g (j)$ .

Step 1 — Apply the separation rule (Fubini for positive terms):

$i = 0 \sum \infty j = 1 \sum \infty 2^{- i} 3^{- j} = (i = 0 \sum \infty (\frac{1}{2})^{i}) \times (j = 1 \sum \infty (\frac{1}{3})^{j})$

Step 2 — First geometric series ( $i$ starts at $0$, ratio $r = \frac{1}{2}$ ):

$S_{1} = \frac{1}{1 - \frac{1}{2}} = 2$

Step 3 — Second geometric series ( $j$ starts at $1$, ratio $r = \frac{1}{3}$ , first term $a = \frac{1}{3}$ ):

$S_{2} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$

Step 4 — Multiply:

$S = 2 \times \frac{1}{2} = 1$

Trap check: A common mistake is to start the $j$ -sum at $0$ instead of $1$, which would give $S_{2} = \frac{1}{1 - 1/3} = \frac{3}{2}$ and a wrong total of $3$. Always read the lower limit carefully.

Verification: Evaluate the inner sum first: $\sum_{j = 1}^{\infty} 2^{- i} 3^{- j} = 2^{- i} \cdot \frac{1}{2}$ . Then $\sum_{i = 0}^{\infty} 2^{- i} \cdot \frac{1}{2} = \frac{1}{2} \cdot 2 = 1$ . Confirmed.

The answer is $1$.

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