Numerical Computation & Estimation, Data Structures, Vector Spaces… Test for GATE DA: 25 Questions with Solutions & Analysis

Attempt the Numerical Computation & Estimation, Data Structures, Vector Spaces… test for GATE DA: 25 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

25 questions · 39 marks · 1.17 minutes. Probability and Statistics: 6 · Programming, Data Structures and Algorithms: 4 · Database Management and Warehousing: 4 · Linear Algebra: 3 · Quantitative Aptitude: 2 · Machine Learning: 2 · Calculus and Optimization: 2 · Verbal Aptitude: 1 · Spatial Aptitude: 1

Free sample questions from Numerical Computation & Estimation, Data Structures, Vector Spaces…

Question 1 · Probability and Statistics · 2025 MCQ

Let

X

be a continuous random variable whose cumulative distribution function (CDF)

F_{X} (x)

, for some

t

, is given as follows:

F_{X} (x) = ⎩ ⎨ ⎧ 0 \frac{x - t}{4 - t} 1 x \leq t t \leq x \leq 4 x \geq 4

If the median of

X

3

, then what is the value of

t

A.
$2$
B.
$1$
C.
$- 1$
D.
$0$

Correct Answer:

Step-by-Step Solution

Insight: The median $m$ of a continuous random variable is the value where the CDF equals 0.5. Set $F (3) = 0.5$ and solve for $t$ .

Exam route: $F (3) = \frac{3 - t}{4 - t} = 0.5$ .

Multiply both sides by $(4 - t)$ : $3 - t = 0.5 (4 - t)$ .

$3 - t = 2 - 0.5 t$ .

$1 = 0.5 t ⟹ t = 2$ .

Learning route: This is a quantile question in reverse. Instead of finding the probability for a given $x$ , we are given the probability (0.5 for the median) and the $x$ value (3), and must find the unknown parameter $t$ . For a continuous, strictly increasing CDF, the median $m$ satisfies $F (m) = 0.5$ . We substitute $x = 3$ into the middle piece of the CDF, set it to 0.5, and solve the resulting linear equation for $t$ . We must also verify that the solution $t = 2$ makes sense (the support becomes $[2, 4]$ , and the median 3 lies inside it).

Question 2 · Probability and Statistics · 2025 MCQ

Let

X = a Z + b

, where

Z

is a standard normal random variable, and

a, b

are two unknown constants. It is given that

E [X] = 1, E [(X - E [X]) Z] = - 2, E [(X - E [X])^{2}] = 4,

where

E [X]

denotes the expectation of random variable

X

. The values of

a, b

are:

A.
$a = - 2, b = 1$
B.
$a = 2, b = - 1$
C.
$a = - 2, b = - 1$
D.
$a = 1, b = 1$

Correct Answer:

Step-by-Step Solution

Insight: The term $E [(X - E [X]) Z]$ is exactly the covariance $C o v (X, Z)$ because $E [Z] = 0$ .

Exam route: Since $Z \sim N (0, 1)$ , $E [Z] = 0$ and $E [Z^{2}] = 1$ . From $X = a Z + b$ , $E [X] = b = 1$ . Then $X - E [X] = a Z$ . The variance condition gives $E [(a Z)^{2}] = a^{2} = 4$ , so $a = \pm 2$ . The covariance condition gives $E [a Z^{2}] = a = - 2$ . Thus $a = - 2, b = 1$ .

Learning route: We are given three expectations involving $X$ and $Z$ . First, we use the linearity of expectation on $X = a Z + b$ . Since $Z$ is standard normal, its mean is 0 and variance is 1, meaning $E [Z] = 0$ and $E [Z^{2}] = 1$ .

Step 1: Find $b$ . $E [X] = E [a Z + b] = a E [Z] + b = a (0) + b = b$ . We are given $E [X] = 1$ , so $b = 1$ .

Step 2: Simplify $X - E [X]$ . Substituting $b = 1$ , we get $X - E [X] = (a Z + 1) - 1 = a Z$ .

Step 3: Use the variance condition. We are given $E [(X - E [X])^{2}] = 4$ . Substituting $X - E [X] = a Z$ , we get $E [(a Z)^{2}] = a^{2} E [Z^{2}] = a^{2} (1) = a^{2}$ . Thus, $a^{2} = 4$ , which means $a = 2$ or $a = - 2$ .

Step 4: Use the covariance condition. We are given $E [(X - E [X]) Z] = - 2$ . Substituting $X - E [X] = a Z$ , we get $E [a Z^{2}] = a E [Z^{2}] = a (1) = a$ . Thus, $a = - 2$ .

Combining these, we get $a = - 2$ and $b = 1$ . This matches the first option.

Question 3 · Probability and Statistics · 2025 MCQ

Consider the cumulative distribution function (CDF) of a random variable

X

F_{X} (x) = ⎩ ⎨ ⎧ 0 \frac{1}{4} (x + 1)^{2} 1 x \leq - 1 - 1 \leq x \leq 1 x \geq 1

The value of

P (X^{2} \leq 0.25)

A.
$0.625$
B.
$0.25$
C.
$0.5$
D.
$0.5625$

Correct Answer:

Step-by-Step Solution

Insight: Rewrite the transformed event $X^{2} \leq 0.25$ as a standard interval $- 0.5 \leq X \leq 0.5$ , then extract the probability using the CDF.

Exam route: $P (- 0.5 \leq X \leq 0.5) = F (0.5) - F (- 0.5)$ .

$F (0.5) = \frac{1}{4} (0.5 + 1)^{2} = \frac{2.25}{4} = 0.5625$ .

$F (- 0.5) = \frac{1}{4} (- 0.5 + 1)^{2} = \frac{0.25}{4} = 0.0625$ .

Probability = $0.5625 - 0.0625 = 0.5$ .

Learning route: This is a transformed event question, recognizable because the condition is on $X^{2}$ rather than $X$ directly. For continuous variables, we first solve the inequality: $X^{2} \leq 0.25 ⟺ - 0.5 \leq X \leq 0.5$ . Then we use the interval probability formula $P (a \leq X \leq b) = F (b) - F (a)$ . We evaluate the correct piece of the piecewise CDF for both bounds. Since both $0.5$ and $- 0.5$ fall in the interval $[- 1, 1]$ , we use the quadratic piece. Subtracting the lower CDF value from the upper gives the exact probability mass in that interval.

Question 4 · Programming, Data Structures and Algorithms · 2025 MCQ

Consider a hash table of size

10

with indices

{0, 1, ..., 9}

, with the hash function

h (x) = 3 x (mod 10),

where linear probing is used to handle collisions. The hash table is initially empty and then the following sequence of keys is inserted into the hash table:

1, 4, 5, 6, 14, 15

. The indices where the keys

14

and

15

are stored are, respectively

A.
$2$ and $5$
B.
$2$ and $6$
C.
$4$ and $5$
D.
$4$ and $6$

Correct Answer:

Step-by-Step Solution

Key idea: This is a hash table tracing question, recognizable because it provides a specific hash function, collision resolution strategy, and a sequence of keys to insert.

Step 1: Understand the hash function and table size. The table has size 10 (indices 0 to 9). The hash function is $h (x) = 3 x (mod 10)$ .

Step 2: Trace the insertion of each key sequentially using linear probing.

Insert 1: $h (1) = 3 (1) (mod 10) = 3$ . Index 3 is empty. Place 1 at index 3.
Insert 4: $h (4) = 3 (4) (mod 10) = 12 (mod 10) = 2$ . Index 2 is empty. Place 4 at index 2.
Insert 5: $h (5) = 3 (5) (mod 10) = 15 (mod 10) = 5$ . Index 5 is empty. Place 5 at index 5.
Insert 6: $h (6) = 3 (6) (mod 10) = 18 (mod 10) = 8$ . Index 8 is empty. Place 6 at index 8.
Insert 14: $h (14) = 3 (14) (mod 10) = 42 (mod 10) = 2$ . Index 2 is occupied (by 4).
Probe 1: $(2 + 1) (mod 10) = 3$ . Index 3 is occupied (by 1).
Probe 2: $(3 + 1) (mod 10) = 4$ . Index 4 is empty. Place 14 at index 4.
Insert 15: $h (15) = 3 (15) (mod 10) = 45 (mod 10) = 5$ . Index 5 is occupied (by 5).
Probe 1: $(5 + 1) (mod 10) = 6$ . Index 6 is empty. Place 15 at index 6.

Step 3: Identify the final indices. Key 14 is at index 4, and key 15 is at index 6.

Answer: D

Question 5 · Programming, Data Structures and Algorithms · 2024 MCQ

Consider performing uniform hashing on an open address hash table with load
factor

α = \frac{n}{m} < 1

, where

n

elements are stored in the table with

m

slots. The
expected number of probes in an unsuccessful search is at most

\frac{1}{1 - α}

.
Inserting an element in this hash table requires at most ______ probes, on average.

A.
$ln (\frac{1}{1 - α})$
B.
$\frac{1}{1 - α}$
C.
$1 + \frac{α}{2}$
D.
$\frac{1}{1 + α}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a probe complexity question, recognizable because it asks for the expected number of probes for a specific hash table operation under the uniform hashing assumption.

Step 1: Recall the mechanics of insertion in open addressing. To insert a new element, the algorithm must find an empty slot in the hash table.

Step 2: Relate insertion to search operations. Finding an empty slot is logically identical to an unsuccessful search, which probes sequentially until it finds an empty slot (indicating the key is not present).

Step 3: Apply the uniform hashing formula. The problem states that the expected number of probes for an unsuccessful search is at most $\frac{1}{1 - α}$ .

Step 4: Conclude the complexity. Since the insertion process requires exactly the same probing sequence as an unsuccessful search, its expected number of probes is also $\frac{1}{1 - α}$ .

Answer: B

Question 6 · Programming, Data Structures and Algorithms · 2026 MSQ

You are given the following Pre-order and In-order traversals of a Binary Tree T with nodes E, F, G, P, Q, R, S.
Pre-order: P Q S E R F G
In-order: S Q E P F R G
Which of the following statements is/are true about the Binary Tree T?

A.
Node P is the root of T
B. The Post-order traversal of T is:
S E Q F G R P
C.
Node Q has only one child
D.
The left subtree of node R contains the node G

Correct Answer:

["A","B"]

Step-by-Step Solution

Insight: Reconstruct the tree using the standard Pre-order and In-order split method.

Exam route: Root is P. Left In-order has 3 elements, Right has 3. Split Pre-order accordingly. Build left and right subtrees. Check options.

Learning route:

Step 1: Root is the first element in Pre-order: P.

Step 2: Find P in In-order. Left of P is S, Q, E (size 3). Right of P is F, R, G (size 3).

Step 3: Split remaining Pre-order (Q, S, E, R, F, G) into Left Pre-order (Q, S, E) and Right Pre-order (R, F, G).

Step 4: Left subtree: Pre (Q, S, E), In (S, Q, E). Root is Q. Left In is S, Right In is E. So Q has left child S, right child E.

Step 5: Right subtree: Pre (R, F, G), In (F, R, G). Root is R. Left In is F, Right In is G. So R has left child F, right child G.

Step 6: Evaluate options.

A) P is root. (True)

B) Post-order is Left Post + Right Post + Root. Left Post: S, E, Q. Right Post: F, G, R. Total: S, E, Q, F, G, R, P. (True)

C) Q has two children (S and E). (False)

D) G is in the right subtree of R. (False)

Question 7 · Database Management and Warehousing · 2026 MCQ

Let R(A,B,C,D,E) be a relational schema with functional dependency set

F = {A \to B C, C D \to E, E \to A}

.
Which of the following statements is correct?

A.
AD, ED and CD are the only candidate keys of R.
B.
AD and ED are the only candidate keys of R.
C.
A, E and CD are the only candidate keys of R.
D.
A and CD are the only candidate keys of R.

Correct Answer:

Step-by-Step Solution

Insight: $D$ appears only on left-hand sides so it must sit in every candidate key, $B$ appears only on right-hand sides so it can never sit in one — then only three combinations remain to close.

Exam route: ${A, D}^{+}$ , ${C, D}^{+}$ and ${E, D}^{+}$ all equal $R$ and each is minimal, so the keys are $A D$ , $C D$ , $E D$ — option A.

Learning route:

Step 1 — the trigger. Every option ends with "the only candidate keys", so the task is exhaustive enumeration plus a minimality proof, not one convenient closure.

Step 2 — classify by appearance in $F = {A \to B C, C D \to E, E \to A}$ .

$A$ : left in $A \to B C$ , right in $E \to A$ — both sides.

$B$ : right only — never in a candidate key. (Whenever $A$ is known, $B$ comes free from $A \to B C$ , and every key here eventually reaches $A$ , directly or through $C D \to E \to A$ ; so deleting $B$ from a key would leave the closure intact, and a key containing $B$ could not be minimal.)

$C$ : left in $C D \to E$ , right in $A \to B C$ — both sides.

$D$ : left only — must be in every candidate key.

$E$ : left in $E \to A$ , right in $C D \to E$ — both sides.

Step 3 — every candidate key is ${D}$ together with a subset of ${A, C, E}$ . Test the three single additions.

${A, D}$ : $A \to B C$ gives ${A, B, C, D}$ ; now $C D \to E$ fires and adds $E$ , giving all of $R$ . Check minimality: $A^{+} = {A, B, C}$ stops there because $C D \to E$ needs $D$ , and $D^{+} = {D}$ . Both proper subsets fail, so $A D$ is a candidate key.

${C, D}$ : $C D \to E$ gives ${C, D, E}$ ; $E \to A$ gives ${A, C, D, E}$ ; $A \to B C$ completes $R$ . Minimality: $C^{+} = {C}$ , $D^{+} = {D}$ . So $C D$ is a candidate key.

${D, E}$ : $E \to A$ gives ${A, D, E}$ ; $A \to B C$ gives all of $R$ . Minimality: $E^{+} = {A, B, C, E}$ — it never acquires $D$ — and $D^{+} = {D}$ . So $E D$ is a candidate key.

Any two- or three-element addition is a proper superset of one of these three, so none is minimal.

Step 4 — the candidate keys are exactly $A D$ , $C D$ , $E D$ . Option A.

Second pass by back-substitution: option B claims only $A D$ and $E D$ , but we just computed $C D^{+} = R$ with both proper subsets failing, so a genuine third key is missing and B is incomplete. Options C and D both include $A$ : $A^{+} = {A, B, C}$ , missing $D$ and $E$ , so $A$ is not a key at all; option C also includes $E$ , whose closure ${A, B, C, E}$ is missing $D$ . Only option A survives both checks.

The tempting wrong path: compute $E^{+}$ , watch $E \to A \to B C$ fill in three attributes, glance at ${A, B, C, E}$ and declare $E$ a key because "it reached almost everything". That produces option C, and its trimmed twin option D. The break is the single line where the closure is compared with $R$ : $D$ is absent, and $D$ is precisely the attribute nothing in $F$ can produce.

Generalisation: an attribute that never appears on a right-hand side must be physically present in every key — never let a closure "almost" cover the relation.

Verification: $A D$ , $C D$ and $E D$ each close to all of $R$ and each loses that property when either attribute is removed; $A$ , $E$ , $C$ alone and every set containing $B$ all fail. Three keys, matching option A exactly.

Question 8 · Database Management and Warehousing · 2026 MSQ

Consider two relations r and s defined on the relational schemas R(A,B) and S(E,C), respectively. A is the primary key of R and E is a foreign key of S referencing A in R.
Which of the following operations will NEVER violate the foreign key constraint?

A.
Inserting records into relation r
B.
Deleting records from relation s
C.
Deleting records from relation r
D.
Inserting records into relation s

Correct Answer:

["A","B"]

Step-by-Step Solution

Insight: Foreign key constraints restrict child inserts/updates and parent deletes/updates, but never restrict parent inserts or child deletes.

Exam route: Identify parent (R) and child (S). Parent insert = safe. Child delete = safe.

Learning route:

Referential integrity dictates that a foreign key in the child table must match an existing primary key in the parent table.

When you insert into the parent (R), you are adding a new valid target. No existing child can reference it yet, so no constraint is violated.

When you delete from the child (S), you are removing a reference. The parent remains intact, and no orphaned records are created in the child, so no constraint is violated.

Conversely, deleting from the parent (R) might leave orphaned records in S, and inserting into the child (S) might reference a non-existent parent in R.

Therefore, inserting into R and deleting from S will NEVER violate the foreign key constraint.

Question 9 · Database Management and Warehousing · 2025 MSQ

A.
aadhaar is a candidate key in the Customer relation
B.
phone can be NULL in the Customer relation
C.
aadhaar is a candidate key in the Person relation
D.
aadhaar can be NULL in the Person relation

Correct Answer:

["C"]

Step-by-Step Solution

Insight: Primary keys cannot be NULL and are always candidate keys; foreign keys can be NULL and thus cannot be candidate keys.

Exam route: Since the question statement is missing, we deduce from standard relational rules and the options. A primary key (aadhaar in Person) cannot be NULL and is a candidate key. A foreign key (aadhaar in Customer) can be NULL, so it is not a candidate key.

Learning route:

Note: The question statement is missing from the prompt, but based on the options, this is a standard question about Person and Customer relations where aadhaar is the primary key of Person and a foreign key in Customer.

Option A: aadhaar is a candidate key in Customer. Since it is a foreign key, it can contain NULL values (unless explicitly constrained). Candidate keys cannot contain NULLs. Thus, A is false.
Option B: phone can be NULL in Customer. Without the statement, this is ambiguous. If phone is a candidate key, it cannot be NULL. If it is just a unique or regular attribute, it can be. We exclude it as it depends on missing text.
Option C: aadhaar is a candidate key in Person. Since aadhaar is the primary key of Person, it is by definition a candidate key. Thus, C is unconditionally true.
Option D: aadhaar can be NULL in Person. Primary keys cannot be NULL (Entity Integrity). Thus, D is false.

The only unconditionally true statement is C.

Question 10 · Linear Algebra · 2025 MCQ

The sum of the elements in each row of $A \in R^{n \times n}$ is $1$ . If $B = A^{3} - 2 A^{2} + A$ , which one of the following statements is correct (for $x \in R^{n}$ )?

A.
The equation $B x = 0$ has no solution
B.
The equation $B x = 0$ has exactly two solutions
C.
The equation $B x = 0$ has infinitely many solutions
D.
The equation $B x = 0$ has a unique solution

Correct Answer:

Step-by-Step Solution

Key idea: The row sum condition directly gives an eigenvalue and its eigenvector, which can be evaluated in the matrix polynomial.

Step 1: The statement "sum of the elements in each row of $A$ is $1$" means that if we multiply $A$ by the all-ones column vector $1 = [1, 1, \dots, 1]^{T}$ , the result is $1$ . Thus, $A 1 = 1 \cdot 1$ .

Step 2: This implies $λ = 1$ is an eigenvalue of $A$ , and $1$ is a corresponding non-zero eigenvector.

Step 3: We are given $B = A^{3} - 2 A^{2} + A$ . We can factor this polynomial as $B = A (A^{2} - 2 A + I) = A (A - I)^{2}$ .

Step 4: Evaluate $B 1$ . Since $(A - I) 1 = A 1 - 1 = 1 - 1 = 0$ , we have $(A - I)^{2} 1 = (A - I) 0 = 0$ .

Step 5: Therefore, $B 1 = A (A - I)^{2} 1 = A 0 = 0$ .

Step 6: Since $1$ is a non-zero vector and $B 1 = 0$ , the homogeneous system $B x = 0$ has a non-trivial solution. Any homogeneous system with a non-trivial solution has infinitely many solutions.

Answer: The equation $B x = 0$ has infinitely many solutions.

Question 11 · Linear Algebra · 2025 MCQ

Let $A \in R^{n \times n}$ be such that $A^{3} = A$ . Which one of the following statements is ALWAYS correct?

A.
$A$ is invertible
B.
Determinant of $A$ is $0$
C.
The sum of the diagonal elements of $A$ is $1$
D.
$A$ and $A^{2}$ have the same rank

Correct Answer:

Step-by-Step Solution

Key idea: The matrix equation $A^{3} = A$ implies that the minimal polynomial of $A$ has distinct linear factors, making $A$ diagonalizable, and forcing a specific relationship between the null spaces of $A$ and $A^{2}$ .

Step 1: Rewrite the given equation as $A^{3} - A = 0$ , which factors to $A (A - I) (A + I) = 0$ .

Step 2: This means the minimal polynomial of $A$ must divide $p (x) = x (x - 1) (x + 1)$ . Since $p (x)$ has distinct linear roots ( $0, 1, - 1$ ), the matrix $A$ is diagonalizable.

Step 3: Evaluate the options using counterexamples or structural properties.

Option A: " $A$ is invertible". Counterexample: $A = 0$ (the zero matrix) satisfies $0^{3} = 0$ , but is not invertible.
Option B: "Determinant of $A$ is $0$". Counterexample: $A = I$ (identity matrix) satisfies $I^{3} = I$ , but $det (I) = 1 \neq = 0$ .
Option C: "The sum of diagonal elements is $1$". Counterexample: $A = 0$ has trace $0$; $A = I$ (for $n = 2$ ) has trace $2$. Neither is always $1$.
Option D: " $A$ and $A^{2}$ have the same rank". Let's prove this structurally.

Step 4: Prove $ker (A) = ker (A^{2})$ .

Clearly, if $A x = 0$ , then $A^{2} x = A (A x) = A (0) = 0$ . So $ker (A) \subseteq ker (A^{2})$ .
Conversely, if $A^{2} x = 0$ , multiply both sides by $A$ to get $A^{3} x = 0$ . Since $A^{3} = A$ , this means $A x = 0$ . So $ker (A^{2}) \subseteq ker (A)$ .

Step 5: Since $ker (A) = ker (A^{2})$ , their dimensions (nullities) are equal: $nullity (A) = nullity (A^{2})$ .

Step 6: By the Rank-Nullity Theorem, $rank (A) = n - nullity (A) = n - nullity (A^{2}) = rank (A^{2})$ . This is ALWAYS correct.

Answer: $A$ and $A^{2}$ have the same rank.

Question 12 · Linear Algebra · 2024 MSQ

Let

R

be the set of real numbers,

U

be a subspace of

R^{3}

and

M \in R^{3 \times 3}

be the
matrix corresponding to the projection on to the subspace

U

.
Which of the following statements is/are TRUE?

A. If $U$ is a 1-dimensional subspace of $R^{3}$ , then the null space of $M$ is a
1-dimensional subspace.
B. If $U$ is a 2-dimensional subspace of $R^{3}$ , then the null space of $M$ is a
1-dimensional subspace.
C.
$M^{2} = M$
D.
$M^{3} = M$

Correct Answer:

["B","C","D"]

Step-by-Step Solution

Key idea: This question tests the fundamental algebraic and geometric properties of orthogonal projection matrices, specifically the Rank-Nullity Theorem and idempotence.

Step 1: Analyze the Null Space dimension (Options A and B).

Let $M$ be the projection matrix onto subspace $U \subseteq R^{3}$ .

The column space of $M$ , denoted $C (M)$ , is exactly the target subspace $U$ . Thus, $rank (M) = dim (U)$ .

By the Rank-Nullity Theorem for a $3 \times 3$ matrix:

$rank (M) + nullity (M) = 3$

$dim (Null (M)) = 3 - dim (U)$

Check Option A: If $dim (U) = 1$ , then $dim (Null (M)) = 3 - 1 = 2$ . Option A claims it is 1-dimensional. This is FALSE.

Check Option B: If $dim (U) = 2$ , then $dim (Null (M)) = 3 - 2 = 1$ . Option B claims it is 1-dimensional. This is TRUE.

Step 2: Analyze Idempotence (Option C).

A matrix $M$ is a projection matrix if and only if it is idempotent, meaning $M^{2} = M$ .

Geometrically, if you project a vector $v$ onto $U$ to get $p$ , projecting $p$ again yields $p$ itself because $p$ is already in $U$ .

Algebraically: $M (M v) = M v ⟹ M^{2} = M$ .

Thus, Option C is TRUE.

Step 3: Analyze Higher Powers (Option D).

Since $M^{2} = M$ , we can compute $M^{3}$ :

$M^{3} = M^{2} \cdot M = M \cdot M = M^{2} = M$

By induction, $M^{k} = M$ for all integers $k \geq 1$ .

Thus, $M^{3} = M$ is TRUE.

Answer: Options B, C, and D are TRUE.

Question 13 · Quantitative Aptitude · 2026 MCQ

Five integers are picked from 0 to 20, with possible repetitions, such that their mean is 12, median is 18, and they have a single mode of 20.
Ignoring permutations, the number of ways to pick these five integers is _____

A.
0
B.
1
C.
2
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: This is a statistical constraint satisfaction question, recognisable because it asks to reconstruct a specific data set given its mean, median, and mode.

Step 1: Use the mean to find the sum. For 5 integers with a mean of 12, the sum is $5 \times 12 = 60$ .

Step 2: Use the median to fix the middle value. Sorting the 5 integers as $x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \leq x_{5}$ , the median is $x_{3}$ . Thus, $x_{3} = 18$ .

Step 3: Use the mode to fix the highest values. The mode is 20, meaning 20 must appear more frequently than any other number. Since $x_{3} = 18$ , the only way to include 20 is if $x_{4} = 20$ and $x_{5} = 20$ .

Step 4: Find the remaining sum. We have $x_{1} + x_{2} + 18 + 20 + 20 = 60$ , which simplifies to $x_{1} + x_{2} = 2$ .

Step 5: Find integer pairs for $x_{1}$ and $x_{2}$ . Since the integers are from 0 to 20 and $x_{1} \leq x_{2} \leq 18$ , the possible pairs summing to 2 are $(0, 2)$ and $(1, 1)$ .

Step 6: Apply the "single mode" constraint. If the pair is $(1, 1)$ , the set is ${1, 1, 18, 20, 20}$ , which has two modes (1 and 20). This violates the single mode condition. If the pair is $(0, 2)$ , the set is ${0, 2, 18, 20, 20}$ , which has a single mode of 20.

Answer: There is exactly 1 valid way to pick these integers.

Question 14 · Quantitative Aptitude · 2024 MCQ

Person 1 and Person 2 invest in three mutual funds A, B, and C. The amounts they
invest in each of these mutual funds are given in the table.

	Mutual fund A	Mutual fund B	Mutual fund C
Person 1	₹10,000	₹20,000	₹20,000
Person 2	₹20,000	₹15,000	₹15,000

At the end of one year, the total amount that Person 1 gets is ₹500 more than
Person 2. The annual rate of return for the mutual funds B and C is 15% each. What
is the annual rate of return for the mutual fund A?

A.
7.5%
B.
10%
C.
15%
D.
20%

Correct Answer:

Step-by-Step Solution

Insight: The difference in total returns between two investors with the same set of funds yields a linear equation in the unknown rate.

Exam route: Write the total return for each person using $R = \sum P_{i} r_{i}$ , subtract the two equations, and solve for $r_{A}$ .

Learning route:

Identify the rates: $r_{B} = 15% = 0.15$ , $r_{C} = 15% = 0.15$ , and let $r_{A} = r$ .
Calculate Person 1's total return:

$R_{1} = 10000 r + 20000 (0.15) + 20000 (0.15) = 10000 r + 3000 + 3000 = 10000 r + 6000$

Calculate Person 2's total return:

$R_{2} = 20000 r + 15000 (0.15) + 15000 (0.15) = 20000 r + 2250 + 2250 = 20000 r + 4500$

Use the given condition that Person 1 gets ₹500 more than Person 2:

$R_{1} - R_{2} = 500$

$(10000 r + 6000) - (20000 r + 4500) = 500$

$- 10000 r + 1500 = 500$

Solve for $r$ :

$10000 r = 1000 ⟹ r = 0.10 = 10%$

The annual rate of return for mutual fund A is 10%.

Question 15 · Machine Learning · 2026 MCQ

For a classification problem, Principal Component Analysis (PCA) has been used to reduce the dimensionality of a feature space from 100 to 10.
Which of the following options is true about the angle

θ

between the first and the tenth principal components?

A.
$θ = 0^{\circ}$
B.
$θ = 9 0^{\circ}$
C.
$9 0^{\circ} < θ \leq 18 0^{\circ}$
D.
$0 < θ < 9 0^{\circ}$

Correct Answer:

Step-by-Step Solution

Insight: This is a geometric property question about PCA.

Exam route: Principal components are eigenvectors of a symmetric covariance matrix, hence they are strictly orthogonal. The angle between any two distinct PCs is exactly 90 degrees.

Learning route:

Step 1: Recall the definition of Principal Components. They are the eigenvectors of the data's covariance matrix.

Step 2: Recall the property of the covariance matrix. It is a real symmetric matrix.

Step 3: Apply the Spectral Theorem. The eigenvectors of a real symmetric matrix corresponding to distinct eigenvalues are mutually orthogonal.

Step 4: Conclude the angle. Since the 1st and 10th principal components are distinct eigenvectors, the angle between them is exactly 90 degrees. Option B is correct.

Question 16 · Machine Learning · 2026 MSQ

Consider that 20 stories of Author X and 10 stories of Author Y were kept together without mentioning the names of the authors. A classifier was then asked to predict the author (X or Y) of each of these stories. Let, out of X’s stories, 6 were classified as that of Y. On the other hand, out of Y’s stories, 2 were classified as that of X.

Considering X and Y as two classes, which of the following statements is/are true?

A.
Accuracy of the classifier is 11/15.
B.
Precision of Class X is higher than the Precision of Class Y.
C.
Recall of Class X is higher than the Recall of Class Y.
D.
Accuracy of the classifier is 14/15.

Correct Answer:

["A","B"]

Step-by-Step Solution

Insight: Construct the confusion matrix by carefully mapping the text to TP, FN, FP, TN for one class, then derive all metrics symmetrically for the other.

Exam route: Total = 30. Correct = (20-6) + (10-2) = 14 + 8 = 22. Accuracy = 22/30 = 11/15. Precision X = 14/(14+2) = 14/16. Precision Y = 8/(8+6) = 8/14. Since 14/16 > 8/14, Precision X > Precision Y. Recall X = 14/20 = 0.7, Recall Y = 8/10 = 0.8, so Recall X < Recall Y.

Learning route:

Define Class X as Positive, Class Y as Negative.
Map counts from the text:

Actual X = 20. Predicted Y = 6 (False Negatives, FN). Thus, Predicted X = 14 (True Positives, TP).
Actual Y = 10. Predicted X = 2 (False Positives, FP). Thus, Predicted Y = 8 (True Negatives, TN).

Calculate Accuracy: $(T P + T N) / Total = (14 + 8) /30 = 22/30 = 11/15$ . (Option A is true, Option D is false).
Calculate Precision for X: $T P / (T P + F P) = 14/ (14 + 2) = 14/16 = 0.875$ .
Calculate Precision for Y (treating Y as Positive): $T P_{Y} = 8$ , $F P_{Y} = 6$ . Precision Y = $8/ (8 + 6) = 8/14 \approx 0.571$ .

Since $0.875 > 0.571$ , Precision of X is higher. (Option B is true).

Calculate Recall for X: $T P / (T P + F N) = 14/ (14 + 6) = 14/20 = 0.7$ .
Calculate Recall for Y: $T P_{Y} / (T P_{Y} + F N_{Y}) = 8/ (8 + 2) = 8/10 = 0.8$ .

Since $0.7 < 0.8$ , Recall of X is lower. (Option C is false).

Question 17 · Calculus and Optimization · 2024 MCQ

Let

f : R \to R

be a function. Note:

R

denotes the set of real numbers.

f (x) = {- x, if x < - 2 \ax^{2} + b x + c, if x \in [- 2, 2] \x, if x > 2

Which ONE of the following choices gives the values of

a, b, c

that make the
function

f

continuous and differentiable?

A.
$a = \frac{1}{4}, b = 0, c = 1$
B.
$a = \frac{1}{2}, b = 0, c = 0$
C.
$a = 0, b = 0, c = 0$
D.
$a = 1, b = 1, c = - 4$

Correct Answer:

Step-by-Step Solution

Insight: Piecewise smoothness requires matching both function values (continuity) and slopes (differentiability) at every boundary point.

Exam route: Set up 4 equations: 2 for continuity at $x = - 2, 2$ and 2 for differentiability at $x = - 2, 2$ . Solve the linear system for $a, b, c$ .

Learning route:

Continuity at $x = - 2$ : $lim_{x \to - 2^{-}} (- x) = f (- 2) ⟹ 2 = 4 a - 2 b + c$ .
Continuity at $x = 2$ : $lim_{x \to 2^{+}} (x) = f (2) ⟹ 2 = 4 a + 2 b + c$ .
Differentiability at $x = - 2$ : LHD = $lim_{x \to - 2^{-}} (- 1) = - 1$ . RHD = $lim_{x \to - 2^{+}} (2 a x + b) = - 4 a + b$ . So $- 4 a + b = - 1$ .
Differentiability at $x = 2$ : LHD = $lim_{x \to 2^{-}} (2 a x + b) = 4 a + b$ . RHD = $lim_{x \to 2^{+}} (1) = 1$ . So $4 a + b = 1$ .
Solve the system: Adding the derivative equations gives $2 b = 0 ⟹ b = 0$ . Subtracting gives $8 a = 2 ⟹ a = 1/4$ .
Substitute $a = 1/4, b = 0$ into the first continuity equation: $4 (1/4) - 0 + c = 2 ⟹ 1 + c = 2 ⟹ c = 1$ .

The values are $a = 1/4, b = 0, c = 1$ .

Question 18 · Calculus and Optimization · 2025 NAT

Let

f : R \to R

be such that

∣ f (x) - f (y) ∣ \leq (x - y)^{2}

for all

x, y \in R

. Then

f (1) - f (0) =

(Answer in integer)

Correct Answer:

0.00

Step-by-Step Solution

Insight: An inequality bounding the difference of function values by a power $> 1$ of the difference in inputs forces the derivative to be zero everywhere.

Exam route: Divide the given inequality by $∣ x - y ∣$ and take the limit as $x \to y$ to show $f^{'} (y) = 0$ , meaning $f$ is constant.

Learning route:

Given $∣ f (x) - f (y) ∣ \leq (x - y)^{2}$ for all $x, y$ .
For $x \neq = y$ , divide both sides by $∣ x - y ∣$ : $\frac{f ( x ) - f ( y )}{x - y} \leq ∣ x - y ∣$ .
Take the limit as $x \to y$ : The right side approaches 0.
By the Squeeze Theorem, $lim_{x \to y} \frac{f ( x ) - f ( y )}{x - y} = 0$ , which means $∣ f^{'} (y) ∣ = 0 ⟹ f^{'} (y) = 0$ .
Since $f^{'} (y) = 0$ for all $y \in R$ , $f (x)$ is a constant function.
Therefore, $f (1) - f (0) = 0$ .

Question 19 · Verbal Aptitude · 2026 MCQ

‘My friend and I parted __ the door __ the cabin that I had rented __ the night.’
Choose the option with the correct sequence of words to fill the blanks.

A.
at; of; for
B.
for; at; of
C.
of; for; in
D.
in; of; for

Correct Answer:

Step-by-Step Solution

Insight: This is a multi-blank preposition question requiring independent analysis of spatial points, possession, and duration.

Exam route: Blank 1: "parted at the door" (specific point). Blank 2: "door of the cabin" (possession/association). Blank 3: "rented for the night" (duration). Sequence: at; of; for.

Learning route:

Step 1: Analyze the first blank. "Parted" happens at a specific physical location (the door). The preposition "at" is used for exact points.

Step 2: Analyze the second blank. The relationship between "door" and "cabin" is one of possession or association. The preposition "of" correctly denotes this relationship.

Step 3: Analyze the third blank. "The night" here refers to a duration of time for which the cabin was rented. The preposition "for" is used to denote a period or duration.

Step 4: Verify the sequence "at; of; for" against the options. Option A matches perfectly.

Question 20 · Spatial Aptitude · 2025 MCQ

In the given figure, the numbers associated with the rectangle, triangle, and ellipse are

1, 2,

and

3

, respectively. Which one among the given options is the most appropriate combination of

P, Q,

and

R

A.
$P = 6; Q = 5; R = 3$
B.
$P = 5; Q = 6; R = 3$
C.
$P = 3; Q = 6; R = 6$
D.
$P = 5; Q = 3; R = 6$

Correct Answer:

Step-by-Step Solution

Insight: This is a Venn diagram region identification problem. The numbers 1, 2, 3 represent the shapes (Rectangle, Triangle, Ellipse). The value for any point P, Q, R is the sum of the IDs of the shapes that strictly contain it.

Exam route: Test each point against the boundaries of the three shapes. Sum the IDs (1 for Rect, 2 for Tri, 3 for Ellipse) for the shapes containing the point. Match the resulting triplet (P, Q, R) to the options.

Learning route:

Identify the shape boundaries from the figure:

Rectangle (1): Bounded region roughly in the center-left.
Triangle (2): Large triangle encompassing the right side.
Ellipse (3): Oval shape overlapping the bottom-right.

Evaluate Point P:

P is located in the central region where all three shapes overlap.
It is inside the Rectangle (1), inside the Triangle (2), and inside the Ellipse (3).
Sum for P = $1 + 2 + 3 = 6$ .

Evaluate Point Q:

Q is located to the right of the Rectangle's right boundary.
It is outside the Rectangle (1).
It is clearly inside the Triangle (2) and inside the Ellipse (3).
Sum for Q = $2 + 3 = 5$ .

Evaluate Point R:

R is located in the lower intersection of the Rectangle and Triangle, but sits just below the lower boundary of the Ellipse.
It is inside the Rectangle (1) and inside the Triangle (2), but outside the Ellipse (3).
Sum for R = $1 + 2 = 3$ .

Match with Options:

We have established $P = 6$ , $Q = 5$ , and $R = 3$ .
Looking at the given options, only Option A ( $P = 6; Q = 5; R = 3$ ) matches this triplet perfectly.

Trap Analysis: The 'boundary illusion' trap occurs when students visually estimate a point's position without strictly checking the boundary lines. For example, a student might think R is inside the Ellipse because it is visually close to it, leading to $R = 1 + 2 + 3 = 6$ . This would incorrectly point to Option C or D. Always trace the exact boundary line (e.g., the bottom curve of the ellipse) to confirm inclusion/exclusion.

Verification: P is deeply inside all three (6). Q is clearly outside the left-most shape (Rectangle), so it cannot be 6. R is below the ellipse's bottom curve, so it cannot include 3. Only Option A satisfies all strict boundary tests.

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