GATE CS 2025_Set1 Question Paper with Solutions: 65 Questions, Answer Key & Section-wise Analysis

GATE CS 2025_Set1 previous year paper: 65 questions with answer key and detailed solutions, section-wise breakdown and free sample questions.

Paper breakdown

65 questions · 100 marks. Engineering Mathematics: 11 · Programming and Data Structures: 7 · Computer Organization and Architecture: 6 · Theory of Computation: 5 · Operating System: 5 · Databases: 5 · Computer Networks: 5 · Algorithms: 5 · Digital Logic: 4 · Compiler Design: 4 · Verbal Aptitude: 3 · Spatial Aptitude: 2 · Quantitative Aptitude: 2 · Analytical Aptitude: 1

Free sample questions from GATE CS 2025_Set1 Question Paper

Question 1 · Engineering Mathematics · 2025_Set1 MCQ

A fair six-faced dice, with the faces labelled ‘1’, ‘2’, ‘3’, ‘4’, ‘5’, and ‘6’, is rolled thrice. What is the probability of rolling ‘6’ exactly once?

A.
$\frac{75}{216}$
B.
$\frac{1}{6}$
C.
$\frac{1}{18}$
D.
$\frac{25}{216}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a Binomial Probability problem, recognizable because we have a fixed number of independent trials (rolling the die 3 times) and we are looking for a specific number of successes (rolling a '6').

Step 1: Identify the parameters of the Bernoulli trial.

Total trials ( $n$ ) = 3.
Success event: Rolling a '6'.
Probability of success ( $p$ ) = $\frac{1}{6}$ .
Probability of failure ( $q$ ) = $1 - p = \frac{5}{6}$ .
Desired number of successes ( $k$ ) = 1.

Step 2: Apply the Binomial Probability Formula.

The probability of getting exactly $k$ successes in $n$ trials is:

$P (X = k) = (k n) p^{k} q^{n - k}$

Step 3: Substitute the values and calculate.

$P (X = 1) = (1 3) (\frac{1}{6})^{1} (\frac{5}{6})^{3 - 1}$

$P (X = 1) = 3 \times \frac{1}{6} \times (\frac{5}{6})^{2}$

$P (X = 1) = 3 \times \frac{1}{6} \times \frac{25}{36} = \frac{75}{216}$

Answer: $\frac{75}{216}$

Question 2 · Engineering Mathematics · 2025_Set1 MCQ

A shop has 4 distinct flavors of ice-cream. One can purchase any number of scoops of any flavor. The order in which the scoops are purchased is inconsequential. If one wants to purchase 3 scoops of ice-cream, in how many ways can one make that purchase?

A.
4
B.
20
C.
24
D.
48

Correct Answer:

Step-by-Step Solution

Key idea: This is a Stars and Bars problem, recognizable by the distribution of identical items (scoops of the same flavor are indistinguishable) into distinct categories (the 4 distinct flavors), with the phrase "order is inconsequential" confirming we are choosing a multiset.

Step 1: Identify the parameters for the Stars and Bars formula.

Number of identical items to distribute: $n = 3$ (scoops).

Number of distinct bins: $k = 4$ (flavors).

Step 2: Apply the standard Stars and Bars formula for non-negative integer solutions.

The number of ways is given by $(k - 1 n + k - 1)$ .

Step 3: Substitute the values and calculate.

$(4 - 1 3 + 4 - 1) = (3 6)$ .

$(3 6) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ .

Answer: B

Question 3 · Engineering Mathematics · 2025_Set1 MCQ

g (.)

is a function from

A

B

f (.)

is a function from

B

C

, and their composition defined as

f (g (.))

is a mapping from

A

C

.

If

f (.)

and

f (g (.))

are onto (surjective) functions, which ONE of the following is TRUE about the function

g (.)

A.
$g (.)$ must be an onto (surjective) function.
B.
$g (.)$ must be a one-to-one (injective) function.
C.
$g (.)$ must be a bijective function, that is, both one-to-one and onto.
D.
$g (.)$ is not required to be a one-to-one or onto function.

Correct Answer:

Step-by-Step Solution

Key idea: This is a function composition property question, recognizable by the given surjectivity of $f$ and $f \circ g$ .

Step 1: Analyze the given conditions. We know $f : B \to C$ is onto, and $f \circ g : A \to C$ is onto.

Step 2: Test if $g$ must be onto. Consider $A = {1}$ , $B = {1, 2}$ , $C = {1}$ . Let $g (1) = 1$ and $f (1) = f (2) = 1$ . Here, $f$ is onto and $f (g (1)) = 1$ is onto, but $g$ is not onto (2 is not in the range of $g$ ). Thus, $g$ is not required to be onto.

Step 3: Test if $g$ must be one-to-one. Consider $A = {1, 2}$ , $B = {1}$ , $C = {1}$ . Let $g (1) = g (2) = 1$ and $f (1) = 1$ . Here, $f$ is onto and $f \circ g$ is onto, but $g$ is not one-to-one.

Step 4: Since $g$ is neither required to be onto nor one-to-one, Option D is the only necessarily true statement.

Answer: Option D.

Question 4 · Programming and Data Structures · 2025_Set1 MSQ

Which of the following statement(s) is/are TRUE for any binary search tree (BST) having $n$ distinct integers?

A.
The maximum length of a path from the root node to any other node is $(n - 1)$ .
B.
An inorder traversal will always produce a sorted sequence of elements.
C.
Finding an element takes $O (lo g_{2} n)$ time in the worst case.
D.
Every BST is also a Min-Heap.

Correct Answer:

["A","B"]

Step-by-Step Solution

Insight: A BST's inorder traversal is always sorted, and its maximum possible depth is $n - 1$ (achieved by a completely skewed tree).

Exam route: Evaluate each option. A is true because a tree with $n$ nodes has at most $n - 1$ edges in any path. B is true by the definition of BST. C is false because worst-case search is $O (n)$ for a skewed tree. D is false because BST ordering (left < parent < right) does not imply heap ordering (parent $\leq$ children).

Learning route:

Option A: A path in a tree with $n$ nodes can have at most $n$ nodes, which means $n - 1$ edges. A right-skewed BST achieves exactly this maximum length. Thus, the maximum length is bounded by and can be $n - 1$ .
Option B: The BST property (left subtree < node < right subtree) guarantees that an inorder traversal (Left, Node, Right) visits elements in strictly ascending order.
Option C: Search time is $O (h)$ , where $h$ is the height. In the worst case (skewed tree), $h = n - 1$ , making the time complexity $O (n)$ , not $O (lo g_{2} n)$ .
Option D: A Min-Heap requires every parent to be $\leq$ its children. A BST only requires left child < parent < right child. For example, a BST with root 10, left child 5, and right child 15 violates the Min-Heap property because 10 is not $\leq$ 5.

Verification: For $n = 3$ , elements ${1, 2, 3}$ inserted as $1, 2, 3$ gives a right-skewed tree. Path length from root (1) to leaf (3) is 2 edges ( $3 - 1 = 2$ ). Inorder is $1, 2, 3$ (sorted). Search for 3 takes 3 steps ( $O (n)$ ). Root 1 is not $\leq$ left child (none), but if we had root 2, left 1, right 3, 2 is not $\leq$ 1, so not a min-heap.

Question 5 · Programming and Data Structures · 2025_Set1 NAT

#include <stdio.h>
void foo(int *p, int x){
*p=x;
}
int main(){
int *z;
int a = 20, b = 25;
z = &a;
foo(z,b);
printf("%d",a);
return 0;
}

The output of the given C program is __________. (Answer in integer)

Correct Answer:

25.00

Step-by-Step Solution

Insight: This is a pointer dereferencing and pass-by-reference simulation question, recognizable by a pointer being passed to a function that modifies the pointed-to value.

Exam route: z holds &a. foo(z, b) passes &a and 25. Inside foo, *p = x modifies a to 25. printf prints 25.

Learning route:

a is initialized to 20, b to 25.
z = &a makes z point to a.
foo(z, b) is called. The arguments passed are the address of a and the value 25.
Inside foo, p receives &a and x receives 25.
*p = x dereferences p (accessing a) and assigns it the value of x (25). Thus, a becomes 25.
Back in main, printf("%d", a) prints the updated value of a, which is 25.

Question 6 · Programming and Data Structures · 2025_Set1 NAT

The height of any rooted tree is defined as the maximum number of edges in the path from the root node to any leaf node.

Suppose a Min-Heap

T

stores 32 keys. The height of

T

is _____________. (Answer in integer)

Correct Answer:

5.00

Step-by-Step Solution

Insight: This is a direct application of the heap height formula. The height of a complete binary tree with $n$ nodes is simply $⌊ lo g_{2} n ⌋$ .

Exam route: The problem states the min-heap stores 32 keys. Using the height formula $h = ⌊ lo g_{2} n ⌋$ , we substitute $n = 32$ . Since $32 = 2^{5}$ , $lo g_{2} 32 = 5$ . The floor of 5 is 5. The height is 5.

Learning route:

Understand the definition: The height of a rooted tree is the maximum number of edges on any path from the root to a leaf.
Recall the structure of a heap: A heap is a complete binary tree. This means all levels except possibly the last are completely full, and the last level is filled from left to right.
Apply the formula: For a complete binary tree with $n$ nodes, the height $h$ is given by $h = ⌊ lo g_{2} n ⌋$ .
Calculate: Here, $n = 32$ . We know that $2^{5} = 32$ , so $lo g_{2} 32 = 5$ .
Conclusion: The height of the min-heap is 5.

Question 7 · Computer Organization and Architecture · 2025_Set1 MCQ

Suppose a program is running on a non-pipelined single processor computer system. The computer is connected to an external device that can interrupt the processor asynchronously. The processor needs to execute the interrupt service routine (ISR) to serve this interrupt. The following steps (not necessarily in order) are taken by the processor when the interrupt arrives:

(i) The processor saves the content of the program counter.
(ii) The program counter is loaded with the start address of the ISR.
(iii) The processor finishes the present instruction.

Which ONE of the following is the CORRECT sequence of steps?

A.
(iii), (i), (ii)
B.
(i), (iii), (ii)
C.
(i), (ii), (iii)
D.
(iii), (ii), (i)

Correct Answer:

Step-by-Step Solution

Key idea: This is an interrupt handling sequence question, testing the exact order of hardware and software actions when an asynchronous interrupt occurs.

Step 1: The processor cannot abort an instruction mid-execution without corrupting the system state. Therefore, it must first finish the currently executing instruction. This is step (iii).

Step 2: Before jumping to the ISR, the processor must save the return address so it can resume the interrupted program later. This is done by saving the content of the Program Counter (PC). This is step (i).

Step 3: Finally, the processor loads the PC with the starting address of the ISR to begin executing the interrupt service routine. This is step (ii).

Step 4: The correct sequence is (iii), (i), (ii).

Answer: A

Question 8 · Computer Organization and Architecture · 2025_Set1 MSQ

The number $- 6$ can be represented as 1010 in 4-bit 2’s complement representation. Which of the following is/are CORRECT 2’s complement representation(s) of $- 6$ ?

A.
1000 1010 in 8-bits
B.
1111 1010 in 8-bits
C.
1000 0000 0000 1010 in 16-bits
D.
1111 1111 1111 1010 in 16-bits

Correct Answer:

["B","D"]

Step-by-Step Solution

Key idea: This is a sign extension question in 2's complement, recognisable because a negative number is given in 4 bits and you must pick its correct wider representations.

Step 1: The sign extension rule for 2's complement says: replicate the MSB (sign bit) into all new higher-order bit positions.

Step 2: The 4-bit representation of $- 6$ is $1010$ . The MSB is $1$ .

Step 3: Extend to 8 bits by prepending four $1$ s: $11111010$ . This matches option B.

Step 4: Extend to 16 bits by prepending twelve $1$ s: $1111111111111010$ . This matches option D.

Step 5: Options A ( $10001010$ ) and C ( $1000000000001010$ ) pad with $0$ s instead of $1$ s. That is the sign-magnitude rule, not 2's complement. For example, $10001010$ in 8-bit 2's complement equals $- 128 + 8 + 2 = - 118$ , not $- 6$ .

Answer: Options B and D.

Question 9 · Computer Organization and Architecture · 2025_Set1 MSQ

A partial data path of a processor is given in the figure, where RA, RB, and RZ are 32-bit registers. Which option(s) is/are CORRECT related to arithmetic operations using the data path as shown?

A.
The data path can implement arithmetic operations involving two registers.
B.
The data path can implement arithmetic operations involving one register and one immediate value.
C.
The data path can implement arithmetic operations involving two immediate values.
D.
The data path can only implement arithmetic operations involving one register and one immediate value.

Correct Answer:

["A","B","C"]

Step-by-Step Solution

Key idea: This is a processor datapath analysis question, recognisable by the diagram showing registers, multiplexers, and an ALU, asking what operations are supported. Why this method applies: We must trace the possible data flows from the inputs (registers and immediate values) through the multiplexers to the ALU inputs. Step 1: Analyze the inputs to Mux_A. The diagram shows Mux_A has two inputs: the register RA and a 32-bit immediate value. Therefore, the first operand to the ALU can be either a register value or an immediate value. Step 2: Analyze the inputs to Mux_B. The diagram shows Mux_B has two inputs: the register RB and a 32-bit immediate value. Therefore, the second operand to the ALU can also be either a register value or an immediate value. Step 3: Evaluate the combinations. Since Mux_A and Mux_B operate independently, we can select any combination of their inputs: - Case 1: Select RA for Mux_A and RB for Mux_B. This implements arithmetic operations involving two registers. (Option A is correct). - Case 2: Select RA for Mux_A and immediate for Mux_B (or vice versa). This implements arithmetic operations involving one register and one immediate value. (Option B is correct). - Case 3: Select immediate for Mux_A and immediate for Mux_B. This implements arithmetic operations involving two immediate values. (Option C is correct). Step 4: Evaluate Option D. Option D states the datapath can only implement operations with one register and one immediate. This is false, as we just proved it can do Reg+Reg and Imm+Imm as well. Answer: Options A, B, and C.

Question 10 · Theory of Computation · 2025_Set1 MCQ

Consider the following context-free grammar

G

, where

S

A

, and

B

are the variables (non-terminals),

a

and

b

are the terminal symbols,

S

is the start variable, and the rules of

G

are described as:

S \to aa B ∣ A bb

A \to a ∣ a A

B \to b ∣ b B

Which ONE of the languages

L (G)

is accepted by

G

A.
$L (G) = {a^{2} b^{n} ∣ n \geq 1} \cup {a^{n} b^{2} ∣ n \geq 1}$
B.
$L (G) = {a^{n} b^{2 n} ∣ n \geq 1} \cup {a^{2 n} b^{n} ∣ n \geq 1}$
C.
$L (G) = {a^{n} b^{n} ∣ n \geq 1}$
D.
$L (G) = {a^{2 n} b^{2 n} ∣ n \geq 1}$

Question 11 · Theory of Computation · 2025_Set1 MSQ

A regular language $L$ is accepted by a non-deterministic finite automaton (NFA) with $n$ states. Which of the following statement(s) is/are FALSE?

A.
$L$ may have an accepting NFA with $< n$ states.
B.
$L$ may have an accepting DFA with $< n$ states.
C.
There exists a DFA with $\leq 2^{n}$ states that accepts $L$ .
D.
Every DFA that accepts $L$ has $> 2^{n}$ states.

Correct Answer:

["D"]

Step-by-Step Solution

Key idea: This is a state complexity bounds question, recognizable by the comparison of NFA and DFA state counts for a regular language $L$ .

Step 1: Analyze the premise. We are given that SOME NFA with $n$ states accepts $L$ . This NFA is not stated to be minimal.

Step 2: Evaluate Option 1. Since the given NFA might have redundant states, there may exist a smaller, minimal NFA for $L$ with $< n$ states. This statement is TRUE.

Step 3: Evaluate Option 2. Similarly, the minimal DFA for $L$ might have fewer states than this specific, possibly bloated, $n$ -state NFA. For example, if $L = {0, 1}^{*}$ , an NFA could be drawn with 5 redundant states ( $n = 5$ ), but the minimal DFA has 1 state ( $< 5$ ). This statement is TRUE.

Step 4: Evaluate Option 3. The subset construction algorithm guarantees that any NFA with $n$ states can be converted into an equivalent DFA with at most $2^{n}$ states. This statement is TRUE.

Step 5: Evaluate Option 4. This claims EVERY DFA for $L$ has $> 2^{n}$ states. This directly contradicts Option 3, which guarantees the existence of at least one DFA with $\leq 2^{n}$ states. Therefore, this statement is definitively FALSE.

Answer: Every DFA that accepts $L$ has $> 2^{n}$ states.

Question 12 · Theory of Computation · 2025_Set1 MCQ

Consider the following two languages over the alphabet

{a, b}

L_{1} = {α β α ∣ α \in {a, b}^{+} AND β \in {a, b}^{+}}

L_{2} = {α β α ∣ α \in {a}^{+} AND β \in {a, b}^{+}}

Which ONE of the following statements is CORRECT?

A.
Both $L_{1}$ and $L_{2}$ are regular languages.
B.
$L_{1}$ is a regular language but $L_{2}$ is not a regular language.
C.
$L_{1}$ is not a regular language but $L_{2}$ is a regular language.
D.
Neither $L_{1}$ nor $L_{2}$ is a regular language.

Correct Answer:

Step-by-Step Solution

Key idea: We must determine the regularity of two languages defined by string patterns. The key is to simplify the pattern definitions by analyzing the constraints on $α$ and $β$ .

Step 1: Analyze $L_{1} = {α β α ∣ α \in {a, b}^{+}, β \in {a, b}^{+}}$ .

$α$ is any non-empty string over ${a, b}$ . The shortest possible $α$ is a single character ('a' or 'b').

$β$ is any non-empty string over ${a, b}$ .

Therefore, any string in $L_{1}$ must have a length of at least $1 + 1 + 1 = 3$ , and it must start and end with the same single character.

Can any string of length $\geq 3$ starting and ending with the same character be formed? Yes. Let the first character be $α$ , the middle part be $β$ , and the last character be $α$ .

Thus, $L_{1}$ is exactly the set of strings of length $\geq 3$ that start and end with 'a', OR start and end with 'b'.

Regular Expression for $L_{1}$ : $a (a + b)^{+} a + b (a + b)^{+} b$ .

Since it can be described by a regular expression, $L_{1}$ is a Regular Language.

Step 2: Analyze $L_{2} = {α β α ∣ α \in {a}^{+}, β \in {a, b}^{+}}$ .

Here, $α$ is restricted to one or more 'a's. So $α = a^{k}$ for some $k \geq 1$ .

The string format is $a^{k} β a^{k}$ .

This requires the number of 'a's at the beginning to exactly match the number of 'a's at the end, with an arbitrary string $β$ in the middle.

This is a classic counting dependency (similar to $a^{n} b a^{n}$ ), which requires unbounded memory. A finite automaton cannot track the arbitrary count $k$ .

Therefore, $L_{2}$ is NOT a Regular Language.

Conclusion: $L_{1}$ is regular, but $L_{2}$ is not regular.

Answer: Option B.

Question 13 · Operating System · 2025_Set1 MCQ

Consider a demand paging memory management system with 32-bit logical address, 20-bit physical address, and page size of 2048 bytes. Assuming that the memory is byte addressable, what is the maximum number of entries in the page table?

A.
$2^{21}$
B.
$2^{20}$
C.
$2^{22}$
D.
$2^{24}$

Question 14 · Operating System · 2025_Set1 NAT

Suppose in a multiprogramming environment, the following C program segment is executed. A process goes into I/O queue whenever an I/O related operation is performed. Assume that there will always be a context switch whenever a process requests for an I/O, and also whenever the process returns from an I/O. The number of times the process will enter the ready queue during its lifetime (not counting the time the process enters the ready queue when it is run initially) is _______. (Answer in integer)

int main()
{
int x=0,i=0;
scanf("%d",&x);
for(i=0; i<20; i++)
{
x = x+20;
printf("%d\n",x);
}
return 0;
}

Question 15 · Operating System · 2025_Set1 MCQ

A computer has two processors,

M_{1}

and

M_{2}

. Four processes

P_{1}, P_{2}, P_{3}, P_{4}

with CPU bursts of 20, 16, 25, and 10 milliseconds, respectively, arrive at the same time and these are the only processes in the system. The scheduler uses non-preemptive priority scheduling, with priorities decided as follows:

•

M_{1}

uses priority of execution for the processes as,

P_{1} > P_{3} > P_{2} > P_{4}

, i.e.,

P_{1}

and

P_{4}

have highest and lowest priorities, respectively.
•

M_{2}

uses priority of execution for the processes as,

P_{2} > P_{3} > P_{4} > P_{1}

, i.e.,

P_{2}

and

P_{1}

have highest and lowest priorities, respectively.

A process

P_{i}

is scheduled to a processor

M_{k}

, if the processor is free and no other process

P_{j}

is waiting with higher priority. At any given point of time, a process can be allocated to any one of the free processors without violating the execution priority rules. Ignore the context switch time. What will be the average waiting time of the processes in milliseconds?

A.
9
B.
8.75
C.
6.5
D.
7.5

Question 16 · Databases · 2025_Set1 MCQ

A schedule of three database transactions

T_{1}

T_{2}

, and

T_{3}

is shown.

R_{i} (A)

and

W_{i} (A)

denote read and write of data item

A

by transaction

T_{i}

i = 1, 2, 3

. The transaction

T_{1}

aborts at the end. Which other transaction(s) will be required to be rolled back?

R_{1} (X) W_{1} (Y) R_{2} (X) R_{2} (Y) R_{3} (Y) A B O R T (T_{1})

A.
Only $T_{2}$
B.
Only $T_{3}$
C.
Both $T_{2}$ and $T_{3}$
D.
Neither $T_{2}$ nor $T_{3}$

Question 17 · Databases · 2025_Set1 MSQ

Consider the following

B^{+}

tree with 5 nodes, in which a node can store at most 3 key values. The value 23 is now inserted in the

B^{+}

tree. Which of the following options(s) is/are CORRECT?

A.
None of the nodes will split.
B.
At least one node will split and redistribute.
C.
The total number of nodes will remain same.
D.
The height of the tree will increase.

Question 18 · Databases · 2025_Set1 MCQ

Consider two relations describing

t e am s

and

pl a y er s

in a sports league:

•

t e am s (t i d, t nam e)

t i d, t nam e

are team-id and team-name, respectively
•

pl a y er s (p i d, p nam e, t i d)

p i d, p nam e

, and

t i d

denote player-id, player-name and the team-id of the player, respectively

Which ONE of the following tuple relational calculus queries returns the name of the players who play for the team having

t nam e

as ‘

M I

’?

A.
${p . p nam e ∣ p \in pl a y er s \land \exists t (t \in t e am s \land p . t i d = t . t i d \land t . t nam e = ‘MI’)}$
B.
${p . p nam e ∣ p \in t e am s \land \exists t (t \in pl a y er s \land p . t i d = t . t i d \land t . t nam e = ‘MI’)}$
C.
${p . p nam e ∣ p \in pl a y er s \land \exists t (t \in t e am s \land t . t nam e = ‘MI’)}$
D.
${p . p nam e ∣ p \in t e am s \land \exists t (t \in pl a y er s \land t . t nam e = ‘MI’)}$

Question 19 · Computer Networks · 2025_Set1 MCQ

Identify the ONE CORRECT matching between the OSI layers and their corresponding functionalities as shown.

OSI Layers                      Functionalities
(a) Network layer            (I) Packet routing
(b) Transport layer          (II) Framing and error handling
(c) Datalink layer            (III) Host to host communication

A.
(a)-(I), (b)-(II), (c)-(III)
B.
(a)-(I), (b)-(III), (c)-(II)
C.
(a)-(II), (b)-(I), (c)-(III)
D.
(a)-(III), (b)-(II), (c)-(I)

Question 20 · Computer Networks · 2025_Set1 MSQ

Consider the 3-way handshaking protocol for TCP connection establishment. Let the three packets exchanged during the connection establishment be denoted as P1, P2, and P3, in order. Which of the following option(s) is/are TRUE with respect to TCP header flags that are set in the packets?

A.
P3: SYN = 1, ACK = 1
B.
P2: SYN = 1, ACK = 1
C.
P2: SYN = 0, ACK = 1
D.
P1: SYN = 1

Question 21 · Computer Networks · 2025_Set1 MCQ

A packet with the destination IP address 145.36.109.70 arrives at a router whose routing table is shown. Which interface will the packet be forwarded to?

A.
E3
B.
E1
C.
E2
D.
E5

Question 22 · Algorithms · 2025_Set1 MCQ

Let $G$ be any undirected graph with positive edge weights, and $T$ be a minimum spanning tree of $G$ . For any two vertices, $u$ and $v$ , let $d_{1} (u, v)$ and $d_{2} (u, v)$ be the shortest distances between $u$ and $v$ in $G$ and $T$ , respectively. Which ONE of the options is CORRECT for all possible $G$ , $T$ , $u$ and $v$ ?

A.
$d_{1} (u, v) = d_{2} (u, v)$
B.
$d_{1} (u, v) \leq d_{2} (u, v)$
C.
$d_{1} (u, v) \geq d_{2} (u, v)$
D.
$d_{1} (u, v) \neq = d_{2} (u, v)$

Question 23 · Algorithms · 2025_Set1 MCQ

Consider the following recurrence relation:

T (n) = 2 T (n - 1) + n 2^{n} for n > 0, T (0) = 1.

Which ONE of the following options is CORRECT?

A.
$T (n) = Θ (n^{2} 2^{n})$
B.
$T (n) = Θ (n 2^{n})$
C.
$T (n) = Θ ((lo g n)^{2} 2^{n})$
D.
$T (n) = Θ (4^{n})$

Question 24 · Algorithms · 2025_Set1 NAT

The pseudocode of a function fun() is given below:

fun(int A[0,…,n-1]){
for i=0 to n-2
for j=0 to n-i-2
if (A[j]>A[j+1])
then swap A[j] and A[j+1]

}

Let

A [0, \dots, 29]

be an array storing 30 distinct integers in descending order. The number of swap operations that will be performed, if the function fun() is called with

A [0, \dots, 29]

as argument, is __________. (Answer in integer)

Correct Answer:

435

Step-by-Step Solution

Key idea: This is a sorting algorithm operation counting question, recognisable because it provides pseudocode and asks for the number of swap operations. The added layer is identifying the algorithm and its behavior on a specific input (reverse sorted).

Step 1: Analyze the pseudocode. The outer loop runs for $i$ from $0$ to $n - 2$ . The inner loop runs for $j$ from $0$ to $n - i - 2$ . Inside, it swaps $A [j]$ and $A [j + 1]$ if $A [j] > A [j + 1]$ .

Step 2: Recognize this as the standard Bubble Sort algorithm.

Step 3: The input array $A$ has $n = 30$ distinct integers in descending order. This is the worst-case input for Bubble Sort.

Step 4: In the worst case, every comparison $A [j] > A [j + 1]$ evaluates to true, so a swap is performed every time.

Step 5: Calculate the total number of comparisons (and thus swaps).

For $i = 0$ , $j$ goes from $0$ to $n - 2$ (which is $n - 1$ iterations).

For $i = 1$ , $j$ goes from $0$ to $n - 3$ (which is $n - 2$ iterations).

...

For $i = n - 2$ , $j$ goes from $0$ to $0$ (which is $1$ iteration).

Step 6: Total swaps = $(n - 1) + (n - 2) + \dots + 1 = \frac{n ( n - 1 )}{2}$ .

Step 7: Substitute $n = 30$ : Total swaps = $\frac{30 \times 29}{2} = 15 \times 29 = 435$ .

Answer: 435

Question 25 · Digital Logic · 2025_Set1 MSQ

Let $X$ be a 3-variable Boolean function that produces output as ‘1’ when at least two of the input variables are ‘1’. Which of the following statement(s) is/are CORRECT, where $a, b, c, d, e$ are Boolean variables?

A.
$X (a, b, X (c, d, e)) = X (X (a, b, c), d, e)$
B.
$X (a, b, X (a, b, c)) = X (a, b, c)$
C.
$X (a, b, X (a, c, d)) = (X (a, b, a) AND X (c, d, c))$
D.
$X (a, b, c) = X (a, X (a, b, c), X (a, c, c))$

Correct Answer:

["B","D"]

Step-by-Step Solution

Insight: $X$ is the 3-variable majority function, which is symmetric and idempotent-friendly under self-composition.

Exam route: Use the identities $X (a, b, a) = a$ and $X (a, b, X (a, b, c)) = X (a, b, c)$ to test the options algebraically or with a strategic counterexample.

Learning route:

The function $X (u, v, w)$ outputs 1 if at least two inputs are 1. This is the majority function: $X (u, v, w) = uv + v w + w u$ .

Note the absorption-like property: $X (u, v, u) = uv + v u + u^{2} = u$ .

Option A: Test with $a = 1, b = 1, c = 0, d = 0, e = 0$ .

LHS: $X (0, 0, 0) = 0 ⟹ X (1, 1, 0) = 1$ .

RHS: $X (1, 1, 0) = 1 ⟹ X (1, 0, 0) = 0$ .

$1 \neq = 0$ , so A is FALSE.

Option B: Let $Y = X (a, b, c)$ . We want to show $X (a, b, Y) = Y$ .

$X (a, b, Y) = ab + bY + Y a = ab + Y (a + b)$ .

Substitute $Y = ab + b c + c a$ :

$Y (a + b) = (ab + b c + c a) (a + b) = a^{2} b + a b^{2} + ab c + a^{2} c + b^{2} c + ab c = ab + ab + ab c + a c + b c + ab c = ab + a c + b c = Y$ .

Thus $X (a, b, Y) = ab + Y = ab + ab + b c + c a = Y$ . TRUE.

Option C: $X (a, b, a) = a$ and $X (c, d, c) = c$ . RHS $= a c$ .

Test $a = 1, b = 1, c = 0, d = 1$ . $X (a, c, d) = X (1, 0, 1) = 1$ .

LHS $= X (1, 1, 1) = 1$ . RHS $= 1 \cdot 0 = 0$ . FALSE.

Option D: $X (a, c, c) = c$ . RHS $= X (a, X (a, b, c), c)$ .

Since $X$ is symmetric, $X (a, Y, c) = X (a, c, Y)$ .

Using the property from B with variables $a, c, b$ : $X (a, c, X (a, c, b)) = X (a, c, b) = X (a, b, c) = Y$ .

Thus RHS $= Y = X (a, b, c)$ . TRUE.

Correct options: B, D.

Question 26 · Digital Logic · 2025_Set1 MCQ

Consider the following four variable Boolean function in sum-of-product form

F (b_{3}, b_{2}, b_{1}, b_{0}) = Σ (0, 2, 4, 8, 10, 11, 12) .

where the value of the function is computed by considering

b_{3} b_{2} b_{1} b_{0}

as a 4-bit binary number, where

b_{3}

denotes the most significant bit and

b_{0}

denotes the least significant bit. Note that there are no don’t care terms. Which ONE of the following options is the CORRECT minimized Boolean expression for

F

A.
$\overline{b_{1}} \overline{b_{0}} + \overline{b_{2}} \overline{b_{0}} + b_{1} \overline{b_{2}} b_{3}$
B.
$\overline{b_{1}} \overline{b_{0}} + \overline{b_{2}} \overline{b_{0}}$
C.
$\overline{b_{2}} \overline{b_{0}} + b_{1} b_{2} b_{3}$
D.
$\overline{b_{0}} \overline{b_{2}} + \overline{b_{3}}$

Correct Answer:

Step-by-Step Solution

Insight: Plot the 7 minterms on a 4-variable K-map; the largest power-of-two rectangles reveal the essential prime implicants.

Exam route: Identify the three essential PIs: $\overset{ˉ}{b_{1}} \overset{ˉ}{b_{0}}$ , $\overset{ˉ}{b_{2}} \overset{ˉ}{b_{0}}$ , and $b_{3} \overset{ˉ}{b_{2}} b_{1}$ .

Learning route:

Minterms in binary ( $b_{3} b_{2} b_{1} b_{0}$ ): 0000, 0010, 0100, 1000, 1010, 1011, 1100.

K-map groups:

Quad ${0, 4, 8, 12}$ : $b_{1} = 0, b_{0} = 0 ⟹ \overset{ˉ}{b_{1}} \overset{ˉ}{b_{0}}$ .
Quad ${0, 2, 8, 10}$ : $b_{2} = 0, b_{0} = 0 ⟹ \overset{ˉ}{b_{2}} \overset{ˉ}{b_{0}}$ .
Pair ${10, 11}$ : $b_{3} = 1, b_{2} = 0, b_{1} = 1 ⟹ b_{3} \overset{ˉ}{b_{2}} b_{1}$ .

Check essentiality:

$m_{4}$ is only covered by $\overset{ˉ}{b_{1}} \overset{ˉ}{b_{0}}$ .
$m_{2}$ is only covered by $\overset{ˉ}{b_{2}} \overset{ˉ}{b_{0}}$ .
$m_{11}$ is only covered by $b_{3} \overset{ˉ}{b_{2}} b_{1}$ .

All three PIs are essential and cover all minterms.

Minimal SOP: $\overset{ˉ}{b_{1}} \overset{ˉ}{b_{0}} + \overset{ˉ}{b_{2}} \overset{ˉ}{b_{0}} + b_{3} \overset{ˉ}{b_{2}} b_{1}$ .

Matches Option A.

Question 27 · Digital Logic · 2025_Set1 NAT

Consider a finite state machine (FSM) with one input

X

and one output

f

, represented by the given state transition table. The minimum number of states required to realize this FSM is ________. (Answer in integer)

Correct Answer:

5.00

Step-by-Step Solution

Insight: To find the minimum number of states, we look for equivalent states using row matching and partition refinement. States with identical next-states and outputs can be merged.

Exam route: Scan the table for identical rows. Merge them. Repeat until no more identical rows exist. Count the remaining unique states.

Learning route:

Step 1: Compare rows for exact matches in Next State and Output columns.

Row B: Next states (D, C), Outputs (0, 0).
Row E: Next states (D, C), Outputs (0, 0).

Rows B and E are identical. Merge E into B.

Row D: Next states (G, A), Outputs (1, 0).
Row H: Next states (G, A), Outputs (1, 0).

Rows D and H are identical. Merge H into D.

Step 2: Update the table with the merged states (replace E with B, and H with D).

Row A: F, B | 0, 0
Row B: D, C | 0, 0
Row C: F, E $\to$ F, B | 0, 0
Row D: G, A | 1, 0
Row F: F, B | 1, 1
Row G: G, H $\to$ G, D | 0, 1

Step 3: Scan the updated table for new identical rows.

Row A: F, B | 0, 0
Row C: F, B | 0, 0

Rows A and C are now identical. Merge C into A.

Step 4: Update the table again (replace C with A).

The remaining unique states are A, B, D, F, G.

Let's verify no more merges are possible:

A: F, B | 0, 0
B: D, A | 0, 0 (since C became A)
D: G, A | 1, 0
F: F, B | 1, 1
G: G, D | 0, 1

All 5 remaining states have distinct output signatures or distinct next-state transitions.

Step 5: The minimum number of states required is 5.

Question 28 · Compiler Design · 2025_Set1 MCQ

Which ONE of the following statements is FALSE regarding the symbol table?

A.
Symbol table is responsible for keeping track of the scope of variables.
B.
Symbol table can be implemented using a binary search tree.
C.
Symbol table is not required after the parsing phase.
D.
Symbol table is created during the lexical analysis phase.

Correct Answer:

Step-by-Step Solution

Key idea: Understand the lifecycle and usage of the symbol table across all compiler phases.

Step 1: Analyze Option A. The symbol table stores information about identifiers, including their scope, type, and memory location. This is TRUE.

Step 2: Analyze Option B. The symbol table can be implemented using various data structures like hash tables, binary search trees, or linked lists, depending on the scope structure. This is TRUE.

Step 3: Analyze Option C. The symbol table is heavily used after the parsing phase. Semantic analysis uses it for type checking, intermediate code generation uses it for address calculation, and the back-end uses it for final memory allocation. Thus, saying it is "not required after the parsing phase" is FALSE.

Step 4: Analyze Option D. The lexical analyzer recognizes identifiers and inserts them into the symbol table. Thus, it is created/populated during the lexical analysis phase. This is TRUE.

Answer: C

Question 29 · Compiler Design · 2025_Set1 MCQ

Which ONE of the following techniques used in compiler code optimization uses live variable analysis?

A.
Run-time function call management
B.
Register assignment to variables
C.
Strength reduction
D.
Constant folding

Question 30 · Compiler Design · 2025_Set1 MSQ

Which of the following statement(s) is/are TRUE while computing First and Follow during top down parsing by a compiler?

A.
For a production $A \to ϵ$ , $ϵ$ will be added to $F i r s t (A)$ .
B.
If there is any input right end marker, it will be added to $F i r s t (S)$ , where $S$ is the start symbol.
C.
For a production $A \to ϵ$ , $ϵ$ will be added to $F o l l o w (A)$ .
D.
If there is any input right end marker, it will be added to $F o l l o w (S)$ , where $S$ is the start symbol.

Question 31 · Verbal Aptitude · 2025_Set1 MCQ

Ravi had ______ younger brother who taught at ______ university. He was widely regarded as ______ honorable man.

Select the option with the correct sequence of articles to fill in the blanks.

A.
a; a; an
B.
the; an; a
C.
a; an; a
D.
an; an; a

Correct Answer:

Step-by-Step Solution

Insight: Articles depend on the specificity of the noun and its phonetic sound, not just its spelling.

Exam route: The first mention of a singular countable noun ('younger brother') takes 'a'. 'University' starts with a consonant sound /j/, so it takes 'a'. 'Honorable' starts with a silent 'h' producing a vowel sound /ɒ/, so it takes 'an'. The sequence is a, a, an.

Learning route:

Step 1: Analyze the first blank. 'Younger brother' is introduced for the first time and is singular/countable, requiring the indefinite article 'a'.

Step 2: Analyze the second blank. 'University' begins with the letter 'U', but phonetically it starts with the consonant sound /j/ (like 'you'). Therefore, it takes 'a', not 'an'.

Step 3: Analyze the third blank. 'Honorable' begins with 'H', but the 'H' is silent. The word starts with the vowel sound /ɒ/. Thus, it requires 'an'.

Combining these gives 'a; a; an'.

Common trap: Students often look at the first letter rather than the sound, leading them to incorrectly choose 'an university' or 'a honorable'.

Verification: Substituting 'a; a; an' into the sentence reads perfectly and follows all phonetic and grammatical rules for articles.

Question 32 · Verbal Aptitude · 2025_Set1 MCQ

The CEO’s decision to downsize the workforce was considered myopic because it sacrificed long-term stability to accommodate short-term gains.

Select the most appropriate option that can replace the word “myopic” without changing the meaning of the sentence.

A.
visionary
B.
shortsighted
C.
progressive
D.
innovative

Correct Answer:

Step-by-Step Solution

Insight: The word "because" provides the exact definition of the target word "myopic" within the sentence itself.

Exam route: The sentence states the decision "sacrificed long-term stability to accommodate short-term gains." This is the definition of lacking foresight. "Shortsighted" is the direct synonym for this concept. "Visionary", "progressive", and "innovative" are positive traits, contradicting the negative context.

Learning route:

Step 1: Identify the logical clue. The word "because" introduces the explanation for why the decision was considered "myopic".

Step 2: Analyze the explanation. The explanation is "sacrificed long-term stability to accommodate short-term gains." This describes a focus on the immediate future while ignoring distant consequences.

Step 3: Predict the meaning. We need a word that means "lacking long-term vision" or "focused only on the immediate".

Step 4: Evaluate options. "Visionary" means thinking about the future with imagination (the opposite). "Shortsighted" means lacking imagination or foresight, matching our prediction perfectly. "Progressive" and "innovative" are positive descriptors, contradicting the negative consequence described.

Question 33 · Verbal Aptitude · 2025_Set1 MCQ

“I put the brown paper in my pocket along with the chalks, and possibly other things. I suppose every one must have reflected how primeval and how poetical are the things that one carries in one’s pocket: the pocket-knife, for instance the type of all human tools, the infant of the sword. Once I planned to write a book of poems entirely about the things in my pocket. But I found it would be too long: and the age of the great epics is past.”

(From G.K. Chesterton’s “A Piece of Chalk”)

Based only on the information provided in the above passage, which one of the following statements is true?

A.
The author of the passage carries a mirror in his pocket to reflect upon things.
B.
The author of the passage had decided to write a poem on epics.
C.
The pocket-knife is described as the infant of the sword.
D.
Epics are described as too inconvenient to write.

Correct Answer:

Step-by-Step Solution

Key idea: This is a detail-extraction and figurative language interpretation question. The correct option must be strictly supported by the text without adding external assumptions or literal misinterpretations.

Step 1: Evaluate Option A. The passage uses "reflected" in the mental sense ("must have reflected how primeval..."), meaning "thought about". It does not imply the physical act of using a mirror. This is a literal interpretation trap.

Step 2: Evaluate Option B. The author planned to write poems about "the things in my pocket", not about epics. The mention of epics is to explain why the pocket-themed book was abandoned (its length would rival great epics, and that age is past).

Step 3: Evaluate Option C. The passage explicitly states: "the pocket-knife, for instance the type of all human tools, the infant of the sword." This directly and accurately matches the option.

Step 4: Evaluate Option D. The author found the book of poems about pocket items would be "too long", not that epics themselves are inconvenient to write.

Answer: C

Question 34 · Spatial Aptitude · 2025_Set1 MCQ

According to the map shown in the figure, which one of the following statements is correct?

Note: The figure shown is representative.

A.
The library is located to the northwest of the canteen.
B.
The hospital is located to the east of the chemistry lab.
C.
The chemistry lab is to the southeast of physics lab.
D.
The classrooms and canteen are next to each other.

Correct Answer:

Step-by-Step Solution

Key idea: This is a map directions and relative positions question, recognisable because it asks to determine the relative location of buildings based on a provided map with a compass rose.

Why this method applies: We can assign approximate grid positions or relative directions (North, South, East, West) to each building based on the map layout and then evaluate each option systematically.

Step 1: Understand the compass and map layout. The compass rose in the bottom right corner indicates standard orientation: North (N) is Up, South (S) is Down, West (W) is Left, and East (E) is Right.

Step 2: Locate the key buildings on the map grid.

Library: Top-Left quadrant.
Physics Lab: Middle-Left (directly below Library).
Canteen: Top-Right quadrant.
Hospital: Middle-Right (directly below Canteen).
Hostels: Bottom-Left (below the horizontal road).
Classrooms: Bottom-most Left (below Hostels).
Chemistry Lab: Bottom-Right (below Hospital).

Step 3: Evaluate Option A. The Library and Canteen are in the same horizontal band (Top). The Library is on the left and Canteen is on the right. Thus, the Library is directly West of the Canteen, not Northwest. Option A is false.

Step 4: Evaluate Option B. The Hospital and Chemistry Lab are in the same vertical band (Right). The Hospital is above the Chemistry Lab. Thus, the Hospital is directly North of the Chemistry Lab, not East. Option B is false.

Step 5: Evaluate Option C. The Physics Lab is in the Middle-Left. The Chemistry Lab is in the Bottom-Right. To move from the Physics Lab to the Chemistry Lab, you must go Right (East) and Down (South). Therefore, the Chemistry Lab is to the Southeast of the Physics Lab. Option C is true.

Step 6: Evaluate Option D. The Classrooms are in the Bottom-Left and the Canteen is in the Top-Right. They are diagonally opposite across the map and separated by roads, so they are not next to each other. Option D is false.

Answer: C

Question 35 · Spatial Aptitude · 2025_Set1 MCQ

A square paper, shown in figure (I), is folded along the dotted lines as shown in the figures (II) and (III). Then a few cuts are made as shown in figure (IV). Which one of the following patterns will be obtained when the paper is unfolded?

Note: The figures shown are representative.

Correct Answer:

Step-by-Step Solution

Key idea: Reverse unfolding using the reflection principle. We start from the final cut state and unfold step-by-step, mirroring the cuts across the fold lines.

Step 1: Analyze the final folded state (Figure IV). It is a right-angled triangle. The cuts are:

A rectangular hole inside the triangle.
A triangular notch on the top edge (hypotenuse of this small triangle).
A rectangular notch on the left edge (a leg of this small triangle).

Step 2: Understand the fold mapping.

The first fold was along the diagonal of the original square.
The second fold was along the altitude to the hypotenuse, creating the small triangle in Figure IV.
The hypotenuse of the small triangle in Figure IV corresponds to the outer perimeter of the original square.
The legs of the small triangle correspond to the diagonal fold lines of the original square.

Step 3: Unfold and reflect.

The rectangular hole is offset from the center (the right-angle vertex). Since the paper has 4 layers here, unfolding will replicate this hole 4 times, symmetrically placed in the four quadrants. The hole's sides are parallel to the hypotenuse (original square's sides), so the unfolded holes will be upright squares/rectangles, not rotated diamonds.
The triangular notch on the hypotenuse (perimeter) will be replicated on all four sides of the original square. Since the cut removes material from the edge, the unfolded representation must show a notch pointing inward (missing material from the boundary).

Step 4: Evaluate the options.

Option A shows notches pointing outward (outside the boundary), which is an incorrect representation of a removed edge piece.
Option B shows diamond-shaped holes, which implies the cut was rotated 45 degrees relative to the sides. This is incorrect.
Option C correctly shows upright square holes and inward-pointing triangular notches on all four edges, matching the reflection of the perimeter cut.
Option D shows a single central hole, which would only occur if the cut was made exactly at the right-angle vertex (the center of the original square).

Answer: C

Question 36 · Quantitative Aptitude · 2025_Set1 MCQ

The average marks obtained by a class in an examination were calculated as 30.8. However, while checking the marks entered, the teacher found that the marks of one student were entered incorrectly as 24 instead of 42. After correcting the marks, the average becomes 31.4. How many students does the class have?

A.
25
B.
28
C.
30
D.
32

Correct Answer:

Step-by-Step Solution

Insight: The increase in the average is caused entirely by the correction in the misrecorded mark, distributed equally among all students.

Exam route: Error in sum = 42 - 24 = 18. Increase in average = 31.4 - 30.8 = 0.6. Number of students = 18 / 0.6 = 30.

Learning route:

Step 1: Identify the error in the total sum. The mark 24 was entered instead of 42. The sum was short by 42 - 24 = 18.

Step 2: Identify the increase in the average. The average went from 30.8 to 31.4, an increase of 31.4 - 30.8 = 0.6.

Step 3: Relate the sum error to the average error. The extra 18 marks were distributed equally among $N$ students, increasing the average by 0.6.

So, $18/ N = 0.6$ .

Step 4: Solve for $N$ . $N = 18/0.6 = 30$ .

Trap warning: Some students might subtract 24 from 42 and get 18, then divide by the initial average 30.8, which is incorrect. The 18 is the total increase in the sum, not a single student's mark.

Verification: Initial sum = 30 * 30.8 = 924. Correct sum = 924 - 24 + 42 = 942. Correct average = 942 / 30 = 31.4. Matches perfectly.

Question 37 · Quantitative Aptitude · 2025_Set1 MCQ

In the diagram, the lines QR and ST are parallel to each other. The shortest distance between these two lines is half the shortest distance between the point P and line QR. What is the ratio of the area of the triangle PST to the area of the trapezium SQRT?

Note: The figure shown is representative.

A.
$\frac{1}{3}$
B.
$\frac{1}{4}$
C.
$\frac{2}{5}$
D.
$\frac{1}{2}$

Correct Answer:

Step-by-Step Solution

Insight: Parallel lines inside a triangle create similar triangles. The ratio of their areas is the square of the ratio of their corresponding heights.

Exam route: Height ratio of $△ P S T$ to $△ P QR$ is $1 : 2$ . Area ratio is $1 : 4$ . Trapezium area is $4 - 1 = 3$ . Ratio is $1/3$ .

Learning route:

Let the shortest distance from $P$ to $QR$ be $H$ .
The problem states the distance between the parallel lines $S T$ and $QR$ is half of this, so it is $H /2$ .
Since $S T$ is parallel to $QR$ and lies between $P$ and $QR$ , the distance from $P$ to $S T$ is $H - H /2 = H /2$ .
$△ P S T$ and $△ P QR$ are similar by the AA test (they share $∠ P$ , and $∠ P S T = ∠ P QR$ due to parallel lines).
The ratio of their corresponding heights is $\frac{H /2}{H} = \frac{1}{2}$ .
The ratio of their areas is the square of the height ratio: $(\frac{1}{2})^{2} = \frac{1}{4}$ .
Let $Area (△ P S T) = 1 x$ and $Area (△ P QR) = 4 x$ .
The trapezium $S QR T$ is the region $△ P QR$ minus $△ P S T$ , so its area is $4 x - 1 x = 3 x$ .
The required ratio is $\frac{Area ( △ P S T )}{Area ( trapezium S QR T )} = \frac{1 x}{3 x} = \frac{1}{3}$ .

Wrong path: Forgetting to square the height ratio, leading to an area ratio of $1 : 2$ . This would make the trapezium area $2 x - 1 x = 1 x$ , giving a ratio of $1/1$ , or mistakenly using the large triangle's area in the denominator to get $1/4$ (Option B).

Question 38 · Analytical Aptitude · 2025_Set1 MCQ

Consider the relationships among P, Q, R, S, and T:

• P is the brother of Q.
• S is the daughter of Q.
• T is the sister of S.
• R is the mother of Q.

The following statements are made based on the relationships given above.

(1) R is the grandmother of S.

(2) P is the uncle of S and T.

(3) R has only one son.

(4) Q has only one daughter.

Which one of the following options is correct?

A.
Both (1) and (2) are true.
B.
Both (1) and (3) are true.
C.
Only (3) is true.
D.
Only (4) is true.

Correct Answer:

Step-by-Step Solution

Insight: Build a generational family tree using standard symbols (squares for males, circles for females, horizontal lines for spouses/siblings, vertical for parent-child) before evaluating any statement.

Exam route:

Step 1: "R is the mother of Q" $\to$ R (Gen 1, Female) $\to$ Q (Gen 2).

Step 2: "P is the brother of Q" $\to$ P (Gen 2, Male) is sibling of Q.

Step 3: "S is the daughter of Q" and "T is the sister of S" $\to$ S, T (Gen 3, Female) are children of Q.

Step 4: Evaluate statements. (1) R is grandmother of S: True (R is mother of Q, Q is parent of S). (2) P is uncle of S and T: True (P is brother of Q, Q is parent of S, T). (3) R has only one son: Unknown (Q's gender is not specified, and R could have other children). (4) Q has only one daughter: False (Q has at least two daughters, S and T).

Step 5: Select "Both (1) and (2) are true."

Learning route:

Step 1: Draw Gen 1: R (Female, circle).

Step 2: Draw Gen 2: P (Male, square) and Q (gender unknown) connected horizontally (siblings), with a vertical line from R.

Step 3: Draw Gen 3: S and T (Females, circles) connected horizontally (sisters), with a vertical line from Q.

Step 4: Trace paths for evaluation. Path R $\to$ Q $\to$ S confirms R is grandmother. Path P $\to$ Q $\to$ S,T confirms P is uncle.

Step 5: Identify "Unknown" traps. "Only one son" requires explicit confirmation of all children and their genders, which is absent.

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