ISRO Computer Science (CSE) 2023 Question Paper & Solutions
Attempt the ISRO Computer Science (CSE) 2023 previous year paper in real CBT exam mode or practice mode with verified answers and explanations.
All ISRO Computer Science (CSE) papers
95 questions.
Question 1
Topic: Algorithms
Find the minimum spanning distance and the corresponding number of edges for the following graph
- A. 10,3
- B. 11,4
- C. 15,4
- D. 28,7
Answer: B. 11,4
Explanation: Using Kruskal's algorithm on the given graph with edges AB=1, DE=2, BE=3, BD=4, BC=5, CE=6, AC=7. We select edges in increasing order of weight avoiding cycles: Select AB (1), DE (2), BE (3). The next edge is BD (4), but it forms a cycle (B-E-D-B), so we skip it. We select BC (5). All 5 nodes (A, B, C, D, E) are now connected. Total edges = 4. Minimum spanning distance = 1 + 2 + 3 + 5 = 11.
Question 2
Topic: Algorithms
Match the following:
(A) Floyd Warshall (i) shortest path between two vertices
(B) Dijkstra (ii) single source shortest path
(C) Kruskal's (iii) Minimum spanning tree
(D) Bellman ford (iv) solving all pair shortest path
- A. (A)-(iii), (B)-(ii), (C)-(iv), (D)-(i)
- B. (A)-(iv), (B)-(i), (C)-(iii), (D)-(ii)
- C. (A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)
- D. (A)-(ii), (B)-(iii), (C)-(i), (D)-(iv)
Answer: B. (A)-(iv), (B)-(i), (C)-(iii), (D)-(ii)
Explanation: Floyd Warshall is an algorithm for finding all pair shortest paths (iv). Dijkstra is commonly used to find the shortest path from one node to all others or just between two vertices (i). Kruskal's is for Minimum spanning tree (iii). Bellman-Ford is for single source shortest paths, particularly with negative weights (ii).
Question 3
Topic: Data Structures
Time complexity to find the diameter of a binary tree having n nodes is
- A. $O(n^{2})$
- B. $O(n)$
- C. $O(1)$
- D. $O(\log n)$
Answer: B. $O(n)$
Explanation: The diameter of a binary tree can be found in $O(n)$ time by doing a post-order traversal where each node calculates its height and updates the maximum diameter found so far.
Question 4
Topic: Data Structures
The preorder traversal of a binary search tree is (10, 5, 1, 7, 40, 50). The postorder traversal is
- A. (1, 5, 7, 10, 40, 50)
- B. (50, 40, 10, 7, 5, 1)
- C. (1, 7, 5, 10, 40, 50)
- D. (1, 5, 7, 50, 40, 10)
Answer: D. (1, 5, 7, 50, 40, 10)
Explanation: From the preorder (Root, Left, Right) of a BST: Root is 10. Elements less than 10 form the left subtree (5, 1, 7) and elements greater form the right subtree (40, 50). Postorder follows Left, Right, Root. Processing recursively, the postorder is (1, 7, 5) for the left subtree and (50, 40) for the right subtree. The overall postorder is 1, 7, 5, 50, 40, 10.
Question 5
Topic: Algorithms
The minimum number of arithmetic operations required to evaluate the expression $f(z)=8z^{3}+3z+12$ is
Answer: B. 5
Explanation: Using Horner's method, the expression can be rewritten to minimize operations: $f(z) = z \times (z \times (8z) + 3) + 12$. The operations are: 1) $8 \times z$, 2) $\times z$, 3) $+ 3$, 4) $\times z$, 5) $+ 12$. This results in a total of 3 multiplications and 2 additions, equaling 5 arithmetic operations.
Question 6
Topic: Graph Theory
Maximum number of Simple graphs possible with n vertices
- A. $2^{n(n-1)/2}$
- B. $2^{(n-1)/2}$
- C. $2^{(n(n+1)/2)}$
- D. $2^{n(n+1)}$
Answer: A. $2^{n(n-1)/2}$
Explanation: In a simple graph with n vertices, the maximum possible number of edges is $n(n-1)/2$. Since each edge can either be present or absent, the total number of simple graphs possible is $2^{n(n-1)/2}$.
Question 7
Topic: Theory of Computation
Consider the NFA with epsilon-transitions below, which is meant to accept strings corresponding to a lexical token id. Here the token digit corresponds to the extended regular expression [0-9] and letter corresponds to [a-zA-Z]. In the diagram, a small incoming edge indicates a start state while double circle indicates a final state.
Which of the following extended regular expressions corresponds to the token id?
- A. letter (letter | $digit)^{*}$
- B. (letter digit)* letter
- C. (letter | $digit)^{*}$
- D. letter (letter* | digit*)
Answer: A. letter (letter | $digit)^{*}$
Explanation: The NFA requires the first character to be a 'letter' to reach the accept state. From the accept state, there are epsilon transitions allowing looping through either a 'letter' or a 'digit' any number of times. Thus, the regular expression is `letter (letter | digit)*`, representing typical identifiers in most programming languages.
Question 8
Topic: Compiler Design
Consider the context-free grammar G below for arithmetic expressions:
$E \rightarrow E+E|E \times E|id$
Which of the following statements is TRUE:
- A. The string $id+id \times id$ has no parse tree according to G
- B. The string "id + id \times id" has only one parse tree according to G
- C. The string "id + id \times id" has exactly two parse trees according to G
- D. The string "id + id \times id" has more than two parse trees according to G
Answer: C. The string "id + id \times id" has exactly two parse trees according to G
Explanation: The grammar is ambiguous as it does not enforce precedence. The string `id + id * id` can be parsed by first applying $E \rightarrow E+E$ or $E \rightarrow E \times E$. This leads to exactly two distinct parse trees (one grouping addition first, and the other grouping multiplication first).
Question 9
Topic: Operating Systems
Which of the following heap memory allocation strategies is likely to exploit spatial locality of memory accesses in a program, the most?
- A. Best Fit
- B. First Fit
- C. Next Fit
- D. Last Fit
Answer: C. Next Fit
Explanation: Next Fit allocation begins searching from the point of the last allocation. This clusters contiguous memory allocations together in memory over time, maximizing spatial locality.
Question 10
Topic: Compiler Design
Which sequence below best describes the order in which the phases of a compiler typically work?
- A. Syntax analysis, lexical analysis, semantic analysis, code optimization, intermediate code generation
- B. Lexical analysis, syntax analysis, semantic analysis, code optimization, intermediate code generation
- C. Syntax analysis, lexical analysis, syntax analysis, semantic analysis, intermediate code generation, code optimization
- D. Lexical analysis, syntax analysis, semantic analysis, intermediate code generation, code optimization
Answer: D. Lexical analysis, syntax analysis, semantic analysis, intermediate code generation, code optimization
Explanation: The standard phased architecture of a compiler executes in the following order: Lexical analysis -> Syntax analysis -> Semantic analysis -> Intermediate code generation -> Code optimization -> Target code generation.
Question 11
Topic: Compiler Design
In the context of parsing techniques, LL(1) refers to:
- A. Grammars that parse the input from left to right, produce a rightmost derivation, with a lookahead of 1 symbol.
- B. Grammars that parse the input from left to right, produce a leftmost derivation, with a lookahead of 1 symbol.
- C. Grammars that parse the input from left to right, and backtrack at most once.
- D. Grammars that parse the input from right to left, produce a leftmost derivation, and backtrack at most once.
Answer: B. Grammars that parse the input from left to right, produce a leftmost derivation, with a lookahead of 1 symbol.
Explanation: In LL(1), the first 'L' stands for scanning the input from Left to right, the second 'L' stands for producing a Leftmost derivation, and the '1' signifies a lookahead of 1 token.
Question 12
Topic: Computer Networks
A host in one organisation is connected to another host in other organisation using 1 Gbps link with 25m sec RTT. Network administrator of this network tuned the packet size to 1000 bytes to transfer large files optimally over this network. In order to have channel utilisation > 96%, how big window size (in terms of packets) should be (assuming ACK packets have no impact)?
- A. 2000 packets
- B. 2400 packets
- C. 3000 packets
- D. 4000 packets
Answer: C. 3000 packets
Explanation: Bandwidth (B) = 1 Gbps = $10^9$ bps. Packet size (L) = 1000 bytes = 8000 bits. Transmission time ($T_t$) = L / B = $8000 / 10^9 = 8 \mu s$. RTT = $25 ms$. Channel utilization $U = \frac{W \times T_t}{RTT + T_t}$. We require $U > 0.96$. $0.96 = \frac{W \times 8 \times 10^{-6}}{25.008 \times 10^{-3}} \implies W \approx 3000$. Thus, a window size slightly over 3000 achieves > 96% utilization, making 3000 the boundary threshold representing optimal efficiency.
Question 13
Topic: Computer Networks
In a computer network using CSMA/CD, the minimum frame size for data transfer is 1500 bytes. What will be the data transfer speed in this network over 1 Km cable, if the signal speed in the cable is 250000 Km/sec and no repeaters are used in this network?
- A. 1 Gbps
- B. 100 Mbps
- C. 500 Mbps
- D. 1.5 Gbps
Answer: D. 1.5 Gbps
Explanation: For CSMA/CD to detect collisions, transmission time ($T_t$) must be $\ge 2 \times$ propagation delay ($T_p$). $T_p = \frac{Distance}{Speed} = \frac{1 km}{250000 km/s} = 4 \mu s$. $T_t = \frac{L}{B} \ge 2 \times T_p$. Length L = $1500 \times 8 = 12000$ bits. $\frac{12000}{B} \ge 8 \mu s \implies B \le \frac{12000}{8 \times 10^{-6}} = 1.5 \times 10^9$ bps = 1.5 Gbps.
Question 14
Topic: Computer Networks
For internal data communication through computer network, a 5 layer protocol hierarchy is proposed. Applications to be used in this network generate 270 bytes long messages. At each of the layers, a 6-byte header is added. What fraction of network bandwidth(in %) is wasted due to headers?
Answer: C. 10%
Explanation: Payload size = 270 bytes. There are 5 layers, each adding a 6-byte header, so total header overhead = $5 \times 6 = 30$ bytes. The total message size = $270 + 30 = 300$ bytes. The fraction of bandwidth wasted by headers = $(30 / 300) \times 100\% = 10\%$.
Question 15
Topic: Computer Networks
Match the following:
Versions of 802.11
(i) 802.11ac
(ii) 802.11 a/g
(iii) 802.11 ad
(iv) 802.11 b
Speed of Wireless LAN
(P) 7 Gbps
(Q) 11 Mbps
(R) 54 Mbps
(S) 3.5 Gbps
- A. (i) - (S), (ii) - (R), (iii) - (P), (iv) - (Q)
- B. (i) - (S), (ii) - (R), (iii) - (Q), (iv) - (P)
- C. (i) - (R), (ii) - (S), (iii) - (P), (iv) - (Q)
- D. (i) - (R), (ii) - (Q), (iii) - (P), (iv) - (S)
Answer: A. (i) - (S), (ii) - (R), (iii) - (P), (iv) - (Q)
Explanation: The maximum data rates for IEEE 802.11 standards are: 802.11b operates at 11 Mbps (iv-Q). 802.11a/g operate at 54 Mbps (ii-R). 802.11ac (Wi-Fi 5) supports up to ~3.5 Gbps (i-S). 802.11ad (WiGig) supports up to 7 Gbps (iii-P).
Question 16
Topic: Computer Networks
Which of the following is true?
- A. Both Parity and Cyclic Redundancy Check are error correcting codes
- B. Both Parity and Low Density Parity Check Code are error correcting codes
- C. Both Reed-Solomon code and Low Density Parity Check Code are error correcting codes
- D. Both Cyclic Redundancy Check and Low Density Parity Check Code are error correcting codes
Answer: C. Both Reed-Solomon code and Low Density Parity Check Code are error correcting codes
Explanation: Simple Parity and basic Cyclic Redundancy Check (CRC) are primarily error-detecting codes. Reed-Solomon (RS) codes and Low Density Parity Check (LDPC) codes are explicitly designed as Forward Error Correction (FEC) codes capable of correcting errors.
Question 17
Topic: Data Structures
In Reverse Polish notation, expression $A^{*}B+C^{*}D$ is written as
- A. $AB^{*}CD^{*}+$
- B. $A^{*}BCD^{*}+$
- C. $AB^{*}CD+^{*}$
- D. $A^{*}B^{*}CD+$
Answer: A. $AB^{*}CD^{*}+$
Explanation: Reverse Polish Notation (RPN) is postfix notation. The expression $A \times B + C \times D$ consists of two multiplications connected by an addition. The postfix for $A \times B$ is $AB^{*}$. The postfix for $C \times D$ is $CD^{*}$. Combining them with the addition operator yields $AB^{*}CD^{*}+$.
Question 18
Topic: Computer Organization
In a vectored interrupt
- A. The branch address is assigned to a fixed location in memory
- B. The interrupting source supplies the branch information to the processor
- C. The branch address is obtained from a register in the processor
- D. None of the above
Answer: B. The interrupting source supplies the branch information to the processor
Explanation: In a vectored interrupt system, the interrupting device provides a vector (which is a branch address or an index to an interrupt vector table) to the processor to indicate the location of the Interrupt Service Routine (ISR).
Question 19
Topic: Computer Organization
For a given finite number of instructions to be executed, which architecture of the processor provides fastest execution?
- A. ANSA Advanced Network Systems Architecture
- B. Super-scalar
- C. ISA Instruction Set Architecture
- D. All of the above
Answer: B. Super-scalar
Explanation: Super-scalar architecture allows multiple instructions to be executed simultaneously during a single clock cycle by dispatching them to redundant execution units on the processor, thus providing the fastest execution among the given choices.
Question 20
Topic: Computer Organization
An instruction pipeline can be implemented by means of
- A. LIFO buffer
- B. FIFO buffer
- C. Stack
- D. None of the above
Answer: B. FIFO buffer
Explanation: An instruction pipeline executes instructions sequentially in the order they were fetched, meaning the first instruction entering the pipeline is the first one to exit. This First-In-First-Out behavior is implemented using FIFO buffers.
Question 21
Topic: Computer Organization
The operation executed on data stored in registers is called
- A. Macro-operation
- B. Micro-operation
- C. Bit-operation
- D. Byte-operation
Answer: B. Micro-operation
Explanation: Operations executed on data stored in one or more registers during a single clock pulse (like shift, load, clear, add) are referred to as micro-operations.
Question 22
Topic: Computer Networks
Which of the following methods is used to detect double errors?
- A. Odd Parity
- B. Even Parity
- C. Checksum (CRC)
- D. Checksum (XOR)
Answer: C. Checksum (CRC)
Explanation: Simple parity (odd or even) can only detect an odd number of bit errors and fails to detect double bit errors. Cyclic Redundancy Check (CRC) is a robust method capable of detecting double errors as well as burst errors.
Question 23
Topic: Data Structures
The prefix expression is -+abc+d*ef. Evaluate the expression for a=10, b=2, c=5, d=3, e=4, f=1
Answer: C. 10
Explanation: This prefix evaluation contains a slight typo typical in ISRO papers where `+abc` represents the sum of three operands: `+ a b c`. The expression is evaluated as: `- (+ a b c) (+ d (* e f))`.
Substituting the values: `+ 10 2 5` = 17.
For the second part: `* 4 1` = 4, `+ 3 4` = 7.
Finally, `- 17 7` = 10.
Question 24
Topic: Algorithms
The asymptotic complexity of 4 functions $f_{1}, f_{2}, f_{3}, f_{4}$ are
$f_{1}(n)=2^{n}$
$f_{2}(n)=n^{(3/2)}$
$f_{3}(n)=n \log n$
$f_{4}(n)=n^{(\log n)}$
Arrange them in increasing order of complexity
- A. $f_{2}, f_{3}, f_{4}, f_{1}$
- B. $f_{1}, f_{2}, f_{3}, f_{4}$
- C. $f_{3}, f_{2}, f_{4}, f_{1}$
- D. $f_{3}, f_{1}, f_{2}, f_{4}$
Answer: C. $f_{3}, f_{2}, f_{4}, f_{1}$
Explanation: Let's compare the growth rates:
1. $f_{3}(n) = n \log n$
2. $f_{2}(n) = n^{1.5}$. (Since $n^{1.5} > n \log n$ for large n, $f_{3} < f_{2}$)
3. $f_{4}(n) = n^{\log n}$. This grows faster than any polynomial because the exponent $\log n$ keeps increasing.
4. $f_{1}(n) = 2^{n}$. Exponential growth is faster than quasi-polynomial $n^{\log n}$.
So the order is $f_{3} < f_{2} < f_{4} < f_{1}$.
Question 25
Topic: Algorithms
What is the maximum number of comparisons needed to sort 6 items using radix sort, if each number is a 3-digit decimal number?
Answer: B. 180
Explanation: In typical ISRO/GATE conventions for radix sort operations/comparisons, it is calculated as: Number of elements $\times$ Number of digits $\times$ Base of the number system.
Here, $n = 6$, $d = 3$, and $k = 10$ (decimal).
Total operations = $6 \times 3 \times 10 = 180$.
Question 26
Topic: Data Structures
What is the Inorder predecessor of 15 in the Binary Search Tree shown below
(Tree contains root 15, left child 6, right child 18. 6 has left child 3, right child 7. 18 has left child 17, right child 20. 3 has left child 2, right child 4. 7 has right child 13. 13 has left child 9)
Answer: D. 13
Explanation: The inorder predecessor of a node in a BST is the maximum value in its left subtree. The left child of 15 is 6. Following the rightmost path in the left subtree: 6 -> 7 -> 13. Node 13 has no right child, making it the maximum value in 15's left subtree. Thus, 13 is the inorder predecessor.
Question 27
Topic: C Programming
Pointer p to an array of integers (of size 2 Bytes) is initialised to 200. What is the value of p+3?
Answer: A. 206
Explanation: Pointer arithmetic scales by the size of the data type it points to. Since the integer size is given as 2 Bytes, adding 3 to the pointer increases the address by $3 \times 2 = 6$ bytes. Therefore, $200 + 6 = 206$.
Question 28
Topic: Data Structures
In a priority queue, insertion and deletion can be done at
- A. Front
- B. Back
- C. Middle
- D. Any position
Answer: D. Any position
Explanation: Unlike a standard queue which strictly follows FIFO (insertion at rear, deletion at front), a priority queue places and removes elements based on their priority. Depending on the underlying implementation (like a sorted linked list or array), insertion or deletion can technically happen at any position to maintain or fetch the highest priority item.
Question 29
Topic: Algorithms
Worst case time complexity of heap sort for n elements
- A. O(nlogn)
- B. O(logn)
- C. $O(n^{2})$
- D. $O(n)$
Answer: A. O(nlogn)
Explanation: Building the heap takes $O(n)$ time. Extracting the maximum element and heapifying takes $O(\log n)$ time, which is done $n$ times. Thus, the worst-case time complexity of heap sort is $O(n \log n)$.
Question 30
Topic: Database Management Systems
Consider a database table T containing two columns X and Y each of type Integer. After the creation of the table, one record (X=1, Y=1) is inserted in the table. Let MX and MY denote the respective maximum values of X and Y among all records in the table at any point in time. Using MX and MY, new records are inserted in the table 128 times with X and Y values being MX+1, 2*MY+1 respectively. It may be noted that each time after the insertion, values of MX and MY change. What will be the output of the following SQL query after the steps mentioned above are carried out?
SELECT Y FROM T WHERE X=7;
Answer: A. 127
Explanation: The table starts with X=1, Y=1 (MX=1, MY=1).
Insert 1: X = 1+1 = 2, Y = 2(1)+1 = 3
Insert 2: X = 2+1 = 3, Y = 2(3)+1 = 7
Insert 3: X = 3+1 = 4, Y = 2(7)+1 = 15
Insert 4: X = 4+1 = 5, Y = 2(15)+1 = 31
Insert 5: X = 5+1 = 6, Y = 2(31)+1 = 63
Insert 6: X = 6+1 = 7, Y = 2(63)+1 = 127
The SQL query SELECT Y FROM T WHERE X=7 will return the Y value when X is 7, which is 127.
Question 31
Topic: Database Management Systems
In SQL the statement select * from R, S is equivalent to
- A. Select * from R natural join S
- B. Select * from R cross join S
- C. Select * from R outer join S
- D. Select * from R inner join S
Answer: B. Select * from R cross join S
Explanation: In SQL, selecting from multiple tables separated by a comma (e.g., `FROM R, S`) without specifying a `WHERE` clause for join conditions evaluates to a Cartesian product, which is functionally equivalent to a cross join.
Question 32
Topic: Database Management Systems
Which of the following SQL query deletes all tuples in the teacher relation for those teachers associated with a department located in the building with name CSE and which is in department relation:
- A. DELETE FROM teacher WHERE dept_name IN 'CSE';
- B. DELETE FROM department WHERE building='CSE';
- C. DELETE FROM teacher WHERE dept_name IN (SELECT dept_name FROM department WHERE building = 'CSE');
- D. None of the above
Answer: C. DELETE FROM teacher WHERE dept_name IN (SELECT dept_name FROM department WHERE building = 'CSE');
Explanation: To delete records from the `teacher` table based on a condition matching the `department` table, a subquery is needed. Option (c) correctly identifies the relevant departments using `SELECT dept_name FROM department WHERE building = 'CSE'` and deletes the associated teachers.
Question 33
Topic: Database Management Systems
Amongst the ACID properties of a transaction, the 'Durability' property requires that the changes made to the database by a successful transaction persist
- A. Except in case of an operating system crash
- B. Except in case of a disk crash
- C. Except in case of a power failure
- D. Always, even if there is a failure of any kind
Answer: D. Always, even if there is a failure of any kind
Explanation: Durability guarantees that once a transaction has successfully committed, its changes are permanently recorded in the database, surviving any subsequent system failures, crashes, or power outages.
Question 34
Topic: Database Management Systems
Fifth Normal form is concerned with
- A. Join dependency
- B. Domain-key
- C. Multivalued dependency
- D. Functional dependency
Answer: A. Join dependency
Explanation: Fifth Normal Form (5NF), also known as Project-Join Normal Form (PJNF), specifically addresses and eliminates redundancy related to join dependencies. It ensures that a relation cannot be losslessly decomposed into smaller tables unless those tables represent candidate keys.
Question 35
Topic: Digital Logic
Which of the following definitions is true
(i) A digital multiplexer takes one input from many inputs and outputs the selected one
(ii) Four select lines are required for 1-16 demultiplexer
(iii) Eight OR gates are required to implement an octal to binary encoder
- A. (i) and (ii)
- B. (ii) and (iii)
- C. (i) alone
- D. none of the above
Answer: A. (i) and (ii)
Explanation: Statement (i) correctly defines a multiplexer. Statement (ii) is correct because $2^{4} = 16$, requiring 4 select lines. Statement (iii) is false because an octal to binary (8-to-3) encoder typically uses only three 4-input OR gates, not eight.
Question 36
Topic: Digital Logic
The time delay obtained through an 8 bit serial register with 400 MHz clock is:
- A. 20 ns
- B. 2.5 $\mu$s
- C. 20 $\mu$s
- D. 2.5 ns
Answer: A. 20 ns
Explanation: The clock frequency is $400$ MHz, so the clock period $T = 1 / (400 \times 10^{6})$ seconds $= 2.5$ ns. An 8-bit serial register requires 8 clock cycles for a bit to pass through. Thus, total delay $= 8 \times 2.5$ ns $= 20$ ns.
Question 37
Topic: Digital Logic
A full adder circuit requires
- A. two inputs and two outputs
- B. two inputs and three outputs
- C. three inputs and two outputs
- D. three inputs and one output
Answer: C. three inputs and two outputs
Explanation: A full adder adds two significant bits and an incoming carry bit, hence it has 3 inputs (A, B, Carry-In). It produces a sum and an outgoing carry, hence it has 2 outputs (Sum, Carry-Out).
Question 38
Topic: Digital Logic
The logical operation of the following circuit is (Circuit with two NAND gates acting as NOT gates feeding into a third NAND gate)
Answer: D. OR
Explanation: The two input NAND gates have their inputs tied together, acting as NOT gates to produce $\overline{A}$ and $\overline{B}$. These are fed into the final NAND gate, resulting in $\overline{\overline{A} \cdot \overline{B}}$. By De Morgan's Law, this simplifies to $A + B$, which is the OR operation.
Question 39
Topic: Digital Logic
A logic circuit that provides a LOW output when both inputs are HIGH or both inputs are LOW is
- A. AND
- B. NAND
- C. XNOR
- D. XOR
Answer: D. XOR
Explanation: An XOR (Exclusive OR) gate outputs 0 (LOW) when both inputs are identical (either both 0s or both 1s) and outputs 1 (HIGH) when the inputs are different.
Question 40
Topic: Digital Logic
The resulting logic circuit obtained when both inputs of NOR and NAND gates are connected together is:
Answer: D. NOT
Explanation: When the inputs of a 2-input NAND gate or a 2-input NOR gate are tied together, the gate inverts the input signal, thereby functioning as a NOT gate.
Question 41
Topic: Engineering Mathematics
Given P is a matrix = $\begin{pmatrix} 3 & 2 & 1 & 4 \\ 4 & 0 & 3 & 1 \\ 6 & 4 & 2 & 8 \\ 2 & 5 & 1 & 3 \end{pmatrix}$
If det |P| denotes the determinant of matrix P, then which of the following is true:
- A. det P is indeterminate
- B. det P is negative
- C. det |P|=0
- D. None of the above
Answer: C. det |P|=0
Explanation: In matrix P, the third row (6, 4, 2, 8) is exactly twice the first row (3, 2, 1, 4). Since two rows are linearly dependent (proportional), the determinant of the matrix is 0.
Question 42
Topic: Discrete Mathematics
Which of the following are true?
(i) In a graph G with 'n' vertices and 'e' edges, sum of degrees of vertices = 2*e
(ii) Eccentricity of a connected graph can never be equal to radius of the graph
(iii) Girth of a graph is the total number of edges in the shortest cycle of the graph
(iv) Graph with equal degree for all vertices is multigraph
- A. (i), (ii), (iii)
- B. (ii), (iii), (iv)
- C. (i), (iii), (iv)
- D. None of the above
Answer: D. None of the above
Explanation: Statement (i) is true (Handshaking lemma). Statement (ii) is false; the radius is defined as the minimum eccentricity among all vertices. Statement (iii) is true. Statement (iv) is false; a graph with equal degrees is a regular graph, not necessarily a multigraph. Since no option perfectly lists just (i) and (iii), (d) is the correct choice.
Question 43
Topic: Discrete Mathematics
If there are five faces and nine vertices in an undirected planar graph, then number of edges is
- A. 14
- B. 16
- C. 12
- D. None of the above
Answer: C. 12
Explanation: Euler's formula for a connected planar graph is $V - E + F = 2$. Given Vertices ($V$) = 9 and Faces ($F$) = 5. Substituting the values: $9 - E + 5 = 2 \implies 14 - E = 2 \implies E = 12$.
Question 44
Topic: Engineering Mathematics
If a bag contains 6 blue balls, 5 white balls and 4 yellow balls and if a ball is drawn out at random, probability of not getting a white ball is
- A. $\frac{2}{3}$
- B. $\frac{1}{3}$
- C. $\frac{1}{5}$
- D. None of the above
Answer: A. $\frac{2}{3}$
Explanation: The total number of balls is $6 + 5 + 4 = 15$. The number of non-white balls (blue or yellow) is $6 + 4 = 10$. The probability of not drawing a white ball is $\frac{10}{15} = \frac{2}{3}$.
Question 45
Topic: Algorithms
The complexity of matrix multiplication of two matrices A and B whose orders are $m\times n$ and $n\times p$ respectively is
- A. $O(m\times p)$
- B. $O(m\times n^{2}\times p)$
- C. $O(m\times n\times p^{2})$
- D. $O(m\times n\times p)$
Answer: D. $O(m\times n\times p)$
Explanation: Matrix A is $m \times n$ and matrix B is $n \times p$. Multiplying them yields a matrix of order $m \times p$. Computing each of the $m \times p$ elements takes $O(n)$ time (multiplying a row by a column). Thus, the overall time complexity is $O(m \times n \times p)$.
Question 46
Topic: Data Structures
From a character string of length m, the number of sub-strings of all lengths that can be formed are:
- A. $m^{2}$
- B. m
- C. $m(m+1)/2$
- D. $m(\log m)$
Answer: C. $m(m+1)/2$
Explanation: For a string of length m, assuming all characters are distinct and excluding the empty substring, the number of substrings of length 1 is m, length 2 is m-1, and so on up to length m which is 1. The sum of the first m natural numbers is $m(m+1)/2$.
Question 47
Topic: Digital Logic
A truth table will need how many rows if there are n variables?
- A. $2^{*}n$
- B. n
- C. 2n
- D. None of the above
Answer: D. None of the above
Explanation: A truth table for n binary variables requires exactly $2^{n}$ rows to account for all possible boolean combinations. Since $2^{n}$ is not explicitly listed in the options (as $2^{*}n$ translates to $2n$), 'None of the above' is the strictly correct choice.
Question 48
Topic: Operating Systems
Which of the following multithreading model is followed in Linux OS?
- A. One User Thread mapping to One Kernel Thread
- B. Many User Threads mapping to One Kernel Thread
- C. One User Thread mapping to Many Kernel Threads
- D. Many User Threads mapping to Many Kernel Threads
Answer: A. One User Thread mapping to One Kernel Thread
Explanation: Linux uses the 1:1 (One-to-One) threading model. The Native POSIX Thread Library (NPTL) maps each user-level thread to a unique kernel-level thread, allowing them to be scheduled independently by the OS kernel.
Question 49
Topic: Operating Systems
Purpose of priority inheritance protocol in synchronization mechanisms is
- A. To prevent priority inversion in systems having two or lesser priorities for a resource
- B. To provide mutual exclusion between threads and prevent data corruption
- C. To prevent priority inversion and ensure fairness in resource allocation
- D. To allow multiple threads to access any resource simultaneously
Answer: C. To prevent priority inversion and ensure fairness in resource allocation
Explanation: The priority inheritance protocol solves the priority inversion problem. When a low-priority task holds a resource required by a high-priority task, the low-priority task temporarily 'inherits' the high priority, ensuring it completes its critical section faster and releases the resource.
Question 50
Topic: Operating Systems
As part of segmentation memory allocation strategy, which of the following assigns segment numbers for various segments of a program like code segment and data segment?
- A. Compiler
- B. Assembler
- C. Loader
- D. Linker
Answer: A. Compiler
Explanation: The compiler (or assembler) divides the user program logically into distinct parts such as the code segment, data segment, and stack segment during compilation, assigning them logical segment numbers.
Question 51
Topic: Operating Systems
Which of the following is false about interrupts?
- A. Interrupts can be triggered by a hardware or by a software
- B. Hardware interrupts may be triggered by sending a signal to CPU through a system bus
- C. Software interrupts may be triggered by executing system calls
- D. Trap is a hardware generated interrupt
Answer: D. Trap is a hardware generated interrupt
Explanation: A trap (or exception) is a software-generated interrupt. It is typically caused either by an error in the program (like division by zero) or a specific software request to the OS via a system call, not by external hardware.
Question 52
Topic: Operating Systems
With respect to Process Schedulers of an operating system mentioned above in the table, which of the following statements is correct?
| Process Scheduler type | Activity |
|---|---|
| (i) Long term scheduler | (P) executes much faster to reduce wastage of CPU time |
| (ii) Medium term scheduler | (Q) controls degree of multiprogramming |
| (iii) Short Term Scheduler | (R) is associated with swapping |
- A. (i) - (R), (ii) - (P), (iii) - (Q)
- B. (i) - (Q), (ii) - (R), (iii) - (P)
- C. (i) - (P), (ii) - (R), (iii) - (Q)
- D. (i) - (R), (ii) - (Q), (iii) - (P)
Answer: B. (i) - (Q), (ii) - (R), (iii) - (P)
Explanation: The Long-term scheduler admits new processes into memory, thus controlling the degree of multiprogramming (Q). The Medium-term scheduler handles swapping processes in and out of main memory (R). The Short-term scheduler (CPU scheduler) must execute extremely fast to constantly allocate the CPU among ready processes (P).
Question 53
Topic: Operating Systems
In a producer-consumer scenario also known as Bounded-Buffer Problem, what would be the most appropriate synchronization primitive to ensure that the consumer waits when the buffer is empty?
- A. Spinlock
- B. Mutex lock
- C. Semaphore
- D. Monitors
Answer: C. Semaphore
Explanation: Counting semaphores are typically used to solve the bounded-buffer problem. Specifically, a semaphore initialized to 0 tracks the number of 'full' slots. The consumer performs a wait/down operation on it, inherently ensuring the consumer blocks when the buffer is completely empty.
Question 54
Topic: Operating Systems
Any attempt by a process to access memory allocated to OS or process of other user results into
- A. Trap to OS
- B. Context Switching
- C. Page fault
- D. Invocation to Scheduler Despatch
Answer: A. Trap to OS
Explanation: When a process attempts an illegal memory access (such as reading memory belonging to the OS kernel or another process), the hardware memory protection unit generates an exception (specifically, a protection fault), which acts as a trap to the Operating System to terminate the offending process.
Question 55
Topic: Theory of Computation
Consider the deterministic finite-state automaton (DFA) below. The alphabet is {a,b}. The state with a small incoming arrow is the initial state, while the double circle state denotes a final state. (Diagram shows initial state transitioning to final state on 'a'. Final state loops on 'b', and transitions to a trap state on 'a'. Initial state transitions to trap state on 'b'. Trap state loops on 'a,b'). Which of the following regular expressions defines the language accepted by the DFA?
- A. $ab^{*}$
- B. $a^{*}b^{*}$
- C. $(ab)^{*}$
- D. $a^{*}b$
Answer: A. $ab^{*}$
Explanation: Based on the transitions in the DFA: The string must start with an 'a' to reach the final state. From the final state, any number of 'b's keeps it in the final state. Any 'b' at the start or an 'a' after the initial 'a' sends the automaton to a trap/dead state. This corresponds to the regular expression $ab^{*}$.
Question 56
Topic: Theory of Computation
Consider the Nondeterministic Finite-State Automaton (NFA) below. States with a small incoming arrow are initial states, while a double circle state denotes a final state. (Diagram shows initial state with a self loop on 'a,b', a transition on 'b' to a second state, and a transition from the second state on 'a,b' to a final state). The number of states in the MINIMAL Deterministic Finite-State Automaton (DFA) for this language is:
Answer: B. 4
Explanation: The given NFA diagram describes the regular expression $(a+b)^{*}b(a+b)$, which represents the language of all strings over {a,b} where the second-to-last symbol is a 'b'. Constructing a minimal DFA for strings where the $k$-th symbol from the end is a specific character requires exactly $2^{k}$ states. Since $k=2$, the minimal DFA requires $2^{2} = 4$ states.
Question 57
Topic: Theory of Computation
Consider the context-free grammar G below. Here S is the starting non-terminal symbol, while a and b are terminal symbols.
$S \rightarrow aSb | T$
$T \rightarrow bT | b$
The language generated by G is:
- A. $\{a^{m}b^{n}:0 \le m < n\}$
- B. $\{a^{m}b^{n}:0 < m < n\}$
- C. $\{a^{m}b^{n}:0 < n \le m\}$
- D. $\{a^{n}b^{n}:0 \le n\}$
Answer: A. $\{a^{m}b^{n}:0 \le m < n\}$
Explanation: The production $T \rightarrow bT | b$ generates the language $b^{k}$ for $k \ge 1$. The production $S \rightarrow aSb | T$ generates strings of the form $a^{m} T b^{m}$ for $m \ge 0$. By substituting T, we get $a^{m} b^{k} b^{m} = a^{m} b^{m+k}$. Since $k \ge 1$, the total number of 'b's ($n = m+k$) is strictly greater than the total number of 'a's ($m$). Therefore, the language is $\{a^{m}b^{n}: 0 \le m < n\}$.
Question 58
Topic: Theory of Computation
Which of the following statements is NOT true?
- A. If a language is recursive its complement is recursive
- B. If a language is recursive its complement is recursively enumerable
- C. If a language and its complement are recursively enumerable it is recursive
- D. If a language is recursively enumerable its complement is also recursively enumerable
Answer: D. If a language is recursively enumerable its complement is also recursively enumerable
Explanation: Recursively enumerable (RE) languages are not closed under complementation. The Halting problem is a classic example of an RE language whose complement is not RE. Statements (a), (b), and (c) are all fundamental theorems in computability.
Question 59
Topic: Theory of Computation
Which of the following statements is FALSE:
- A. The intersection of a regular language and a context-free language is context-free
- B. The intersection of a regular language and context-free language is regular
- C. The union of two context-free languages is context-free
- D. The union of two regular languages is regular
Answer: B. The intersection of a regular language and context-free language is regular
Explanation: The intersection of a context-free language (CFL) and a regular language is guaranteed to be a CFL, but it is not necessarily regular. For instance, the intersection of $L_{CFL} = \{a^n b^n | n \ge 0\}$ and $L_{Regular} = (a+b)^*$ results in $L_{CFL}$, which is context-free but not regular.
Question 60
Topic: Theory of Computation
Consider the context-free grammar G below. Here S is the starting non-terminal symbol, while a and b are terminal symbols.
$S \rightarrow aaSb | T$
$T \rightarrow Tb | a$
Which of the following statements is true about the language L(G) generated by G?
- A. aabbaabb belongs to L(G) but aabb does not
- B. aaaaabbb belongs to L(G) but aaaabb does not
- C. aaaabb belongs to L(G) but aabbaabb does not
- D. aaabb belongs to L(G) but aaaaabbb does not
Answer: B. aaaaabbb belongs to L(G) but aaaabb does not
Explanation: The production $T \rightarrow Tb | a$ generates $ab^{k}$ (an 'a' followed by $k \ge 0$ 'b's). The production $S \rightarrow aaSb | T$ generates $a^{2m} T b^{m}$ ($m \ge 0$). Substituting T, we get $a^{2m} a b^{k} b^{m} = a^{2m+1} b^{k+m}$. Thus, L(G) consists of strings with an ODD number of 'a's followed by some number of 'b's. The string 'aaaaabbb' has 5 'a's (odd) and belongs to L(G), whereas 'aaaabb' has 4 'a's (even) and does not belong.
Question 61
Topic: Artificial Intelligence
Which of the following is not a connective of First Order Logic with reference to AI (Artificial Intelligence)?
Answer: C. <>
Explanation: In standard First Order Logic, the primary connectives include AND (often represented as $\wedge$ or 'A' in the provided text), OR ($\vee$ or 'V'), NOT ($\neg$ or $\sim$), and implications. The symbol '<>' typically represents 'not equal to' in various programming languages, but it is not a standard logical connective in AI or First Order Logic.
Question 62
Topic: Artificial Intelligence
A hybrid Bayesian network contains
- A. Only continuous variables
- B. Only discrete variables
- C. Both discrete and discontinuous variables
- D. Both continuous and discrete variables
Answer: D. Both continuous and discrete variables
Explanation: A Bayesian network that incorporates both continuous (real-valued) and discrete (categorical) random variables within the same model is referred to as a hybrid Bayesian network.
Question 63
Topic: Big Data
Which of the following scenario may not be a good fit for HDFS in Big data paradigm?
- A. HDFS is not suitable for scenarios requiring multiple/simultaneous writes to the same file
- B. HDFS is suitable for storing data related to applications requiring low latency data access
- C. HDFS is suitable for storing data related to applications requiring high latency data access
- D. None of the above
Answer: B. HDFS is suitable for storing data related to applications requiring low latency data access
Explanation: HDFS (Hadoop Distributed File System) is optimized for batch processing and high throughput of large datasets, which naturally incurs high latency. Therefore, using HDFS for applications that require low-latency data access (like traditional relational databases) is a fundamentally poor fit. Option (b) incorrectly claims it is suitable for low latency, highlighting the scenario that is not a good fit.
Question 64
Topic: Java Technologies
A standard Java API for monitoring and managing applications is
Answer: C. JMX
Explanation: JMX stands for Java Management Extensions. It is a standard Java technology and API that supplies tools for managing and monitoring applications, system objects, devices, and service-oriented networks.
Question 65
Topic: Data Mining / Machine Learning
Which of the following clustering technique is used by K-Means Algorithm:
- A. Hierarchical Technique
- B. Partitional technique
- C. Divisive
- D. Agglomerative
Answer: B. Partitional technique
Explanation: K-Means is a centroid-based, partitional clustering algorithm. It works by partitioning a dataset into exactly 'K' non-overlapping, distinct clusters without imposing a hierarchical structure.
Question 66
Topic: Databases / Big Data
Which type of data storage system Cassandra is?
- A. Distributed
- B. Centralized
- C. Parallel
- D. Dumb
Answer: A. Distributed
Explanation: Apache Cassandra is a highly scalable, distributed NoSQL database management system designed to handle large amounts of data across many commodity servers, providing high availability with no single point of failure.
Question 67
Topic: Cloud Computing
The cloud infrastructure solely for a single organisation within its premises is
- A. public cloud
- B. private cloud
- C. common cloud
- D. hybrid cloud
Answer: B. private cloud
Explanation: A private cloud is a cloud computing environment dedicated entirely to a single organization. It can be hosted either on-premises (within the organization's own data center) or by a third-party service provider, but the resources are not shared with other organizations.
Question 68
Topic: Cloud Computing
Temporarily renting the capacity to handle spikes in load is called
- A. Cloud sharing
- B. Cloud computing
- C. Cloud bursting
- D. Cloud loading
Answer: C. Cloud bursting
Explanation: Cloud bursting is a deployment model in which an application runs in a private cloud or local data center and 'bursts' into a public cloud when the demand for computing capacity spikes, allowing the organization to temporarily rent external capacity.
Question 69
Topic: Operating Systems
State whether true or false :
(i) Virtualisation facilitates multiple OS simultaneously on the same hardware
(ii) In dual boot, multiple OS can run simultaneously
- A. (i) True (ii) False
- B. (i) False (ii) True
- C. (i) True (ii) True
- D. (i) False (ii) False
Answer: A. (i) True (ii) False
Explanation: Statement (i) is true: Virtualization allows a hypervisor to run multiple isolated operating systems concurrently on a single physical machine. Statement (ii) is false: In a dual-boot setup, multiple operating systems are installed on the disk, but only one can be booted and run at any given time.
Question 70
Topic: Operating Systems / Virtualization
In which of the virtualization techniques, hypervisor modifies the guest OS?
- A. Full virtualization
- B. Para virtualization
- C. Both (a) and (b)
- D. None of the above
Answer: B. Para virtualization
Explanation: In para-virtualization, the guest operating system is explicitly modified to be aware that it is running in a virtualized environment. This allows the guest OS to communicate directly with the hypervisor (using hypercalls) for better performance, unlike full virtualization where the guest OS remains unmodified.
Question 71
Topic: Computer Networks / Security
Which of the following is the cause of Ping of death issue related to ICMP packets?
- A. Buffer overflow
- B. Divide by ZERO
- C. Missing input sanitisation
- D. Privilege escalation
Answer: A. Buffer overflow
Explanation: The 'Ping of Death' attack involves sending a malicious, oversized ICMP echo request (ping) packet to a target machine. When the target attempts to reassemble the oversized packet fragments, it exceeds the maximum IP packet size (65,535 bytes), leading to a buffer overflow that can crash or freeze the system.
Question 72
Topic: Cyber Security
Match the following and select the correct option:
(i) Encryption (P) Manipulation of identification of a host (through NIC identifier)
(ii) MAC Cloning (Q) Determining version and name of OS and OS utilities
(iii) ARP Spoofing (R) Protecting the confidentiality
(iv) Finger printing (S) Poisoning the IP to MAC address mapping
- A. (i) - (R), (ii) - (P), (iii) - (S), (iv) - (Q)
- B. (i) - (P), (ii) - (R), (iii) - (S), (iv) - (Q)
- C. (i) - (R), (ii) - (S), (iii) - (P), (iv) - (Q)
- D. (i) - (R), (ii) - (Q), (iii) - (S), (iv) - (P)
Answer: A. (i) - (R), (ii) - (P), (iii) - (S), (iv) - (Q)
Explanation: Encryption is used for protecting confidentiality (R). MAC Cloning involves spoofing or manipulating the hardware identification (NIC identifier) of a host (P). ARP Spoofing is the technique of poisoning the local network's IP-to-MAC address resolution cache (S). Fingerprinting is the process of probing a system to determine its OS and version details (Q).
Question 73
Topic: Cyber Security
Match the following in the context of Information Security which are closely associated
(i) Ingress filtering (P) Data leakage prevention
(ii) Egress filtering (Q) Hiding identity of systems
(iii) NAT (R) Keep track of TCP/IP connections
(iv) Stateful firewall (S) Malicious traffic prevention
- A. (i) - (S), (ii) - (P), (iii) - (Q), (iv) - (R)
- B. (i) - (P), (ii) - (R), (iii) - (S), (iv) - (Q)
- C. (i) - (S), (ii) - (R), (iii) - (P), (iv) - (Q)
- D. (i) - (R), (ii) - (Q), (iii) - (S), (iv) - (P)
Answer: A. (i) - (S), (ii) - (P), (iii) - (Q), (iv) - (R)
Explanation: Ingress filtering checks incoming packets to prevent malicious traffic and spoofing from entering the network (S). Egress filtering monitors outbound traffic, serving as a layer of data leakage prevention (P). Network Address Translation (NAT) hides the internal IP identity of systems from the outside world (Q). A Stateful firewall actively tracks the state and context of active TCP/IP connections (R).
Question 74
Topic: Cryptography
Asymmetric encryption is not suitable for
- A. Confidentiality
- B. Authentication
- C. Key Exchange
- D. Prevention of Denial of Service
Answer: D. Prevention of Denial of Service
Explanation: Asymmetric (public-key) encryption relies on complex mathematical operations, making it computationally heavy. It is excellent for authentication, secure key exchange, and confidentiality of small data payloads. However, its high processing overhead makes it completely unsuitable for mitigating Denial of Service (DoS) attacks; in fact, forcing a server to perform heavy asymmetric decryptions is a common DoS vector.
Question 75
Topic: Cyber Security
CAPTCHA is used to provide protection from
- A. Zero day attack
- B. Buffer overflow
- C. Automated scripted attack
- D. Man in the middle attack
Answer: C. Automated scripted attack
Explanation: CAPTCHA stands for 'Completely Automated Public Turing test to tell Computers and Humans Apart'. It is primarily used on websites to prevent bots and automated scripts from spamming forms, scraping data, or launching brute-force attacks.
Question 76
Topic: Software Engineering
The MCDC software testing technique stands for
- A. Multiple Conditions Decision Coverage
- B. Multiple Coverage Decision Conditions
- C. Modified Condition Decision Coverage
- D. None of the above
Answer: C. Modified Condition Decision Coverage
Explanation: MC/DC stands for Modified Condition/Decision Coverage. It is a structural coverage metric used extensively in software testing, especially for safety-critical systems, to ensure that every condition within a decision independently affects the decision's outcome.
Question 77
Topic: Software Engineering
Regression testing focuses on validating _________.
- A. module structure
- B. module interfaces
- C. software changes
- D. None of the above
Answer: C. software changes
Explanation: Regression testing is the process of re-running functional and non-functional tests to ensure that previously developed and tested software still performs after a software change (such as an update, patch, or bug fix) has been introduced.
Question 78
Topic: Software Engineering
HIPO chart is also known as
- A. UML diagram
- B. Structure chart
- C. DFD
- D. None of the above
Answer: B. Structure chart
Explanation: HIPO (Hierarchical Input Process Output) is a systems analysis design aid and documentation technique from the 1970s. It visually represents the hierarchical structure of a system's modules and is widely considered a form of or synonymous with a structure chart.
Question 79
Topic: Software Engineering
McCabe's cyclomatic complexity number of a program control graph 'G' with e edges, n nodes and p disconnected paths is defined as
- A. $n-e+2p$
- B. $e-n+2+p$
- C. $e-n+2p$
- D. $n-e+p$
Answer: C. $e-n+2p$
Explanation: Thomas McCabe's Cyclomatic Complexity formula for a control flow graph is $V(G) = E - N + 2P$, where E is the number of edges, N is the number of nodes, and P is the number of connected components (or disconnected subgraphs/paths).
Question 80
Topic: Database Management Systems
Entity Relationship model shows the
- A. Static view
- B. Functional view
- C. Dynamic view
- D. All of the above
Answer: A. Static view
Explanation: An Entity-Relationship (ER) model provides a conceptual abstraction of a database's structure, defining the entities, their attributes, and the relationships between them. Since it does not capture states, behaviors, or data flow over time, it represents the purely static view of the system.
Question 81
Topic: Aptitude
Which option replaces the question mark?
Answer: A. 1
Explanation: Based on the visual analogy in the provided image: Paint is to a Paintbrush as a Spool of thread is to a Needle (Option 1).
Question 82
Topic: Aptitude
Which option replaces the question mark?
Answer: D. D
Explanation: Based on the visual pattern: the main figure rotates 90 degrees clockwise (so vertical stripes become horizontal), and the small corner shape shifts from the top-left to the top-right while changing from white to black.
Question 83
Topic: Aptitude
Which option replaces the question mark?
Answer: A. A
Explanation: Applying the rules from the image to the sequence of shapes. The 'Crossed circle' operator makes the first item half size, and the 'Green Target' operator toggles the shading of the first item. The other items remain unchanged, leading to option A.
Question 84
Topic: Aptitude
You are a retail clerk who typically works Monday through Friday. You arrive at work and notice the new schedule is posted, and your manager has given you a Saturday shift. It happens to be your son's birthday on Saturday, and you have a party planned. What is the most appropriate response?
- A. Speak with your manager and explain you have a birthday party planned for your son on Saturday and ask if they can find someone else or if you can find someone to cover your shift
- B. Ask your coworkers if they can cover your shift because you have a birthday party for your son planned
- C. Pretend like you did not see the schedule since you always work Monday through Friday and do not show up on Saturday
- D. Do not say anything to anyone, and plan on calling in sick on Saturday
Answer: A. Speak with your manager and explain you have a birthday party planned for your son on Saturday and ask if they can find someone else or if you can find someone to cover your shift
Explanation: Directly communicating with the manager about the scheduling conflict is the most professional, transparent, and responsible course of action.
Question 85
Topic: Aptitude
Which of the following best completes the passage below?
In a survey of job applicants, two-fifths admitted to being at least a little dishonest. However, the survey may underestimate the proportion of job applicants who are dishonest, because ________?
- A. some dishonest people taking the survey might have claimed on the survey to be honest
- B. some generally honest people taking the survey might have claimed on the survey to be dishonest
- C. some people who claimed on the survey to be at least a little dishonest may be very dishonest
- D. some people who claimed on the survey to be dishonest may have been answering honestly
Answer: A. some dishonest people taking the survey might have claimed on the survey to be honest
Explanation: If dishonest applicants lie on the survey and claim to be honest, they will not be counted in the 'two-fifths' figure, logically causing the survey to underestimate the actual number of dishonest people.
Question 86
Topic: Aptitude
Eight friends P, Q, R, S, T, V, W and Y are sitting around a square table.
1. Out of eight, four persons are sitting at the corners of the table and the other four are sitting at the mid points of each side of the table.
2. Persons at the corners are facing the centre while the persons at the mid - points of side are facing outside.
3. S is third to the right of P. P is facing the centre.
4. Y is not sitting beside P or S.
5. T is third to the right of R.
6. R is not sitting at the mid - point of any side of the table.
7. R is also not beside Y.
8. There is only one person between P and V.
9. Q is not sitting beside V.
How many people are there between T and Q?
Answer: A. 1
Explanation: Following the seating arrangement rules logically establishes the circular order: P, W, V, S, R, Q, Y, T. Between T and Q, there is exactly 1 person (Y) in one direction (and 5 in the other, but 1 is the available valid option).
Question 87
Topic: Aptitude
Which option replaces the question mark?
Answer: B. 2
Explanation: Based on the visual dot patterns on the dominoes provided in the image, Option 2 best completes the logical sequence.
Question 88
Topic: Aptitude
Which one will replace the question mark?
0 6 6
2 4 14
10 24 ?
Answer: C. 38
Explanation: Based on numerical grid logic often found in such puzzles, applying row/column mathematical operators determines the missing value as 38.
Question 89
Topic: Aptitude
Which option replaces the question mark?
Answer: C. C
Explanation: Following the visual flowchart transformations mapped to the input geometric shape.
Question 90
Topic: Aptitude
Fact 1: Pictures can tell a story.
Fact 2: All storybooks have pictures.
Fact 3: Some storybooks have words.
If the first three statements are facts, which of the following statements must also be a fact?
I: Pictures can tell a story better than words can.
II: The stories in storybooks are very simple.
III: Some storybooks have both words and pictures.
- A. I only
- B. II only
- C. III only
- D. None of the statements is a known fact
Answer: C. III only
Explanation: Since all storybooks have pictures (Fact 2) and some storybooks have words (Fact 3), it logically follows that the storybooks that have words must also have pictures. Therefore, some storybooks have both (Statement III is true). Statements I and II introduce unsupported opinions.
Question 91
Topic: Aptitude
Statement: Exporters in the capital are alleging that commercial banks are violating a Reserve Bank of India directive to operate a post-shipment export credit denominated in foreign currency at international rates from January this year.
Courses of Action:
I. The officers concerned in the commercial banks are to be suspended.
II. The RBI should be asked to stop giving such directives to commercial banks.
- A. Only I follow
- B. Only II follows
- C. Either I or II follows
- D. Neither I nor II follows
Answer: D. Neither I nor II follows
Explanation: Action I is too extreme as the violation is currently just an 'allegation' and requires investigation before suspension. Action II is illogical, as stopping regulatory directives entirely just because they are being violated is not a proper administrative solution.
Question 92
Topic: Aptitude
Find out the wrong number in the series.
7, 8, 18, 57, 228, 1165, 6996
Answer: D. 228
Explanation: The pattern is: $7 \times 1 + 1 = 8$; $8 \times 2 + 2 = 18$; $18 \times 3 + 3 = 57$; $57 \times 4 + 4 = 232$. The given number is 228 instead of 232. Continuing the correct pattern: $232 \times 5 + 5 = 1165$, which matches the next term. Thus, 228 is incorrect.
Question 93
Topic: Aptitude
Directions : The graph given below represents the variations in earnings in rupees over a week. Study the graph and answer questions given below.
The difference in earnings was large between
- A. Sunday-Monday
- B. Tuesday-Wednesday
- C. Thursday-Friday
- D. None of these
Answer: A. Sunday-Monday
Explanation: Reading the graph, earnings on Sunday are ~100 Rs and on Monday they jump to ~500 Rs, creating a massive difference of 400 Rs, which is the steepest single-day slope shown.
Question 94
Topic: Aptitude
Study the following table carefully and answer the questions given below it:
(Table data provided in source)
What was the percentage (approximate) increase in production of D type of toys from 1983 to 1985?
Answer: C. 19
Explanation: From the table, the production of D type toys in 1983 was 105. In 1985, it was 125. The increase is $125 - 105 = 20$. The percentage increase is $(20 / 105) \times 100 \approx 19.04\%$, which is approximately 19.
Question 95
Topic: Aptitude
What is the maximum area of the rectangle with perimeter $620~mm$?
- A. $24,025~mm^{2}$
- B. $22,725~mm^{2}$
- C. $24,000~mm^{2}$
- D. $24,075~mm^{2}$
Answer: A. $24,025~mm^{2}$
Explanation: A rectangle with a fixed perimeter has its maximum area when it is a square. Perimeter = $4a = 620~mm \implies a = 155~mm$. The maximum area is $155 \times 155 = 24,025~mm^{2}$.