Computer Science Previous Year Questions (page 7)

Explanation

Objects in an object-oriented database contain not only data but also methods for processing the data – correct. Object-oriented databases store computational instructions in the same place as the data – correct. Object-oriented databases are more adept at handling structured (analytical) data than relational databases – incorrect. Object-oriented databases store more types of data than relational databases and access that data faster – correct. So, option (c) is correct.

Explanation

wc -l <a >b &: wc is the word count program, -l counts lines, <a takes input from a, >b writes output to b, and & runs the command in the background. So it counts the number of lines in a, writes the result to b, and does it in the background.

Explanation

The cron daemon is the standard tool for running commands on a pre-determined schedule. It starts when the system boots and runs as long as the system is up. Crontabs for individual users are stored under /var/spool/cron; there is one crontab file per user. So statement (d) is not correct.

Explanation

In UNIX, file access permissions are given as three octal digits: the first for the owner, the second for the group, the third for others. Each digit is rwx: 7 = rwx, 6 = rw-, 5 = r-x, 4 = r--, 3 = -wx, 2 = -w-, 1 = --x, 0 = none. 641 → owner can read and write, group members can read, others can execute only. 100 → owner can execute; all other access is prohibited. So both statements are correct.

Explanation

Zombie Process : The UNIX system assumes that if child process terminate, the parent collects some data about the child, As soon as this collection is made, the system releases information bits about the child process and removes its PID from the process table. The UNIX is forced to keep the child's Process ID (PID) and termination data in the process table indefinitely. Such a terminated process whose status is not collected called a zombie process.

Explanation

Parent and child share only the share memory segments everything else (like stack, heap etc) is duplicated.

Explanation

UNIX file system: Super block contains the number of i-nodes, the number of disk blocks, and the start of the list of free disk blocks.Correct An i-node contains accounting information as well as enough information to locate all the disk blocks that holds the file’s data.Correct All the files and directories are stored in data blocks.Correct Each i-node is 256-bytes long.Incorrect.It is 128 byte long.

Explanation

KILL (SIGKILL) cannot be blocked, caught or ignored, so the statement "KILL is a blockable signal" is not correct.

Explanation

With location transparency the user does not need to know where the data is stored, but still needs to know how the data is fragmented; so a SQL query needs to specify fragments. Fragmentation transparency is the highest level, where the user does not even need to know that the database is partitioned.

Explanation

SQ₁: S tuples with D < 7 have S.C = 4, 2, 3, 1. Joining with R.A: S.C = 1 matches R tuples with A = 1 (R.B = 2, 2, 3, 4) and S.C = 3 matches <3, 1, 4> (R.B = 1). Grouping by R.B gives groups 1, 2, 3, 4, so M = 4. SQ₂: S.B = 1 has distinct D {5, 3} (count 2), S.B = 2 has {7, 4} (count 2), S.B = 3 has {4} (count 1). The third group is not taken because COUNT(DISTINCT S.D) > 1 fails, so N = 2.

Explanation

Semi-join strategies are techniques for query processing in distributed database system. In which only the joining attributes are sent from one site to another and then only the required rows are returned.

Explanation

If we let A represent availability, then the simplest formula for availability A = uptime/(uptime + downtime) MTBF is mean time between failures A = MTTR is mean time to repair A = MTBF/(MTBF + MTTR) So MTBF = 30 days = 30 × 24 = 720 hr A = 720/(720 + 12) = 98.3%

Explanation

I, II and III all return exactly those tuples of R whose B value appears in S, whereas IV returns every (A, R.B) pair of R as long as S is non-empty. Example: R = {(10, 1), (20, 2)}, S = {(2, 30), (3, 40)}. (I) R ⋈ S = {(20, 2, 30)}, so π_A,B(R ⋈ S) = {(20, 2)}. (II) π_B(S) = {(2), (3)}, so R ⋈ π_B(S) = {(20, 2)}. (III) π_A(R) × π_B(S) = {(10, 2), (10, 3), (20, 2), (20, 3)}, and R ∩ that = {(20, 2)}. (IV) π_A,R.B(R × S) = {(10, 1), (20, 2)}. So I, II and III are the same query.

Explanation

D₂: R₁ ∩ R₂ = {A, D}, and AD → B, so AD is a key of R₁ (A, B, D); the decomposition is lossless. D₁ is not always lossless since no FDs are given: if B → A and C → A, then the common attribute A is not a key of either table, so the decomposition is lossy. Hence only D₂.

Explanation

The channel capacity of a discrete memoryless channel is defined by C = max over p(x) of I(x; y) = max over p(x) of [H(y) – H(y|x)] = max over p(x) of [H(x) – H(x|y)]. max H(x|y) does not define the channel capacity.

Explanation

Transmission rate of a circuit = frame rate × number of bits in a slot = 4000 × 8 bps = 32 Kbps. So, option (c) is correct.

Explanation

Multiplexing is of three types: (1) Frequency division multiplexing (FDM) – analog signals. (2) Wavelength division multiplexing (WDM) – analog signals (used in optical fiber). (3) Time division multiplexing (TDM) – digital signals; a single frame is made with interleaved digital signals from n devices. So (1), (2) and (4) are true and (3) is false.

Explanation

Throughput of pure ALOHA is S = G × e⁻²ᴳ, where G is the load (frames per frame-transmission time). Frame transmission time = 200 bits / 200 kbps = 1 ms. The system produces 500 frames per second = 0.5 frames per ms, so G = 1/2. S = 0.5 × e⁻¹ ≈ 0.184 (the maximum throughput of pure ALOHA). So the answer is (b).

Explanation

There are five levels defined along the continuum of the model and according to the SEI. (1) Initial : The starting point for use of a new repeat process (2) Repeatable : The process is at least documented sufficiently such that repeating the same steps may be attempted. (3) Defined : The process is defined as a standard business process. (4) Managed : This process is quantitatively managed in accordance with agreed upon metrics (5) Optimizing : Process management includes deliberate process improvement.

Explanation

(i) Corrective maintenance fixes errors observed when the software is in use – (B). (ii) Adaptive maintenance changes the software to make it adaptable to a new environment – (C). (iii) Perfective maintenance implements new or changed user requirements – (D). (iv) Preventive maintenance performs activities to reduce complexity and prevent future errors – (A).

Explanation

LOC = 500 FP × 50 LOC/FP = 25000 LOC = 25 KLOC. E = a(KLOC)ᵇ = 2.5 × (25)¹ = 62.5 person months.

Explanation

A software program that infers and manipulates existing knowledge in order to generate new knowledge is known as Inference engine. A data dictionary is a collection of descriptions of the data objects or items in a data model for the benefit of programmers and others who need to refer to them. Control strategies are specific action plans for bringing a process back into control. Reference is what relates words to the world of objects on whose condition the truth of sentences hinges and it is natural to wonder what sorts of relations underlie the reference relation-to wonder, that is, what constitutes the mechanism of reference.

Explanation

(i) Coupling measures how dependent modules are on each other – (D). (ii) Cohesion reflects how well a module's elements work toward a single purpose – (C). (iii) Scalability means the system can grow (add functionality/workload) without redesign – (B). (iv) Readable code is easy to inspect and understand – (A).

Explanation

Software safety is a software quality assurance activity that focuses on the identification and assessment to potential hazards that may affect software negatively and cause an entire system to fail. If hazards can be identified early in the software process, software design factures can be specified that will either eliminate or control potential hazards.

Explanation

HTML is not programming language Programming Language means where writing logic is possible which is not supported by HTML.

Explanation

Hermite curves smoothly interpolate between key points. Bezier curves approximate their control points and lie entirely within the convex hull of the control points. The number of control points determines the degree of the Bezier polynomial, so IV is false. So the answer is option (d).

Explanation

Language requirements for ADTs: a syntactic unit in which to encapsulate the type definition, and some primitive operations built into the language processor. Ada provides packages that simulate ADTs, C++ provides data abstraction through classes, and Java's data abstraction is similar to C++. C++, Ada, Java 5.0 and C# 2005 support parameterized ADTs. So (1), (2) and (3) are true. Hence, the answer is (d).

Explanation

Assigning one object reference variable to another copies only the reference (address), not the object. Both variables then refer to the same object; no distinct copy of the object is created. So a copy of the reference is created.

Explanation

Each constraint is satisfied by the points of a closed half space of Rⁿ, and the set of feasible solutions is the intersection of all these half spaces, which is a convex polyhedron. So the feasible region has the convexity property. The answer is (b).

Explanation

The revised simplex method is an improvement over the simplex method: it is computationally more efficient (fewer computations), it automatically generates the inverse of the current basis matrix, and fewer entries are needed in each table, since the simplex method updates the entire tableau while only a small part is used. So answer is option (d).

Explanation

(326.4)₈ = (3 × 8² + 2 × 8¹ + 6 × 8⁰ + 4 × 8⁻¹)₁₀ = (214.5)₁₀. (326.4)₈ = (011 010 110.100)₂ = (1101 0110.1000)₂ = (D6.8)₁₆.

Explanation

In 1's complement there are two representations of zero: –0 (11111111) and +0 (00000000). In 2's complement there is only one representation of 0 (00000000), so 2's complement is the most efficient for performing arithmetic operations on numbers.

Explanation

Grouping the 1s of the K-map: row AB gives AB; the cells with A = 1 and D = 1 give AD; the cells with A = 1 and C = 1 give AC; and cell A'BCD is covered by BCD. Therefore f = AB + AC + AD + BCD.

Explanation

G1 = (x + y)'. G2 = (x + (x + y)')' = x'(x + y) = x'y. G3 = (y + (x + y)')' = y'(x + y) = xy'. f = x'y + xy' = x ⊕ y, i.e. EXCLUSIVE-OR.

Explanation

Half adder adds 2 bits; full adder adds 3 bits; a controlled inverter transmits a binary word or its 1's complement; a binary adder adds two binary numbers. So A-iii, B-i, C-iv, D-ii. Hence, answer is (d).

Explanation

Using a Venn diagram: A – B ⊆ A ⊕ B ⊆ A ∪ B, and |A ∪ B| = |A| + |B| – |A ∩ B| ≤ |A| + |B|. Therefore |A – B| ≤ |A ⊕ B| ≤ |A ∪ B| ≤ |A| + |B|.

Explanation

~(~p∧q) ∧ ~(~p∧~q) = (p ∨ ~q) ∧ (p ∨ q) = p ∨ (~q ∧ q) = p. So the function is p ∨ (p ∧ r) = p(1 + r) = p.

Explanation

MVI A, 35H: A = 35H. MOV B, A: B = 35H. STC: carry = 1. CMC: carry = 0. RAR: rotate A right through carry, A = 0001 1010 (1AH). XRA B: (0001 1010) XOR (0011 0101) = 0010 1111 = 2FH.

Explanation

A queue can be implemented using two stacks S1 and S2: enqueue pushes onto S1; for dequeue, if S2 is not empty pop from it, otherwise first pop all elements of S1 and push them into S2, then pop from S2. So S₁ is correct. Similarly, a stack can be implemented using two queues, so S₂ is also correct. Hence, the answer is option (c).

Explanation

Evaluating from the right: + 3 3 = 6; ↑ 3 6 = 3⁶ = 729; + 3 729 = 732; + 3 732 = 735; * 735 3 = 2205.

Explanation

Although we have a tail pointer, it is of no use for deletion because there is no previous pointer. To delete the last node we need a pointer to the second-last node so that it can become the new last node; finding it requires traversing the whole list, which takes O(n) time.

Explanation

The root is ↔, its left subtree is ~(p ∧ q) and its right subtree is (~p ∧ ~q). Prefix: ↔~∧pq∧~p~q Postfix: pq∧~p~q~∧↔ Option (b) is the answer.

Explanation

Printing the elements in order means printing them in sorted order; inorder traversal of a BST gives the sorted order and visits each node once, taking O(n) time. For a balanced BST T(n) = 2T(n/2) + c and for a skewed tree T(n) = T(n – 1) + c; both solve to O(n).

Explanation

In a 10-bit palindrome the first 5 bits can be chosen freely and the last 5 bits are fixed by them. So there are 2⁵ palindromes out of 2¹⁰ strings. Probability = 2⁵/2¹⁰ = 1/2⁵ = 1/32.

Explanation

x * 3 = 1.5, which is non-zero (true), and 3 is true, so x * 3 && 3 = 1. Then 1 || (j | k) = 1 (logical OR returns 1). The output is 1.

Explanation

Basically polymorphism means having many forms. In C++, polymorphism requires Inheritance, Virtual functions and references, because polymorphism occurs when there is a hierarchy of classes and they are related by inheritance which include virtual functions and references.

Explanation

To minimize average response time the jobs must run in shortest-job-first order: C(3), E(3 < x < 5), D(5), B(6), A(9).

Explanation

P1 arrives at time 1 and runs. At time 3, P2 arrives with a shorter remaining time (10 < 18), so there is a switch to P2 (1). P2 runs to completion at time 13 since no shorter job arrives, then there is a switch to P1 (2). P1 (remaining 18) runs to completion since its remaining time is less than P3's burst time (30), then there is a switch to P3 (3), which runs to completion. So 3 context switches are required.