Q3. UGC NET Dec 2018 To overcome difficulties in Readers-Writers problem, which of the following statement/s is/are true? 1. Writers are given exclusive access to shared objects. 2. Readers are given exclusive access to shared objects. 3. Both Readers and Writers are given exclusive access to shared objects. Choose the correct answer from the code […]

### Category: UGC NET CS

## Suppose a system has 12 instances | UGC NET Dec 2018

Q2. UGC NET Dec 2018 : Suppose a system has 12 instances of some resource with n processes competing for that resource. Each process may require 4 instances of the resources. The maximum value of n for which the system never enters into deadlock is a) 3 b) 4 c) 5 d) 6 Sol. What […]

## Data warehouse | UGC NET Dec 2018

Q1. UGC NET Dec 2018 Data warehouse contains _____ data that is never found in operational environment. a) Summary b) Encoded c) Encrypted d) Scripted Sol. A data warehouse is a large collection of business data used to help an organization make decisions. In data warehouse summaries are very valuable because they are pre-compute long operations […]

## UGC NET CS Paper 2 June 2012

UGC NET CS Paper 2 June 2012 1. The postfix expression AB + CD – * can be evaluated using a (A) Stack (B) Tree (C) Queue (D) Linked list Answer: (A)

## UGC NET November 2017 Paper II

## Net 52

CBSE NET Q. Convert the following infix expression into its equivalent post fix expression (A + B^ D) / (E – F) + G (A) ABD^ + EF – / G+ (B) ABD + ^EF – / G+ (C) ABD + ^EF / – G+ (D) ABD^ + EF / – G+ Ans: (A) Explanation: […]

## Net 51

CBSE NET JUNE 2015 Paper II Q. The in-order and pre-order Traversal of binary Tree are dbeafcg and abdecfg respectively. The post-order Traversal is______. A)dbefacg B)debfagc C)dbefcga D)debfgca Ans: (D) Explanation: Inorder (Left, Root, Right),Preorder (Root, Left, Right),Postorder (Left, Right, Root) Given, in-order=dbeafcg, pre-order=abdecfg. Than, binary tree from in-order and pre-order become- Post-order from above […]

## Net 50

CBSE NET JUNE 2015 Paper II Q. A disk drive has 100 cylinders,numbered 0 to 99. Disk request come to the disk driver for cylinders 12,26,24,4,42,8,50 in that order. The driver is currently serving a request at cylinder 24. A seek takes 6msec per cylinder moved. How much seek time is needed for the shortest […]

## Net 49

CBSE NET Q. A LRU page replacement is used with four page frames and eight pages. How many page faults will occur with the reference string 0172327103. if the four frames are initially empty. A) 6 B) 7 C) 5 D) 8. Ans: (B) Explanation: LRU (Least recently used) algorithm. Here, F shows page faults […]

## Net 48

CBSE NET Q. Consider an array A[20, 10], assume 4 words per memory cell and the base address of array A is 100. What is the address of A[11, 5] ? Assume row major storage. (A) 560 (B) 565 (C) 570 (D) 575 Ans: (A) Explanation: Formula for calculating address of element in row major […]

## Net 47

CBSE NET Q. An analog signal has a bit rate of 6000 bps and a baud rate of 2000 baud. How many data elements are carried by each signal element ? (A) 0.336 bits/baud (B) 3 bits/baud (C) 120,00,000 bits/baud (D) None of the above Ans: (B) Explanation: A same signal has bit rate of […]

## Net 46

CBSE NET Q. In a demand paging memory system, page table is held in registers. The time taken to service a page fault is 8 m.sec. if an empty frame is available or if the replaced page is not modified, and it takes 20 m.secs., if the replaced page is modified. What is the average […]

## Net 45

CBSE NET DECEMBER 2012 PAPER II Q. Given an empty stack, after performing push (1), push (2), Pop, push (3), push (4), Pop, Pop, push (5), pop, what is the value of the top of the stack? (A) 4 (B) 3 (C) 2 (D) 1 Ans: (D) Explanation: See the image below:

## Net 44

CBSE NET JUNE 2012 PAPER II Q. Consider the following page trace : 4,3, 2, 1, 4, 3, 5, 4, 3, 2, 1, 5 Percentage of page fault that would occur if FIFO page replacement algorithm is used with number of frames for the JOB m = 4 will be (A) 8 (B) 9 (C) […]

## Net 43

CBSE NET JUNE 2012 PAPER II Q. Page Shift Keying (PSK) Method is used to modulate digital signal at 9600 bps using 16 level. Find the line signals and speed (i.e. modulation rate). (A) 2400 bauds (B) 1200 bauds (C) 4800 bauds (D) 9600 bauds Ans: (A) Explanation: To calculate baud rate, modulation have to […]

## Net 42

CBSE NET JUNE 2012 PAPER II Q. The post order traversal of a binary tree is DEBFCA. Find out the preorder traversal. (A) ABFCDE (B) ADBFEC (C) ABDECF (D) None of the above Ans: (C) Explanation: DFD – Inorder (Left,Root,Right), Preorder (Root,Left,Right), Postorder (Left,Right,Root). So, POSTORDER to BINARY TREE will like this. Than BINARY TREE […]

## Net 41

CBSE NET JUNE 2012 PAPER II Q. The postfix expression AB + CD – * can be evaluated using a (A) Stack (B) Tree (C) Queue (D) Linked list Ans: (A) Explanation: Stack method is used to convert infix to postfix and postfix to infix. For example, see image, to learn how to convert infix […]

## Net 40

CBSE NET DECEMBER 2013 PAPER II Q. Consider a disk queue with request for input/output to block on cylinders 98, 183, 37, 122, 14, 124, 65, 67 in that order. Assume that disk head is initially positioned at cylinder 53 and moving towards cylinder number 0. The total number of head movements using Shortest Seek […]

## Net 39

CBSE NET DECEMBER 20133 PAPER II OPERATING SYSTEM Q. Consider a pre-emptive priority based scheduling algorithm based on dynamically changing priority. Larger priority number implies higher priority. When the process is waiting for CPU in the ready queue (but not yet started execution), its priority changes at a rate a = 2. When it starts […]

## Net 38

CBSE NET Q. Consider the following transportation problem. The transportation cost in the initial basic feasible solution of the above transportation problem using Vogel’s Approximation method is: (A) 1450 (B) 1465 (C) 1480 (D) 1520 Ans. (B) Explanation: In VAM, Row difference and column difference is calculated. To calculate row difference, difference of 2 smallest […]

## Net 37

CBSE NET June 2011 Q. The number of different trees with 8 nodes is (A) 256 (B) 255 (C) 248 (D) None of the above Ans :- (C) Explanation:- Use formula 2n-n. Here the value of n=8. So, the formula would become 28-8=256-8=248.

## Net 36

CBSE NET December 2010 PAPER II Q. A binary tree with 27 nodes has _______ null branches. (A) 54 (B) 27 (C) 26 (D) None of the above Ans :- (D) Explanation:- A binary tree with n nodes has n+1 null branches. So a binary tree with 27 nodes will have 28 null branches.

## Net 35

CBSE NET December 2010 PAPER II Q. What is the maximum number of nodes in a B-tree of order 10 of depth 3 (root at depth 0) ? (A) 111 (B) 999 (C) 9999 (D) None of the above Ans :- (C) Explanation:- The formula for calculating the maximum number of nodes in a B-tree […]

## Net 34

CBSE NET December 2015 PAPER II Q.Which of the following property/ies a Group G must hold, in order to be an Abelian group? (a) The distributive property (b) The commutative property (c) The symmetric property Codes: (A) (a) and (b) (B) (b) and (c) (C) (a) only (D) (b) only Ans :- (D) Explanation:- A […]

## Net 33

CBSE NET July 2016 PAPER II Q. How many committees of five people can be chosen from 20 men and 12 women such that each committee contains at least three women? (A) 75240(B) 52492(C) 41800(D) 9900 Ans :- (A) Explanation:- At least 3 women means, minimum 3 women should be in 5 people group. So, […]

## Net 32

CBSE NET July 2016 PAPER II Q. What is the value returned by the function f given below when n=100 ? int f (int n) { if (n==0) then return n; else return n + f(n-2); } (A) 2550 (B) 2556 (C) 5220 (D) 5520 Ans :- (A) Explanation:- It’s a recursive function. Its gives […]

## Net 31

CBSE NET July 2016 PAPER II Q. In a Positive edge triggered JK flip-flop, if J and K both are high then the output will be _______ on the rising edge of the clock. (A) No change (B) Set (C) Reset (D) Toggle Ans :- (D) Explanation:- State table of JK Flip Flop – J […]

## Net 29

CBSE NET July 2016 PAPER II DATA STRUCTURE Q. Which of the following logic expressions is incorrect ? (A) 1⊕0 = 1 (B) 1⊕1⊕1 = 1 (C) 1⊕1⊕0 = 1 (D) 1⊕1 = 0 Ans :- (C) Explanation:- In XOR gate, in case of 2 inputs, both the input should be different and in case […]

## Net 30

CBSE NET July 2016 PAPER II Q. The Simplified form of a Boolean equation (AB’+AB’C+AC)(A’C’ +B’) is (A) 0 (B) 1 (C) -1 (D) -2 Ans :- (A) Explanation:- (AB’+AB’C+AC)(A’C’ +B’) On simplifying above equation, (AB’A’C’+AB’CA’C’+ACA’C’+AB’B’+AB’CB’+ACB’) (0+0+0+AB’+AB’C+ACB’) Because multiplication of element with its complement is 0. (AB’+AB’C) AB’(1+C) AB’

## Net 28

CBSE NET January 2017 PAPER II OPERATING SYSTEM Q. There are three processes P1, P2 and P3 sharing a semaphore for synchronising a variable. Initial value of semaphore is one. Assume that negative value of semaphore tells us how many processes are waiting in queue. Processes access the semaphore in following order:(a) P2 needs to […]

## Net 26

CBSE NET JUNE 2013 PAPER III Q. An operating system using banker’s algorithm for deadlock avoidance has ten dedicated devices (of same type) and has three processes P1, P2 and P3 with maximum resource requirements of 4, 5 and 8 respectively. There are two states of allocation of devices as follows : State 1 Processes […]

## Net 27

CBSE NET January 2017 PAPER II DATA STRUCTURE Q. The seven elements A, B, C, D, E, F and G are pushed onto a stack in reverse order, i.e., starting from G. The stack is popped five times and each element is inserted into a queue. Two elements are deleted from the queue and pushed […]

## Net 25

CBSE NET DECEMBER 2008 PAPER II Which level is called as “defined” in capability maturity model? A) level 0B) level 3C) level 4D) level 1 Ans:- BExplanation:For explanation click here.

## Net 24

CBSE NET JUNE 2009 PAPER II Capability Maturity Model is meant for: A) ProductB) ProcessC) Product and processD) None of the above Ans:- B Explanation:For explanation click here.

## Net 23

CBSE NET JUNE 2012 PAPER III Key process areas of CMM level 4 are also classified by a process which is (A) CMM level 2 (B) CMM level 3 (C) CMM level 5 (D) All of the above Ans:- B Explanation:For explanation click here.

## Net 22

CBSE NET DECEMBER 2012 PAPER II The maturity levels used to measure a process are (A) Initial, Repeatable, Defined, Managed, Optimized.(B) Primary, Secondary, Defined,Managed, Optimized. (C) Initial, Stating, Defined, Managed, Optimized. (D) None of the above Ans:- A Explanation:For explanation click here.

## Net 21

CBSE NET JUNE 2014 PAPER III Which one of the following is not a key process area in CMM level 5 ? (A) Defect prevention (B) Process change management(C) Software product engineering (D) Technology change management Ans:- C Explanation:For explanation click here.

## Net 20

CBSE NET JUNE 2014 PAPER II KPA in CMM stands forA) Key Process AreaB) Key Product AreaC) Key Principal AreaD) Key Performance Area Ans:- A Explanation:For explanation click here.

## Net 17

CBSE NET JUNE 2013 PAPER II Cyclomatic complexity of a flow graph G with n vertices and e edges is A)V(G)=e+n-2B)V(G)=e-n+2C)V(G)=e+n+2D)V(G)=e-n-2 Ans: (B) Explanation:To solve above problem, first remember these 3 rules to compute the cyclomatic complexity.1. The number of regions correspond to the cyclomatic complexity.2. Cyclomatic complexity V(G) for a flow graph G, is defined […]

## Net 18

CBSE NET JUNE 2012 PAPER III Which one of the following statements is incorrect ? (A) The number of regions corresponds to the cyclomatic complexity.(B) Cyclometric complexity for a flow graph G is V(G) = N – E + 2, where E is the number of edges and N is the number of nodes in […]