UGC-NET Computer Science Neural Networks, Artificial Intelligence, Fuzzy Logic and Predicate Logic Previous Year Questions

UGC-NET COMPUTER SCIENCE Software Engineering Previous Year Questions

UGC-NET COMPUTER SCIENCE Software Engineering Previous Year Questions

UGC-NET COMPUTER SCIENCE Software Engineering Previous Year Questions

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ugc net computer science december 2008 question papers

UGC-NET Computer Science UNIX Questions with Explanation

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q : 1 Which of the following is incorrect related to UNIX?
(A) "$ chmod +r emp1" and "$ chmod a+r emp1" are equivalent
(B) "$ chmod +x emp1" and "$ chmod a+x emp1" are equivalent
(C) "$ chmod +w emp1" and "$ chmod a+w emp1" are equivalent
(D) None of the above

Answer: (C)

chmod: Changing File Permissions
The default security feature provided by UNIX write protects a file from all except the owner of the file. Generally, all users have read access by default, but it could be different on each system. We can change the default permissions by umask command. 

Structure of chmod command contains three components:
1. Category of user (owner, group or others)
2. Operation to be performed (assign or remove a permission)
3. Permission type (read, write, execute)

Abbreviations used by chmod:

chmod Usage

Hence option (C) is incorrect.

UGC-NET Computer Science Data Structures Questions with Explanation

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q : 6 A full binary tree with n leaves contains

(A) n nodes
(B) log₂n nodes
(C) 2n –1 nodes
(D) 2ⁿ nodes

Answer : (C)

Explanation :

A full binary tree is one in which each node has exactly two children. Like on given below is a full binary tree with 15 nodes:

No. of internal nodes are given by 'No. of External or Leaf nodes - 1'.
In our question we have to find total no. of nodes in the tree. We are given that total no. of leaf nodes are 'n'. Total no. of nodes in tree = Total No. of Leaf Nodes + Total No. of Internal Nodes
=> Total no. of nodes in tree = n + n - 1 = 2n-1.
We can examine this by putting the no. leaf nodes of the above tree in given formula which is 8.
Total no. nodes in tree = 2 * 8 - 1 = 16 - 1 = 15.

Hence correct option is (C).

This question was taken from (UGC-NET Computer Science December 2014 Paper 2).

UGC-NET Computer Science Data Structures Questions with Explanation

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q : 4 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

(UGC-NET Computer Science December 2014)


Answer: (A)

Explanation:

In computers memory the array elements are always stored linearly whether they are 1-Dimensional or 2-Dimensional. Like this is an example of 2-Dimensional array stored in memory linearly:

As we can see element A11 is stored at location '0' in memory and it is known as its offset.
Step 1:
Now to calculate the offset in case of Row major order we will use the formula:
Offset = (Row No. * Total No. of Columns) + Column No.

Now to calculate the offset in case of Column major order we will use the formula:
Offset = (Column No. * Total No. of Rows) + Row No.

'Row No.' is the row number of the location whose address we need to calculate.
'Column No.' is the column number of the location whose address we need to calculate.
'Total No. of Columns' is the total number of columns in the given matrix.
'Total No. of Rows' is the total number of rows in the given matrix.

Step 2:

Multiply the offset calculated with the size of each memory word.

Step 3:

Add the Base address to calculate the effective address.

Given matrix is A[20][10], Size of memory word is 4 bytes, address of the location to find A[11][5], Base address is 100.

Step 1:
Offset = (11 * 10) + 5 = 115.
Step 2:
115 * 4 = 460.
Step 3:
100 + 460 = 560.

So, correct option is (A) 560.

This Question was taken from UGC-NET Computer Science December 2014.

UGC-NET Computer Science Operating Systems Questions with Explanation

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q : 1 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 access time to service a page fault assuming that the page to be replaced is modified 70% of the time ?

(A) 11.6 m.sec.

(B) 16.4 m.sec.
(C) 28 m.sec.

(D) 14 m.sec.

((UGC-NET Computer Science December 2014 Paper 2))

Answer: (B)

Explanation:

Time to serve  page fault if frame is empty or page replaced is not modified = 8 m.sec

Time to serve  page fault if page is modified = 20 m.sec

Frequency of Page modification  = 70% = 70/100 = 0.7

Frequency when page is not modified or frame is empty = (100-70)% = 30% = 30/100 = 0.3

Average access time = (0.3 * 8) + (0.7 * 20) = 2.4 + 14 = 16.4 m.sec

Hence correct option is (B) 16.4 m.sec.

This question was taken from UGC-NET Computer Science December 2014 Paper 2.

UGC-NET Computer Science Discrete Mathematics Questions with Explanation

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q : 1 Consider a set A = {1, 2, 3, …….., 1000}. How many members of A shall be divisible by
3 or by 5 or by both 3 and 5 ?

(UGC-NET Computer Science December 2014 Paper 2)

(A) 533
(B) 599
(C) 467
(D) 66

Answer: (C)

Explanation:

Given Set is A = {1, 2, 3, …….., 1000}.
The members of 'A' which are divisible by 3 are given by {3,6,9,12...,999} these are total 333 elements.
The members of 'A' which are divisible by 5 are given by {5,10,15,20,...,1000} these are total 200 elements.
The members of 'A' which are divisible by both 3 and 5 i.e. divisible by 15 are given by {15,30,45,...,990} these are total 66 elements.

So, we have to find How many members of A shall be divisible by
3 or by 5 or by both 3 and 5?
i.e. No. of elements divisible by 3 +  No. of elements divisible by 5 -  No. of elements divisible by 15. 
(Union operation n(X union Y)=n(X)+n(Y) - n(X intersection Y))

=> 333 + 200 - 66=533 - 66 = 467.

This question was taken from UGC-NET Computer Science December 2014 Paper 2.

UGC-NET Computer Science C Language Programming Questions with Explanation

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q:1 What is the result of the following expression?
        (1 & 2) + (3 & 4)
(A) 1
(B) 3
(C) 2
(D) 0

Answer: (D)

(UGC-NET Computer Science December 2012)

Explanation:

In this question the operator '&' is bitwise AND operator So, it will work on individual bits.
Binary of 1 in 8 bits will be : 00000001
Binary of 2 in 8 bits will be : 00000010
Binary of 3 in 8 bits will be : 00000011
Binary of 4 in 8 bits will be : 00000100

   1 : 00000001
   2 : 00000010
1&2: 00000000

   3 : 00000011
   4 : 00000100

3&4: 00000000

1&2 results in 0 in decimal number system.
3&4 results in 0 in decimal number system.

(1 & 2) + (3 & 4) = 0+0=0.
So, correct option is (D).

This question was from UGC-NET Computer Science December 2012.

UGC-NET Computer Science Data Structures Questions with Explanation

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q:3 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

(UGC-NET Computer Science December 2012)

Answer: (D)

Explanation:

An empty stack will look like this:

After performing Push(1):

1

1 will be on TOP.

After performing Push(2):

2
1

2 will be on TOP.

After performing POP:

1

Again 1 will be on TOP.

After performing Push(3), Push(4):

4
3
1

After performing POP, POP:

1

1 will be on TOP.

After performing Push(5), POP again 1 will be on TOP of the Stack. Hence option (D) is Correct.

This question was taken from UGC-NET Computer Science December 2012.

UGC-NET Computer Science Data Structures Questions with Explanation

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q: 2  A hash function f  defined as f(key) = key mod 7, with linear probing it is used to insert the key 37,38,72,48,98,11,56 into a table index from 0 to 6. What will be the locations of 11 :

(A) 3
(B) 4
(C) 5
(D) 6

(UGC-NET Computer Science December 2004, December 2009)

Answer: (C)

Explanation: According to question f(key) = key mod 7. So, key will be inserted at after the result of key % 7. It will be 0,1,2,3,4,5 or 6. Because if we divide a number with 7 these are the possible remainders. Now in our given question the keys to be inserted are in order 37,38,72,48,98,11,56. Beginning with  37 it will be inserted at 37 % 7 = 2 i.e. location 2 in the table.
Next is 38, it will be inserted at 38 % 7 = 3 i.e. location 3 in the table.
Next is 72, it will be inserted at 72 % 7 = 2 but this location is already occupied by 37 So, by using linear probing we will examine next location which is 3 but it is also occupied by 38 So, we will examine next location which is 4 and it is empty hence 72 will be inserted at location 4 in the table.
Next is 48, it will be inserted at 48 % 7 = 6 i.e. location 6 in the table.
Next is 98, it will be inserted at 98 % 7 = 0 i.e. location 0 in the table.
Next is 11, it will be inserted at 11 % 7 = 4 but this location is already occupied by 72 So, we will examine next location which is 5 and also empty so 11 will be inserted at location 5 in the table.
Next is 56, it will be inserted at 56 % 7 = 0 but this location is already occupied So by linear probing method it will be inserted at next location i.e. 1.
The table will look like this after all keys are inserted.

98	56	37	38	72	11	48
0	1	2	3	4	5	6

UGC-NET Computer Science Data Structures Questions with Explanation

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q:1 The number of different trees with 8 nodes is

(A) 256
(B) 255
(C) 248
(D) None of these

(June 2011 Paper 2)

Answer: (C) 248

Explanation: The number of different trees with 'n' nodes is given by the formula 2 ⁿ - n. So, the number of trees with 8 nodes will be 2⁸ - 8 i.e. 256 - 8 = 248. Hence (C) option is correct.