# UGC NET COMPUTER SCIENCE SOLVED PAPERS 2014-16 - UGC NET Computer Science Paper 2 December 2015

>>>>>>>>UGC NET Computer Science Paper 2 December 2015

• A

Prototyping  • B

Iterative  • C

Time-boxing  • D

Glass-boxing  • Option : D
• Explanation :
Prototyping, iterative, are well known software process model. In time boxing model, development is done iteratively as in the iterative enhancement model but in time boxing model, each iteration is done in a timebox of fixed duration. Glass boxing model is a software testing pattern. So, option (D) is correct

• A

4960  • B

2600  • C

23751  • D

8855  • Option : B
• Explanation :
x ≥ 1, y ≥ 2, z ≥ 3, we have to subtract these constraints from total number of choices: i.e. 29-(1+2+3+0) = 23 (23 + 4 - 1)C23 = 2600 So, option (B) is correct.

• A

32 GB  • B

64 GB  • C

16 GB  • D

1 GB  • Option : C
• Explanation :
block = 210
block pointer size = 4B
entries possible in block = 210/22 = 256
direct pointer gives = 10 * 256 = 10 blocks
single indirect gives = 256 * 256 = 28 blocks
double indirect gives = 256 * 256 * 256 = 216 blocks
triple indirect gives = 256 * 256 * 256 * 256 = 224 blocks
total = 10 blocks + 28 blocks + 216 blocks + 224 blocks
= 16843018 blocks = 16843018 * 1024 = 17247250432 ≈ 16 GB

• A

M-Banking  • B

E-Banking  • C

O-Banking  • D

C-Banking  • Option : B
• Explanation :
E-banking uses electronic means to transfer funds directly from one account to another rather than by cheque or cash. E-banking is faster then other modes of banking and it saves time as well as resources. It is much safer then cash and cheque and payment gateway is present on the same platform where the market place available. In this mode merchant and bank have their own agreement of privacy and transaction rates. So, option (B) is correct.

• A

2 3 4 6 7 13 15 17 18 18 20  • B

20 18 18 17 15 13 7 6 4 3 2  • C

15 13 20 4 7 17 18 2 3 6 18  • D

2 4 3 13 7 6 15 17 20 18 18  • Option : D
• Explanation :
In inorder traversal first we traverse left node then root node and then right node: In the following tree we first go to the leftmost node then its root after that right i.e. 2 4 3 13 7 6 15 17 20 18 18. In rest of the option inorder property is violating. So, option (D) is correct.