# UGC-NET CS 2017 Nov – II

 Question 1
If the time is now 4 O’clock, what will be the time after 101 hours from now ?
 A 9 O’clock B 8 O’clock C 5 O’clock D 4 O’clock
Clocks    UGC-NET CS 2017 Nov - II
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Question 1 Explanation:
We know that there are 24 hours in a day, so we will take mod for 101 hours ie (101 mod 24) = 5. So residue is 5. On adding these remaining 5 hours to the presenet time will give 9 O'CLOCK time.
 Question 2
Let m=(313)4 and n=(322)4. Find the base 4 expansion of m+n.
 A (635)4 B (32312)4 C (21323)4 D (1301)4
Digital Logic & Number representation    Number Representation    UGC-NET CS 2017 Nov - II
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Question 2 Explanation:
We have m=(313)4 and n=(322)4 convert m and n into decimal:
m = 3*42 + 1*41 + 3*40
m = 48 + 4 + 3
m = 55.
Now n=3*42 + 2*41 +2*40
n = 48 + 8 +2
n = 58.
m + n = 55 + 58
m + n = 113
Now we have to convert 113 in to base 4:
ie step 1-113 % 4 = 1
113 / 4 = 28
step 2- 28 % 4 = 0
28 / 4 = 7
step 3- 7 % 4  = 3
7 / 4  = 1
step 4- 1 % 4  = 1
1 / 4  --> will not divide it in quant.
So we have to stop here.
The answer will be Residue from step 4 to step 1 inorder ie 1301 Ans-(1301)4
 Question 3
Find the boolean product A⊙B of the two matrices.
 A (1) B (2) C (3) D (4)
Linear Algebra    UGC-NET CS 2017 Nov - II
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 Question 4
How many distinguishable permutations of the letters in the word BANANA are there ?
 A 720 B 120 C 60 D 360
Permutation and Combination    UGC-NET CS 2017 Nov - II
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Question 4 Explanation:
In BANANA we have six letters in total but here we have some duplicate letters too so we have to deal with it and have to remove those duplicate case. B - 1 A - 3 N - 2 So total no of words possible is factorial(6) ie 6! but we must remove duplicate words: ie- (6!/(2!*3!)) which gives 60 So 60 distinguishable permutation of the letters in BANANA. So, option (A) is correct.
 Question 5
Consider the graph given below : Use Kruskal’s algorithm to find a minimal spanning tree for the graph. The List of the edges of the tree in the order in which they are choosen is? (1) AD, AE, AG, GC, GB, BF (2) GC, GB, BF, GA, AD, AE (3) GC, AD, GB, GA, BF, AE (4) AD, AG, GC, AE, GB, BF
 A (1) B (1),(2) C (1),(2),(3) D (1),(2),(3),(4)
Graph Theory    UGC-NET CS 2017 Nov - II
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Question 5 Explanation:
Using Kruskal's algorithm for minimal spanning tree: First select the minimum weight edge ie 2 we have two options GC and AD Now select Next edge of minimum weight: for GC we have to select GB or AD and so on select minimum weight keep in mind no vertex left to visit. Here all option are matching with minimum spanning tree. So, option (D) is correct.
 Question 6
The Boolean function with the Karnaugh map is:
 A (A+C).D+B B (A+B).C+D C (A+D).C+B D (A+C).B+D
Digital Logic & Number representation    UGC-NET CS 2017 Nov - II    Logic functions and Minimization
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Question 6 Explanation:
To find the boolean function here we will use K-map Select neighbouring Minterms as max as possible in powers of 2 ie 2 ,4,8 and so on. Here we have selected group of minterms: First we will take 8 minterms group which will give us B. Now take 4 minterms group it will give us CD. Another 4 minterms group will give us AD. We have the expression B + CD + AD. which can be reduced as B + D ( A + C ) So, option (A) is correct.
 Question 7
The Octal equivalent of the binary number 1011101011 is :
 A 7353 B 1353 C 5651 D 5657
Number Representation    UGC-NET CS 2017 Nov - II
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Question 7 Explanation:
First we will make pair of 3 bits from LSB : ie- 1 011 101 011 now convert these bits into decimal and that will be 1 - 1 011 - 3 101 - 5 011 - 3 so the answer will be 1353. So, option (B) is correct.
 Question 8
Let P and Q be two propositions, ¬ (P ↔ Q) is equivalent to: (I) P ↔ ¬ Q (II) ¬ P ↔ Q (III) ¬ P ↔ ¬ Q (IV) Q → P
 A Only (I) and (II) B Only (II) and (III) C Only (III) and (IV) D None of the above
Propositional and First Order Logic.    UGC-NET CS 2017 Nov - II
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Question 8 Explanation:
We know that ¬(P ↔ Q) = (¬P ↔ Q) = (P ↔ ¬Q) = ¬(¬P ↔ ¬Q) Only statement (I) and (II) are correct. So, option (A) is correct.
 Question 9
‘ptrdata’ is a pointer to a data type. The expression *ptrdata++ is evaluated as (in C++) :
 A *(ptrdata++) B (*ptrdata)++ C *(ptrdata)++ D Depends on compiler
C Pointer Basics    C Advanced Pointer    UGC-NET CS 2017 Nov - II
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Question 9 Explanation:
Here ++ (unary operator) have high precedence than * (unary operator). So ++ will be evaluated first then * will be in action. So A is the correct option. So, option (A) is correct.
 Question 10
The associativity of which of the following operators is Left to Right, in C++ ?
 A Unary Operator B Logical not C Array element access D addressof
C++ Misc    UGC-NET CS 2017 Nov - II
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Question 10 Explanation:
Array element access has left to right associativity in C++. So, option (C) is correct.
There are 50 questions to complete.

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