UGC-NET CS 2017 Nov – II

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.

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

Here we have to follow two basics: 1- Matrics multiplication rule- For multiplication of two matrics, column in first matrics must be equal to rows in second matrics. 2- We have to find boolean product of two matrics so, we must know boolean product ie No we have A= and B= Now the product AB= AB= So, option (A) is correct.

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.

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.

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.

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.

We know that ¬(P ↔ Q) = (¬P ↔ Q) = (P ↔ ¬Q) = ¬(¬P ↔ ¬Q) Only statement (I) and (II) are correct. So, option (A) is correct.

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.

Array element access has left to right associativity in C++. So, option (C) is correct.