HSSC PGT Computer Science Solution of Question Paper March 2016

Ans 40: (C) 12,20,25

Solution: In a Poset, Maximal element is that which is not related to any other element.

Here Operation is Divisibility.

2 divides 4,10,12,20

5 divides 10, 20, 25

4 divides 12,20

10 divides 20

but 12,20,25 does not divide any other element in the set. Hence it is Maximal element.

Extra Information: Minimal element is which is not divisible by any other element in the set.

Here 2 and 5 are minimal elements

Ans 41: (D) Partial Order Relation

Solution: 1. Reflexive: Since A is a subset of A. Hence Reflexive

2. Symmetric: Since A is a subset of B and B is a subset of A is possible only if A=B.

So its not symmetric but its anti symmetric.

3. Transitive: Since if A is a subset of B and B is a subset of C implies A is a subset of C.

Hence it is transitive

So It is Partial order Relation

Ans 42: (C)

Solution: Since the operation given is Multiplication.

So, Multiplicative inverse of -i

=1/-i

=-1/i

=i²/i

=i

Ans 43: (D) -2³¹ to 2³¹-1

Solution: Here 1 bit will be used to represent sign.

Remaining bits=31

So range of numbers will be 2³¹

But Zero is included in positive numbers.

So range of positive numbers will be 1 less than 2³¹

Finally range is: -2³¹ to 2³¹-1

Ans 44: 2n-1

Solution: Full binary tree has either 0 or 2 children

So n leaves means total 2n-1 nodes

Ans 45: (D) 1 or 3

Solution: In a full binary tree, only one vertex, namely, the root is of even degree (namely 2).

Each of the other (n-1) vertices is of odd degree (namely 1 or 3.)

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