UGC NET Computer Science 2018 Previous Year Questions (PYQs) – Page 1 of 3

A statement in mathematical logic is a declarative sentence which has a definite truth value, either true or false.

$(i)$ There will be snow in January.
This is a declarative sentence. So, it is a statement.

$(ii)$ What is the time now?
This is a question. So, it is not a statement.

$(iii)$ Today is Sunday.
This is a declarative sentence. So, it is a statement.

$(iv)$ You must study Discrete Mathematics.
This is a command or advice. So, it is not a statement.

Therefore, only $(i)$ and $(iii)$ are statements.

Equivalence is represented by

$p\Leftrightarrow q$

So,

$(a)\rightarrow(iv)$

Contrapositive of $p\Rightarrow q$ is

$\sim q\Rightarrow \sim p$

So,

$(b)\rightarrow(iii)$

Converse of $p\Rightarrow q$ is

$q\Rightarrow p$

So,

$(c)\rightarrow(ii)$

Implication is represented by

$p\Rightarrow q$

So,

$(d)\rightarrow(i)$

Therefore, correct matching is

$(a)-(iv),\ (b)-(iii),\ (c)-(ii),\ (d)-(i)$

Total balls are

$6+4=10$

Number of ways to select $4$ balls from $10$ balls is

${}^{10}C_4$

We need exactly $2$ red balls and $2$ green balls.

Number of ways to select $2$ red balls from $6$ red balls is

${}^{6}C_2$

Number of ways to select $2$ green balls from $4$ green balls is

${}^{4}C_2$

So, required probability is

$\dfrac{{}^{6}C_2 \times {}^{4}C_2}{{}^{10}C_4}$

$=\dfrac{15 \times 6}{210}$

$=\dfrac{90}{210}$

$=\dfrac{3}{7}$

Let Bus, Train and Automobile be represented by $B,T,A$ respectively.

Given,

$n(B)=30$

$n(T)=35$

$n(A)=100$

$n(B\cap T)=15$

$n(B\cap A)=15$

$n(T\cap A)=20$

$n(B\cap T\cap A)=5$

Using the formula,

$n(B\cup T\cup A)=n(B)+n(T)+n(A)-n(B\cap T)-n(B\cap A)-n(T\cap A)+n(B\cap T\cap A)$

Now,

$n(B\cup T\cup A)=30+35+100-15-15-20+5$

$=165-50+5$

$=120$

Therefore, $120$ people completed the survey form.

Statement $(i)$ is true.

NAND and NOR gates are functionally complete gates.

So, every logic network can be represented using only NAND gates or only NOR gates.

Statement $(ii)$ is true.

Boolean expressions and logic networks can be represented using labelled acyclic digraphs.

Statement $(iii)$ is false.

Any two Boolean algebras with the same number of atoms are isomorphic.

Statement $(iv)$ is false.

In a finite Boolean algebra, every non-zero element can be uniquely expressed as a join of atoms.

Therefore, only $(i)$ and $(ii)$ are true.

Given Boolean expression is

$\overline{A}\cdot B+A\cdot \overline{B}+A\cdot B$

Now, group the last two terms:

$= \overline{A}\cdot B+A(\overline{B}+B)$

Since,

$\overline{B}+B=1$

So,

$= \overline{A}\cdot B+A$

Using Boolean identity,

$A+\overline{A}B=A+B$

Therefore,

$\overline{A}\cdot B+A\cdot \overline{B}+A\cdot B=A+B$

In PERT/CPM network, a merge event is an event where two or more activities come together.

It represents the joining of two or more events or activities.

Therefore, the merge event represents joining.

Given objective function is

$z=-2x_1-3x_2$

To maximize $z=-2x_1-3x_2$, we need to minimize the negative loss.

The main lower constraint is

$x_1+x_2\geq 2$

Now check the feasible corner points on $x_1+x_2=2$:

If $x_2=0$, then

$x_1=2$

So, point is $(2,0)$.

At $(2,0)$,

$z=-2(2)-3(0)$

$z=-4$

If $x_1=0$, then

$x_2=2$

So, point is $(0,2)$.

At $(0,2)$,

$z=-2(0)-3(2)$

$z=-6$

Since $-4>-6$, maximum value is $-4$.

Also, $(2,0)$ satisfies the other constraints:

$2x_1+x_2=2(2)+0=4\leq 10$

$x_1+x_2=2+0=2\leq 8$

Therefore, optimal solution is

$x_1=2,\ x_2=0,\ z=-4$

In computers, subtraction is generally performed using $2$'s complement method.

For subtracting $B$ from $A$, the computer takes the $2$'s complement of $B$ and adds it to $A$.

This makes subtraction possible using the same binary adder circuit used for addition.

Therefore, subtraction in computers is generally carried out by $2$'s complement.