UGC NET Computer Science 2019 Previous Year Questions (PYQs) – Page 1 of 20

In divisibility poset, least element means an element which divides every element.

Here, $3$ does not divide $5$, so $3$ is not the least element.

Similarly, no other element divides all elements of the set.

So, least element does not exist.

Greatest element means an element which is divisible by every element.

Here, $45$ is not divisible by $24$.

Also, no other element is divisible by all elements of the set.

So, greatest element does not exist.

Therefore, there does not exist a greatest element and a least element.

Let the number of balls in four distinguishable bins be $x_1,x_2,x_3,x_4$.

Then,

$x_1+x_2+x_3+x_4=8$

Here, balls are indistinguishable and bins are distinguishable.

So, the number of non-negative integral solutions is

${}^{n+r-1}C_{r-1}$

Here, $n=8$ and $r=4$.

So,

${}^{8+4-1}C_{4-1}={}^{11}C_{3}$

$=\dfrac{11 \times 10 \times 9}{3 \times 2 \times 1}$

$=165$

Total length of bit string is $10$.

Let $A$ be the set of bit strings starting with $1$.

So, first bit is fixed and remaining $9$ bits are free.

$n(A)=2^9=512$

Let $B$ be the set of bit strings ending with $00$.

So, last two bits are fixed and remaining $8$ bits are free.

$n(B)=2^8=256$

Now, strings which start with $1$ and end with $00$:

First bit is fixed as $1$ and last two bits are fixed as $00$.

Remaining $7$ bits are free.

$n(A \cap B)=2^7=128$

Therefore,

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

$=512+256-128$

$=640$

Given,

$V=6$

Each vertex has degree $4$.

So, sum of degrees is

$6 \times 4=24$

By handshaking theorem,

$2E=24$

$E=12$

For a connected planar graph, Euler’s formula is

$V-E+R=2$

Substitute $V=6$ and $E=12$.

$6-12+R=2$

$R=8$

In a complete bipartite graph $K_{m,n}$, vertices are divided into two parts.

A Hamilton circuit must visit every vertex exactly once and return to the starting vertex.

In a bipartite graph, every cycle alternates between the two parts.

So, for a Hamilton circuit to include all vertices, both parts must have the same number of vertices.

Therefore,

$m=n$

Also, a cycle needs at least $2$ vertices in each part.

So,

$m,n \ge 2$

Hence, $K_{m,n}$ has a Hamilton circuit when

$m=n,\ m,n \ge 2$

Given statement is

$[(p \vee q)\wedge[p \to q]]$

We know that

$p \to q \equiv \neg p \vee q$

So,

$[(p \vee q)\wedge[p \to q]]=(p \vee q)\wedge(\neg p \vee q)$

Now truth table se false cases:

When $p=0,\ q=0$, expression false hoti hai.
Corresponding maxterm is

$p \vee q$

When $p=1,\ q=0$, expression false hoti hai.
Corresponding maxterm is

$\neg p \vee q$

Therefore, PCNF is

$(p \vee q)\wedge(\neg p \vee q)$

Hence, correct PCNF is $(p \vee q)\wedge(\neg p \vee q)$.

A standard deck has

$13$ hearts

and

$52-13=39$ non-heart cards.

To guarantee at least $3$ hearts, consider the worst case.

First, we may select all $39$ non-heart cards.

Then, we may select only $2$ hearts.

So, cards selected without guaranteeing $3$ hearts:

$39+2=41$

Now, the next card must be a heart.

So, required number of cards is

$41+1=42$

First,

$p \to q \equiv \neg p \vee q$

So,

$(a) \rightarrow (iv)$

Now,

$\neg p \to q \equiv p \vee q$

So,

$(b) \rightarrow (iii)$

Now,

$q \to \neg p \equiv \neg q \vee \neg p$

Therefore,

$\neg(q \to \neg p)=\neg(\neg q \vee \neg p)$

Using De Morgan’s law,

$\neg(\neg q \vee \neg p)=q \wedge p$

$=p \wedge q$

So,

$(c) \rightarrow (i)$

Now,

$\neg(p \to q)=\neg(\neg p \vee q)$

Using De Morgan’s law,

$\neg(\neg p \vee q)=p \wedge \neg q$

So,

$(d) \rightarrow (ii)$

Therefore, the correct matching is

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

Given matrix is

$A=\left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 0\\1 & 1 & 0\end{array}\right]$

From the matrix, the relation contains:

$1 \to 1,\ 1 \to 3,\ 2 \to 2,\ 3 \to 1,\ 3 \to 2$

Now, since $1 \to 3$ and $3 \to 2$, therefore $1 \to 2$ must be added.

Also, since $3 \to 1$ and $1 \to 3$, therefore $3 \to 3$ must be added.

So, the transitive closure matrix is

$\left[\begin{array}{ccc}1 & 1 & 1\\0 & 1 & 0\\1 & 1 & 1\end{array}\right]$

Hence, the correct answer is Option 1.

Given objective function is

$\text{Max } Z=2x_1-x_2+2x_3$

In the objective function, the positive coefficient of $x_3$ is $2$.

Taking $x_3$ as the entering variable, we apply the minimum ratio test.

From the constraints:

$2x_1+x_2\le 10$

$x_1+2x_2-2x_3\le 20$

$x_1+2x_3\le 5$

Only the third constraint has positive coefficient of $x_3$.

So,

$2x_3=5$

$x_3=\dfrac{5}{2}$

At the first iteration,

$x_1=0,\ x_2=0,\ x_3=\dfrac{5}{2}$

Now,

$Z=2x_1-x_2+2x_3$

$Z=2(0)-0+2\left(\dfrac{5}{2}\right)$

$Z=5$