UGC NET Computer Science Boolean Algebra Previous Year Questions (PYQs) – Page 1 of 2
Study Stuff for MCA Examinations
| List-I (Boolean Expression) | List-II (Dual of Boolean Expression) |
| A. $x(y+0)$ | I. $(x+0)(y\cdot z)$ |
| B. $\bar{x}+(\bar{y}+z)$ | II. $x+y\cdot1$ |
| C. $\bar{x}(y+0)$ | III. $\bar{x}+\bar{y}\cdot1$ |
| D. $x+(\bar{y}+z)$ | IV. $(x+0)(y\cdot \bar{z})$ |
Solution
Dual is obtained by interchanging:
$+$ ↔ $\cdot$
$0$ ↔ $1$
Applying dual rule:
A → III
B → IV
C → II
D → I
A. Storage allocation
B. Checking type compatibility
C. Suppressing duplicate error messages
Choose the correct answer from the options given below:
Solution
It helps in storage allocation.
It helps in checking type compatibility.
It also helps in avoiding or suppressing duplicate error messages for the same identifier.
So, all A, B and C are correct.
How many different Boolean functions of degree $n$ are there?
Solution
For a Boolean function of $n$ variables, the number of possible input combinations is
$2^n$
For each input combination, the output can be either $0$ or $1$.
So, total number of Boolean functions is
$2^{2^n}$
Solution
$(\overline{P}\land Q)\lor(P\land \overline{Q})\lor(P\land Q)$
Take common $P$ from last two terms:
$=(\overline{P}\land Q)\lor P(\overline{Q}\lor Q)$
Since,
$\overline{Q}\lor Q=1$
So,
$=(\overline{P}\land Q)\lor P$
Using identity,
$P\lor(\overline{P}\land Q)=P\lor Q$
Therefore,
$(\overline{P}\land Q)\lor(P\land \overline{Q})\lor(P\land Q)=P\lor Q$
Solution
Solution
Solution
$AB+A\overline{B}+\overline{A}C+AC$
Take $A$ common from first two terms:
$A(B+\overline{B})+\overline{A}C+AC$
Since,
$B+\overline{B}=1$
So,
$A(1)+\overline{A}C+AC$
$=A+\overline{A}C+AC$
Now,
$A+AC=A$
So,
$A+\overline{A}C$
Using identity $A+\overline{A}C=A+C$
Therefore, the simplified expression is:
$A+C$
The final expression does not contain $B$, so it is unaffected by the value of $B$.
Consider the Boolean expression given by
$F=(X+Y+Z)(X'+Y)(Y'+Z)$
Which of the following Boolean expressions is/are equivalent to $F'$?
A. $(X'+Y'+Z')(X+Y')(Y+Z')$
B. $(XY'+Z')$
C. $(X+Z')(Y'+Z')$
D. $(XY'+YZ'+X'Y'Z')$
Solution
$X:a\lor(b\land(a\lor c))=(a\lor b)\land(a\lor c)$
$Y:a\land(b\lor(a\land c))=(a\land b)\lor(a\land c)$
$a\lor(b\land c)=(a\lor b)\land c$ is satisfied if
Solution
$X:a\lor(b\land(a\lor c))=(a\lor b)\land(a\lor c)$
This follows the distributive law:
$a\lor(b\land d)=(a\lor b)\land(a\lor d)$
Here,
$d=(a\lor c)$
So,
$a\lor(b\land(a\lor c))=(a\lor b)\land(a\lor(a\lor c))$
$=(a\lor b)\land(a\lor c)$
Therefore, $X$ is true.
Now check $Y$:
$Y:a\land(b\lor(a\land c))=(a\land b)\lor(a\land c)$
Using distributive law:
$a\land(b\lor d)=(a\land b)\lor(a\land d)$
Here,
$d=(a\land c)$
So,
$a\land(b\lor(a\land c))=(a\land b)\lor(a\land a\land c)$
$=(a\land b)\lor(a\land c)$
Therefore, $Y$ is true.
Hence, both $X$ and $Y$ are true.
Hence, the correct answer is $\boxed{(c)}$
Which of the following statements are true?
$(i)$ Every logic network is equivalent to one using just NAND gates or just NOR gates.
$(ii)$ Boolean expressions and logic networks correspond to labelled acyclic digraphs.
$(iii)$ No two Boolean algebras with $n$ atoms are isomorphic.
$(iv)$ Non-zero elements of finite Boolean algebras are not uniquely expressible as joins of atoms.
Choose the correct answer from the code given below:
Solution
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.
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