Boolean Algebra Questions and Answers

Q: Which WFF is in CNF form ?

The formula in CNF is (p &or; q) &and; (¬r &or; p).A formula is in conjunctive normal form when it is a conjunction (AND) of clauses, each clause being a disjunction (OR) of literals, and a literal is a variable or a negated variable. (p &or; q) and (¬r &or; p) are both clauses of literals, joined by &and;, so the formula qualifies.p → q uses the implication connective, which CNF does not allow until it is rewritten as ¬p &or; q. p &or; (q &and; r) has an AND inside an OR, the shape of disjunctive form; its CNF is (p &or; q) &and; (p &or; r). In ¬(p &and; q) &or; r the negation applies to a compound expression rather than to a single variable.

Q: The minimal expression of A + A'BA is

The minimal expression is A.The second term contains both A and its complement, which makes it 0:\[ \begin{aligned} A + A'BA &= A + (A'A)B \\ &= A + 0 \cdot B \\ &= A + 0 = A \end{aligned} \]A + B would be the answer to the different expression A + A'B, which is the likely trap. AB and B do not follow, since the term A'BA contributes nothing for any input. A quick check: if A = 1 the expression is 1, and if A = 0 every term contains A or is A, so it is 0; the output always equals A.

Q: De-Morgan's Law states that

De Morgan's law states (AB)' = A' + B'.The complement of a product is the sum of the complements, and the complement of a sum is the product of the complements: (A + B)' = A'B'.The other options complement only one variable or none: (A + B)' = A'B leaves B uncomplemented, (AB)' = A' + B misses B', and (AB)' = A + B complements nothing.De Morgan's laws let any AND-OR expression be rewritten with NAND or NOR gates, which is why they matter in digital design.

Q: Which one of the following is NOT a valid identity ?

The identity that is not valid is (x+y)&oplus;z = x&oplus;(y+z).Take x = 1, y = 1, z = 0. The left side is (1 + 1)&oplus;0 = 1&oplus;0 = 1, while the right side is 1&oplus;(1 + 0) = 1&oplus;1 = 0. The two sides differ, so OR and XOR cannot be regrouped this way.The others hold: XOR is associative; when xy = 0 the two inputs are never both 1, so XOR equals OR; and x&oplus;y is the complement of XNOR, (xy + x'y')'.

Q: Total number of boolean functions possible over n boolean variables are

The number of Boolean functions is \(2^{2^{n}}\).With n Boolean variables there are \(2^{n}\) rows in the truth table. A function assigns 0 or 1 to each row independently, so there are\[ 2 \times 2 \times \cdots \times 2 = 2^{2^{n}} \]functions. For n = 2 that is 16. \(n^{2^{n}}\) and \(2^{n!}\) count the wrong things, and since only one expression is right, 'More than one of these' is wrong.

Q: Which one of the following is not equivalent to p ↔ q ?

The expression not equivalent to p ↔ q is (¬ p &and; q) &or; (p &and; ¬ q).That expression is true exactly when p and q differ, which is exclusive OR, the negation of p ↔ q.The others are equivalent. (¬p &or; q) is p → q and (p &or; ¬q) is q → p, so their conjunction is p ↔ q, and option B writes the same thing with q → p. (¬p &and; ¬q) &or; (p &and; q) is true exactly when p and q are equal, which is again p ↔ q.

Q: Which Boolean law states A + 0 = A ?

A + 0 = A is the identity law.The identity laws say that ORing a variable with 0, or ANDing it with 1, leaves it unchanged: A + 0 = A and A · 1 = A. 0 is the identity element for OR, as 1 is for AND.The annulment (null) laws are A + 1 = 1 and A · 0 = 0. The complement laws are A + A' = 1 and A · A' = 0. The idempotent laws are A + A = A and A · A = A.

Q: The Boolean expression A(A+B) simplifies to

A(A+B) simplifies to A.Multiply out and use the idempotent and absorption laws:\[ \begin{aligned} A(A+B) &= AA + AB \\ &= A + AB \\ &= A(1 + B) = A \end{aligned} \]This is the absorption law. A check with a truth table agrees: when A = 0 the expression is 0, and when A = 1 it is 1 whatever B is. A+B and B depend on B, so they are wrong, and AB is 0 when A = 1 and B = 0, where the expression is 1.

11.

Which WFF is in CNF form ?

A.p → q
B.(p ∨ (q ∧ r))
C.¬(p ∧ q) ∨ r
D.(p ∨ q) ∧ (¬r ∨ p)

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: D

The formula in CNF is (p ∨ q) ∧ (¬r ∨ p).

A formula is in conjunctive normal form when it is a conjunction (AND) of clauses, each clause being a disjunction (OR) of literals, and a literal is a variable or a negated variable. (p ∨ q) and (¬r ∨ p) are both clauses of literals, joined by ∧, so the formula qualifies.

p → q uses the implication connective, which CNF does not allow until it is rewritten as ¬p ∨ q. p ∨ (q ∧ r) has an AND inside an OR, the shape of disjunctive form; its CNF is (p ∨ q) ∧ (p ∨ r). In ¬(p ∧ q) ∨ r the negation applies to a compound expression rather than to a single variable.

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12.

The minimal expression of A + A'BA is

A.A
B.A + B
C.AB
D.B

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: A

The minimal expression is A.

The second term contains both A and its complement, which makes it 0:

\[ \begin{aligned} A + A'BA &= A + (A'A)B \\ &= A + 0 \cdot B \\ &= A + 0 = A \end{aligned} \]

A + B would be the answer to the different expression A + A'B, which is the likely trap. AB and B do not follow, since the term A'BA contributes nothing for any input. A quick check: if A = 1 the expression is 1, and if A = 0 every term contains A or is A, so it is 0; the output always equals A.

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13.

De-Morgan's Law states that

A.(A + B)' = A' * B
B.(AB)' = A' + B'
C.(AB)' = A' + B
D.(AB)' = A + B

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: B

De Morgan's law states (AB)' = A' + B'.

The complement of a product is the sum of the complements, and the complement of a sum is the product of the complements: (A + B)' = A'B'.

The other options complement only one variable or none: (A + B)' = A'B leaves B uncomplemented, (AB)' = A' + B misses B', and (AB)' = A + B complements nothing.

De Morgan's laws let any AND-OR expression be rewritten with NAND or NOR gates, which is why they matter in digital design.

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14.

Which one of the following is NOT a valid identity ?

A.(x⊕y)⊕z = x⊕(y⊕z)
B.(x+y)⊕z = x⊕(y+z)
C.x⊕y = x + y if xy=0
D.x⊕y = (xy+x'y')'

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: B

The identity that is not valid is (x+y)⊕z = x⊕(y+z).

Take x = 1, y = 1, z = 0. The left side is (1 + 1)⊕0 = 1⊕0 = 1, while the right side is 1⊕(1 + 0) = 1⊕1 = 0. The two sides differ, so OR and XOR cannot be regrouped this way.

The others hold: XOR is associative; when xy = 0 the two inputs are never both 1, so XOR equals OR; and x⊕y is the complement of XNOR, (xy + x'y')'.

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15.

Total number of boolean functions possible over n boolean variables are

A.\(2^{2^{n}}\)
B.\(n^{2^{n}}\)
C.\(2^{n!}\)
D.More than one of these

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: A

The number of Boolean functions is \(2^{2^{n}}\).

With n Boolean variables there are \(2^{n}\) rows in the truth table. A function assigns 0 or 1 to each row independently, so there are

\[ 2 \times 2 \times \cdots \times 2 = 2^{2^{n}} \]

functions. For n = 2 that is 16. \(n^{2^{n}}\) and \(2^{n!}\) count the wrong things, and since only one expression is right, 'More than one of these' is wrong.

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16.

Which one of the following is not equivalent to p ↔ q ?

A.(¬ p ∨ q) ∧ (p ∨ ¬ q)
B.(¬p∨q) ∧ (q → p)
C.(¬ p ∧ q) ∨ (p ∧ ¬ q)
D.(¬ p ∧ ¬ q) ∨ (p ∧ q)

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: C

The expression not equivalent to p ↔ q is (¬ p ∧ q) ∨ (p ∧ ¬ q).

That expression is true exactly when p and q differ, which is exclusive OR, the negation of p ↔ q.

The others are equivalent. (¬p ∨ q) is p → q and (p ∨ ¬q) is q → p, so their conjunction is p ↔ q, and option B writes the same thing with q → p. (¬p ∧ ¬q) ∨ (p ∧ q) is true exactly when p and q are equal, which is again p ↔ q.

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17.

Which Boolean law states A + 0 = A ?

A.Identity law
B.Complement law
C.Annulment law
D.Idempotent law

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: A

A + 0 = A is the identity law.

The identity laws say that ORing a variable with 0, or ANDing it with 1, leaves it unchanged: A + 0 = A and A · 1 = A. 0 is the identity element for OR, as 1 is for AND.

The annulment (null) laws are A + 1 = 1 and A · 0 = 0. The complement laws are A + A' = 1 and A · A' = 0. The idempotent laws are A + A = A and A · A = A.

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18.

The Boolean expression A(A+B) simplifies to

A.A+B
B.AB
C.A
D.B

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: C

A(A+B) simplifies to A.

Multiply out and use the idempotent and absorption laws:

\[ \begin{aligned} A(A+B) &= AA + AB \\ &= A + AB \\ &= A(1 + B) = A \end{aligned} \]

This is the absorption law. A check with a truth table agrees: when A = 0 the expression is 0, and when A = 1 it is 1 whatever B is. A+B and B depend on B, so they are wrong, and AB is 0 when A = 1 and B = 0, where the expression is 1.

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19.

The dual of a Boolean expression is obtained by interchanging

A.Boolean sums and Boolean products
B.Boolean sums and Boolean products or interchanging 0's and 1's
C.Boolean sums and Boolean products and interchanging 0's & 1's
D.Interchanging 0's and 1's

Exam Year(s)

UGC NET - 2013,

Answer: C

The dual of a Boolean expression is obtained by interchanging Boolean sums and Boolean products and also interchanging 0's and 1's. For example, the dual of x(y + 0) is x + (y . 1).

Swapping only sums and products leaves out the constants, and swapping only the constants leaves the operators unchanged. The option that says 'or' also does not give the dual, since both swaps have to be made together.

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20.

The sum of products expansion for the function \(F(x, y, z) = (x + y)\bar{z}\) is given as

A.\(\bar{x}\bar{y}z + xy\bar{z} + \bar{x}y\bar{z}\)
B.\(xyz + xy\bar{z} + x\bar{y}\bar{z}\)
C.\(x\bar{y}\bar{z} + \bar{x}\bar{y}\bar{z} + xy\bar{z}\)
D.\(xy\bar{z} + x\bar{y}\bar{z} + \bar{x}y\bar{z}\)

Exam Year(s)

UGC NET - 2013,

Answer: D

Expand F = (x + y) z' . Multiplying out gives x z' + y z'. Each term lacks a variable, so it is expanded to full minterms by multiplying with (y + y') and (x + x').

\[ xz' = xyz' + xy'z' \]\[ yz' = xyz' + x'yz' \]

Combining and removing the repeated term xyz' gives xyz' + xy'z' + x'yz', the last option. The others contain minterms such as x'y'z or xyz, in which z is present, or x'y'z', in which x and y are both false, so they cannot make F equal to 1.

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