Boolean algebra borrows its symbols from arithmetic and almost none of its rules: A · A is A, not A squared, and A + A·B collapses straight back to A. That mismatch is what a good distractor exploits, because a half-remembered law feels exactly as convincing as a correct one. Two moves settle almost every question below: substitute a single row of values, or name the law being applied. Attempt each question before reading its explanation. KnowledgeGate's question bank holds more than 60 Boolean algebra MCQs, with further practice in Digital Electronics MCQs and Digital Logic.
1. Reset the notation before solving
Use this compact working key:
+means OR.·,*, or juxtaposition means AND.A prime or overbar means NOT.
Every input is a binary value, either 0 or 1.
So 1 + 1 = 1 in Boolean OR, not 2. For A=1 and B=0,
(A · B)′ = (1 · 0)′ = 0′ = 1
and
A′ + B′ = 0 + 1 = 1.
This row illustrates De Morgan's law. A complete identity, however, must be proved algebraically or checked across every truth-table row. Boolean algebra sits alongside number systems, K-maps and sequential circuits in the wider CS Fundamentals category.
2. Duality and De Morgan's law
Question 1: form the dual without changing the variables
Asked in UGC NET 2013.
The dual of a Boolean expression is obtained by interchanging:
(a) the Boolean Sum (OR, +) and Boolean Product (AND, ·) operators only, keeping the constants 0's and 1's unchanged
(b) the Boolean Sum and Boolean Product operators, or alternatively the constants 0's and 1's: either one substitution by itself
(c) the Boolean Sum and Boolean Product operators and the constants 0's and 1's simultaneously, while every variable stays unchanged
(d) only the constants 0's and 1's, keeping the operators unchanged
Answer: (c) the Boolean Sum and Boolean Product operators and the constants 0's and 1's simultaneously, while every variable stays unchanged. Duality simultaneously swaps + ↔ · and 0 ↔ 1; it leaves every variable unchanged. For example, the dual of A + 0 is A · 1. Both simplify to A, illustrating the dual identity laws. Options (a), (b) and (d) each leave part of the required swap incomplete.
Question 2: complement a product correctly
Asked in TPSC 2025.
According to De Morgan's Law, which expression correctly equals (A · B)′?
(a) (A + B)′ = A′ · B′
(b) (A · B)′ = A′ + B′
(c) (A · B)′ = A′ + B
(d) (A · B)′ = A + B′
Answer: (b) (A · B)′ = A′ + B′. Break the bar and change AND to OR. With A=1, B=0, the left side is (1·0)′=1, while the right side is 0+1=1. With A=1, B=1, both sides are 0. The last check exposes expressions that complement only one input.
3. Absorption, distributivity and associative structure
Question 3: distinguish absorption from idempotence
Asked in UGC NET 2011.
The absorption law in Boolean algebra says that
(a) X + X = X
(b) X · X = X
(c) x + x · y = x
(d) None of the above
Answer: (c) x + x · y = x. Factor and simplify:
x + xy = x(1+y) = x·1 = x.
For x=0, y=1, both sides are 0. For x=1, y=0, both are 1. Options (a) and (b) are also true, but they are the idempotent laws, not the named absorption law.
Question 4: collapse a product of sums
Asked in Indian Space Research Organization 2011.
In Boolean algebra, rule (X+Y)(X+Z) =
(a) Y+XZ
(b) X+YZ
(c) XY+Z
(d) XZ+Y
Answer: (b) X+YZ. Expand the left side:
(X+Y)(X+Z) = XX + XZ + XY + YZ
= X + XZ + XY + YZ
= X + YZ.
Here XX=X, while X+XZ=X and X+XY=X by absorption. For X=0, Y=1, Z=1, the original is (1)(1)=1, and X+YZ=0+1=1.
Question 5: find the non-associative connective
Asked in UGC NET 2004.
An example of a connective which is not associative is:
(a) AND
(b) OR
(c) EX-OR
(d) NAND
Answer: (d) NAND. Write NAND as ↑ and take A=1, B=1, C=0:
(A↑B)↑C = 0↑0 = 1,
but
A↑(B↑C) = 1↑1 = 0.
One counterexample disproves associativity. AND, OR and XOR are associative; NAND is not.
4. Truth tables and paired Boolean laws
Question 6: name the gate table
Asked in DSSSB 2021.
The input-output relationship of the binary variable for each gate can be represented in tabular form by a __________.
(a) symbol table
(b) decision table
(c) truth table
(d) characteristic table
Answer: (c) truth table. A truth table is an exhaustive input-output map. A two-input AND gate has these rows:
A | B | A·B |
|---|---|---|
0 | 0 | 0 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
There are four rows because two binary inputs have 2² possible assignments.
Question 7: verify both named laws
Asked in BEL 2023.
Which of the following pair is CORRECT?
I. Associative law : (x*y)*z = x*(y*z)
II. Distributive law : x*(y+z) = (x*y)+(x*z)
(a) Only II
(b) Both I and II
(c) Only I
(d) Neither I nor II
Answer: (b) Both I and II. Statement I is associativity of AND. Statement II is AND distributing over OR. With x=1, y=0, z=1, Statement I gives 0 on both sides. Statement II gives 1*(0+1)=1 and (1*0)+(1*1)=0+1=1. This substitution illustrates the laws; their identity status does not rest on one row.
5. Solve the system, then reject the false property
Question 8: test simultaneous Boolean equations
Asked in GATE 2000.
The simultaneous equations on the Boolean variables x, y, z and w are:
x + y + z = 1
xy = 0
xz + w = 1
xy + z'w' = 0
They have the following solution for x, y, z and w, respectively.
(a) 0 1 0 0
(b) 1 1 0 1
(c) 1 0 1 1
(d) 1 0 0 0
Answer: (c) 1 0 1 1. Test the restrictive equations first. Option (a) fails xz+w=1 because it gives 0. Option (b) fails xy=0 because it gives 1. Option (d) also makes xz+w=0. For Option (c), (x,y,z,w)=(1,0,1,1):
x+y+z=1xy=0xz+w=1xy+z'w'=0+0·0=0
Remember that Boolean + is OR.
Question 9: disprove one alleged property
Asked in GATE 2026.
Which one of the following options is not a property of Boolean Algebra?
(+ denotes OR, . denotes AND, ′ denotes NOT)
(a) a+b=b+a
(b) a.a′=1
(c) a+a′=1
(d) a.b=b.a
Answer: (b) a.a′=1. The correct complement law is a·a′=0, not 1. If a=0, then 0·1=0; if a=1, then 1·0=0. Options (a) and (d) are commutative laws. Option (c) is the OR complement law.
6. Use parity for XOR and recognise XNOR
Question 10: locate the incorrect XOR expression
Asked in UGC NET 2016.
Which of the following logical expressions is incorrect?
(a) 1 ⊕ 0 = 1
(b) 1 ⊕ 1 ⊕ 1 = 1
(c) 1 ⊕ 1 ⊕ 0 = 1
(d) 1 ⊕ 1 = 0
Answer: (c) 1 ⊕ 1 ⊕ 0 = 1. XOR tests odd parity: its output is 1 only when the number of 1s is odd. Check each option: 1⊕0=1; 1⊕1⊕1=0⊕1=1; 1⊕1⊕0=0⊕0=0, not 1; and 1⊕1=0.
Question 11: decide whether a custom operator is associative
Asked in UGC NET 2013.
A Boolean operator s is defined as follows:
1 s 1 = 1, 1 s 0 = 0 , 0 s 1 = 0, 0 s 0 = 1
What will be the truth value of the expression (x s y) s z = x s (y s z)
(a) Always false
(b) Always true
(c) Sometimes true
(d) True when x, y, z are all true
Answer: (b) Always true. The operator s is XNOR because it returns 1 when its inputs are equal. Write x s y = ¬(x⊕y). Then both (x s y) s z and x s (y s z) reduce to x⊕y⊕z. For 000, both grouped sides are 0. For 111, both are 1. The equality is always true.
7. Finish with a full Boolean simplification
Question 12: simplify the bracket before applying the final literal
Asked in BEL 2007.
The Boolean expression Y = (A + B̅ + A̅ B) C̅ is given by
(a) A C̅
(b) B C̅
(c) C̅
(d) AB
Answer: (c) C̅. First use A+A̅B=A+B:
A+B̅+A̅B = A+B+B̅ = A+1 = 1.
Therefore Y=1·C̅=C̅. Check A=0, B=0, C=1: the bracket is 1 and Y=0. With the same A and B but C=0, Y=1. Both results match C̅.
The short version
Use a three-line revision routine:
Rewrite the notation key from memory.
Solve all 12 items again without looking at the answers.
Record each miss under one law or one truth-table row.
Translate, simplify, verify: that is the whole method. Zero to Hero provides a broad CS fundamentals route, while GATE Guidance by Sanchit Sir provides structured GATE-oriented preparation.
