Boolean Algebra Solved Examples – Practice Problems with Solutions

This page provides a comprehensive set of solved problems covering all aspects of Boolean algebra. Practice these to build confidence for exams and interviews.

Category 1: Direct Simplification

Problem 1: Simplify F = XY + XY\'Z + XY\'Z\'

Problem 2: Simplify F = A\'B + ABC\' + ABC

Problem 3: Simplify F = (A + B)(A + C)(B + C)

Actually, apply consensus: AB + AC + BC. Here BC is the consensus of... No, let\'s check if this can be simplified further. By consensus theorem (with respect to variable A): AB + A\'C + BC = AB + A\'C. But we don\'t have A\' here. So AB + AC + BC cannot be reduced further without knowing which variable to use.

Wait — let me reconsider: (A+B)(A+C)(B+C). By consensus theorem for POS: (A+B)(A\'+C)(B+C) = (A+B)(A\'+C). But our middle term is (A+C), not (A\'+C). So this is different.

Let me just expand properly:

This is the minimum form.

Problem 4: Simplify F = A\'B\'C + A\'BC + AB\'C + ABC

Problem 5: Simplify F = A\'B + AB\' + AB + A\'B\'

The function always outputs 1 (tautology).

Category 3: Complement Finding

Problem 9: Find the complement of F = AB\' + A\'B

Note: F = A XOR B, so F\' = A XNOR B ✓

Problem 10: Find complement of F = (X+Y)(X\'+Z)

Category 4: Canonical Form Conversion

Problem 11: Express F = A + B\'C in canonical SOP (3 variables)

Problem 12: Convert F = Σm(0, 1, 3, 5, 7) to POS

For 3 variables, total minterms = 8 (0 to 7). Missing from SOP: 2, 4, 6

Category 5: Implementation Questions

Problem 13: Implement F = A\'B + AB\' using only NAND gates

But we need A\' and B\' first:

Total: 5 NAND gates.

Actually, there\'s a better way:

Total: 4 NAND gates! (This is the standard XOR using NAND.)

Problem 14: How many 2-input gates are needed for F = AB + CD + EF?

Category 6: Word Problems

Problem 15: Design a circuit where the output is HIGH only when the majority of three inputs are HIGH.

This is the majority function for 3 inputs A, B, C.

Truth table:

SOP: F = A\'BC + AB\'C + ABC\' + ABC

Simplification:

Verification: AB covers minterms 6,7. AC covers minterms 5,7. BC covers minterms 3,7. Union: {3,5,6,7} ✓

Quick Reference: Simplification Checklist

• [ ] Factor common variables
• [ ] Apply complement law (X + X\' = 1)
• [ ] Apply absorption (A + AB = A)
• [ ] Apply consensus theorem
• [ ] Check for De Morgan\'s opportunities
• [ ] Verify with truth table

Interview Questions

Q1: Simplify (A + B + C)(A + B + C\')(A + B\' + C)

Combine first two: (A+B+C)(A+B+C\') = A+B (they differ only in C). Then F = (A+B)(A+B\'+C). Expand: AA + AB\' + AC + AB + BB\' + BC = A + AB\' + AC + AB + BC = A + BC. So F = A + BC.

Q2: If F = Σm(1,2,3,5,7), find the simplified SOP using algebraic methods.

Write minterms: A\'B\'C + A\'BC\' + A\'BC + AB\'C + ABC. Group: A\'B\'C + AB\'C = B\'C. A\'BC\' + A\'BC = A\'B. A\'BC + ABC = BC. So F = B\'C + A\'B + BC = A\'B + C(B\'+B) = A\'B + C. Final: F = A\'B + C.

Q3: What is the minimum number of 2-input NAND gates needed to implement XOR?

4 NAND gates. The construction is: N1=NAND(A,B), N2=NAND(A,N1), N3=NAND(B,N1), F=NAND(N2,N3). This is more efficient than the 5-gate version that creates A\' and B\' separately.

Q4: Prove that A + A\'B = A + B without using truth table.

A + A\'B = (A + A\')(A + B) [OR distributes over AND] = 1·(A + B) = A + B. Alternatively: A + A\'B = A·1 + A\'B = A(1+B) + A\'B = A + AB + A\'B = A + B(A+A\') = A + B.

Q5: How do you determine if two Boolean expressions are equivalent?

Three methods: (1) Simplify both to the same minimal form, (2) Construct truth tables and compare all outputs, (3) Convert both to canonical form and compare minterm lists. Method 2 is most reliable; method 1 might miss equivalence if simplification gets stuck.