Level 1 — RecognitionBoolean Algebra & Logic Gates

Boolean Algebra & Logic Gates

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Level: 1 (Recognition) Time Limit: 20 minutes Total Marks: 30

Section A — Multiple Choice (1 mark each) [12 marks]

Q1. The decimal value of the binary number $1011_2$ is: (a) 9 (b) 11 (c) 13 (d) 15

Q2. How many distinct values can be represented using one byte? (a) 8 (b) 16 (c) 128 (d) 256

Q3. The hexadecimal representation of $11011010_2$ is: (a) $DA$ (b) $AD$ (c) $BA$ (d) $D2$

Q4. The octal representation of $110101_2$ is: (a) $65$ (b) $56$ (c) $61$ (d) $47$

Q5. In 8-bit two's complement, the representation of $-5$ is: (a) $11111010$ (b) $11111011$ (c) $10000101$ (d) $00000101$

Q6. Which gate outputs $1$ only when its two inputs are different? (a) AND (b) NOR (c) XOR (d) XNOR

Q7. According to De Morgan's theorem, $\overline{A + B}$ is equal to: (a) $\bar{A} + \bar{B}$ (b) $\bar{A}\cdot\bar{B}$ (c) $A\cdot B$ (d) $\overline{A}\cdot B$

Q8. Which of the following is a universal gate? (a) AND (b) OR (c) NAND (d) XOR

Q9. The expression $A\bar{B} + AB$ simplifies to: (a) $A$ (b) $B$ (c) $\bar{A}$ (d) $AB$

Q10. A minterm-based (canonical) representation of a function typically produces which form? (a) POS (b) SOP (c) NAND-only (d) K-map

Q11. In a 4-variable Karnaugh map, the total number of cells is: (a) 4 (b) 8 (c) 16 (d) 32

Q12. The Quine–McCluskey method is primarily used to: (a) convert decimal to binary (b) minimise Boolean functions algorithmically (c) measure propagation delay (d) build truth tables

Section B — Matching (1 mark each) [6 marks]

Q13. Match each gate (Column X) to its Boolean output (Column Y).

Column X	Column Y
(i) NAND	(P) $A\oplus B$
(ii) NOR	(Q) $\overline{A\cdot B}$
(iii) XOR	(R) $\overline{A+B}$
(iv) XNOR	(S) $\overline{A\oplus B}$

Write the correct pairings (e.g. i–?, ii–?, iii–?, iv–?).

Q14. Match the law (Column X) to its expression (Column Y).

Column X	Column Y
(i) Commutative	(P) $A(B+C)=AB+AC$
(ii) Associative	(Q) $A+B = B+A$
(iii) Distributive	(R) $(A+B)+C = A+(B+C)$

Write the correct pairings.

Section C — True/False WITH Justification (2 marks each) [12 marks]

(1 mark for correct T/F, 1 mark for correct justification.)

Q15. "The NOR gate alone can be used to build any logic circuit." — True or False? Justify.

Q16. " $A + \bar{A} = 0$ is a valid Boolean identity." — True or False? Justify.

Q17. "In two's complement, an 8-bit register can represent numbers from $-128$ to $+127$ ." — True or False? Justify.

Q18. "Don't-care conditions in a K-map must always be treated as 0." — True or False? Justify.

Q19. "De Morgan's theorem states $\overline{A\cdot B} = \bar{A}+\bar{B}$ ." — True or False? Justify.

Q20. "Propagation delay is the time taken for a change at a gate's input to produce a change at its output." — True or False? Justify.

End of Paper

Answer keyMark scheme & solutions

Section A

Q1 — (b) 11. $1011_2 = 8+0+2+1 = 11$ . (1)

Q2 — (d) 256. One byte = 8 bits, $2^8 = 256$ distinct values. (1)

Q3 — (a) DA. Group into nibbles: $1101=D$ , $1010=A$ → $DA$ . (1)

Q4 — (a) 65. Group into triples from right: $110\,101 = 6\,5 = 65_8$ . (1)

Q5 — (b) 11111011. $+5 = 00000101$ ; invert → $11111010$ ; add 1 → $11111011$ . (1)

Q6 — (c) XOR. XOR = 1 iff inputs differ. (1)

Q7 — (b) $\bar{A}\cdot\bar{B}$ . De Morgan: NOR-of-OR = AND-of-complements. (1)

Q8 — (c) NAND. NAND is functionally complete (universal). (1)

Q9 — (a) $A$ . $A\bar{B}+AB = A(\bar{B}+B)=A(1)=A$ . (1)

Q10 — (b) SOP. Sum of minterms is the canonical SOP form. (1)

Q11 — (c) 16. $2^4 = 16$ cells. (1)

Q12 — (b) minimise Boolean functions algorithmically. (1)

Section B

Q13 (¼ each — round to marks: all-or-award per correct pair, total 1)

i–Q (NAND $=\overline{A\cdot B}$ )
ii–R (NOR $=\overline{A+B}$ )
iii–P (XOR $=A\oplus B$ )
iv–S (XNOR $=\overline{A\oplus B}$ )

Q14

i–Q (Commutative $A+B=B+A$ )
ii–R (Associative)
iii–P (Distributive) (1)

Section C

Q15 — True. (T:1) NOR is a universal gate; NOT, AND, OR can all be constructed from NOR alone, so any circuit can be built. (Justify:1)

Q16 — False. (T:1) The complement law gives $A+\bar{A}=1$ , not $0$ (an element ORed with its complement always yields 1). (Justify:1)

Q17 — True. (T:1) 8-bit two's complement range is $-2^{7}$ to $2^{7}-1 = -128$ to $+127$ . (Justify:1)

Q18 — False. (T:1) Don't-cares may be treated as either 0 or 1, whichever gives greater simplification. (Justify:1)

Q19 — True. (T:1) De Morgan: complement of a product equals the sum of the complements, $\overline{AB}=\bar A+\bar B$ . (Justify:1)

Q20 — True. (T:1) Propagation delay is precisely the input-to-output response time of a gate. (Justify:1)

[
  {"claim":"1011 binary equals 11","code":"result = (int('1011',2)==11)"},
  {"claim":"one byte gives 256 values","code":"result = (2**8==256)"},
  {"claim":"11011010 binary equals hex DA","code":"result = (format(int('11011010',2),'X')=='DA')"},
  {"claim":"110101 binary equals octal 65","code":"result = (oct(int('110101',2))[2:]=='65')"},
  {"claim":"-5 in 8-bit twos complement is 11111011","code":"result = (format((-5)&0xFF,'08b')=='11111011')"},
  {"claim":"8-bit twos complement range is -128 to 127","code":"result = ((-2**7==-128) and (2**7-1==127))"},
  {"claim":"A*~B + A*B simplifies to A","code":"A,B=symbols('A B'); result = (simplify_logic(Or(And(A,Not(B)),And(A,B)))==A)"}
]