Prashnikaप्रश्निका

Boolean Algebra - ISC Class 12 Computer Science Questions with Answers, Page 7

189 past-paper questions on Boolean Algebra from ISC Class 12 Computer Science papers (2026-2017), newest first, in full. Questions 121-140 are on this page, 20 to a page. Tap "Show answer" under a question to see its answer.

Practise these questions with filters
2022 · 1 mark · MCQOpen: The law used:

Choose the correct option.

The law used:
  • (i)Distributive Law
  • (ii)De Morgan Law
  • (iii)Associative Law
  • (iv)Idempotent Law
Show the case
Given the Boolean expression $F = (P + R) \cdot (P \cdot Q + Q \cdot R')$, identify:
Show answer

Answer

AI

Correct option: (ii)

Answer: (ii) De Morgan Law Complementing the expression uses De Morgan's law to turn sums into products and vice versa.
2022 · 2 marks · MCQOpen: The basic logic gate that represents the simplification of the Boolean…

Choose the correct option.

The basic logic gate that represents the simplification of the Boolean expression $A \cdot (A' + B) \cdot (A + B)$ is:
  • (a)OR gate
  • (b)NOT gate
  • (c)AND gate
  • (d)None of the above
Show answer

Answer

AI

Correct option: (c)

Answer: (c) AND gate $A(A'+B) = AB$, and $AB(A+B) = AB$. The result is A.B, an AND gate.
2022 · 2 marks · MCQOpen: The reduced expression of the Boolean function given above is:

Choose the correct option.

The reduced expression of the Boolean function given above is:
  • (i)$ACD' + B'D' + BD$
  • (ii)$(A + C' + D') \cdot (B' + D') \cdot (A + C')$
  • (iii)$C'D' + AC' + B'D'$
  • (iv)$(C + D') \cdot (B' + D') \cdot (A + B + D)$
Show the case
Reduce the given Boolean function $F(A,B,C,D) = \sum(0,2,4,8,9,10,12,13)$ by using 4-variable Karnaugh map and answer the following questions:
Show answer

Answer

AI

Correct option: (iii)

Answer: (iii) $C'D' + AC' + B'D'$ The three quads give $AC'$ (8,9,12,13), $B'D'$ (0,2,8,10) and $C'D'$ (0,4,8,12).
2022 · 1 mark · MCQOpen: The dual of the Boolean equation is:

Choose the correct option.

The dual of the Boolean equation $(X + Y) \cdot 1 = X + Y$ is:
  • (a)$X + Y + 0$
  • (b)$X \cdot Y + 0 = X \cdot Y$
  • (c)$(X \cdot Y) + 1 = X \cdot Y$
  • (d)$(X + Y) + 0 = X \cdot Y$
Show answer

Answer

AI

Correct option: (b)

Answer: (b) $X \cdot Y + 0 = X \cdot Y$ The dual swaps + with . and 0 with 1, so $(X+Y) \cdot 1 = X+Y$ becomes $X \cdot Y + 0 = X \cdot Y$.
2022 · 1 mark · MCQOpen: What will be the least number of groups and their types formed for reduction?

Choose the correct option.

What will be the least number of groups and their types formed for reduction?
  • (i)6 pairs
  • (ii)2 quad and 2 pairs
  • (iii)1 quad and 3 pairs
  • (iv)3 quads
Show answer

Answer

AI

Correct option: (iv)

Answer: (iv) 3 quads The map groups are 8,9,12,13 ; 0,2,8,10 ; 0,4,8,12: three quads cover all minterms.
2022 · 2 marks · MCQOpen: The proposition is a:

Choose the correct option.

The proposition $\sim(a \wedge b) \vee (\sim a => b)$ is a:
  • (a)Contradiction
  • (b)Contingency
  • (c)Tautology
  • (d)Implication
Show answer

Answer

AI

Correct option: (c)

Answer: (c) Tautology ~a => b is a + b, so the expression is $(a' + b') + a + b$, which is always 1.
2022 · 1 mark · MCQOpen: What will be the least number of groups and their types formed for reduction?

Choose the correct option.

What will be the least number of groups and their types formed for reduction?
  • (i)6 pairs
  • (ii)3 quads
  • (iii)1 quad and 3 pairs
  • (iv)2 quad and 3 pairs
Show the case
Reduce the given Boolean function $F(A,B,C,D) = \pi(3,4,5,6,7,11,13,15)$ by using 4-variable Karnaugh map and answer the following questions:
Show answer

Answer

AI

Correct option: (ii)

Answer: (ii) 3 quads The maxterm groups are 4,5,6,7 ; 5,7,13,15 ; 3,7,11,15: three quads cover all zeros.
2022 · 2 marks · MCQOpen: The reduced expression for the Boolean expression is:

Choose the correct option.

The reduced expression for the Boolean expression $F(X,Y,Z) = \sum(0,1,2,3,4,5,6,7)$ is:
  • (a)$XY' + X'Y$
  • (b)1
  • (c)0
  • (d)None of the above
Show answer

Answer

AI

Correct option: (b)

Answer: (b) 1 All 8 minterms of three variables are present, so the function is always 1.
2022 · 1 mark · MCQOpen: The maximum input combinations for the above truth table will be:

Choose the correct option.

The maximum input combinations for the above truth table will be:
  • (i)24
  • (ii)16
  • (iii)8
  • (iv)4
Show the case
A school intends to select candidates for an Inter school competition as per the criteria given below: • The student has participated in an earlier competition and is very creative Or • The student is very creative and has excellent general awareness, but has not participated in any competition earlier Or • The student has excellent general awareness and has won prize in an inter-house competition The inputs are:
Inputs
AParticipated in a competition earlier
BIs very creative
CWon prize in an inter house competition
DHas excellent general awareness
(In all the above cases 1 indicates yes and 0 indicates no). Output: X [1 indicates yes and 0 indicates no for all cases].
Show answer

Answer

AI

Correct option: (ii)

Answer: (ii) 16 With 4 inputs there are 2^4 = 16 input combinations.
2020 · 1 mark · Short answerOpen: Find the dual for the Boolean equation: .

Answer the following in short.

Find the dual for the Boolean equation: $AB' + BC' + 1 = 1$.
Show answer

Answer

AI
To get the dual, interchange $+$ with $\cdot$ and $0$ with $1$; the variables and their complements stay unchanged. Given: $AB' + BC' + 1 = 1$ Dual: $(A + B') \cdot (B + C') \cdot 0 = 0$
2020 · 1 mark · Short answerOpen: Minimize: using Boolean laws.

Answer the following in short.

Minimize: $F = XY + (XZ)' + XY'Z$ using Boolean laws.
Show answer

Answer

AI
$F = XY + (XZ)' + XY'Z$ $= XY + X' + Z' + XY'Z$ (De Morgan's law on $(XZ)'$) $= (X' + XY) + Z' + XY'Z$ (regrouping) $= X' + Y + Z' + XY'Z$ (using $A + A'B = A + B$, i.e. $X' + XY = X' + Y$) $= X' + Y + Z' + Y'Z$ (using $X' + XY'Z = X' + Y'Z$) $= X' + Y + Z' + Z$ (using $Y + Y'Z = Y + Z$) $= X' + Y + 1$ (complement law $Z + Z' = 1$) $= 1$ (null law) So $F = 1$ (verified with the boolean tool: the expression is a tautology).
2020 · 1 mark · Short answerOpen: State the properties of zero in Boolean algebra.

Answer the following in short.

State the properties of zero in Boolean algebra.
Show answer

Answer

AI
Properties of zero (0) in Boolean algebra: 1. Identity law (OR with 0): $A + 0 = A$ 2. Null law (AND with 0): $A \cdot 0 = 0$ 3. Complement of zero: $0' = 1$, and $A \cdot A' = 0$ (a variable ANDed with its complement gives 0).
2020 · 1 mark · Short answerOpen: State whether the following proposition is a tautology, contradiction or a…

Answer the following in short.

State whether the following proposition is a tautology, contradiction or a contingency: $F = (P \Rightarrow Q) \lor (Q \Rightarrow \sim P)$
Show answer

Answer

AI
$F = (P \Rightarrow Q) \lor (Q \Rightarrow \sim P)$ $P \Rightarrow Q = P' + Q$ and $Q \Rightarrow \sim P = Q' + P'$ $F = P' + Q + Q' + P' = P' + 1 = 1$ Truth table:
PQP=>Q~PQ=>~PF
001111
011111
100011
111001
The final column is 1 for every combination of P and Q, so the proposition is a Tautology.
2020 · 1 mark · Short answerOpen: Convert the Boolean expression into its cardinal form.

Answer the following in short.

Convert the Boolean expression $F(X,Y,Z) = X'Y'Z + X'YZ' + XYZ$ into its cardinal form.
Show answer

Answer

AI
$F(X,Y,Z) = X'Y'Z + X'YZ' + XYZ$ Each term contains all three variables, so it is a minterm: $X'Y'Z = 001 = m_1$, $X'YZ' = 010 = m_2$, $XYZ = 111 = m_7$. Cardinal form: $F(X,Y,Z) = \Sigma(1, 2, 7)$ (verified with the boolean tool). Equivalently, in maxterm form: $F(X,Y,Z) = \pi(0, 3, 4, 5, 6)$.
2020 · 1 mark · Short answerOpen: Find the complement of the following Boolean expression using De Morgan’s law:

Answer the following in short.

Find the complement of the following Boolean expression using De Morgan’s law: $F(P, Q, R) = P + (Q' \cdot R)$
Show answer

Answer

AI
$F = P + (Q' \cdot R)$ $F' = [P + (Q' \cdot R)]' = P' \cdot (Q' \cdot R)'$ (De Morgan's law) $= P' \cdot (Q'' + R') = P' \cdot (Q + R')$ (De Morgan's law, involution law) So $F' = P'(Q + R') = P'Q + P'R'$ (verified with the boolean tool).
2020 · 2 marks · Short answerOpen: Convert the following expression to its cardinal SOP form:

Answer the following in short.

Convert the following expression to its cardinal SOP form: $F(P,Q,R) = P'Q'R + P'QR + PQ'R' + PQR'$
Show answer

Answer

AI
$F(P,Q,R) = P'Q'R + P'QR + PQ'R' + PQR'$ Each term already contains all three variables, so it is a minterm. Writing each in binary ($P'Q'R = 001$, $P'QR = 011$, $PQ'R' = 100$, $PQR' = 110$): $P'Q'R = m_1$, $P'QR = m_3$, $PQ'R' = m_4$, $PQR' = m_6$ Cardinal (sum of minterms) form: $F(P,Q,R) = \Sigma(1, 3, 4, 6)$
2020 · 1 mark · DrawingOpen: Draw the logic gate diagram for the reduced expression. Assume that the…

Answer the following for the given Boolean function.

Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs.

Draw: Logic gate diagram for the reduced expression

Show the case
Given the Boolean function $F(P, Q, R, S) = \pi (5, 7, 8, 10, 12, 14, 15)$:
Show answer

Answer

AI
Logic gate diagram for $F = (P' + S)(P + Q' + S')(Q' + R' + S')$: one 2-input OR gate with inputs (P', S) and two 3-input OR gates with inputs (P, Q', S') and (Q', R', S') feed a 3-input AND gate whose output is F. The complemented inputs P', Q', R', S' are taken directly as inputs.
Diagram for this answer
2020 · 2 marks · Short answerOpen: Simplify the following expression using Boolean laws:

Answer the following in short.

Simplify the following expression using Boolean laws: $F = [ (X' + Y) \cdot (Y' + Z) ]' + (X' + Z)$
Show answer

Answer

AI
$F = [(X' + Y) \cdot (Y' + Z)]' + (X' + Z)$ $= (X' + Y)' + (Y' + Z)' + (X' + Z)$ (De Morgan's law) $= XY' + YZ' + X' + Z$ (De Morgan's law, involution law) $= (X' + XY') + (Z + YZ')$ (regrouping) $= (X' + Y') + (Z + Y)$ (using $A + A'B = A + B$) $= (Y + Y') + X' + Z$ (regrouping) $= 1 + X' + Z$ (complement law) $= 1$ (null law) So $F = 1$ (verified with the boolean tool: the expression is a tautology).

Questions on other pages on Boolean Algebra

Other Computer Science chapters