Answer the following using a truth table.
Using a truth table, state whether the following proposition is a Tautology, Contradiction or Contingency:
$\sim(P =>Q) <=> (\sim P \lor Q)$
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Boolean Algebra - ISC Class 12 Computer Science Questions with Answers, Page 9
189 past-paper questions on Boolean Algebra from ISC Class 12 Computer Science papers (2026-2017), newest first, in full. Questions 161-180 are on this page, 20 to a page. Tap "Show answer" under a question to see its answer.
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Answer the following in short.
If $F(A, B, C) = A'.B'.C' + A'.B.C'$ then find $F'$ using De Morgan’s Law.
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Answer
AI$F(A,B,C) = A'.B'.C' + A'.B.C'$
By De Morgan's Law, $F' = (A'.B'.C')'.(A'.B.C')'$
Applying De Morgan's Law again to each term: $(A'.B'.C')' = A+B+C$ and $(A'.B.C')' = A+B'+C$
$F' = (A+B+C).(A+B'+C)$
(this can be further simplified using the distributive law: $(A+C)+B.B' = (A+C)+0 = A+C$, so $F' = A+C$)
Answer the following in short.
Find the dual of: $X.Y + X.Y' = X + 0$
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Answer
AIThe dual of an expression is obtained by interchanging AND (.) with OR (+) and interchanging 0 with 1, keeping the variables unchanged.
$X.Y + X.Y' = X + 0$
Dual: $(X+Y).(X+Y') = X.1$
Answer the following in short.
Write the canonical POS expression of: $F(P, Q) = \Pi( 0,2 )$
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Answer
AI$F(P,Q)=\pi(0,2)$ means F=0 for minterms 0 (P=0,Q=0) and 2 (P=1,Q=0); each maxterm is $(P+Q)$ for row 0 and $(P'+Q)$ for row 2 (complement the variable that is 1 in that row).
Canonical POS: $F(P,Q) = (P+Q).(P'+Q)$
Answer the following using a truth table.
Using the truth table, state whether the following proposition is a tautology, contingency or a contradiction:
$\sim( A \land B ) \lor ( \sim A \implies B )$
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Answer
AI$\sim(A \land B) \lor (\sim A \implies B)$
Truth table (verified with the boolean tool):
The final column is 1 (True) for every combination of A and B. Hence the given proposition is a Tautology.
| A | B | A.B | (A.B)' | A' | A'=>B | (A.B)'+(A'=>B) |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 1 | 1 |
Answer the following using a truth table.
Verify the following proposition with the help of a truth table:
$( P \land Q ) \lor ( P \land \sim Q ) = P$
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Answer
AI$(P \land Q) \lor (P \land \sim Q) = P$
Truth table (verified with the boolean tool):
The column for $(P \land Q) \lor (P \land \sim Q)$ is identical to the column for $P$ in every row. Hence the proposition is verified: $(P \land Q) \lor (P \land \sim Q) = P$.
| P | Q | P.Q | Q' | P.Q' | (P.Q)+(P.Q') | P |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 1 | 1 |
Reduce the following using a Karnaugh map.
Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).
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Answer
AI$F(A,B,C,D) = \pi(3,4,5,6,7,10,11,14,15)$: the 0s are plotted on a 4-variable K-map and grouped (POS form, verified with the boolean tool):
Quad (10, 11, 14, 15): $(A' + C')$
Quad (4, 5, 6, 7): $(A + B')$
Quad (3, 7, 11, 15): $(C' + D')$
Reduced expression: $F = (A' + C') \cdot (A + B') \cdot (C' + D')$
Answer the following.
Convert the following expression into its canonical POS form:
$F(X, Y, Z) = (X+Y' ) \cdot (Y'+Z)$
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Answer
AI$F(X,Y,Z) = (X+Y').(Y'+Z)$
Each term is expanded to include the missing variable by ORing it with (variable.variable'):
$(X+Y') = (X+Y'+Z.Z') = (X+Y'+Z).(X+Y'+Z')$
$(Y'+Z) = (Y'+Z+X.X') = (X+Y'+Z).(X'+Y'+Z)$
Combining and removing the repeated term $(X+Y'+Z)$:
Canonical POS: $F(X,Y,Z) = (X+Y'+Z).(X+Y'+Z').(X'+Y'+Z)$
(verified with the boolean tool)
Reduce the following using a Karnaugh map.
Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).
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Answer
AI$F(A,B,C,D)=\Sigma(0,2,4,8,9,10,12,13)$. Plotting on a 4-variable K-map and grouping (verified with the boolean tool):
Quad {8,9,12,13}: $AC'$
Quad {0,2,8,10}: $B'D'$
Quad {0,4,8,12}: $C'D'$
Minimal SOP: $F = AC' + B'D' + C'D'$
Answer the following.
If $F(A, B, C) = A' (BC' + B'C)$, then find $F'$
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Answer
AI$F(A,B,C) = A'.(B.C' + B'.C)$
By De Morgan's law: $F' = A + (B.C'+B'.C)'$
$(B.C'+B'.C)' = (B.C')'.(B'.C)' = (B'+C).(B+C') = B'.B + B'.C' + C.B + C.C' = B'C' + BC$
So $F' = A + BC + B'C'$
(verified with the boolean tool)
Reduce the following using a Karnaugh map.
Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs.
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Answer
AILogic gate diagram for $F = AC' + B'D' + C'D'$: three 2-input AND gates (AC', B'D', C'D') feeding one 3-input OR gate.
Answer the following.
Find the dual of:
$(A'+B) \cdot (1+B') = A'+B$
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Answer
AI$(A'+B).(1+B') = A'+B$
To find the dual: interchange $+$ with $\cdot$ and interchange $0$ with $1$, keeping all variables and their complements unchanged.
Dual of LHS: $(A'.B)+(0.B')$
Dual of RHS: $A'.B$
Dual: $(A'.B)+(0.B') = A'.B$
(verified with the boolean tool - both sides are equivalent, since $0.B'=0$)
Reduce the following using a Karnaugh map.
Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs.
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Answer
AILogic gate diagram for $F = (A' + C') \cdot (A + B') \cdot (C' + D')$: three 2-input OR gates feeding one 3-input AND gate.
Answer the following.
Simplify the following expression, using Boolean laws:
$A \cdot ( A' + B ) \cdot C \cdot ( A + B )$
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Answer
AI$A.(A'+B).C.(A+B)$
$A.(A'+B) = A.A' + A.B = 0 + AB = AB$ (Distributive law, then Complement law)
Expression becomes: $AB.C.(A+B) = ABC.(A+B)$
$ABC.(A+B) = ABC.A + ABC.B = ABC + ABC = ABC$ (Distributive law, then Idempotent law)
Final simplified expression (verified with the boolean tool): $F = ABC$
Answer the following using a truth table.
State the Commutative law and prove it with the help of a truth table.
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Answer
AICommutative Law: The order of the operands does not change the result of the OR or AND operation.
For OR: $A+B = B+A$
For AND: $A \cdot B = B \cdot A$
Proof by truth table (for OR; AND follows the same way):
The columns for $A+B$ and $B+A$ are identical in every row, so $A+B=B+A$. A similar truth table for $A.B$ and $B.A$ gives identical columns, proving $A.B=B.A$. Hence the Commutative law is verified.
| A | B | A+B | B+A |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 |
Answer the following using a truth table.
A training institute intends to give scholarships to its students as per the criteria given below :
• The student has excellent academic record but is financially weak.
OR
• The student does not have an excellent academic record and belongs to a backward class.
OR
• The student does not have an excellent academic record and is physically impaired.
The inputs are:
(In all the above cases 1 indicates yes and 0 indicates no).
Output : X [1 indicates yes, 0 indicates no for all cases]
Draw the truth table for the inputs and outputs given above and write the SOP expression for $X(A,F,C,I)$.
| INPUTS | |
| A | Has excellent academic record |
| F | Financially sound |
| C | Belongs to a backward class |
| I | Is physically impaired |
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Answer
AILet X=1 mean the student gets the scholarship. From the criteria: $X = A.F' + A'.C + A'.I$, where $A.F'$ means 'excellent record (A=1) but financially weak (F=0, i.e. NOT financially sound)', $A'.C$ means 'not an excellent record and belongs to a backward class', $A'.I$ means 'not an excellent record and is physically impaired'.
Truth table (columns A, F, C, I; verified with the boolean tool):
SOP expression: $X(A,F,C,I) = A.F' + A'.C + A'.I$
| A | F | C | I | X |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 0 | 1 |
| 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 | 0 |
Answer the following using a truth table.
A school intends to select candidates for an Inter-School Essay 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:
(In all the above cases 1 indicates yes and 0 indicates no).
Output : X [1 indicates yes, 0 indicates no for all cases]
Draw the truth table for the inputs and outputs given above and write the POS expression for $X(A, B, C, D)$.
| INPUTS | |
| A | participated in a competition earlier |
| B | is very creative |
| C | won prize in an inter-house competition |
| D | has excellent general awareness |
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Reduce the following using a Karnaugh map.
Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).
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Given the Boolean function $F(A, B, C, D) = \Sigma (2, 3, 4, 5, 6, 7, 8, 10, 11)$.
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Reduce the following using a Karnaugh map.
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
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Answer
AILogic gate diagram for $F = A'B + B'C + AB'D'$: three AND gates with inputs (A', B), (B', C) and (A, B', D') feed one 3-input OR gate whose output is F. The complemented inputs A', B', D' are taken directly as inputs.
Answer the following question.
If $(\sim P \implies Q)$ then write its inverse.
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Answer
AIGiven $\sim P \Rightarrow Q$. Inverse: negate both the antecedent and the consequent.
Inverse: $\sim(\sim P) \Rightarrow \sim Q$, i.e. $P \Rightarrow \sim Q$.