Answer the following.
Minterm
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Given that P = 0, Q = 1, R = 0, S = 0, write its:
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Answer
AIFor P = 0, Q = 1, R = 0, S = 0:
Minterm = P' • Q • R' • S'
Boolean Algebra - ISC Class 12 Computer Science Questions with Answers, Page 2
189 past-paper questions on Boolean Algebra from ISC Class 12 Computer Science papers (2026-2017), newest first, in full. Questions 21-40 are on this page, 20 to a page. Tap "Show answer" under a question to see its answer.
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Study the Assertion and Reason and choose the correct option.
Assertion: Its inverse in statement form is If you do not submit your assignment on time, then you will not get full marks for submission.
Reason: A conditional statement $P \Rightarrow Q$ can be expressed as $\sim P \vee Q$
$P$ = You submit your assignment on time
$Q$ = You get full marks for submission
If $P \Rightarrow Q$ then,
- (a)Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
- (b)Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
- (c)Assertion is true and Reason is false.
- (d)Both Assertion and Reason are false.
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Answer
AICorrect option: (b)
Answer: (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
Assertion: The inverse of $P \Rightarrow Q$ is $\sim P \Rightarrow \sim Q$, i.e. 'If you do not submit your assignment on time, then you will not get full marks for submission' — this correctly states the inverse, so the Assertion is true.
Reason: $P \Rightarrow Q \equiv \sim P \vee Q$ is also a true statement, but it explains how a conditional can be written as a disjunction — it does not explain why the given statement is the inverse of $P \Rightarrow Q$. Hence Reason does not correctly explain the Assertion.
Answer the following.
What is Gray code in Karnaugh map?
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Answer
AIGray code in a Karnaugh map is a unit-distance binary ordering (00, 01, 11, 10) used to label adjacent rows and columns such that consecutive cells differ by exactly one binary bit. This allows adjacent cells to be combined using Boolean reduction laws.
Answer the following.
Consider the two propositions given below:
A = You use ecofriendly methods
B = Pollution is reduced
If A implies to B, then write its Contrapositive statement.
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Answer
AIContrapositive statement:
If pollution is not reduced, then you do not use ecofriendly methods.
Symbolic form: ~B => ~A
Study the Assertion and Reason and choose the correct option.
Assertion: If Sujata is in the merit list, then she is a topper ($Y \Rightarrow X$).
Reason: Inverse is formed when both antecedent and consequent are negated.
Study the given propositions and the statements marked Assertion and Reason that follow it. Choose the correct option on the basis of your analysis.
$X$ – Sujata is a topper
$Y$ – Sujata is in the merit list
- (a)Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
- (b)Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
- (c)Assertion is true and Reason is false.
- (d)Both Assertion and Reason are false.
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Answer
AICorrect option: (b)
Answer: (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
The Reason correctly defines an Inverse (formed by negating both antecedent and consequent). However the Assertion 'Y⇒X' is actually the Converse of the original proposition X⇒Y, not its Inverse, so the Reason (about Inverse) does not correctly explain the Assertion, even though both statements are individually true.
Choose the correct option.
The complement of the Boolean expression $(P + Q') \cdot (R' + P)$ is:
- (a)$(P \cdot Q') + (R' \cdot P)$
- (b)$(P' \cdot Q) + (R \cdot P')$
- (c)$(P' \cdot Q) \cdot (R \cdot P')$
- (d)$(P + Q') \cdot (R' + P)$
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Answer
AICorrect option: (b)
Answer: (b) (P'.Q) + (R.P')
Complement of (P + Q').(R' + P) = (P + Q')' + (R' + P)' [De Morgan's law], applying De Morgan again inside each bracket = (P'.Q) + (R.P').
Answer the following using a truth table.
Post pandemic, to encourage the tourism industry in India, the Ministry of Tourism started a policy in which a tourist would be allowed to book a resort at a rebate, if any one of the following criteria matches.
● The tourist has an AADHAR card and has no criminal record.
● The tourist is the government employee and has repeated the resort visit in a span of six months.
● The tourist has an AADHAR card and has repeated the resort visit in a span of six months.
Inputs:
- A: The tourist has an AADHAR CARD
- C: The tourist has a criminal record
- G: The tourist is the government employee
- V: The tourist has repeated the resort visit in a span of six months
Output:
- F: The tourist would be allowed to book the resort at rebate. [1 indicates Yes and 0 indicates No].
Draw the truth table for inputs and outputs given above and write the Cardinal Sum of product expression for $F(A, C, G, V)$.
No answer yet.
Answer the following.
Write the minterms in canonical form for the Boolean Function X (A, B), from the truth table given below:
| A | B | X |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
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Answer
AIFrom the truth table, X=1 for A=0,B=0 (minterm $A'B'$) and A=1,B=1 (minterm $AB$).
Canonical (SOP) form: $X(A,B) = A'B' + AB = \Sigma(0,3)$
Choose the correct option.
The complement of the Boolean expression $(A \cdot B') + (B' \cdot C)$ is:
- (a)$(A + B') \cdot (B' + C)$
- (b)$(A' \cdot B) + (B \cdot C')$
- (c)$(A' + B) \cdot (B + C')$
- (d)$(A \cdot B') + (B \cdot C')$
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Answer
AICorrect option: (c)
Answer: (c) $(A' + B) \cdot (B + C')$
By De Morgan's law, the complement of $(A \cdot B') + (B' \cdot C)$ is $(A \cdot B')' \cdot (B' \cdot C)' = (A' + B) \cdot (B + C')$.
Reduce the following using a Karnaugh map.
$F(P, Q, R, S) = (P + Q + R + S) \cdot (P + Q + R + S') \cdot (P + Q + R' + S) \cdot (P + Q' + R + S) \cdot (P + Q' + R + S') \cdot (P + Q' + R' + S) \cdot (P + Q' + R' + S') \cdot (P' + Q + R + S) \cdot (P' + Q + R + S')$
(a)[4.0]
Reduce the Boolean function $F(P, Q, R, S) = (P + Q + R + S) \cdot (P + Q + R + S') \cdot (P + Q + R' + S) \cdot (P + Q' + R + S) \cdot (P + Q' + R + S') \cdot (P + Q' + R' + S) \cdot (P + Q' + R' + S') \cdot (P' + Q + R + S) \cdot (P' + Q + R + S')$ by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).
(b)[1.0]
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 (b)
Official answer keyLogic gate diagram for $F = (Q+R) \cdot (P+Q') \cdot (P+S)$, drawn from the three OR-gate outputs feeding a 3-input AND gate.
Three 2-input OR gates: OR gate 1 has inputs Q and R, giving output (Q+R). OR gate 2 has inputs P and S, giving output (P+S). OR gate 3 has inputs P and Q', giving output (P+Q'). The three OR-gate outputs are wired into the three inputs of a 3-input AND gate, whose output is labelled F(P,Q,R,S) = (Q+R).(P+Q').(P+S).
Choose the correct option.
Commutative law states that:
- (a)$A \cdot (A + B) = A$
- (b)$(A \cdot B) \cdot C = A \cdot (B \cdot C)$
- (c)$A + (B + C) = (A + B) + C$
- (d)$(A + B) = (B + A)$
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Answer
AICorrect option: (d)
Answer: (d) (A+B) = (B+A)
The Commutative law states that changing the order of operands does not change the result: A+B = B+A (and similarly A.B = B.A).
Study the Assertion and Reason and choose the correct option.
Assertion: s2 is converse of s1
Reason: Three-sided polygon must be a triangle.
Study the given propositions and the statements marked Assertion and Reason that follow it. Choose the correct option on the basis of your analysis.
p = I am a triangle
q = I am a three-sided polygon
s1 = $p \rightarrow q$
s2 = $q \rightarrow p$
- (a)Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
- (b)Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
- (c)Assertion is true and Reason is false.
- (d)Assertion is false and Reason is true.
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Answer
Official answer keyCorrect option: (b)
Answer: (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
Given $s1: p \rightarrow q$, its converse is $q \rightarrow p$, which is exactly $s2$, so the Assertion ('s2 is the converse of s1') is true - this is a structural/definitional fact about conditionals. The Reason ('a three-sided polygon must be a triangle') is also true, but it is a fact of geometry, unrelated to why $s2$ is termed the converse of $s1$ - so it does not explain the Assertion.
Study the Assertion and Reason and choose the correct option.
Assertion: If it is not a Sunday, then it is not a holiday. ($Q' \Rightarrow P'$)
Reason: Inverse is formed when antecedent and consequent are interchanged.
Study the given propositions and the statements marked, Assertion and Reason that follow it. Choose the correct option on the basis of your analysis.
P – It is a holiday
Q – It is a Sunday
- (a)Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
- (b)Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
- (c)Assertion is true and Reason is false.
- (d)Both Assertion and Reason are false.
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Answer
AICorrect option: (c)
Answer: (c) Assertion is true and Reason is false.
$Q' \Rightarrow P'$ is the contrapositive of $P \Rightarrow Q$, and a contrapositive is always logically equivalent to the original conditional, so the Assertion is a valid, true transformation. However, the Reason is false: the Inverse of a conditional is formed by negating both the antecedent and consequent ($\sim P \Rightarrow \sim Q$), not by interchanging them - interchanging antecedent and consequent gives the Converse, not the Inverse.
Answer the following from the logic gate diagram.
From the logic gate diagram given below:
(a)[4.0]
Derive Boolean expression for (1), (2) and R. Reduce the derived expression.
(b)[1.0]
Name the logic gate that represents the reduced expression.
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Answer (a)
Checked answer.
(1) $= (A \oplus B)' = A \cdot B + A' \cdot B'$, (2) $= A + B$, $R = (A \cdot B)' = A' + B'$
- The first gate is an XNOR gate with inputs A and B: (1) $= (A \oplus B)' = A \cdot B + A' \cdot B'$
- The second gate is an OR gate with inputs A and B: (2) $= A + B$
- The last gate is a NAND gate with inputs (1) and (2): $R = [(A \cdot B + A' \cdot B') \cdot (A + B)]'$
- Expanding: $(A \cdot B + A' \cdot B') \cdot (A + B) = A \cdot B + A \cdot B \cdot B + A \cdot A' \cdot B' + A' \cdot B \cdot B'$
- $A \cdot A' = 0$ and $B \cdot B' = 0$ (Complement law), and $A \cdot B + A \cdot B = A \cdot B$ (Idempotent law), so the product is $A \cdot B$
- $R = (A \cdot B)' = A' + B'$ (De Morgan's law)
Answer (b)
Checked answer.
$R = (A \cdot B)'$, so the reduced expression represents a NAND gate.
Choose the correct option.
The complement of the reduced expression of $F(A, B) = \Sigma (0, 1, 2, 3)$ is:
- (a)1
- (b)$A \cdot B$
- (c)0
- (d)$A' + B'$
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Answer
Official answer keyCorrect option: (c)
Answer: (c) 0
For 2 variables A and B there are only 4 possible minterms (0,1,2,3), and $F(A,B)=\Sigma(0,1,2,3)$ includes all of them, so the reduced expression is $F=1$ (always true). The complement of this reduced expression is therefore $1'=0$.
Study the Assertion and Reason and choose the correct option.
Assertion: The contrapositive of $p' \Rightarrow q$ is $q' \Rightarrow p$.
Reason: Contrapositive is the conditional statement, obtained after interchanging antecedent and consequent.
- (a)Both A and R are true, and R is the correct explanation of A.
- (b)Both A and R are true, but R is not the correct explanation of A.
- (c)A is true, but R is false.
- (d)A is false, but R is true.
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Answer
AICorrect option: c
Answer: (c) A is true, but R is false.
For p' => q, the contrapositive negates both sides and swaps them: ~q => ~(p') = q' => p, exactly as the assertion states, so A is true. R is false because it describes only interchanging the antecedent and consequent without negating them - that is the definition of the converse, not the contrapositive.
Answer the following questions.
Given the Boolean function $F(P,Q,R,S) = \Sigma(2,3,5,7,8,10,11,12,13,15)$.
(a)[3.0]
Reduce the above expression by using a 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).
(b)[2.0]
Draw the logic gate diagram for the reduced expression using NAND gate only. Assume that the variables and their complements are available as inputs.
Draw: Logic gate diagram for reduced expression using only NAND gates
Must show: NAND gates
Show answer
Answer (b)
AINAND-only (NAND-NAND) implementation of $F = PR'S' + QS + Q'R$: one NAND gate for each of the three product terms, and a 3-input NAND gate combining them, since $[(PR'S')' \cdot (QS)' \cdot (Q'R)']' = PR'S' + QS + Q'R$ by De Morgan's law. Variables and their complements (P, P', Q, Q', R, R', S, S') are assumed available as inputs, as stated.
Answer the following using a truth table.
By using truth table, verify if the following proposition is valid or not.
No answer yet.
Differentiate between the following.
Differentiate between Tautology and Contradiction.
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Answer
AIA Tautology is a compound proposition that is TRUE for every possible combination of truth values of its component propositions (the result column of its truth table is all 1's). A Contradiction is a compound proposition that is FALSE for every possible combination of truth values of its component propositions (the result column of its truth table is all 0's).
Answer the following using a truth table and a Karnaugh map.
A food delivery app offers free home delivery to its customers who meet any of the following criteria.
• The order is above ₹ 1000 and payment is made through UPI
OR
• Food is ordered from a partner restaurant and payment is made through UPI
OR
• The customer uses the app for the first time and places order above ₹ 1000
The inputs are:
(In all the above cases, 1 indicates YES, 0 indicates NO)
Output: D - Denotes free home delivery [1 indicates YES and 0 indicates NO in all cases]
| INPUTS | |
|---|---|
| A | Order is above ₹ 1000 |
| U | Payment is done through UPI |
| P | Food is ordered from a partner restaurant |
| F | Customer uses the app for the first time |
(i)[5.0]
Draw a truth table for the inputs and the outputs given above. Write the SOP expression for D(A, U, P, F).
(ii)(a)[2.5]
Reduce the above expression D(A, U, P, F) by using 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs).
(ii)(b)[2.5]
Draw the logic gate diagram for the reduced expression using NAND gates only. Assume that the variables and their complements are available as inputs.
Draw: logic gate diagram using NAND gates only for the reduced expression
Show answer
Answer
AI3(i): | A | U | P | F | D |
D(A,U,P,F) = Σm(6,7,9,11,12,13,14,15)
SOP expression:
D(A,U,P,F) = A'UPF' + A'UPF + AU'P'F + AU'PF + AUP'F' + AUP'F + AUPF' + AUPF
3(ii): K-map (variables A, U, P, F) grouping the eight 1s (minterms 6, 7, 9, 11, 12, 13, 14, 15) into three quads (verified with the boolean tool):
Quad 1: cells 12, 13, 14, 15 (A=1, U=1) → term A.U
Quad 2: cells 9, 11, 13, 15 (A=1, F=1) → term A.F
Quad 3: cells 6, 7, 14, 15 (U=1, P=1) → term U.P
Reduced (minimal) SOP: D(A,U,P,F) = A.U + A.F + U.P
With NAND gates only (NAND-NAND): one 2-input NAND gate for each term and a 3-input NAND gate combining them, since $[(AU)' \cdot (AF)' \cdot (UP)']' = AU + AF + UP$ by De Morgan's law.
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |