Boolean Algebra - ISC Class 12 Computer Science Questions with Answers, Page 5

189 past-paper questions on Boolean Algebra from ISC Class 12 Computer Science papers (2026-2017), newest first, in full. Questions 81-100 are on this page, 20 to a page. Tap "Show answer" under a question to see its answer.

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2024 · 1 mark · MCQOpen: The complement of the Boolean expression is:

Choose the correct option.

The complement of the Boolean expression $(P' \cdot Q) + (R \cdot S')$ is:

(a)$(P' + Q) \cdot (R' + S)$
(b)$(P + Q') \cdot (R' + S)$
(c)$(P' + Q) \cdot (R + S')$
(d)$(P + Q') \cdot (R + S')$

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Answer

Correct option: (b)

Answer: (b) $(P + Q') \cdot (R' + S)$ By De Morgan's laws: $[(P' \cdot Q) + (R \cdot S')]' = (P' \cdot Q)' \cdot (R \cdot S')' = (P+Q')\cdot(R'+S)$.

2024 · 2 marks · Short answerOpen: Write the canonical form of the cardinal terms, and for .

Answer the following.

Write the canonical form of the cardinal terms, $m_3$ and $M_5$ for $F (A, B, C, D)$.

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Answer

For F(A,B,C,D) (4 variables), an uncomplemented variable in a minterm corresponds to bit 1 and a complemented variable corresponds to bit 0 (minterm index read directly as the binary value ABCD); for a maxterm the convention is reversed - an uncomplemented variable corresponds to bit 0 and a complemented variable corresponds to bit 1. $m_3$: 3 = 0011 in binary (A B C D), so A=0, B=0, C=1, D=1: $m_3 = A'B'CD$ $M_5$: 5 = 0101 in binary (A B C D), so A=0, B=1, C=0, D=1 (bit 1 -> complemented literal, bit 0 -> uncomplemented literal): $M_5 = A + B' + C + D'$

2024 · 1 mark · MCQOpen: According to the Principle of duality, the Boolean equation will be equivalent…

Choose the correct option.

According to the Principle of duality, the Boolean equation $(A + B') \cdot (A + 1) = A + B'$ will be equivalent to:

(a)$(A' + B) \cdot (A' + 1) = A' + B$
(b)$(A \cdot B') + (A \cdot 0) = A \cdot B'$
(c)$(A' \cdot B) + (A' \cdot 1) = A' \cdot B$
(d)$(A' \cdot B) + (A' \cdot 0) = A' \cdot B$

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Answer

Correct option: (b)

Answer: (b) $(A \cdot B') + (A \cdot 0) = A \cdot B'$ Applying the Principle of Duality: interchange every $+$ with $\cdot$, and every $1$ with $0$ (variables and complements stay unchanged). $(A+B')\cdot(A+1)=A+B'$ becomes $(A\cdot B')+(A\cdot 0)=A\cdot B'$.

2024 · 1 mark · One wordOpen: Write the canonical SOP expression for

Answer the following.

Write the canonical SOP expression for $F (A, B) = A \Leftrightarrow B$

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Answer

$F(A,B) = A \Leftrightarrow B$ is true (1) when A and B have the same value, i.e. when $A=B=0$ or $A=B=1$. Canonical SOP: $F(A,B) = A'B' + AB$

2024 · 10 marks · Case basedOpen: A Football Association coach analyses the criteria for a win/draw of his team…

Answer the following using a truth table and a Karnaugh map.

A Football Association coach analyses the criteria for a win/draw of his team depending on the following conditions. • If the Centre and Forward players perform well but Defenders do not perform well. OR • If Goalkeeper and Defenders perform well but the Centre players do not perform well. OR • If all perform well. The inputs are:

INPUTS
C	Centre players perform well
D	Defenders perform well
F	Forward players perform well
G	Goalkeeper perform well

(In all the above cases, 1 indicates yes and 0 indicates no.) Output: X - Denotes the win/draw criteria [1 indicates win/draw and 0 indicates defeat in all cases]

(i)[5.0]

Draw the truth table for the inputs and outputs given above and write the SOP expression for $X(C, D, F, G)$.

(ii)(a)[2.5]

Reduce the above expression $X (C, D, F, G)$ 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. Assume that the variables and their complements are available as inputs.

Draw: logic gate diagram for the reduced expression

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Answer

3(i): Truth table for $X(C,D,F,G)$:

C	D	F	G	X
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	1
0	1	1	0	0
0	1	1	1	1
1	0	0	0	0
1	0	0	1	0
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	1

From the three given conditions: Condition 1 (Centre, Forward well; Defenders not well): $C \cdot D' \cdot F$ Condition 2 (Goalkeeper, Defenders well; Centre not well): $C' \cdot D \cdot G$ Condition 3 (all perform well): $C \cdot D \cdot F \cdot G$ SOP expression: $X(C,D,F,G) = C \cdot D' \cdot F + C' \cdot D \cdot G + C \cdot D \cdot F \cdot G$ 3(ii): Plotting the minterms of $X = CD'F + C'DG + CDFG$, i.e. $\Sigma(5,7,10,11,15)$, on a 4-variable K-map (rows CD: 00,01,11,10; columns FG: 00,01,11,10) gives three pairs (no quad or octet is possible for this function): Pair (10,11): $C=1,D=0,F=1$, G varies $\Rightarrow CD'F$ Pair (5,7): $C=0,D=1,G=1$, F varies $\Rightarrow C'DG$ Pair (7,15): $D=1,F=1,G=1$, C varies $\Rightarrow DFG$ Reduced (minimal) SOP: $X(C,D,F,G) = CD'F + C'DG + DFG$ Logic gate diagram: three 3-input AND gates feeding one 3-input OR gate.

2024 · 2 marks · Short answerOpen: Find the complement of the following expression and reduce it by using Boolean…

Answer the following.

Find the complement of the following expression and reduce it by using Boolean laws. $P \cdot (P + Q) \cdot Q \cdot (Q + R')$

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Answer

Let E = P.(P+Q).Q.(Q+R') By absorption law, P.(P+Q) = P, so E = P.Q.(Q+R') Again by absorption law, Q.(Q+R') = Q, so E = P.Q Complement of E: E' = (P.Q)' = P' + Q' (De Morgan's law) Verification by taking complement of the original expression directly: E' = [P.(P+Q).Q.(Q+R')]' = P' + (P+Q)' + Q' + (Q+R')' (De Morgan's law) = P' + P'.Q' + Q' + Q'.R (De Morgan's law again) = P' + Q' + Q'.R (absorption: P' + P'.Q' = P') = P' + Q' (absorption: Q' + Q'.R = Q') Reduced complement: E' = P' + Q'

2023 · 5 marks · Case basedOpen: Given the Boolean function . Reduce the above expression by using 4-variable…

Answer the following.

Given the Boolean function $F(A,B,C,D) = \Sigma(2, 3, 6, 7, 8, 10, 12, 14, 15)$.

(a)[4.0]

Reduce the above expression 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

Must show: variables and their complements as inputs

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Answer (b)

Logic gate diagram for $F = A'C + BC + AD'$: three 2-input AND gates (A', C), (B, C) and (A, D') feed one 3-input OR gate whose output is F.

2023 · 1 mark · MCQOpen: If then its contra positive will be:

Answer the following.

If $(\sim p \Rightarrow \sim q)$ then its contra positive will be:

(a)$p \Rightarrow q$
(b)$q \Rightarrow p$
(c)$\sim q \Rightarrow p$
(d)$\sim p \Rightarrow q$

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Answer

Correct option: (b)

Answer: (b) $q \Rightarrow p$ The contrapositive of $a \Rightarrow b$ is $\sim b \Rightarrow \sim a$; here $a = \sim p$, $b = \sim q$, so it is $\sim(\sim q) \Rightarrow \sim(\sim p)$, i.e. $q \Rightarrow p$.

2023 · 2 marks · Short answerOpen: Simplify the following expression using Boolean laws:

Answer the following.

Simplify the following expression using Boolean laws: $F = PQ + ( P + Q ) \cdot ( P + PR ) + Q$

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Answer

$F = PQ + (P+Q)\cdot(P+PR) + Q$ $P+PR = P(1+R) = P$ (since $1+R=1$) So $(P+Q)\cdot(P+PR) = (P+Q)\cdot P = P$ (Absorption law: $P\cdot(P+Q)=P$) $F = PQ + P + Q$ $PQ+P = P$ (Absorption law) $F = P + Q$ (verified with the boolean tool) Final simplified expression: $F = P+Q$

2023 · 3 marks · Truth tableOpen: Verify if the following proposition is valid:

Answer the following using a truth table.

Verify if the following proposition is valid: $( P \Rightarrow Q ) \wedge ( P \Rightarrow R ) = P \Rightarrow ( Q \wedge R )$

No answer yet.

2023 · 1 mark · MCQOpen: The dual of is:

Answer the following.

The dual of $(X' + 1) \cdot (Y' + 0) = Y'$ is:

(a)$X \cdot 0 + Y \cdot 1 = Y$
(b)$X' \cdot 1 + Y' \cdot 0 = Y'$
(c)$X' \cdot 0 + Y' \cdot 1 = Y'$
(d)$(X' + 0) + (Y' + 1) = Y'$

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Answer

Correct option: (c)

Answer: (c) $X' \cdot 0 + Y' \cdot 1 = Y'$ For the dual, interchange $+$ and $\cdot$ and interchange 0 and 1, leaving the variables and their complements unchanged.

2023 · 1 mark · MCQOpen: The law which states is:

Choose the correct option.

The law which states $a + (b \cdot c) = (a + b) \cdot (a + c)$ is:

(a)Associative Law
(b)Distributive Law
(c)Involution Law
(d)Commutative Law

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Answer

Correct option: (b)

Answer: (b) Distributive Law The identity $a+(b\cdot c)=(a+b)\cdot(a+c)$ is the (OR-over-AND) form of the Distributive Law, which distributes one operation over the other.

2023 · 1 mark · One wordOpen: Write the canonical sum of product form of the function .

Answer the following.

Write the canonical sum of product form of the function $y(A, B) = A + B$.

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Answer

The canonical SOP form lists every minterm where the output is 1. For $y=A+B$, the output is 1 for $(A,B)=(0,1),(1,0),(1,1)$, i.e. minterms $A'B, AB', AB$: $y(A,B) = A'B + AB' + AB$

2023 · 5 marks · Truth tableOpen: A shopping mall allows customers to shop using cash or credit card of any…

Answer the following.

A shopping mall allows customers to shop using cash or credit card of any nationalised bank. It awards bonus points to their customers on the basis of criteria given below: • The customer is an employee of the shopping mall and makes the payment using a credit card OR • The customer shops items which carry bonus points and makes the payment using a credit card with a shopping amount of less than ₹10,000/- OR • The customer is not an employee of the shopping mall and makes the payment not through a credit card but in cash for the shopping amount above ₹10,000/- The inputs are:

INPUTS
C	Payment through a credit card
A	Shopping amount is above ₹10,000/-
E	The customer is an employee of the shopping mall
I	Item carries a bonus point

(In all the above cases, 1 indicates yes and 0 indicates no.) Output: X [1 indicates bonus point awarded, 0 indicates bonus point not awarded for all cases] Draw the truth table for the inputs and outputs given above and write the POS expression for $X (C, A, E, I)$.

No answer yet.

2023 · 5 marks · Case basedOpen: Given the Boolean function . Reduce the above expression by using 4-variable…

Reduce the following using a Karnaugh map.

Given the Boolean function $F(A, B, C, D) = \pi(0, 1, 3, 5, 6, 7, 9, 11, 13, 14, 15)$.

(a)[4.0]

Reduce the above expression 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

Must show: variables and their complements as inputs

Show answer

Answer (b)

Logic gate diagram for $F=(A+B+C)\cdot(B'+C')\cdot D'$: a 3-input OR gate (A, B, C) and a 2-input OR gate (B', C') feed a 3-input AND gate together with D'.

2023 · 2 marks · Short answerOpen: Write the maxterm and minterm for the function when, and .

Answer the following.

Write the maxterm and minterm for the function $F(A, B, C, D)$ when, $A=1, B=1, C=0$ and $D=1$.

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Answer

For $A=1, B=1, C=0, D=1$: reading ABCD as binary $1101 = 13$ in decimal. Minterm ($m_{13}$): uncomplemented literal for 1, complemented for 0: $m_{13} = ABC'D$ Maxterm ($M_{13}$): complemented literal for 1, uncomplemented for 0 (reverse convention): $M_{13} = A'+B'+C+D'$

2023 · 5 marks · Case basedOpen: Given the Boolean function . Reduce the above expression by using 4-variable…

Answer the following.

Given the Boolean function $F(A,B,C,D) = \pi(0, 1, 2, 4, 5, 8, 10, 11, 14, 15)$.

(a)[4.0]

Reduce the above expression 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

Must show: variables and their complements as inputs

Show answer

Answer (b)

$F = (A + C)(A' + C')(B + D)$ is drawn with three 2-input OR gates whose outputs feed one 3-input AND gate.

2023 · 3 marks · Short answerOpen: From the logic diagram given below, write the Boolean expression for (1) and…

Answer the following.

From the logic diagram given below, write the Boolean expression for (1) and (2). Also, derive the Boolean expression (F) and simplify it.

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Answer

From the diagram: gate (1) is a NAND gate with inputs X and Y, gate (2) is a NOR gate with inputs Y and Z, and their outputs go into an OR gate giving F. (1) $= (X \cdot Y)'$ (2) $= (Y + Z)'$ $F = (X \cdot Y)' + (Y + Z)' = X' + Y'$

1. Output of the NAND gate: (1) $= (X \cdot Y)'$
2. Output of the NOR gate: (2) $= (Y + Z)'$
3. The OR gate adds them: $F = (X \cdot Y)' + (Y + Z)'$
4. By De Morgan's laws: $F = X' + Y' + Y' \cdot Z'$
5. By the absorption law ($Y' + Y'Z' = Y'$): $F = X' + Y'$
6. By De Morgan's law this can also be written $F = (X \cdot Y)'$, so $F = X' + Y'$

2023 · 1 mark · MCQOpen: According to De Morgan’s law will be equal to:

Answer the following.

According to De Morgan’s law $(a + b + c')'$ will be equal to:

(a)$a' + b' + c'$
(b)$a' + b' + c$
(c)$a' \cdot b' \cdot c'$
(d)$a' \cdot b' \cdot c$

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Answer

Correct option: (d)

Answer: (d) $a' \cdot b' \cdot c$ By De Morgan's law the complement of a sum is the product of the complements: $(a + b + c')' = a' \cdot b' \cdot (c')' = a' \cdot b' \cdot c$.

2023 · 1 mark · MCQOpen: The complement of the Boolean expression is:

Choose the correct option.

The complement of the Boolean expression $(P \cdot Q)' + R'$ is:

(a)$(P + Q) \cdot R$
(b)$PQR$
(c)$(P' + Q') \cdot R'$
(d)$(P' + Q') \cdot R$

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Answer

Correct option: (b)

Answer: (b) $PQR$ By De Morgan's laws, complementing $(P\cdot Q)'+R'$ gives $[(P\cdot Q)']' \cdot [R']' = (P\cdot Q)\cdot R = PQR$ (verified with the boolean tool).

Boolean Algebra - ISC Class 12 Computer Science Questions with Answers, Page 5

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Questions on other pages on Boolean Algebra

Other Computer Science chapters

C	D	F	G	X
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	1
0	1	1	0	0
0	1	1	1	1
1	0	0	0	0
1	0	0	1	0
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	1

C	D	F	G	X
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	1
0	1	1	0	0
0	1	1	1	1
1	0	0	0	0
1	0	0	1	0
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	1

C	D	F	G	X
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	1
0	1	1	0	0
0	1	1	1	1
1	0	0	0	0
1	0	0	1	0
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	1