Answer the following using a truth table.
Draw a truth table with a 3 input combination which outputs 1 if there are odd number of 0’s. Also derive an SOP expression for the output. Reduce the expression using Karnaugh Map.
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Boolean Algebra - ISC Class 12 Computer Science Questions with Answers, Page 8
189 past-paper questions on Boolean Algebra from ISC Class 12 Computer Science papers (2026-2017), newest first, in full. Questions 141-160 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 using a truth table.
Draw the truth table to prove the proportional logic expression:
$(X \Rightarrow Y) \land (Y \Rightarrow X) = X \Leftrightarrow Y$
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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
Show the case
Given the Boolean function: $F(A,B,C,D) = \Sigma (0, 1, 2, 3, 4, 5, 8, 9, 10, 11, 12, 13, 14)$.
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Answer
AILogic gate diagram for $F = B' + C' + AD'$: one 2-input AND gate takes A and D'; its output, together with B' and C', feeds a 3-input OR gate whose output is F. The complemented inputs B', C', D' are taken directly as inputs.
Answer the following in short.
State Associative law and prove it with the help of a truth table.
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Answer
AIAssociative law: the way in which the variables are grouped in an OR (or AND) operation does not change the result.
$A + (B + C) = (A + B) + C$ and $A \cdot (B \cdot C) = (A \cdot B) \cdot C$
Truth table proof:
The columns for $A+(B+C)$ and $(A+B)+C$ are identical, and so are the columns for $A.(B.C)$ and $(A.B).C$, so the Associative law is proved.
| A | B | C | B+C | A+(B+C) | A+B | (A+B)+C | B.C | A.(B.C) | A.B | (A.B).C |
|---|---|---|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Simplify the expression and draw its circuit.
Simplify the following Boolean expression and draw the gate for the reduced expression:
$F = A'B + AB'C + A$
Draw: Gate for the reduced expression
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Answer
AI$F = A'B + AB'C + A$
$= A + A'B + AB'C$ (rearranging)
$= (A + B) + AB'C$ (using $A + A'B = A + B$)
$= A + B$ (absorption law, since $A + AB'C = A$)
Reduced expression: $F = A + B$ (verified with the boolean tool).
Gate for the reduced expression: a single 2-input OR gate with inputs A and B, whose output is F.
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
Show the case
Given the Boolean function: $F(A, B, C, D) = \pi ( 3, 4, 6, 9, 11, 12, 13, 14, 15 )$.
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Answer
AILogic gate diagram for $F = (A' + D')(B' + D)(B + C' + D')$: two 2-input OR gates with inputs (A', D') and (B', D), and one 3-input OR gate with inputs (B, C', D'), feed a 3-input AND gate whose output is F. The complemented inputs A', B', C', D' are taken directly as inputs.
Answer the following for the given Boolean function.
Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).
Show the case
Given the Boolean function $F(P, Q, R, S) = \pi (5, 7, 8, 10, 12, 14, 15)$:
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Answer the following in short.
Find the dual of: $(A' + 0) \cdot (B' + 1) = A'$
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Answer
AITo get the dual, interchange $+$ with $\cdot$ and $0$ with $1$; the variables and their complements stay unchanged.
Given: $(A' + 0) \cdot (B' + 1) = A'$
Dual: $(A' \cdot 1) + (B' \cdot 0) = A'$
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 the following using a truth table.
A company intends to develop a device to show the high status power load for a household invertor depending on the criteria given below:
• If Air conditioner and Geyser are on
OR
• If Air conditioner is off, but Geyser and Refrigerator are on
OR
• If Geyser is off, but Air conditioner and Water purifier are on
OR
• When all are on
The inputs are:
(In all the above cases 1 indicates yes and 0 indicates no.)
Output : X [1 indicates high power, 0 indicates low power for all cases]
Draw the truth table for the inputs and outputs given above and write the SOP expression for $X(A,G,R,W)$.
| INPUTS | |
|---|---|
| A | Air conditioner is on |
| G | Geyser is on |
| R | Refrigerator is on |
| W | Water purifier is on |
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Answer the following in short.
Define Proposition. How does tautology differ from contradiction?
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Answer
AIA proposition is a declarative statement that is either true or false but not both, e.g. 'The Sun rises in the east'.
Tautology: a compound proposition that is always true, i.e. its truth table has 1 (True) in every row of the result column, e.g. $P \lor \sim P$.
Contradiction: a compound proposition that is always false, i.e. its truth table has 0 (False) in every row of the result column, e.g. $P \land \sim P$.
(A proposition that is sometimes true and sometimes false is called a contingency.)
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 $F(A, B, C, D) = \Sigma (0, 2, 3, 6, 8, 10, 11, 14, 15)$:
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Answer
AILogic gate diagram for $F = AC + B'C + B'D' + CD'$: four 2-input AND gates with inputs (A, C), (B', C), (B', D') and (C, D') feed one 4-input OR gate whose output is F. The complemented inputs B' and D' are taken directly as inputs.
Answer the following using a truth table.
The owner of a company pays bonus to his salesmen as per the criteria given below:
• If the salesman works overtime for more than 4 hours but does not work on off days/holidays.
OR
• If the salesman works when festival sales are on and updates showroom arrangements.
OR
• If the salesman works on an off day/holiday when the festival sales are on.
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 $X(O,F,H,U)$.
| INPUTS | |
|---|---|
| O | Works overtime for more than 4 hours |
| F | Festival sales are on |
| H | Working on an off day/holiday |
| U | Updates showroom arrangements |
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Write the proposition asked for.
Converse
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If A= “It is cloudy” and B= “It is raining”, then write the proposition for:
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Answer
AIConverse ($B \Rightarrow A$): "If it is raining, then it is cloudy."
Answer the following in short.
Simplify the following expression, using Boolean laws:
$( X + Z ) . ( X.Y + Y.Z' ) + X.Z + Y$
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Answer
AI$(X+Z).(X.Y+Y.Z') + X.Z + Y$
$X.Y+Y.Z' = Y.(X+Z')$ (Distributive law - factor out Y)
$(X+Z).(X.Y+Y.Z') = (X+Z).Y.(X+Z') = Y.(X+Z).(X+Z')$
$(X+Z).(X+Z') = X+Z.Z' = X+0 = X$ (Distributive law, then Complement law $Z.Z'=0$, then Identity law)
So $(X+Z).(X.Y+Y.Z') = Y.X = XY$
Expression becomes: $XY + XZ + Y$
$XY+Y = Y$ (Absorption law), so $XY+XZ+Y = (XY+Y)+XZ = Y+XZ$
Final simplified expression (verified with the boolean tool): $F = Y + XZ$
Answer the following for the given Boolean function.
Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).
Show the case
Given the Boolean function $F(A, B, C, D) = \Sigma ( 0, 2, 3, 4, 5, 8, 10, 11, 12, 13 )$.
No answer yet.
Answer the following for the given Boolean function.
Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).
Show the case
Given the Boolean function: $F(P, Q, R, S) = \pi ( 0, 1, 2, 8, 9, 11, 13, 15 )$.
No answer yet.
Write the proposition asked for.
Contrapositive
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Answer
AITaking the underlying proposition as "If A, then B" (i.e. "If it is cloudy, then it is raining"):
Contrapositive ($\sim B \Rightarrow \sim A$): "If it is not raining, then it is not cloudy."
Answer the following for the given Boolean function.
Draw the logic gate diagram for the reduced expression using only NAND gates. Assume that the variables and their complements are available as inputs.
Draw: Logic gate diagram of the reduced expression using only NAND gates
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Answer
AINAND-only (NAND-NAND) implementation of $F = BC' + B'C + C'D'$: one 2-input NAND gate for each product term, and a 3-input NAND gate combining them. By De Morgan's law, $[(BC')' \cdot (B'C)' \cdot (C'D')']' = BC' + B'C + C'D'$.
Answer the following for the given Boolean function.
Draw the logic gate diagram for the reduced expression using only NOR gates. Assume that the variables and their complements are available as inputs.
Draw: Logic gate diagram of the reduced expression using only NOR gates
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Answer
AINOR-only (NOR-NOR) implementation of $F = (P'+S') \cdot (P+Q+S) \cdot (Q+R)$: one NOR gate for each sum term, and a 3-input NOR gate combining them. By De Morgan's law, $[(P'+S')' + (P+Q+S)' + (Q+R)']' = (P'+S') \cdot (P+Q+S) \cdot (Q+R)$.
