Question 13 Marks
Construct the simplified circuit for the following circuit.
Answer
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Questions
Solve the Following Question.(3 Marks)
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20 questions · timed · auto-graded
Question 23 Marks
Using truth table prove that : $p \leftrightarrow q=(p \wedge q) \vee(\sim p \wedge \sim q)$
Answer
View full question & answer→1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
A | B | ||||||
P | q | $p \leftrightarrow q$ | $p \wedge q$ | $\sim p$ | $\sim q$ | $\begin{gathered}\sim p \wedge \\ \sim q\end{gathered}$ | $A \vee B$ |
T T F F
| T F T F | T F F T | T F F F | F F T T
| F T F T | F F F T | T F F T |
By column number 3 and 8
p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Question 33 Marks
If $p:$ It is a day time, $q:$ It is warm, write the compound statement in verbal form denoted by -
(a) $p \wedge \sim q$
(b) $\sim p \rightarrow q$
(c) $q \leftrightarrow p$
(a) $p \wedge \sim q$
(b) $\sim p \rightarrow q$
(c) $q \leftrightarrow p$
Answer
View full question & answer→Given, p : It is a day time.
q : It is warm.
a) p ∧ ∼ q : It is a day time but it is not warm.
b) ∼p → q : It is not a day time then it is warm.
c) q ↔ p : It is warm if and only it is day time.
Question 43 Marks
Construct the new switching circuit for the following circuit with only one switch by simplifying the given circuit.
Answer
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Question 53 Marks
State the converse, inverse and contrapositive of the conditional statement : 'If a sequence is bounded, then it is convergent'.
Answer
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Question 63 Marks
Write the truth values of the following statements :
(a) 2 is a rational number and $\sqrt{2}$ is an irrational number.
(b) $2+3=5$ or $\sqrt{2}+\sqrt{3}=\sqrt{5}$
(a) 2 is a rational number and $\sqrt{2}$ is an irrational number.
(b) $2+3=5$ or $\sqrt{2}+\sqrt{3}=\sqrt{5}$
Answer
View full question & answer→(1) Let p : 2 is rational number, q : $\sqrt{2}$ is an irrational number. . So given compound statement is p ^ q. Truth value of of p is T and that q is T. Hence, truth value p ^ q = T ^ T =T.
(2) Let p = 2 + 3 = 5 which is T.
q = $\sqrt{2}+\sqrt{3}=\sqrt{5}$ which is F
∴ Truth value of given statment is T ∨ F = T.
Question 73 Marks
Write the negations of the following statements :
(a) If diagonals of a parallelogram are perpendicular, then it is a rhombus.
(b) Mangoes are delicious, but expensive.
(c) A person is rich if and only if he is a software engineer.
(a) If diagonals of a parallelogram are perpendicular, then it is a rhombus.
(b) Mangoes are delicious, but expensive.
(c) A person is rich if and only if he is a software engineer.
Answer
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Question 83 Marks
Write converse, inverse and contrapositive of the following conditional statement: "If an angle is a right angle then its measure is $90^{\circ}$."
Answer
View full question & answer→Converse: If the measure of an angle is 90° then it is a right angle
Inverse : If an angle is not a right angle then its measure is not 90°.
Contrapositive: If the measure of an angle is not 90° then it is not a right angle.
Question 93 Marks
Using the truth table, prove the following logical equivalence : $p \leftrightarrow q=(p \wedge q) \vee(\sim p \wedge \sim q)$
Answer
View full question & answer→1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
A | B | ||||||
P | q | $p \leftrightarrow q$ | $p \wedge q$ | $\sim p$ | $\sim q$ | $\begin{gathered}\sim p \wedge \\ \sim q\end{gathered}$ | $A \vee B$ |
T T F F
| T F T F | T F F T | T F F F | F F T T
| F T F T | F F F T | T F F T |
By column number 3 and 8
p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Question 103 Marks
Construct the switching circuit for the statement $(p \wedge q) \vee(\sim p) \vee(p \wedge \sim q)$
Answer
View full question & answer→Let
$p =$ Switch $S_1$ is closed
$q =$ Switch $S_2$ is closed
$\sim p=$ switch $\quad S_1^{\prime}$ and $\sim q \equiv S_2^{\prime}$
Question 113 Marks
Using truth tables, examine whether the statement pattern $(p \wedge q) \vee(p \wedge r)$ is a tautology, contradiction or contingency.
Answer
View full question & answer→No of rows = 2n=23 =8
No. of columns = m+ n=3+3= 6
| P | q | r | $p \wedge q$ | $p \wedge r$ | $\begin{aligned} & (p \wedge q) \vee(p \wedge \\ & r)\end{aligned}$ |
| T | T | T | T | T | T |
| T | T | F | T | F | T |
| T | F | T | F | T | T |
| T | F | F | F | F | F |
| F | T | T | F | F | F |
| F | T | F | F | F | F |
| F | F | T | F | F | F |
| F | F | F | F | F | F |
In the last column, the truth values of the statement is neither all T nor all F. Hence, it is neither a tautology nor a contradiction i.e. it is a contingency.
Question 123 Marks
Without using truth table show that : $\sim(p \vee q) \vee(\sim p \wedge q) \equiv \sim p$
Answer
View full question & answer→~(p v q)v(~p ∧ q)
≡~(p v q)v~(p ∨ ~q) by De Morgan's Law
≡~[(p ∨ q) ∧ (p ∨ ~q)] by De Morgan's Law
≡~{[(p ∨ q) ∧ p] ∨ [(p ∨ q)∧ ~q)]} by Distributive Law
≡ ~{[p] ∨ [(p ∨ q) ∧ ~q]} by Absorption Law
≡ ~{[p] ∨ [(p∧ ~q) ∨ (q ∧ ~q)]} by Distributive Law
≡~{[p] ∨ [(p ∧ ~q) ∨ F]} by Complement Law
≡~{[p] ∨ [(p ∧ ~q)]} by Identity Law
≡~p ∧ (~p ∨ q) by De Morgan's Law
≡ ~p by Absorption Law
Question 133 Marks
Examine whether the following logical statement pattern is tautology, contradiction or contingency.
$[(p \rightarrow q) \wedge q] \rightarrow p$
$[(p \rightarrow q) \wedge q] \rightarrow p$
Answer
View full question & answer→Consider the statement pattern : [(p → q) ∧ q ] → p
No. of rows = 2n = 2 × 2 = 4
No. of column = m + n = 3 + 2 = 5
Thus the truth table of the given logical statement:
[(p → q) ∧ q] → p
| P | q | $p \rightarrow q$ | $(p \rightarrow q) \wedge q$ | $[(p \rightarrow q) \wedge q] \rightarrow p$ |
| T | T | T | T | T |
| T | F | F | F | T |
| F | T | T | T | F |
| F | F | T | F | T |
The entries in the last column of the above truth table are neither all T nor all F.
∴ [(p → q) ∧ q] → p is contingency.
Question 143 Marks
Construct the switching circuit for the following statement :$[p \vee(\sim p \wedge q)] \vee[(\sim q \wedge r) \vee \sim p]$
Answer
View full question & answer→Let, $p :$ Switch $S_1$ is closed.
$\sim p :$ Switch $S_1$ is open.
$q :$ Switch $S_2$ is closed.
$\sim q :$ Switch $S_2$ is open
$r :$ Switch $S_3$ is closed.
$\sim r :$ Switch $S_3$ is open.
Now,
$\sim p :$ Switch $S_1$ is open.
$q :$ Switch $S_2$ is closed.
$\sim q :$ Switch $S_2$ is open
$r :$ Switch $S_3$ is closed.
$\sim r :$ Switch $S_3$ is open.
Now,
Question 153 Marks
Discuss the statement pattern, using truth table : $\sim(\sim p \wedge \sim q) \vee q$
Answer
View full question & answer→Consider the statement pattern: ∼ (∼ p ∧ ∼ q) ∨ q
Thus the truth table of the given logical statement: ~(~p ∧ ~q) ∨ q
| P | q | $\sim p$ | $\sim q$ | $\sim p \wedge \sim q$ | $\sim(\sim p$$\wedge \sim q)$ | $\sim(\sim p$$\wedge \sim q) \vee q$ |
| T | T | F | F | F | T | T |
| T | F | F | T | F | T | T |
| F | T | T | F | F | T | T |
| F | F | T | T | T | F | F |
The above statement is contingency.
Question 163 Marks
Using truth table examine whether the following statement pattern is tautology, contradiction or contingency.
$(p \wedge \sim q) \leftrightarrow(p \rightarrow q)$
$(p \wedge \sim q) \leftrightarrow(p \rightarrow q)$
Answer
View full question & answer→| p | q | $\sim q$ | $p \wedge \sim q$ | $p \rightarrow q$ | $(p \wedge \sim q) \leftrightarrow(p \rightarrow q)$ |
| T | T | F | F | T | F |
| T | F | T | T | F | F |
| F | T | F | F | T | F |
| F | F | T | F | T | F |
All the entries in the last column of the above truth table are F.
$(p \wedge \sim q) \leftrightarrow(p \rightarrow q)$ is is a contradiction
Question 173 Marks
Without using truth table show that:
$p \leftrightarrow q=(p \wedge q) \vee(\sim p \wedge \sim q)$
$p \leftrightarrow q=(p \wedge q) \vee(\sim p \wedge \sim q)$
Answer
View full question & answer→L.H.S = p ↔ q
≡ (p → q) ∧ (q → p) ........(Biconditional Law)
≡ (∼ p ∨ q) ∧ (∼ q ∨ p) ........(Conditional Law)
≡ [∼ p ∧ (∼ q ∨ p)] ∨ [q ∧ (∼ q ∨ p)] ....(Distributive Law)
≡ [(∼ p ∧ ∼ q)] ∨ (∼ p ∧ p)] ∨ [(q ∧ ∼ q) ∨ (q ∧ p)] .........(Distributive Law)
≡ [(∼ p ∧ ∼ q) ∨ F] ∨ [F ∨ (q ∧ p)] ........(Complement Law)
≡ (∼ p ∧ ∼ q) ∨ (q ∧ p) .......(Identity Law)
≡ (∼ p ∧ ∼ q) ∨ (p ∧ q) ........(Commutative Law)
≡ (p ∧ q) ∨ (∼ p ∧ ∼ q) ........(Commutative Law)
≡ R.H.S.
Question 183 Marks
Using truth table, examine whether the following statement pattern is a tautology, a contradiction or a contingency: $(p \vee q) \vee r \leftrightarrow p \vee(q \vee r)$
Answer
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Question 193 Marks
Express the following switching circuit in symbolic form of logic. Construct its switching table and write your conclusion from it :
Answer
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Question 203 Marks
Write the converse, inverse and contrapositive of the following statement: "If it rains then the match will be cancelled."
Answer
View full question & answer→Let
p : It rains,
q : the match will be cancelled.
The symbolic form of the given statement is p → q.
Converse: q → p
i.e., If the match is cancelled then it rains.
Inverse: ~p → ~q
i.e., If it does not rain then the match will not be cancelled.
Contrapositive: ~q → ~p
i.e. If the match is not cancelled then it does not rain.