01Medium115×since 2008Q4348The Boolean expression (p∧∼q)⇒(q∨∼p)(p \wedge \sim q) \Rightarrow (q \vee \sim p)(p∧∼q)⇒(q∨∼p) is equivalent to :Aq⇒pq \Rightarrow pq⇒pBp⇒qp \Rightarrow qp⇒qC∼q⇒p\sim q \Rightarrow p∼q⇒pDp⇒ ∼qp \Rightarrow \, \sim qp⇒∼qCheck answerSkip
02Medium115×since 2008Q4349Which of the following Boolean expressions is not a tautology?A(p ⇒\Rightarrow⇒ q) ∨\vee∨ (∼\sim∼ q ⇒\Rightarrow⇒ p)B(q ⇒\Rightarrow⇒ p) ∨\vee∨ (∼\sim∼ q ⇒\Rightarrow⇒ p)C(p ⇒\Rightarrow⇒ ∼\sim∼ q) ∨\vee∨ (∼\sim∼ q ⇒\Rightarrow⇒ p)D(∼\sim∼ p ⇒\Rightarrow⇒ q) ∨\vee∨ (∼\sim∼ q ⇒\Rightarrow⇒ p)Check answerSkip
03Easy115×since 2008Q4350The Boolean expression (p⇒q)∧(q⇒∼p)(p \Rightarrow q) \wedge (q \Rightarrow \sim p)(p⇒q)∧(q⇒∼p) is equivalent to :A∼\sim∼ qBqCpD∼\sim∼ pCheck answerSkip
04Medium115×since 2008Q4351The compound statement (P∨Q)∧(∼P)⇒Q(P \vee Q) \wedge ( \sim P) \Rightarrow Q(P∨Q)∧(∼P)⇒Q is equivalent to :AP∨QP \vee QP∨QBP∧∼QP \wedge \sim QP∧∼QC∼(P⇒Q)\sim (P \Rightarrow Q)∼(P⇒Q)D∼(P⇒Q)⇔P∧∼Q\sim (P \Rightarrow Q) \Leftrightarrow P \wedge \sim Q∼(P⇒Q)⇔P∧∼QCheck answerSkip
05Hard115×since 2008Q4352If the truth value of the Boolean expression ((p∨q)∧(q→r)∧(∼r))→(p∧q)\left( {\left( {p \vee q} \right) \wedge \left( {q \to r} \right) \wedge \left( { \sim r} \right)} \right) \to \left( {p \wedge q} \right)((p∨q)∧(q→r)∧(∼r))→(p∧q) is false, then the truth values of the statements p, q, r respectively can be :AT F TBF F TCT F FDF T FCheck answerSkip
06Easy115×since 2008Q4353Consider the two statements : (S1) : (p →\to→ q) ∨\vee∨ (∼\sim∼ q →\to→ p) is a tautology . (S2) : (p ∧\wedge∧ ∼\sim∼ q) ∧\wedge∧ (∼\sim∼ p ∧\wedge∧ q) is a fallacy. Then :Aonly (S1) is true.Bboth (S1) and (S2) are false.Cboth (S1) and (S2) are true.Donly (S2) is true.Check answerSkip
07Easy115×since 2008Q4354The statement (p ∧\wedge∧ (p →\to→ q) ∧\wedge∧ (q →\to→ r)) →\to→ r is :Aa tautologyBequivalent to p →\to→ ∼\sim∼ rCa fallacyDequivalent to q →\to→ ∼\sim∼ rCheck answerSkip
08Easy115×since 2008Q4355The Boolean expression (p ∧\wedge∧ q) ⇒\Rightarrow⇒ ((r ∧\wedge∧ q) ∧\wedge∧ p) is equivalent to :A(p ∧\wedge∧ q) ⇒\Rightarrow⇒ (r ∧\wedge∧ q)B(q ∧\wedge∧ r) ⇒\Rightarrow⇒ (p ∧\wedge∧ q)C(p ∧\wedge∧ q) ⇒\Rightarrow⇒ (r ∨\vee∨ q)D(p ∧\wedge∧ r) ⇒\Rightarrow⇒ (p ∧\wedge∧ q)Check answerSkip
09Hard115×since 2008Q4356Let *, ▢ ∈\in∈{∧\wedge∧, ∨\vee∨} be such that the Boolean expression (p * ∼\sim∼ q) ⇒\Rightarrow⇒ (p ▢ q) is a tautology. Then :A* = ∨\vee∨, ▢ = ∨\vee∨B* = ∧\wedge∧, ▢ = ∧\wedge∧C* = ∧\wedge∧, ▢ = ∨\vee∨D* = ∨\vee∨, ▢ = ∧\wedge∧Check answerSkip
10Easy115×since 2008Q4414The contrapositive of the statement "If I reach the station in time, then I will catch the train" is :AIf I will catch the train, then I reach the station in time.BIf I do not reach the station in time, then I will not catch the train.CIf I will not catch the train, then I do not reach the station in time.DIf I do not reach the station in time, then I will catch the train.Check answerSkip