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JEE • MATHEMATICS • MATHEMATICAL REASONING

Mathematical reasoning practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 211-232 of 232 questions

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211 Easy
211 of 232
Analyzing +4 | -1
Let p,q,rp, q, r be three statements such that the truth value of ∼(∼p∨∼q)→(∼q∨r)\sim(\sim p \vee \sim q) \rightarrow(\sim q \vee r) is FF. Then the truth values of p,q,rp, q, r are respectively:
View answer and explanation
Correct option: D
T, T, F
The correct answer is D (T, T, F) because when p and q are true and r is false, the given expression evaluates to false, satisfying the condition.
212 Easy
212 of 232
Analyzing +4 | -1
Negation of the Boolean statement (p∨q)⇒((∼r)∨p)(p \vee q) \Rightarrow((\sim r) \vee p) is equivalent to
View answer and explanation
Correct option: C
(∼p)∧q∧r(\sim p) \wedge q \wedge r
Option C ((∼p)∧q∧r\sim p) \wedge q \wedge r) is correct because it represents the negation of the given Boolean statement after applying De Morgan's laws and logical equivalences.
213 Easy
213 of 232
Understanding +4 | -1
The contrapositive of the statement "I go to school if it does not rain" is
View answer and explanation
Correct option: B
If I do not go to school, it rains.
Option B (If I do not go to school, it rains) is correct because it is the contrapositive of the given statement.
214 Medium
214 of 232
Understanding +4 | -1
The negation of the compound proposition p∨(∼p∨q)p \vee \left( { \sim p \vee q} \right) is
View answer and explanation
Correct option: A
(p∧∼q)∧∼p\left( {p \wedge \sim q} \right) \wedge \sim p
Option A is correct because it correctly applies De Morgan's law to negate the given compound proposition.
215 Easy
215 of 232
Analyzing +4 | -1
Consider the following statements :

P : Suman is brilliant

Q: Suman is rich.

R\mathrm{R} : Suman is honest

the negation of the statement

"Suman is brilliant and dishonest if and only if suman is rich" can be equivalently expressed as :
View answer and explanation
Correct option: D
∼Q↔P∧∼R\sim \mathrm{Q} \leftrightarrow \mathrm{P} \wedge \sim \mathrm{R}
Option D (∼Q↔P∧∼R\sim Q \leftrightarrow P \wedge \sim R) is correct because the negation of "Suman is brilliant and dishonest if and only if Suman is rich" is equivalent to ∼Q↔P∧∼R\sim Q \leftrightarrow P \wedge \sim R.
216 Easy
216 of 232
Understanding +4 | -1
Contrapositive of the statement “If two numbers are not equal, then their squares are not equal” is
View answer and explanation
Correct option: B
If the squares of two numbers are equal, then the numbers are equal.
Option B (If the squares of two numbers are equal, then the numbers are equal) is correct because it is the contrapositive of the given statement.
217 Easy
217 of 232
Analyzing +4 | -1
Let Δ∈{∧,∨,⇒,⇔}\Delta \in\{\wedge, \vee, \Rightarrow, \Leftrightarrow\} \quad be such that (p∧q)Δ((p∨q)⇒q)(p \wedge q) \Delta((p \vee q) \Rightarrow q) is a tautology. Then Δ\Delta is equal to :
View answer and explanation
Correct option: C
⇒\Rightarrow
Option C (⇒\Rightarrow) is correct because the truth table for (p∧q)⇒((p∨q)⇒q)(p \wedge q) \Rightarrow ((p \vee q) \Rightarrow q) results in all true values, indicating that it is a tautology.
218 Easy
218 of 232
Application +4 | -1
   −[p∧(q⇒∼r)]=……\,\,\,-[p \wedge(q \Rightarrow \sim r)]=\ldots \ldots
View answer and explanation
Correct option: B
∼p∨(q∧r)\sim p \vee(q \wedge r)
Option B (∼p∨(q∧r)\sim p \vee (q \wedge r)) is correct because by applying logical equivalences and De Morgan's law, the given expression simplifies to this form.
219 Easy
219 of 232
Analyzing +4 | -1
For any two statements pp and qq, ∼(p∨q)∨(∼p∧q)\sim \left( {p \vee q} \right) \vee \left( { \sim p \wedge q} \right) is logically equivalent to
View answer and explanation
Correct option: B
∼p\sim p
Option B (∼p\sim p) is correct because the given expression simplifies to ∼p\sim p using logical equivalences.
220 Easy
220 of 232
Understanding +4 | -1
If p: Nidhi is intelligent; q\mathrm{q} : Nidhi will pass, then the statement Nidhi is intelligent but will not pass is written in symbols as
View answer and explanation
Correct option: C
p∧∼qp\wedge \sim q
Option C (p∧∼qp\wedge \sim q) is correct because it accurately represents "Nidhi is intelligent but will not pass" in symbolic form.
221 Easy
221 of 232
Understanding +4 | -1
The statement p∧∼q→rp \wedge \sim q \rightarrow r is logically equivalent to
View answer and explanation
Correct option: A
∼p∨q∨r\sim p \vee q \vee r
Option A (∼p∨q∨r\sim p \vee q \vee r) is correct because it is logically equivalent to the given statement p∧∼q→rp \wedge \sim q \rightarrow r after applying logical equivalences.
222 Easy
222 of 232
Analyzing +4 | -1
Which of the following Boolean expression is a tautology?
View answer and explanation
Correct option: D
(p∧q)→(p→q)(p \wedge q) \to (p \to q)
The correct answer is D ((p∧q)→(p→q)(p \wedge q) \to (p \to q)) because its truth table always results in true, making it a tautology.
223 Medium
223 of 232
Analyzing +4 | -1
The compound statement (∼(P∧Q))∨((∼P)∧Q)⇒((∼P)∧(∼Q))(\sim(P \wedge Q)) \vee((\sim P) \wedge Q) \Rightarrow((\sim P) \wedge(\sim Q)) is equivalent to
View answer and explanation
Correct option: D
((∼P)∨Q)∧((Q)∨P)((\sim P) \vee Q) \wedge((Q) \vee P)
Option D (((∼P)∨Q)∧((Q)∨P)(\sim P) \vee Q) \wedge((Q) \vee P)) is correct because the given expression simplifies to this logical statement using logical equivalences.
224 Easy
224 of 232
Understanding +4 | -1
The statement p⇔qp \Leftrightarrow q is equal to -
View answer and explanation
Correct option: C
(∼p∨q)∧(p∨∼q)(\sim p \vee q) \wedge(p \vee \sim q)
Option C ((∼p∨q)∧(p∨∼q)\sim p \vee q) \wedge(p \vee \sim q)) is correct because p⇔qp \Leftrightarrow q is equivalent to (p→q)∧(q→p)(p \rightarrow q) \wedge (q \rightarrow p), which expands to the given expression.
225 Easy
225 of 232
Understanding +4 | -1
the contrapositive of "If two triangles are identical, then these are similar" is
View answer and explanation
Correct option: A
If two triangles are not similar, then these are not identical
Option A (If two triangles are not similar, then these are not identical) is correct because it is the contrapositive of the given conditional statement.
226 Easy
226 of 232
Analyzing +4 | -1
The proposition ∼(p∨∼q)∨∼(p∨q)\sim(p \vee \sim q) \vee \sim(p \vee q) is logically equivalent to
View answer and explanation
Correct option: C
∼p\sim p
Option C (∼p\sim p) is correct because by applying De Morgan's laws and simplification, the expression reduces to ∼p\sim p.
227 Easy
227 of 232
Understanding +4 | -1
p∧(q∧r)p \wedge \left( {q \wedge r} \right) is logically equivalent to
View answer and explanation
Correct option: B
(p∧q)∧r\left( {p \wedge q} \right) \wedge r
Option B ((p∧q)∧r\left( {p \wedge q} \right) \wedge r) is correct because the conjunction is associative.
228 Easy
228 of 232
Understanding +4 | -1
The contrapositive of p→(∼q→∼r)p \rightarrow(\sim q \rightarrow \sim r) is
View answer and explanation
Correct option: A
(∼q∧r)→∼p(\sim q \wedge r) \rightarrow \sim p
Option A is correct because it accurately represents the contrapositive of the given logical statement, derived by negating and reversing the implication.
229 Easy
229 of 232
Understanding +4 | -1
Which of the following statement has the truth value 'F'?
View answer and explanation
Correct option: A
A quadratic equation has always a real root
Option A is correct because a quadratic equation can have complex roots, thus not always a real root.
230 Medium
230 of 232
Application +4 | -1
The negation of (p∨q)∧(p∨∼r)(p \vee q) \wedge(p \vee \sim r) is
View answer and explanation
Correct option: C
(−p∧−q)∨(−p∧r)(-p \wedge-q) \vee(-p \wedge r)
Option C ((∼p∧∼q)∨(∼p∧r)\sim p \wedge \sim q) \vee (\sim p \wedge r)) is correct because the negation of (p∨q)∧(p∨∼r)(p \vee q) \wedge (p \vee \sim r) is found by applying De Morgan's laws.
231 Hard
231 of 232
Analyzing +4 | -1
The number of choices for Δ∈{∧,∨,⇒,⇔},\Delta \in \left\{ { \wedge , \vee , \Rightarrow , \Leftrightarrow } \right\}, such that (pΔq)⇒((pΔ∼q)∨((∼p)Δq))\left( {p\Delta q} \right) \Rightarrow \left( {\left( {p\Delta \sim q} \right) \vee \left( {\left( { \sim p} \right)\Delta q} \right)} \right) is a tautology, is
View answer and explanation
Correct option: B
2
The correct answer is 2 because only two logical connectives, ∨\vee and ⇒\Rightarrow, result in a tautology when substituted for Δ\Delta in the given expression.
232 Easy
232 of 232
Understanding +4 | -1
The negation of the Boolean expression x↔ ∼yx \leftrightarrow \, \sim y is equivalent to
View answer and explanation
Correct option: B
(x∧y)∨(∼x∧∼y)(x \wedge y) \vee ( \sim x \wedge \sim y)
Option B ((x∧y)∨(∼x∧∼y)(x \wedge y) \vee (\sim x \wedge \sim y)) is correct because it represents the negation of the biconditional statement using logical equivalences.
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