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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 151-180 of 232 questions

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151 Medium
151 of 232
Understanding +4 | -1
The negation of ' x∈Rx \in \mathrm{R} if and only if ∣x∣≥0|x| \geq 0 ' is ...... .
View answer and explanation
Correct option: C
{x∈R\{x \in \mathrm{R} and ∣x∣≥0}|x| \geq 0\} or {∣x∣≥0\{|x| \geq 0 and x∉R}x \notin \mathrm{R}\}
Option C is correct because it accurately represents the contrapositive of the given conditional statement by negating and reversing the hypothesis and conclusion.
152 Easy
152 of 232
Understanding +4 | -1
Negation of " 3 is an odd number and 7 is a rational number is -
View answer and explanation
Correct option: A
3 is not an odd number and 7 is not a rational number
Option D (3 is not an odd number or 7 is not a rational number) is correct because the negation of 'p and q' is 'not p or not q'.
153 Easy
153 of 232
Application +4 | -1
The negation of the Boolean expression ∼s∨(∼r∧s)\sim s \vee ( \sim r \wedge s) is equivalent to
View answer and explanation
Correct option: A
s∧rs \wedge r
Option A (s∧rs \wedge r) is correct because the negation simplifies to this expression using De Morgan's laws and Boolean algebra identities.
154 Easy
154 of 232
Understanding +4 | -1
The Negation of the statement (p∧q)→r(p \wedge q) \rightarrow r is -
View answer and explanation
Correct option: C
(p∧q)∧∼r(p \wedge q) \wedge \sim r
Option C ((p∧q)∧∼r(p \wedge q) \wedge \sim r) is correct because the negation of a conditional statement (a -> b) is a AND (not b).
155 Medium
155 of 232
Understanding +4 | -1
Consider the following statements :

PP : Suman is brilliant

Q:Q: Suman is rich

RR : Suman is honest

The negation of the "Suman is brilliant and dishonest if and only if Suman is rich" can be expressed as
View answer and explanation
Correct option: C
∼(Q↔(P∧∼R))\sim(Q \leftrightarrow(P \wedge \sim R))
Option C (∼(Q↔(P∧∼R))\sim(Q \leftrightarrow(P \wedge \sim R))) is correct because it accurately represents the negation of the given statement using logical symbols.
156 Easy
156 of 232
Analyzing +4 | -1
If 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) \to \left( {p \wedge q} \right)} \right) is false, then the truth values of statements pp, qq and rr respectively can be
View answer and explanation
Correct option: C
T F F
Option C (T F F) is correct because it satisfies the condition where the given Boolean expression evaluates to false.
157 Easy
157 of 232
Understanding +4 | -1
The converse of (p∧(∼q))⇒r(p \wedge(\sim q)) \Rightarrow r is
View answer and explanation
Correct option: C
(∼p∨q)⇒∼r(\sim p \vee q) \Rightarrow \sim r
The correct answer is C because the converse of a conditional statement involves swapping the hypothesis and conclusion, and then applying De Morgan's law and the implication rule.
158 Easy
158 of 232
Analyzing +4 | -1
Among the statements :

(S1) ((p∨q)⇒r)⇔(p⇒r)((p \vee q) \Rightarrow r) \Leftrightarrow(p \Rightarrow r)

(S2) ((p∨q)⇒r)⇔((p⇒r)∨(q⇒r))((p \vee q) \Rightarrow r) \Leftrightarrow((p \Rightarrow r) \vee(q \Rightarrow r))
View answer and explanation
Correct option: C
neither (S1) nor (S2) is a tautology
Neither (S1) nor (S2) is a tautology because we can find truth values for p, q, and r that make the statements false. A tautology must be true for all possible truth values.
159 Medium
159 of 232
Understanding +4 | -1
The statement p→(q→p)p \rightarrow(q \rightarrow p) is equivalent to
View answer and explanation
Correct option: B
p→(p∨q)p \rightarrow(p \vee q)
Option B (p→(p∨q)p \rightarrow (p \vee q)) is correct because both p→(q→p)p \rightarrow (q \rightarrow p) and p→(p∨q)p \rightarrow (p \vee q) are tautologies.
160 Medium
160 of 232
Analyzing +4 | -1
A relation RR on the set of non-zero complex numbers is defined by z1Rz2⇔z1−z2z1+z2{z_1}R{z_2} \Leftrightarrow \frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}} is real, then RR is
View answer and explanation
Correct option: D
Equivalence
Option D (Equivalence) is correct because the relation R is reflexive, symmetric, and transitive, satisfying the conditions for an equivalence relation.
161 Easy
161 of 232
Analyzing +4 | -1
Given that water freezes below zero degree Celsius. Consider the following statements :
p:p: Water froze this morning, qq: This morning temperature was below 00C{0^0}\mathrm{C}
Which of the following is the correct?
View answer and explanation
Correct option: A
pp and qq are logically equivalent
Option A (pp and qq are logically equivalent) is correct because the statements express the same condition: water freezes if and only if the temperature is below zero.
162 Easy
162 of 232
Application +4 | -1
The region represented by the inequality ∣2Z−3i∣<∣3Z−2i∣|2 Z-3 i|<|3 Z-2 i| is

1) the unit disc with its centre at Z=0Z=0

2) the exterior of the unit circle with its centre at Z=0Z=0

3) the interior of a square of side 2 units with its centre at Z=0Z=0

4) the exterior of a square of side 2 units with its centre at Z=0Z=0
View answer and explanation
Correct option: A
2
Option A (2) is correct because the inequality represents the region outside the unit circle centered at the origin.
163 Easy
163 of 232
Application +4 | -1
If p,q,rp,q,r have truth values T,F,TT,F,T respectively, which of the following is true?
View answer and explanation
Correct option: D
q→(p∧r)q \to \left( {p \wedge r} \right)
Option D (q→(p∧r)q \to (p \wedge r)) is correct because with the given truth values, it evaluates to True, while the other options evaluate to False.
164 Medium
164 of 232
Analyzing +4 | -1
If p,q,r,sp, q, r, s and tt are simple statements, such that p:p: A number is terminating.
q:q: A number is recurring.
rr : A number is a rational number.
ss : A number is an irrational number.
tt : A number is a real number.
Then, the compound statement, 'A number is an irrational number if and only if it is neither non-terminating nor not recurring or rational if and only if it is terminating or if non-terminating, then must be recurring and if the number is real, it can be rational or irrational', is represented by
View answer and explanation
Correct option: B
[s↔−(p∨q)]∨[r↔{p∨(−p→q)}]∧[t→(r∨s)][s \leftrightarrow-(p \vee q)] \vee[r \leftrightarrow\{p \vee(-p \rightarrow q)\}] \wedge[t \rightarrow(r \vee s)]
The correct answer is B because it accurately translates the given compound statement into logical symbols, correctly representing the relationships between terminating, recurring, rational, irrational, and real numbers.
165 Medium
165 of 232
Analyzing +4 | -1
∼(p∨q)∨(∼p∧q)\sim(p \vee q) \vee(\sim p \wedge q) is logical equivalent to
View answer and explanation
Correct option: A
∼p\sim p
Option A (∼p\sim p) is correct because the given expression simplifies to ∼p\sim p using logical equivalences and properties of boolean algebra.
166 Easy
166 of 232
Understanding +4 | -1
The contrapositive of the statement "If it is raining, then I will not come", is :
View answer and explanation
Correct option: D
If I will come, then it is not raining.
Option D (If I will come, then it is not raining) is correct because it is the contrapositive of the given statement.
167 Medium
167 of 232
Understanding +4 | -1
(∼p∨q)(\sim p \vee q) is logically equal to -
View answer and explanation
Correct option: A
p→qp \rightarrow q
Option A (p→qp \rightarrow q) is correct because (∼p∨q)(\sim p \vee q) is the logical equivalent of p→qp \rightarrow q.
168 Easy
168 of 232
Analyzing +4 | -1
If p⇒(∼p∨q)p \Rightarrow \left( { \sim p \vee q} \right) is false, the truth value of pp and qq are respectively
View answer and explanation
Correct option: D
T,F
Option D (T,F) is correct because for p⇒(∼p∨q)p \Rightarrow (\sim p \vee q) to be false, pp must be true and (∼p∨q)(\sim p \vee q) must be false, implying qq is false.
169 Easy
169 of 232
Understanding +4 | -1
The contra positive of (∼p∧q)→(q∧∼r)(\sim p \wedge q) \rightarrow(q \wedge \sim r) is
View answer and explanation
Correct option: C
∼r∨(p∨∼q)\sim r \vee(p \vee \sim q)
Option C (∼r∨(p∨∼q)\sim r \vee(p \vee \sim q)) is correct because the contrapositive of p→qp \rightarrow q is ∼q→∼p\sim q \rightarrow \sim p.
170 Easy
170 of 232
Understanding +4 | -1
Let p,q,rp, q, r be the following:

pp : Rohit is healthy

qq : Rohit is wealthy

rr : Rohit is wise.

The statement p∨q→∼rp \vee q \rightarrow \sim r means
View answer and explanation
Correct option: A
If Rohit is healthy or wealthy then Rohit is not wise
Option A (If Rohit is healthy or wealthy then Rohit is not wise) is correct because p∨qp \vee q represents "Rohit is healthy or wealthy" and ∼r\sim r represents "Rohit is not wise", so p∨q→∼rp \vee q \rightarrow \sim r means "If Rohit is healthy or wealthy, then Rohit is not wise".
171 Easy
171 of 232
Analyzing +4 | -1
If the compound statement p→(∼p∨q)p \rightarrow(\sim p \vee q) is false, then the truth values of pp and qq are respectively
View answer and explanation
Correct option: B
T, F
Option B (T, F) is correct because p→(∼p∨q)p \rightarrow (\sim p \vee q) is false only when p is true and (∼p∨q)(\sim p \vee q) is false, which occurs when q is false.
172 Medium
172 of 232
Understanding +4 | -1
The contrapositive of the statement "I go to college if its not a holiday
View answer and explanation
Correct option: A
If I do not go to the college then its a holiday
Option A (If I do not go to the college then its a holiday) is correct because it represents the contrapositive of the given statement.
173 Easy
173 of 232
Understanding +4 | -1
Consider the statement “The match will be played only if the weather is good and ground is not wet”.
Select the correct negation from the following
View answer and explanation
Correct option: C
The match will be played and weather is not good or ground is wet.
Option C (The match will be played and weather is not good or ground is wet) is correct because it represents the negation of the given conditional statement.
174 Easy
174 of 232
Analyzing +4 | -1
Which one of the following statements is not a tautology?
View answer and explanation
Correct option: D
(p∨q)→(p∨(∼q))(p \vee q) \to (p \vee ( \sim q))
Option D ((p∨q)→(p∨(∼q))(p \vee q) \to (p \vee ( \sim q))) is correct because its truth table does not always result in 'True', meaning it's not a tautology.
175 Easy
175 of 232
Understanding +4 | -1
The negation of the Boolean expression p∨(∼p∧q)p \vee ( \sim p \wedge q) is equivalent to
View answer and explanation
Correct option: B
∼p∧∼q\sim p \wedge \sim q
Option B (∼p∧∼q\sim p \wedge \sim q) is correct because it represents the negation of the given expression using De Morgan's laws and simplification.
176 Easy
176 of 232
Understanding +4 | -1
If pp is any statement, then which of the following is a contradiction?
View answer and explanation
Correct option: B
p∧∼pp \wedge \sim p
Option B (p∧∼pp \wedge \sim p) is a contradiction because it is always false, regardless of the truth value of pp.
177 Easy
177 of 232
Understanding +4 | -1
The conditional statement of 'You will get a sweet dish after the dinner' is
View answer and explanation
Correct option: A
if you take the dinner, then you will get a sweet dish
Option A (if you take the dinner, then you will get a sweet dish) is the correct conditional statement of the given statement.
178 Easy
178 of 232
Analyzing +4 | -1
Let r∈{p, q, ∼p, ∼q}r \in \{ p,\,q,\, \sim p,\, \sim q\} be such that the logical statement r∨(∼p)  ⇒(p∧q)∨rr \vee \left( { \sim p} \right)\,\, \Rightarrow \left( {p \wedge q} \right) \vee r is a tautology. Then rr is equal to:
View answer and explanation
Correct option: C
∼p\sim p
Option C (∼p\sim p) is correct because it makes the logical statement a tautology, meaning it is always true regardless of the truth values of p and q.
179 Easy
179 of 232
Understanding +4 | -1
The logically equivalent preposition of p⇔qp \Leftrightarrow q is
View answer and explanation
Correct option: A
(p⇒q)∧(q⇒p)(p \Rightarrow q) \wedge(q \Rightarrow p)
Option A ((p⇒q)∧(q⇒p)(p \Rightarrow q) \wedge(q \Rightarrow p)) is correct because it accurately represents the logical equivalence of the biconditional statement p⇔qp \Leftrightarrow q.
180 Medium
180 of 232
Understanding +4 | -1
The contrapositive of the statement
'If 7 is greater than 5 , then 8 is greater than 6 ', is
View answer and explanation
Correct option: C
if 8 is not greater than 6 , then 7 is not greater than 5
The correct answer is 'if 8 is not greater than 6, then 7 is not greater than 5' because it correctly applies the contrapositive transformation: if not q, then not p.
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