Mathematics Objective Questions and Answers: Mathematical Reasoning
Review structured MCQ sets for JEE Mathematics Mathematical Reasoning. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Download Mathematical Reasoning MCQs with Answers
Access the complete set of multiple-choice questions for Mathematical Reasoning below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.
Multiple Choice Questions
(a) Three plus two equals six
(b) Three plus two equals five
(c) Tomorrow is Tuesday
(d) There are 380 days in a year
Show Answer & Explanation
Answer: (c) Tomorrow is Tuesday
Explanation:
1. (a), (b) and (d) are definitely false, true and false respectively.
2. 'Tomorrow is Tuesday' is true or false depending on the day, so it is not a statement.
Teacher's Note:
1. A statement must be either true or false, but not both; questions, commands and exclamations are not statements.
2. A sentence whose truth depends on an unknown (such as 'tomorrow is Friday') is not a statement.
(a) False because the sun rises in the east
(b) True
(c) Cannot be determined
(d) It is false because 2 + 2 = 4
Show Answer & Explanation
Answer: (b) True
Explanation:
1. The hypothesis 'the sun rises in the west' is false and the conclusion '2 + 2 = 6' is false.
2. A conditional with a false hypothesis is true (F ⇒ F = T).
Teacher's Note:
1. Substitute the given truth values step by step, innermost brackets first.
2. A conditional p ⇒ q is false only when p is true and q is false.
(a) A number is odd implies that its cube is odd.
(b) A number is odd only if its cube is odd.
(c) If the cube of a number is not odd, then the number is not odd.
(d) For the cube of a number to be odd, it is necessary that the number is odd.
Show Answer & Explanation
Answer: (d) For the cube of a number to be odd, it is necessary that the number is odd.
Explanation:
1. p is 'odd number ⇒ odd cube'.
2. (a) and (b) say the same thing; (c) is its contrapositive.
3. (d) says 'odd cube ⇒ odd number', the converse, which is a different statement.
Teacher's Note:
1. 'p ⇒ q' can be read as: if p then q; p is sufficient for q; q is necessary for p; p only if q.
2. 'q only if p' and 'q is sufficient for p' describe q ⇒ p, which is the converse.
(a) Necessary is: "pass the exam"; sufficient is: "getting a good college"
(b) Necessary is: "getting a good college"; sufficient is: "pass the exam"
(c) Necessary is: "not getting a good college"; sufficient is: "pass the exam"
(d) Necessary is: "not pass the exam"; sufficient is: "getting a good college"
Show Answer & Explanation
Answer: (b) Necessary is: "getting a good college"; sufficient is: "pass the exam"
Explanation:
1. In p ⇒ q, p is sufficient for q and q is necessary for p.
2. Here p : pass the exam (sufficient) and q : get a good college (necessary).
Teacher's Note:
1. 'p ⇒ q' can be read as: if p then q; p is sufficient for q; q is necessary for p; p only if q.
2. 'q only if p' and 'q is sufficient for p' describe q ⇒ p, which is the converse.
(a) p ∧ q
(b) ~(p ∧ q)
(c) ~p ∧ ~q
(d) ~p ∧ q
Show Answer & Explanation
Answer: (c) ~p ∧ ~q
Explanation:
1. The second factor ~p ∧ ~q implies ~p, hence ~p ∨ q is true whenever the second factor is true.
2. So the whole expression equals ~p ∧ ~q.
Teacher's Note:
1. Use p ⇒ q = ~p ∨ q and p ⇔ q = (p ⇒ q) ∧ (q ⇒ p) to reduce every expression to ∧, ∨, ~.
2. Absorption laws p ∧ (p ∨ q) = p and p ∨ (p ∧ q) = p shorten many simplifications.
(a) 36 is divisible by 2
(b) 36 is divisible by 3
(c) 36 is divisible by 6
(d) 36 is divisible by 18
Show Answer & Explanation
Answer: (d) 36 is divisible by 18
Explanation:
1. The components are: 36 is divisible by 2, by 3, and by 6.
2. '36 is divisible by 18' is not one of them.
Teacher's Note:
1. A statement must be either true or false, but not both; questions, commands and exclamations are not statements.
2. A sentence whose truth depends on an unknown (such as 'tomorrow is Friday') is not a statement.
(a) a tautology
(b) a fallacy
(c) p
(d) q
Show Answer & Explanation
Answer: (a) a tautology
Explanation:
1. ~q ∧ q = F and p ∨ ~p = T.
2. F ∨ T = T, so the statement is a tautology.
Teacher's Note:
1. A tautology is true for every combination of truth values and a contradiction (fallacy) is false for every combination.
2. Simplify with the laws of algebra of statements, or build a short truth table to be certain.
(a) 2 + 3 ≠ 5 and 8 < 10
(b) 2 + 3 = 5 or 8 < 10
(c) 2 + 3 ≠ 5 or 8 ≥ 10
(d) None of these
Show Answer & Explanation
Answer: (c) 2 + 3 ≠ 5 or 8 ≥ 10
Explanation:
1. ~(p ∧ q) = ~p ∨ ~q.
2. ~p : 2 + 3 ≠ 5 and ~q : 8 ≥ 10, so the negation is: 2 + 3 ≠ 5 or 8 ≥ 10.
Teacher's Note:
1. Use De Morgan's laws: ~(p ∧ q) = ~p ∨ ~q and ~(p ∨ q) = ~p ∧ ~q.
2. The negation of p ⇒ q is p ∧ ~q, and the negation of p ⇔ q is (p ∧ ~q) ∨ (~p ∧ q).
(a) A is not in class X and B is in class XII
(b) A is not in class X and B is not in class XII
(c) Either A is not in class X or B is not in class XII
(d) None
Show Answer & Explanation
Answer: (b) A is not in class X and B is not in class XII
Explanation:
1. ~(p ∨ q) = ~p ∧ ~q.
2. So: A is not in class X and B is not in class XII.
Teacher's Note:
1. Use De Morgan's laws: ~(p ∧ q) = ~p ∨ ~q and ~(p ∨ q) = ~p ∧ ~q.
2. The negation of p ⇒ q is p ∧ ~q, and the negation of p ⇔ q is (p ∧ ~q) ∨ (~p ∧ q).
(a) p ≡ true, q ≡ true
(b) p ≡ true, q ≡ false
(c) p ≡ false, q ≡ false
(d) p ≡ false, q ≡ true
Show Answer & Explanation
Answer: (b) p ≡ true, q ≡ false
Explanation:
1. A conditional is false only when the hypothesis is true and the conclusion is false.
Teacher's Note:
1. Substitute the given truth values step by step, innermost brackets first.
2. A conditional p ⇒ q is false only when p is true and q is false.
(a) ~p ⇒ ~q
(b) ~q ⇒ p
(c) ~q ⇒ ~p
(d) q ⇒ ~p
Show Answer & Explanation
Answer: (a) ~p ⇒ ~q
Explanation:
1. 'a man is not happy' is ~p and 'he is not rich' is ~q.
2. So the statement is ~p ⇒ ~q.
Teacher's Note:
1. 'p ⇒ q' can be read as: if p then q; p is sufficient for q; q is necessary for p; p only if q.
2. 'q only if p' and 'q is sufficient for p' describe q ⇒ p, which is the converse.
(a) T, T
(b) T, F
(c) F, T
(d) F, F
Show Answer & Explanation
Answer: (b) T, F
Explanation:
1. A conditional is false only if p is true and ~p ∨ q is false.
2. With p = T, ~p ∨ q = q, so q = F.
3. Hence p = T, q = F.
Teacher's Note:
1. Substitute the given truth values step by step, innermost brackets first.
2. A conditional p ⇒ q is false only when p is true and q is false.
(a) ~p ⇒ q
(b) ~q ⇒ p
(c) q ⇒ ~p
(d) None
Show Answer & Explanation
Answer: (c) q ⇒ ~p
Explanation:
1. Contrapositive of X ⇒ Y is ~Y ⇒ ~X.
2. With X = p and Y = ~q: ~(~q) ⇒ ~p = q ⇒ ~p.
Teacher's Note:
1. The contrapositive of p ⇒ q is ~q ⇒ ~p and it is always logically equivalent to p ⇒ q.
2. The converse q ⇒ p and the inverse ~p ⇒ ~q are equivalent to each other but not to the original statement.
(a) p, q, r are all false
(b) p, q, r are all true
(c) p, q are true and r is false
(d) p is true and q, r are false
Show Answer & Explanation
Answer: (b) p, q, r are all true
Explanation:
1. A conjunction is true only when every part is true, so p, q and r must all be true.
Teacher's Note:
1. Substitute the given truth values step by step, innermost brackets first.
2. A conditional p ⇒ q is false only when p is true and q is false.
(a) p
(b) q
(c) ~q
(d) ~p
Show Answer & Explanation
Answer: (b) q
Explanation:
1. By absorption, (~p ∧ q) ∨ q = q.
Teacher's Note:
1. Use p ⇒ q = ~p ∨ q and p ⇔ q = (p ⇒ q) ∧ (q ⇒ p) to reduce every expression to ∧, ∨, ~.
2. Absorption laws p ∧ (p ∨ q) = p and p ∨ (p ∧ q) = p shorten many simplifications.
(a) Neither I will become PM nor I will open more IITs.
(b) I will not become PM or I will not open more IITs.
(c) I will become PM and I will not open more IITs.
(d) Either I will not become PM or I will not open more IITs.
Show Answer & Explanation
Answer: (c) I will become PM and I will not open more IITs.
Explanation:
1. ~(p ⇒ q) = p ∧ ~q.
2. So: I will become PM and I will not open more IITs.
Teacher's Note:
1. Use De Morgan's laws: ~(p ∧ q) = ~p ∨ ~q and ~(p ∨ q) = ~p ∧ ~q.
2. The negation of p ⇒ q is p ∧ ~q, and the negation of p ⇔ q is (p ∧ ~q) ∨ (~p ∧ q).
(a) p ⇒ q and q ⇏ p
(b) p ⇏ q and q ⇒ p
(c) p ⇔ q
(d) p ⇒ q and ~q ⇒ p
Show Answer & Explanation
Answer: (c) p ⇔ q
Explanation:
1. For positive x and y, squaring is increasing, so x < y implies x² < y².
2. Also x² < y² implies (y − x)(y + x) > 0, so y > x since y + x > 0.
3. Both implications hold, so p ⇔ q.
Teacher's Note:
1. 'p ⇒ q' can be read as: if p then q; p is sufficient for q; q is necessary for p; p only if q.
2. 'q only if p' and 'q is sufficient for p' describe q ⇒ p, which is the converse.
(a) If two integers a and b are such that b³ − a² is a composite number then a divides b.
(b) If a divides b then a, b are integers and b³ − a² is a composite number.
(c) If b³ − a² is a composite number then a, b are integers and a divides b.
(d) If a, b are integers and a does not divide b then b³ − a² is not a composite number.
Show Answer & Explanation
Answer: (a) If two integers a and b are such that b³ − a² is a composite number then a divides b.
Explanation:
1. The converse of 'H ⇒ C' is 'C ⇒ H'.
2. Hypothesis H : a and b are integers and a divides b. Conclusion C : b³ − a² is composite.
3. The converse is option (a); (d) is the inverse.
Teacher's Note:
1. The converse of p ⇒ q is q ⇒ p: just swap the hypothesis and the conclusion.
2. Do not confuse the converse with the contrapositive, which negates both parts as well as swapping them.
(a) There is no irrational number which when added to a particular irrational number results in a rational number.
(b) The sum of two irrational numbers is always irrational.
(c) There exist at least one pair of irrational numbers whose sum is irrational.
(d) We do not always get a rational number when we add two irrational numbers.
Show Answer & Explanation
Answer: (b) The sum of two irrational numbers is always irrational.
Explanation:
1. P : there exist irrational x, y in S with x + y rational.
2. Negation: for all irrational x, y in S, x + y is not rational, that is, the sum of any two irrational numbers is always irrational.
Teacher's Note:
1. The negation of 'for all' is 'there exists ... not', and the negation of 'there exists' is 'for all ... not'.
2. Negate the property inside the quantifier as well.
(a) p ∧ (~p) is a fallacy
(b) p ∨ (~p) is a tautology
(c) (p ⇒ q) ⇔ (~q ⇒ ~p) is a contradiction
(d) ~(~p) ⇔ p is a tautology
Show Answer & Explanation
Answer: (c) (p ⇒ q) ⇔ (~q ⇒ ~p) is a contradiction
Explanation:
1. (a), (b) and (d) are standard true facts.
2. (p ⇒ q) ⇔ (~q ⇒ ~p) is a tautology, because a conditional and its contrapositive are equivalent; calling it a contradiction is incorrect.
Teacher's Note:
1. A tautology is true for every combination of truth values and a contradiction (fallacy) is false for every combination.
2. Simplify with the laws of algebra of statements, or build a short truth table to be certain.
(a) (S₁ ∧ S₂) ⇒ S₃
(b) (~S₁ ∧ S₂) ⇒ S₃
(c) (S₂ ∧ ~S₂) ⇒ S₃
(d) (S₁ ∧ S₂) ⇒ (~S₃)
Show Answer & Explanation
Answer: (b) (~S₁ ∧ S₂) ⇒ S₃
Explanation:
1. 'f is continuous' is ~S₁ and 'differentiable' is S₂, so the hypothesis is ~S₁ ∧ S₂.
2. The conclusion is S₃, so the statement is (~S₁ ∧ S₂) ⇒ S₃.
Teacher's Note:
1. 'p ⇒ q' can be read as: if p then q; p is sufficient for q; q is necessary for p; p only if q.
2. 'q only if p' and 'q is sufficient for p' describe q ⇒ p, which is the converse.
(a) a tautology
(b) a fallacy
(c) can't say
(d) equivalent to p ⇒ q
Show Answer & Explanation
Answer: (a) a tautology
Explanation:
1. If ~p is true then p = F, and p ⇔ ~q requires ~q = F, so q = T.
2. Whenever the antecedent is true, q is true, so the implication is a tautology.
Teacher's Note:
1. A tautology is true for every combination of truth values and a contradiction (fallacy) is false for every combination.
2. Simplify with the laws of algebra of statements, or build a short truth table to be certain.
(a) The negation of "There exists a capital for every state in India" is "there exists a state in India which does not have its capital"
(b) The negation of "There is a tree in this campus all of whose leaves are green" is "For all trees in this campus, there is a leaf which is not green"
(c) The negation of "There is a tree in this campus with at least one brown leaf" is "no tree in this campus has no brown leaf"
(d) The negation of "For every real number x, either x > 1 or x < −1" is "there exists a real number x such that −1 ≤ x ≤ 1"
Show Answer & Explanation
Answer: (c) The negation of "There is a tree in this campus with at least one brown leaf" is "no tree in this campus has no brown leaf"
Explanation:
1. (a), (b) and (d) apply the quantifier rules correctly.
2. In (c) the correct negation is 'no tree in this campus has a brown leaf'. The statement given, 'no tree has no brown leaf', means every tree has a brown leaf, which is not the negation.
Teacher's Note:
1. The negation of 'for all' is 'there exists ... not', and the negation of 'there exists' is 'for all ... not'.
2. Negate the property inside the quantifier as well.
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