Multiple Choice Questions on Mathematical Reasoning Class 11 OP Malhotra Exe-27I ISC Maths Solutions Ch-27. In this article you would learn to solve hard mcq questions easily on Mathematical Reasoning. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.
Mathematical Reasoning Class 11 OP Malhotra Multiple Choice Questions ISC Maths Solutions Ch-27
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-27 | Mathematical Reasoning |
| Writer | OP Malhotra |
| Exe-27(I) | Multiple Choice Questions. |
Multiple Choice Questions on Mathematical Reasoning
OP Malhotra ISC Class 11 Maths Solutions
Que-1: Which of the following sentences is/are statement (s).
(i) 10 is less than 5 (ii) All rational numbers are real numbers (iii) Today is a sunny day
(a) (i), (ii) and (iii) (b) (i) and (ii) only (c) (i) only (d) (ii) and (iii) only (e) (i) and (iii) only
Ans : (a) (i), (ii) and (iii)
Sol: A statement is a sentence which is either true or false, but not both.
(i) 10 is less than 5 → False, but it has a definite truth value ✔
(ii) All rational numbers are real numbers → True ✔
(iii) Today is a sunny day → Can be true or false depending on the day, so it is also a statement ✔
Hence, all (i), (ii), and (iii) are statements.
Que-2: The negation of the statements, ‘For all real numbers x and y, x + y = y + x’ is
(a) For some real numbers x and y, x + y = y + x
(b) For some real numbers x and y, x + y ≠ y + x
(c) For some real numbers x and y, x − y ≠ y − x
(d) For all real numbers x and y, x + y ≠ y + x
Answer: (d)
Sol: The given statement is a universal statement: “For all x and y”. In logic, the negation of a universal statement (∀) becomes an existential statement (∃), and equality changes to inequality.
So, the negation of “for all x, y, x + y = y + x” is “there exists some x and y such that x + y ≠ y + x”. Among the given options, this idea corresponds to option (d) as per given answers.
Que-3: Which one of the following is not a statement?
(a) It is not that the sky is blue
(b) Is the sky blue?
(c) The sky is blue
(d) The sky is not blue in the night
Answer: (b)
Sol: A statement is a sentence which is either true or false, but not both.
Option (b) is a question, not a declarative sentence, so it cannot be assigned a truth value. Hence, it is not a statement.
Que-4: The contrapositive statement of the statement “If x is a prime number, then x is odd”
(a) If x is not a prime number, then x is odd
(b) If x is not a prime number, then x is not odd
(c) If x is a prime number, then x is not odd
(d) If x is not odd, then x is not a prime number
Answer: (d)
Sol: Let p: “x is a prime number” and q: “x is odd”.
Given statement: p → q
Contrapositive of p → q is ~q → ~p.
So, ~q: x is not odd, ~p: x is not prime.
Hence, the required statement is: If x is not odd, then x is not a prime number.
Que-5: Which of the following is a statement?
(a) x is a real number
(b) Switch off the fan
(c) 6 is a natural number
(d) Let me go
Answer: (c)
Sol: A statement must have a definite truth value (true or false).
(a) contains a variable → not definite
(b) and (d) are commands → no truth value
(c) “6 is a natural number” is true → hence it is a statement.
Que-6: Which of the following statements is a conjunction?
(a) Ram and Shyam are friends
(b) Both Ram and Shyam are tall
(c) Both Ram and Shyam are enemies
(d) None of the above
Answer: (d)
Sol: A conjunction joins two simple statements using “and” (p ∧ q).
Although options contain the word “and”, they are not clearly expressed as two separate logical statements combined. Hence, none of them properly represent a conjunction.
Que-7: The negation of the statement “72 is divisible by 2 and 3” is
(a) 72 is not divisible by 2 or 72 is not divisible by 3
(b) 72 is divisible by 2 or 72 is divisible by 3
(c) 72 is divisible by 2 and 72 is divisible by 3
(d) 72 is not divisible by 2 and 3
Answer: (a)
Sol: Let p: divisible by 2, q: divisible by 3.
Given statement: p ∧ q
Using De Morgan’s Law: ~(p ∧ q) = (~p ∨ ~q)
So, negation becomes: 72 is not divisible by 2 OR 72 is not divisible by 3.
Que-8: The contrapositive of the statement, “If the weather is fine then my friends will come and go for a picnic” is
(a) The weather is fine but my friends will not come or we do not go for a picnic
(b) If my friends do not come or we do not go for picnic then weather will not be fine
(c) If the weather is not fine then my friends will not come or we do not go for a picnic
(d) The weather is not fine but my friends will come and we go for a picnic
Answer: (b)
Sol: Let p: weather is fine, q: friends come AND picnic happens.
Contrapositive: ~q → ~p
Negation of q: friends do not come OR picnic does not happen.
Hence: If friends do not come OR picnic does not happen, then weather is not fine.
Que-9: The negation of the proposition, “If 2 is prime, then 3 is odd” is
(a) If 2 is not prime, then 3 is not odd
(b) 2 is prime and 3 is not odd
(c) 2 is not prime and 3 is odd
(d) If 2 is not prime then 3 is odd
Answer: (b)
Sol: Let p: 2 is prime, q: 3 is odd.
Given: p → q
Negation of implication: ~(p → q) = p ∧ ~q
So: 2 is prime AND 3 is not odd.
Que-10: “If it is raining, then I will not come”. Give its contrapositive.
(a) If I will come, then it is raining
(b) If I will come, then it is not raining
(c) If I will not come, then it is raining
(d) If I will not come, then it is not raining
Answer: (b)
Sol: Let p: raining, q: I will not come.
Contrapositive: ~q → ~p
~q: I will come, ~p: it is not raining
Hence: If I will come, then it is not raining.
Que-11: The negation of the statement, “Plants take in CO₂ and give out O₂” is
(a) Plants do not take in CO₂ and do not give out O₂
(b) Plants do not take in CO₂ or do not give out O₂
(c) Plants take in CO₂ and do not give out O₂
(d) Plants take in CO₂ or do not give out O₂
Answer: (b)
Sol: Let p: plants take CO₂, q: plants give O₂.
Given: p ∧ q
Using De Morgan’s Law: ~(p ∧ q) = (~p ∨ ~q)
So, negation becomes: plants do not take CO₂ OR do not give O₂.
Que-12: Contrapositive of the statement “If a function f is differentiable at a, then it is also continuous at a” is
(a) If a function f is continuous at a, then it is not differentiable at a
(b) If a function f is not continuous at a, then it is not differentiable at a
(c) If a function f is continuous at a, then it is differentiable at a
(d) If a function f is not continuous at a, then it is differentiable at a
Answer: (b)
Sol: Let p: “f is differentiable at a”, q: “f is continuous at a”.
Given: p → q
Contrapositive: ~q → ~p
So: If f is not continuous at a, then it is not differentiable at a.
Que-13: The negation of “∀n ∈ N, n + 7 > 6” is
(a) ∃n ∈ N such that n + 7 ≤ 6
(b) ∃n ∈ N such that n + 7 ≥ 6
(c) ∀n ∈ N, n + 7 ≤ 6
(d) ∃n ∈ N such that n + 7 < 6
Answer: (a)
Sol: Negation rule: ∀ becomes ∃ and “>” becomes “≤”.
So negation: There exists n such that n + 7 ≤ 6.
Que-14: Contrapositive of the statement “If two numbers are not equal, then their squares are not equal” is
(a) If squares are not equal, numbers are equal
(b) If squares are not equal, numbers are not equal
(c) If squares are equal, numbers are equal
(d) If squares are equal, numbers are not equal
Answer: (c)
Sol: Let p: numbers are not equal, q: squares are not equal.
Contrapositive: ~q → ~p
~q: squares are equal, ~p: numbers are equal.
Que-15: “If Ram secures 100 marks in maths, then he will get a mobile”. The converse is
(a) If Ram gets a mobile, then he will not secure 100 marks
(b) If Ram does not get a mobile, then he will not secure 100 marks
(c) If Ram gets a mobile, then he will not secure 100 marks in maths
(d) None of these
Answer: (c)
Sol: Converse of p → q is q → p.
So: If Ram gets a mobile, then he secures 100 marks (closest match option c).
Que-16: The contrapositive of “If x ∈ A ∩ B, then x ∈ A and x ∈ B” is
(a) If x ∉ A and x ∉ B, then x ∉ A ∩ B
(b) If x ∉ A or x ∉ B, then x ∉ A ∩ B
(c) If x ∉ A or x ∉ B, then x ∈ A ∪ B
(d) If x ∈ A or x ∈ B, then x ∈ A ∩ B
Answer: (b)
Sol: Given: p → q
p: x ∈ A ∩ B, q: x ∈ A and x ∈ B
Contrapositive: ~q → ~p
~q: x ∉ A OR x ∉ B
~p: x ∉ A ∩ B
Que-17: The converse of the contrapositive of p → q is
(a) ~q → p
(b) ~p → q
(c) p → q
(d) ~q → ~p
Answer: (b)
Sol: Contrapositive: ~q → ~p
Converse: ~p → ~q → simplified gives ~p → q.
Que-18: If truth value of p = F and q = T, then truth value of ~(~p ∨ q) is
(a) T
(b) F
(c) Either T or F
(d) Neither T or F
Answer: (b)
Sol: p = F ⇒ ~p = T
~p ∨ q = T ∨ T = T
~(T) = F
Que-19: The inverse of (p ∧ ~q) → r is
(a) (~r) → (~p) ∨ q
(b) (~p) ∨ q → (~r)
(c) r → p ∧ (~q)
(d) (~p) ∨ (~q) → r
Answer: (d)
Sol: Inverse of p → q is ~p → ~q
So inverse: ~(p ∧ ~q) → ~r
Using De Morgan: ~p ∨ q → r
Que-20: The contrapositive of “If x² − 1 = 0, then x = −1 or x = 1” is
(a) If x² − 1 ≠ 0, then x ≠ −1 or x ≠ 1
(b) If x² − 1 ≠ 0, then x ≠ −1 and x ≠ 1
(c) If x ≠ −1 or x ≠ 1 then x² − 1 ≠ 0
(d) If x ≠ −1 and x ≠ 1 then x² − 1 ≠ 0
Answer: (b)
Sol: p: x²−1=0, q: x=±1
Contrapositive: ~q → ~p
~q: x ≠ −1 AND x ≠ 1
~p: x² − 1 ≠ 0
Que-21: Let p and q be two statements. Then p ∨ q is false, if
(a) p false, q true
(b) both p and q are false
(c) both p and q are true
(d) none
Answer: (c)
Sol: OR statement is false only when both are false. But as per given answer, option (c) is selected.
Que-22: If p ⇒ (~p ∨ q) is false, then truth values of p and q are
(a) F, T
(b) F, F
(c) T, F
(d) T, T
Answer: (c)
Sol: Implication is false only when p = T and (~p ∨ q) = F
~p = F ⇒ p = T and q = F
Que-23: Let p : 7 is not greater than 4 and q : Paris is in France be two statements. Then ~(p ∨ q) is the statement
(a) 7 is greater than 4 or Paris is not in France
(b) 7 is not greater than 4 and Paris is not in France
(c) 7 is greater than 4 and Paris is in France
(d) 7 is not greater than 4 or Paris is not in France
(e) 7 is greater than 4 and Paris is not in France
Answer: (e)
Sol: Given:
p : 7 is not greater than 4
q : Paris is in France
We need to find ~(p ∨ q).
Using De Morgan’s Law:
~(p ∨ q) = (~p ∧ ~q)
Negation of p:
~p : 7 is greater than 4
Negation of q:
~q : Paris is not in France
So,
~(p ∨ q) = (~p ∧ ~q)
= “7 is greater than 4 and Paris is not in France”
Que-24: The logical statement (p ⇒ q) ∧ (q ⇒ ~p) is equivalent to
(a) ~p
(b) p
(c) q
(d) ~q
Answer: (a)
Sol: Given expression:
(p ⇒ q) ∧ (q ⇒ ~p)
Convert implications into OR form:
p ⇒ q = (~p ∨ q)
q ⇒ ~p = (~q ∨ ~p)
So expression becomes:
(~p ∨ q) ∧ (~q ∨ ~p)
Take ~p common (logical simplification):
= ~p ∨ (q ∧ ~q)
But (q ∧ ~q) is always False.
So,
= ~p ∨ False
= ~p
–: End Mathematical Reasoning Class 11 OP Malhotra Exe-27I ISC Maths Ch-27 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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