Get the most accurate ISC Solutions for Class 11 Mathematics Chapter 27 Mathematical Reasoning here. Updated for the 2026-27 academic session, these solutions are based on the latest ISC textbooks for Class 11 Mathematics. Our expert-created answers for Class 11 Mathematics are available for free download in PDF format.
Detailed Chapter 27 Mathematical Reasoning ISC Solutions for Class 11 Mathematics
For Class 11 students, solving ISC textbook questions is the most effective way to build a strong conceptual foundation. Our Class 11 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 27 Mathematical Reasoning solutions will improve your exam performance.
Class 11 Mathematics Chapter 27 Mathematical Reasoning ISC Solutions PDF
S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(D)
Question 1. Let p, q, r and s represent simple statements. Assume that p is false, q is true, r is false, and s is true. Determine the truth value of each statement expressed below :
(i) \( q \wedge r \)
(ii) \( r \vee p \)
(iii) \( p \wedge s \)
(iv) \( p \vee s \)
(v) \( \sim q \)
(vi) \( q \vee s \)
(vii) \( \sim r \)
(viii) \( s \wedge q \)
(ix) \( r \wedge p \)
Answer:
Given that statement \( p \) is false, \( q \) is true, \( r \) is false, and \( s \) is true.
(i) For \( q \wedge r \): Since \( q \) is True and \( r \) is False, the compound statement \( q \wedge r \) is False. This is because a conjunction (AND) statement is only true if all its parts are true.
(ii) For \( r \vee p \): Since \( r \) is False and \( p \) is False, the compound statement \( r \vee p \) is False. An disjunction (OR) statement is only false if all its parts are false.
(iii) For \( p \wedge s \): Since \( p \) is False and \( s \) is True, the compound statement \( p \wedge s \) is False. One false component makes the entire AND statement false.
(iv) For \( p \vee s \): Since \( p \) is False and \( s \) is True, the compound statement \( p \vee s \) is True. Because at least one component (s) is true, the OR statement is true.
(v) For \( \sim q \): Since \( q \) is True, its negation \( \sim q \) is False. The negation flips the truth value of the original statement.
(vi) For \( q \vee s \): Since \( q \) is True and \( s \) is True, the compound statement \( q \vee s \) is True. When both parts of an OR statement are true, the whole statement is true.
(vii) For \( \sim r \): Since \( r \) is False, its negation \( \sim r \) is True. This flips the original false truth value to true.
(viii) For \( s \wedge q \): Since \( s \) is True and \( q \) is True, the compound statement \( s \wedge q \) is True. Both statements being true makes their conjunction true.
(ix) For \( r \wedge p \): Since \( r \) is False and \( p \) is False, the compound statement \( r \wedge p \) is False. Any false component makes an AND statement false.
In simple words: We are given what is true or false for p, q, r, and s. Then, we figure out if new statements made by combining them with "and" (\( \wedge \)), "or" (\( \vee \)), or "not" (\( \sim \)) are true or false. Remember, "and" needs both parts to be true, "or" needs at least one part to be true, and "not" just flips the answer.
🎯 Exam Tip: When evaluating compound statements, clearly write down the truth value of each simple statement first. Then, apply the truth table rules for \( \wedge \) (AND), \( \vee \) (OR), and \( \sim \) (NOT) step-by-step to avoid errors.
Question 2. Let a, b, c and d represent simple statements. Assume that \( a \wedge d \) is true, \( b \wedge c \) is false, and \( \sim c \) is false.
(i) What is the truth value of a?
(ii) What is the truth value of d?
(iii) What is truth value of c?
(iv) What is the truth value of b?
Answer:
We are given the following information:
1. \( a \wedge d \) is true.
2. \( b \wedge c \) is false.
3. \( \sim c \) is false.
Let's find the truth values step-by-step:
From (3), \( \sim c \) is false.
\( \implies \) This means the statement \( c \) itself must be true. (Because if NOT c is false, then c has to be true).
From (1), \( a \wedge d \) is true.
\( \implies \) For an AND statement to be true, both individual statements must be true. So, \( a \) is true, and \( d \) is true. (If either a or d were false, \( a \wedge d \) would be false).
From (2), \( b \wedge c \) is false.
\( \implies \) We already know \( c \) is true. So, we have \( b \wedge \text{True} \) is false.
\( \implies \) For \( b \wedge \text{True} \) to be false, \( b \) must be false. (If b were true, then True \( \wedge \) True would be True).
So, the truth values are:
(i) The truth value of \( a \) is true.
(ii) The truth value of \( d \) is true.
(iii) The truth value of \( c \) is true.
(iv) The truth value of \( b \) is false.
In simple words: We work backwards from the given facts. If "not c" is false, then "c" must be true. If "a and d" is true, then both "a" and "d" must be true. If "b and c" is false, and we know "c" is true, then "b" must be false to make the "and" statement false.
🎯 Exam Tip: Always deduce truth values logically in a clear, step-by-step manner. Start with negations, then conjunctions (AND), and finally disjunctions (OR) if available, as they often simplify the problem quickly.
Question 3. Assume that two given statements p and q are both true and indicate whether or not you would expect each of the following statements to be true :
(i) \( p \wedge q \)
(ii) \( p \vee q \)
(iii) \( p \vee (\sim q) \)
(iv) \( (\sim p) \vee (\sim q) \)
Answer:
We are given that statement \( p \) is true and statement \( q \) is true.
(i) For \( p \wedge q \): Since \( p \) is True and \( q \) is True, the statement \( p \wedge q \) is True. (Both parts are true, so the AND statement is true).
(ii) For \( p \vee q \): Since \( p \) is True and \( q \) is True, the statement \( p \vee q \) is True. (At least one part is true, so the OR statement is true).
(iii) For \( p \vee (\sim q) \): First, find \( \sim q \). Since \( q \) is True, \( \sim q \) is False. Now, we have \( p \vee (\sim q) \) which is True \( \vee \) False. This statement is True. (Because p is true, the OR statement is true).
(iv) For \( (\sim p) \vee (\sim q) \): First, find \( \sim p \). Since \( p \) is True, \( \sim p \) is False. Next, find \( \sim q \). Since \( q \) is True, \( \sim q \) is False. Now, we have \( (\sim p) \vee (\sim q) \) which is False \( \vee \) False. This statement is False. (Both parts are false, so the OR statement is false).
In simple words: We assume both 'p' and 'q' are true. Then we check different combinations: "p and q" is true; "p or q" is true; "p or not q" is true (because p is true); and "not p or not q" is false (because both "not p" and "not q" are false).
🎯 Exam Tip: Remember the core definitions: \( \wedge \) (AND) is true only if *all* parts are true; \( \vee \) (OR) is true if *at least one* part is true; \( \sim \) (NOT) reverses the truth value. Break down complex statements into simpler components.
Question 4. Construct truth tables for :
\( (\sim p) \wedge q \)
Answer:
| \( p \) | \( q \) | \( \sim p \) | \( (\sim p) \wedge q \) |
|---|---|---|---|
| T | T | F | F |
| T | F | F | F |
| F | T | T | T |
| F | F | T | F |
🎯 Exam Tip: When constructing truth tables, ensure you list all possible combinations of truth values for the simple statements (2^n combinations for n statements). Calculate intermediate steps like \( \sim p \) before the final compound statement.
Question 5. Construct truth tables for :
\( (\sim p) \wedge (\sim q) \)
Answer:
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( (\sim p) \wedge (\sim q) \) |
|---|---|---|---|---|
| T | T | F | F | F |
| T | F | F | T | F |
| F | T | T | F | F |
| F | F | T | T | T |
🎯 Exam Tip: Pay close attention to De Morgan's laws: \( \sim (p \vee q) \equiv (\sim p) \wedge (\sim q) \) and \( \sim (p \wedge q) \equiv (\sim p) \vee (\sim q) \). Understanding these can help verify your truth table results for negations of conjunctions/disjunctions.
Question 6. Construct truth tables for :
\( \sim (p \wedge q) \)
Answer:
| \( p \) | \( q \) | \( p \wedge q \) | \( \sim (p \wedge q) \) |
|---|---|---|---|
| T | T | T | F |
| T | F | F | T |
| F | T | F | T |
| F | F | F | T |
🎯 Exam Tip: Always calculate the truth value of the expression inside the parentheses or brackets first, then apply any negations (~) to that result. This ensures the correct order of operations in logic.
Question 7. Construct truth tables for :
\( p \wedge (\sim q) \)
Answer:
| \( p \) | \( q \) | \( \sim q \) | \( p \wedge (\sim q) \) |
|---|---|---|---|
| T | T | F | F |
| T | F | T | T |
| F | T | F | F |
| F | F | T | F |
🎯 Exam Tip: Ensure that the column for \( \sim q \) is correctly derived by negating the values in the \( q \) column. A mistake in an intermediate column will lead to an incorrect final truth table.
Question 8. Construct truth tables for :
\( \sim [p \vee (\sim q)] \)
Answer:
| \( p \) | \( q \) | \( \sim q \) | \( p \vee (\sim q) \) | \( \sim [p \vee (\sim q)] \) |
|---|---|---|---|---|
| T | T | F | T | F |
| T | F | T | T | F |
| F | T | F | F | T |
| F | F | T | T | F |
🎯 Exam Tip: When dealing with nested logical operations, always work from the innermost part outwards. For \( \sim [p \vee (\sim q)] \), first resolve \( (\sim q) \), then \( [p \vee (\sim q)] \), and finally apply the outermost negation \( \sim \).
Question 9. Construct truth tables for :
\( \sim (\sim p \wedge \sim q) \)
Answer:
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( \sim p \wedge \sim q \) | \( \sim (\sim p \wedge \sim q) \) |
|---|---|---|---|---|---|
| T | T | F | F | F | T |
| T | F | F | T | F | T |
| F | T | T | F | F | T |
| F | F | T | T | T | F |
🎯 Exam Tip: This expression is a classic example of De Morgan's Law: \( \sim (\sim p \wedge \sim q) \equiv p \vee q \). If you construct the truth table for \( p \vee q \) separately, you'll find the final columns match, which is a good way to check your work.
Question 10. Construct truth tables for :
\( (p \wedge q) \vee (\sim p \wedge q) \)
Answer:
| \( p \) | \( q \) | \( p \wedge q \) | \( \sim p \) | \( \sim p \wedge q \) | \( (p \wedge q) \vee (\sim p \wedge q) \) |
|---|---|---|---|---|---|
| T | T | T | F | F | T |
| T | F | F | F | F | F |
| F | T | F | T | T | T |
| F | F | F | T | F | F |
🎯 Exam Tip: Notice that this expression is equivalent to \( q \). You can factor out \( q \) using the distributive law: \( (p \wedge q) \vee (\sim p \wedge q) \equiv (p \vee \sim p) \wedge q \). Since \( (p \vee \sim p) \) is always True, the expression simplifies to True \( \wedge q \), which is \( q \). This kind of simplification can help understand and check your truth table results.
Question 11. Construct truth tables for :
\( p \wedge (q \vee r) \)
Answer:
| \( p \) | \( q \) | \( r \) | \( q \vee r \) | \( p \wedge (q \vee r) \) |
|---|---|---|---|---|
| T | T | T | T | T |
| T | T | F | T | T |
| T | F | T | T | T |
| T | F | F | F | F |
| F | T | T | T | F |
| F | T | F | T | F |
| F | F | T | T | F |
| F | F | F | F | F |
🎯 Exam Tip: For expressions with three simple statements (like p, q, r), remember there are 8 rows in the truth table. Systematically list all combinations to ensure no case is missed, perhaps by alternating T/F for r, then TT/FF for q, then TTTT/FFFF for p.
Question 12. Construct truth tables for :
\( (\sim p \wedge \sim q) \vee (p \wedge \sim q) \)
Answer:
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( \sim p \wedge \sim q \) | \( p \wedge \sim q \) | \( (\sim p \wedge \sim q) \vee (p \wedge \sim q) \) |
|---|---|---|---|---|---|---|
| T | T | F | F | F | F | F |
| T | F | F | T | F | T | T |
| F | T | T | F | F | F | F |
| F | F | T | T | T | F | T |
🎯 Exam Tip: Similar to Question 10, this expression also simplifies: \( (\sim p \wedge \sim q) \vee (p \wedge \sim q) \equiv (\sim p \vee p) \wedge \sim q \). Since \( (\sim p \vee p) \) is a tautology (always True), the expression simplifies to True \( \wedge \sim q \), which is \( \sim q \). Use this to quickly verify the final column of your truth table.
Question 13. Construct truth tables for :
\( (p \vee q) \vee (r \wedge \sim q) \)
Answer:
| \( p \) | \( q \) | \( r \) | \( \sim q \) | \( p \vee q \) | \( r \wedge \sim q \) | \( (p \vee q) \vee (r \wedge \sim q) \) |
|---|---|---|---|---|---|---|
| T | T | T | F | T | F | T |
| T | T | F | F | T | F | T |
| T | F | T | T | T | T | T |
| T | F | F | T | T | F | T |
| F | T | T | F | T | F | T |
| F | T | F | F | T | F | T |
| F | F | T | T | F | T | T |
| F | F | F | T | F | F | F |
🎯 Exam Tip: For complex expressions with multiple operators and statements, create a column for each intermediate sub-expression. For example, for \( (p \vee q) \vee (r \wedge \sim q) \), you need columns for \( \sim q \), \( p \vee q \), \( r \wedge \sim q \), and finally the full expression. This systematic approach reduces errors.
Question 14. Let p be "Ananya is beautiful,” and let q be “Ananya is 165 centimetres tall.”
(i) Under what conditions is the statement, “Ananya is beautiful and 165 centimetres tall.” true?
(ii) Under what conditions is the statement, “Ananya is beautiful and 165 centimetres tall,” false?
(iii) Under what conditions is the statement, “Ananya is beautiful or 165 centimetres tall,” true?
(iv) Under what conditions is the statement, “Ananya is beautiful or 165 centimetres tall,” false?
Answer:
Let \( p \) represent "Ananya is beautiful."
Let \( q \) represent "Ananya is 165 centimetres tall."
(i) The statement "Ananya is beautiful and 165 centimetres tall" corresponds to \( p \wedge q \). This statement is true only if both \( p \) is true (Ananya is beautiful) and \( q \) is true (Ananya is 165 cm tall). Both conditions must be met for an "AND" statement to be true.
(ii) The statement "Ananya is beautiful and 165 centimetres tall" corresponds to \( p \wedge q \). This statement is false if Ananya is not beautiful OR not 165 cm tall, or both. In simpler terms, if either \( p \) is false, or \( q \) is false, or both are false, the entire "AND" statement becomes false.
(iii) The statement "Ananya is beautiful or 165 centimetres tall" corresponds to \( p \vee q \). This statement is true if Ananya is beautiful OR 165 cm tall, or both. Only one of the conditions needs to be true for the "OR" statement to be true.
(iv) The statement "Ananya is beautiful or 165 centimetres tall" corresponds to \( p \vee q \). This statement is false only if Ananya is neither beautiful nor 165 cm tall. For an "OR" statement to be false, both \( p \) must be false and \( q \) must be false. This is the only way it can be false.
In simple words: When we say "AND", both parts must be true for the whole sentence to be true. When we say "OR", only one part needs to be true for the whole sentence to be true. So, for "Ananya is beautiful AND 165 cm tall" to be true, she must be both. For "Ananya is beautiful OR 165 cm tall" to be true, she just needs to be one of those things (or both).
🎯 Exam Tip: Clearly define the simple statements (e.g., p, q) before analyzing compound statements. Remember that "and" (\( \wedge \)) requires all parts to be true for the compound statement to be true, while "or" (\( \vee \)) requires at least one part to be true.
Free study material for Mathematics
ISC Solutions Class 11 Mathematics Chapter 27 Mathematical Reasoning
Students can now access the ISC Solutions for Chapter 27 Mathematical Reasoning prepared by teachers on our website. These solutions cover all questions in exercise in your Class 11 Mathematics textbook. Each answer is updated based on the current academic session as per the latest ISC syllabus.
Detailed Explanations for Chapter 27 Mathematical Reasoning
Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 11 Mathematics chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 11 students who want to understand both theoretical and practical questions. By studying these ISC Questions and Answers your basic concepts will improve a lot.
Benefits of using Mathematics Class 11 Solved Papers
Using our Mathematics solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 11 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 27 Mathematical Reasoning to get a complete preparation experience.
FAQs
The complete and updated OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Exercise 27 (D) is available for free on StudiesToday.com. These solutions for Class 11 Mathematics are as per latest ISC curriculum.
Yes, our experts have revised the OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Exercise 27 (D) as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Mathematics concepts are applied in case-study and assertion-reasoning questions.
Toppers recommend using ISC language because ISC marking schemes are strictly based on textbook definitions. Our OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Exercise 27 (D) will help students to get full marks in the theory paper.
Yes, we provide bilingual support for Class 11 Mathematics. You can access OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Exercise 27 (D) in both English and Hindi medium.
Yes, you can download the entire OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Exercise 27 (D) in printable PDF format for offline study on any device.