Ex 9.1 Class 9 Maths Ganita Manjari Solutions | Sid Classes
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 9 Exercise 9.1 Propositions and their Converses. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough makes formulating converses, checking truth values, constructing mathematical justifications, and identifying counterexamples entirely clear.
🔑 Key Concepts & Definitions:
- Proposition: An 'If-then' statement of the form \(P \implies Q\).
- Converse: Reversing the hypothesis and conclusion, i.e., \(Q \implies P\).
- Counterexample: A specific example that proves a conditional statement or conjecture is false.
NCERT Class 9 Maths Ganita Manjari Chapter 9 Exercise 9.1 Solutions
Question 1
If two lines are parallel, then the corresponding angles formed by a transversal are equal.
Solution:
Converse: If the corresponding angles formed by a transversal with two lines are equal, then the two lines are parallel.
Truth Value & Justification: Both the original statement and its converse are true. This is a fundamental property of parallel lines and transversals in Euclidean geometry (corresponding angles axiom and its converse).
Question 2
If a quadrilateral is a square, then all its angles are equal.
Solution:
Converse: If all angles of a quadrilateral are equal, then the quadrilateral is a square.
Truth Value & Justification: The original proposition is true (each angle is \(90^\circ\)). However, the converse is false.
Counterexample: A rectangle has all four angles equal to \(90^\circ\), but it is not necessarily a square because its adjacent sides can be unequal.
Question 3
Given any \(\triangle ABC\), angle bisectors of B and C meet at incentre I, extended to E and F. Proposition: If \(AB = AC\), then \(IE = IF\).
Solution:
Converse: If \(IE = IF\), then \(AB = AC\).
Truth Value & Justification: Both the proposition and its converse are true. In an isosceles triangle (\(AB = AC\)), the angle bisectors and distances to the sides exhibit symmetry leading to \(IE = IF\), and vice versa.
Question 4
If \(x = y\), then \(a + x = a + y\), where \(x, y\) and \(a\) are any three numbers.
Solution:
Converse: If \(a + x = a + y\), then \(x = y\).
Truth Value & Justification: Both the proposition and its converse are true, representing standard algebraic equality axioms (adding or subtracting the same quantity from both sides).
Question 5
If \(a\) and \(b\) are perfect squares, then \(ab\) is a perfect square.
Solution:
Converse: If \(ab\) is a perfect square, then \(a\) and \(b\) are perfect squares.
Truth Value & Justification: The original statement is true (if \(a = m^2\) and \(b = n^2\), then \(ab = (mn)^2\)). The converse is false.
Counterexample: Let \(a = 2\) and \(b = 8\). Then \(ab = 16\), which is a perfect square (\(4^2\)), but neither 2 nor 8 are perfect squares.
Question 6
If \(x = y\), then \(x^2 = y^2\) (where \(x, y\) are real numbers).
Solution:
Converse: If \(x^2 = y^2\), then \(x = y\).
Truth Value & Justification: The original statement is true. The converse is false.
Counterexample: Let \(x = 2\) and \(y = -2\). Then \(x^2 = 4\) and \(y^2 = 4\) (so \(x^2 = y^2\)), but \(2 \neq -2\).
Question 7
If \(x = y\), then \(x^3 = y^3\).
Solution:
Converse: If \(x^3 = y^3\), then \(x = y\).
Truth Value & Justification: Both the proposition and its converse are true because the cube function for real numbers is one-to-one.
Question 8
If \(n\) is divisible by 24, then it is divisible by both 4 and 6.
Solution:
Converse: If \(n\) is divisible by both 4 and 6, then it is divisible by 24.
Truth Value & Justification: The original statement is true (since 4 and 6 are factors of 24). The converse is false.
Counterexample: Let \(n = 12\). 12 is divisible by both 4 and 6, but it is not divisible by 24.
Question 9
If \(n\) is divisible by 60, then it is divisible by both 5 and 12.
Solution:
Converse: If \(n\) is divisible by both 5 and 12, then it is divisible by 60.
Truth Value & Justification: Both the proposition and its converse are true because 5 and 12 are coprime, so any number divisible by both must be divisible by their product, 60.
Question 10
If \(n\) is the square of a prime number, then it has exactly 3 factors.
Solution:
Converse: If \(n\) has exactly 3 factors, then it is the square of a prime number.
Truth Value & Justification: Both the proposition and its converse are true. For a prime \(p\), \(p^2\) has factors \(1, p, p^2\) (exactly 3 factors), and any number with exactly 3 factors must be the square of a prime.
Question 11
If \(n\) is a product of two unequal prime numbers, then it has exactly 4 divisors.
Solution:
Converse: If \(n\) has exactly 4 divisors, then it is a product of two unequal prime numbers.
Truth Value & Justification: The original statement is true (\(n = pq\) has divisors \(1, p, q, pq\)). The converse is false.
Counterexample: Let \(n = 8 = 2^3\). Its divisors are \(1, 2, 4, 8\) (exactly 4 divisors), but 8 is not a product of two unequal primes (it is the cube of a prime).
Question 12
If \(n\) and \(n+3\) have no factors in common, then \(n\) is not a multiple of 3.
Solution:
Converse: If \(n\) is not a multiple of 3, then \(n\) and \(n+3\) have no factors in common.
Truth Value & Justification: Both the proposition and its converse are true. If \(n\) is a multiple of 3, then \(n+3\) is also a multiple of 3, meaning they share 3 as a factor. Conversely, if they share any common factor greater than 1, that factor must divide their difference \((n+3) - n = 3\), meaning the factor is 3, so \(n\) must be a multiple of 3.
Question 13
Find counterexamples to the following claims:
(i) All numbers of the form \(4n^2 + 1\) are prime.
(ii) All numbers of the form \(n^2 + n + 11\) are prime.
(iii) All numbers of the form \(4^n + 3\) are prime.
Solution:
(i) For \(4n^2 + 1\): Let \(n = 4\). Then \(4(4)^2 + 1 = 4(16) + 1 = 65 = 5 \times 13\) (composite).
(ii) For \(n^2 + n + 11\): Let \(n = 11\). Then \(11^2 + 11 + 11 = 121 + 22 = 143 = 11 \times 13\) (composite).
(iii) For \(4^n + 3\): Let \(n = 4\). Then \(4^4 + 3 = 256 + 3 = 259 = 7 \times 37\) (composite).
Question 14
Find counterexamples to the following statements:
(i) If \(n\) is a prime number, then \(2^n - 1\) is a prime number.
(ii) If \(n\) is an even number, then \(2^n + 1\) is a prime number.
Solution:
(i) Let \(n = 11\) (prime). Then \(2^{11} - 1 = 2048 - 1 = 2047 = 23 \times 89\) (composite).
(ii) Let \(n = 6\) (even). Then \(2^6 + 1 = 64 + 1 = 65 = 5 \times 13\) (composite).
Question 15
Consider: 'If a number is divisible by 8, then it is divisible by both 2 and 4.' (i) Justify. (ii) Is it enough to check divisibility by 2 and 4 to ensure divisibility by 8? Why or why not?
Solution:
(i) Justification: Since \(8 = 2 \times 4\), any multiple of 8 contains both 2 and 4 as factors, making it divisible by both.
(ii) Sufficiency: No, it is not enough.
Counterexample: Consider 12. 12 is divisible by both 2 and 4, but 12 is not divisible by 8.
Question 16
Express the relationship between 'a number is divisible by 3' and 'sum of the digits is a multiple of 3' using 'If-then' sentences.
Solution:
* Statement: If a number is divisible by 3, then the sum of its digits is a multiple of 3.
* Converse: If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.
Question 17
Constructing quadrilaterals of type Q using two thin sticks as diagonals.
(i) If type Q requires equal-length diagonals: (a) Should sticks be equal length? Why? (b) Does placement matter?
(ii) If type Q is defined by having equal diagonals: answers to (a) and (b)?
Solution:
(i)
(a) Yes, the two sticks must be of equal length because quadrilaterals of type Q are specified to have equal-length diagonals.
(b) Yes, how the sticks are put together (such as intersecting at right angles or midpoints) determines the specific quadrilateral category (e.g., square, rectangle, or isosceles trapezoid).
(ii)
(a) Yes, the sticks must still be of equal length since any quadrilateral satisfying type Q must have equal diagonals.
(b) Yes, relative placement and intersection angles still dictate the specific geometric shape and properties of type Q.
Frequently Asked Questions (FAQs)
Q1: How do you write the converse of a conditional statement?
A: Interchange the hypothesis and the conclusion of the 'If-then' statement.
Q2: What is the purpose of a counterexample?
A: A single counterexample is sufficient to prove that a mathematical proposition or converse is false.
Quick Navigation:
Last updated according to the latest NCERT Ganita Manjari syllabus.