NCERT Solutions for Class 9 Maths Ganita Manjari Part 2 Chapter 9 – Propositions and Their Converses
The NCERT Solutions for Class 9 Maths Ganita Manjari Part 2 Chapter 9 – Propositions and Their Converses help students understand how mathematical statements are formed, verified, and related to their converses. This chapter introduces students to propositions written in the “if–then” form and explains how to determine whether a proposition and its converse are true or false.
Through examples involving numbers, divisibility, prime numbers, algebra, triangles, and quadrilaterals, students learn how to justify mathematical statements and use counterexamples to disprove incorrect claims. These solutions provide clear, step-by-step explanations to help students strengthen their logical reasoning and prepare effectively for Class 9 Maths examinations.
NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 9 Exercise Set 9.1
1. If two lines are parallel, then the corresponding angles formed by a transversal are equal.
Answer:
If two parallel lines are cut by a transversal, each pair of corresponding angles is equal.
2. If a quadrilateral is a square, then all its angles are equal.
Answer:
If a quadrilateral is a square, then all four of its angles are equal, and each angle measures 90°.
3. Given any △ABC, Iet us bisect the angles at B and C. The bisectors meet at the incentre I of the triangle.
Now extend the bisectors beyond I till they meet the opposite sides at E and F respectively, as shown.
Proposition: If AB = AC, then IE = IF.
Answer:
Proposition: If AB = AC, then IE = IF.
Given:
In △ABC, AB = AC. BI and CI are the angle bisectors of ∠B and ∠C respectively. BI and CI are extended to meet AC at E and AB at F respectively.
To prove:
IE = IF
Proof:
Since AB = AC, △ABC is an isosceles triangle.
Therefore,
∠ABC = ∠BCA
(Angles opposite equal sides are equal)
Since BI and CI bisect ∠B and ∠C respectively,
∠IBC = ∠BCI
Therefore, in △BIC,
IB = IC
(Sides opposite equal angles are equal)
Now consider △BIF and △CIE:
- IB = IC
- ∠IBF = ∠ICE
(Halves of equal angles ∠B and ∠C) - ∠BIF = ∠CIE
(Vertically opposite angles)
Therefore,
△BIF ≅ △CIE
(ASA congruence criterion)
Hence,
IE = IF
(Corresponding parts of congruent triangles)
Hence proved.
4. If x = y, then a + x = a + y, where x, y and a are any three numbers.
This proposition and its converse are routinely used while solving equations.
Answer:
Proposition
If x = y, then a + x = a + y, where a, x, and y are any numbers.
This follows from the addition property of equality: adding the same number to both sides of an equation keeps the two sides equal.
Converse
If a + x = a + y, then x = y.
Subtracting a from both sides gives:
x = y
Therefore, both the proposition and its converse are true, which is why they are commonly used while solving equations.
5. If a and b are perfect squares, then ab is a perfect square.
Answer:
If a and b are perfect squares, then ab is a perfect square.
Converse:
If ab is a perfect square, then a and b are perfect squares.
Checking the proposition:
The proposition is true.
Justification:
Let
a = m² and b = n²,
where m and n are integers.
Then,
ab = m²n² = (mn)²
Therefore, ab is a perfect square.
Checking the converse:
The converse is false.
Counterexample:
Take a = 2 and b = 8.
Then,
ab = 2 × 8 = 16
Since 16 is a perfect square, ab is a perfect square. However, neither 2 nor 8 is a perfect square.
Therefore, the converse is false.
6. If x = y, then x2 = y2.
Answer:
If x = y, then x² = y².
Converse:
If x² = y², then x = y.
Checking the proposition:
The proposition is true.
Justification:
If x = y, then multiplying both sides by the same number gives:
x × x = y × y
Therefore,
x² = y²
So, the proposition is true.
Checking the converse:
The converse is false.
Counterexample:
Take x = 2 and y = −2.
Then,
x² = 4 and y² = 4
So, x² = y², but x ≠ y.
Therefore, the converse is false.
7. If x = y, then x3 = y3.
Answer:
If x = y, then x³ = y³.
Converse:
If x³ = y³, then x = y.
Checking the proposition:
The proposition is true.
Justification:
If x = y, then multiplying equal quantities by themselves three times gives:
x × x × x = y × y × y
Therefore,
x³ = y³
So, the proposition is true.
Checking the converse:
The converse is also true.
Justification:
If x³ = y³, then taking the cube root of both sides gives:
x = y
Therefore, both the proposition and its converse are true.
8. If n is divisible by 24, then it is divisible by both 4 and 6.
Answer:
Converse:
If n is divisible by both 4 and 6, then n is divisible by 24.
Checking the proposition:
The proposition is true.
Justification:
If n is divisible by 24, then
n = 24k
for some integer k.
Since 24 = 4 × 6,
n = 4(6k)
and
n = 6(4k)
Therefore, n is divisible by both 4 and 6.
Checking the converse:
The converse is false.
Counterexample:
Take n = 12.
12 is divisible by both 4 and 6, but 12 is not divisible by 24.
Therefore, the converse is false.
9. If n is divisible by 60, then it is divisible by both 5 and 12.
Answer:
Proposition:
If n is divisible by 60, then n is divisible by both 5 and 12.
Converse:
If n is divisible by both 5 and 12, then n is divisible by 60.
Checking the proposition:
The proposition is true.
Justification:
If n is divisible by 60, then
n = 60k
for some integer k.
Since 60 = 5 × 12,
n = 5(12k)
and
n = 12(5k)
Therefore, n is divisible by both 5 and 12.
Checking the converse:
The converse is also true.
Justification:
Since 5 and 12 are co-prime,
LCM(5, 12) = 60.
Therefore, any number divisible by both 5 and 12 must also be divisible by 60.
Hence, both the proposition and its converse are true.
10. If n is the square of a prime number, then it has exactly 3 factors.
Answer:
Let
n = p²
where p is a prime number.
The factors of p² are:
1, p, p²
Thus, n has exactly 3 factors.
Checking the converse:
The converse is also true.
Justification:
If n has exactly 3 factors, then its factors must be of the form:
1, p, p²
where p is prime.
Therefore,
n = p²
Hence, n is the square of a prime number.
Therefore, both the proposition and its converse are true.
11. If n is a product of two unequal prime numbers, then it has exactly 4 divisors.
Answer:
Proposition:
If n is a product of two unequal prime numbers, then n has exactly 4 divisors.
Converse:
If n has exactly 4 divisors, then n is a product of two unequal prime numbers.
Checking the proposition:
The proposition is true.
Justification:
Let
n = pq
where p and q are unequal prime numbers.
The divisors of n are:
1, p, q, pq
Thus, n has exactly 4 divisors.
Checking the converse:
The converse is false.
Counterexample:
Take n = 8.
The divisors of 8 are:
1, 2, 4, 8
So, 8 has exactly 4 divisors. However,
8 = 2 × 2 × 2
and it cannot be written as the product of two unequal prime numbers.
Therefore, the converse is false.
12. If n and n + 3 have no factors in common, then n is not a multiple of 3.
Answer:
12. If n and n + 3 have no factors in common, then n is not a multiple of 3.
Proposition:
If n and n + 3 have no common factor other than 1, then n is not a multiple of 3.
Converse:
If n is not a multiple of 3, then n and n + 3 have no common factor other than 1.
Checking the proposition:
The proposition is true.
Justification:
Suppose n were a multiple of 3.
Then,
n = 3k
for some integer k.
So,
n + 3 = 3k + 3 = 3(k + 1)
Thus, both n and n + 3 would be divisible by 3, which means they would have 3 as a common factor.
This contradicts the given condition.
Therefore, n is not a multiple of 3.
Checking the converse:
The converse is also true.
Justification:
Any common factor of n and n + 3 must also divide their difference:
(n + 3) − n = 3
So, their common factor can only be 1 or 3.
Since n is not divisible by 3, 3 cannot be a common factor.
Therefore, n and n + 3 have no common factor other than 1.
Hence, both the proposition and its converse are true.
13. There are no known ‘neat’ expressions that generate only primes! Find counterexamples to the following claims.
(i) AII numbers of the form 4n2 + 1 are prime.
(ii) AII numbers of the form n2 + n + 11 are prime.
(iii) AII numbers of the form 4n + 3 are prime.
Answer:
(i) All numbers of the form 4n² + 1 are prime.
Take n = 4.
4n² + 1
= 4(4²) + 1
= 4 × 16 + 1
= 65
= 5 × 13
Since 65 is not prime, the claim is false.
(ii) All numbers of the form n² + n + 11 are prime.
Take n = 10.
n² + n + 11
= 10² + 10 + 11
= 100 + 10 + 11
= 121
= 11 × 11
Since 121 is not prime, the claim is false.
(iii) All numbers of the form 4n + 3 are prime.
4n + 3 = 44 + 3 = 256 + 3 = 259
But 259 = 7 × 37
Therefore, 259 is not prime. Counterexample: n = 4.
14. Find counterexamples to the fouowing statements.
(i) If n is a prime number then 2n – 1is a prime number.
(ii) If n is an even number, then 2n + 1 is a prime number.
Answer:
(i) Take n = 11, then
2n – 1 = 211 – 1 = 2048 – 1 = 2047
But 2047 = 23 × 89
Therefore, 2047 is not a prime number.
Counterexample: n = 11.
(ii) Take n = 6, then
2n + 1 = 26 + 1 = 64 + 1 = 65
But 65 = 5 × 13
Therefore, 65 is not a prime number.
Counterexample: n = 6.
15. Consider the statement: ‘If a number is divisible by 8, then it is divisible by both Z and 4’.
(i) Justify the statement.
(ii) RecaII the divisibility shortcuts that we have studied for di-fferent numbers. To check whether a given number is divisible by 8, is it enough to check whether it is divisille by 2 and 4? Why or why not?
Answer:
(i) Justification:
Since 8 = 2 × 4, every number divisible by 8 is also divisible by both 2 and 4.
For example, 24 is divisible by 8.
24 ÷ 2 = 12
24 ÷ 4 = 6
Hence, the statement is true.
(ii) No, it is not enough to check whether a number is divisible by both 2 and 4.
For example, 12 is divisible by both 2 and 4:
12 ÷ 2 = 6
12 ÷ 4 = 3
But 12 is not divisible by 8 because 12 ÷ 8 is not a whole number.
Therefore, being divisible by both 2 and 4 does not guarantee that a number is divisible by 8.
16. Recall that a shortcut to check whether a given number is divisible by 3 is to add the digits of the number and check if the sum is a multiple of 3. Express the relationship betlveen ‘a number is divisible by 3’ and ‘sum of the digits is a multiple of 3’ using ‘If – then’ sentences.
Answer:
Proposition:
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.
Therefore, both the proposition and its converse are true.
17. We have identified different types of quadrilaterals – squares, rectangles, parallelograms, rhombi, kites and trapezia. One can identify more types (e.g., we could create a category of quadrilaterals that have equal-length opposite sides).
Suppose we have identified a category of quadrilaterals called Q, and we have to construct a quadrilateral of this type. For this, we are to use two thin sticks, put them together as diagonals so that the quadrilateral obtained by joining their endpoints is of type Q (see the Fig. 9.2).
(i) Suppose Q satisfies the following property.
If a quadrilateral is of type Q, then it has equal-length diagonals.
(a) Should the two sticks be of equal length? Why or why not?
(b) WiIl it matter how the two sticks are put together?
(ii) Instead of the property mentioned above, suppose Q satisfies the following property.
If a quadrilateral has equal diagonals, then it is of type Q. What will be your answers to (a) and (b) now?
Answer:
(i) If a quadrilateral is of type Q, then it has equal-length diagonals.
(a) Should the two sticks be of equal length? Why or why not?
Yes.
Since every quadrilateral of type Q has equal-length diagonals, the two sticks representing the diagonals must be of equal length.
If the sticks have unequal lengths, the quadrilateral obtained cannot be of type Q.
(b) Will it matter how the two sticks are put together?
Yes, it may matter.
The given statement only tells us that:
Q ⇒ equal diagonals
It does not say that every quadrilateral having equal diagonals is of type Q.
Therefore, merely taking two equal sticks does not guarantee that the quadrilateral formed is of type Q. Their arrangement may also have to satisfy other properties of Q.
(ii) If a quadrilateral has equal diagonals, then it is of type Q.
(a) Should the two sticks be of equal length?
Yes, to guarantee that the quadrilateral formed is of type Q, we should take two equal-length sticks.
Then the quadrilateral will have equal diagonals, and by the given statement, it must be of type Q.
(b) Will it matter how the two sticks are put together?
No.
Since the statement says:
equal diagonals ⇒ Q
once the two sticks are of equal length, the resulting quadrilateral will have equal diagonals and hence will be of type Q, irrespective of how the sticks intersect, provided they form a quadrilateral.
So, in this case, equal-length sticks are sufficient to ensure that the quadrilateral is of type Q.
NCERT Solutions for Class 9 Maths Ganita Manjari Part 1 - All Chapters
| Chapter | Chapter name | Exercise sets |
|---|---|---|
| 1 | Orienting Yourself: The Use of Coordinates | 2 sets + End of Chapter |
| 2 | Introduction to Linear Polynomials | 6 sets + End of Chapter |
| 3 | The World of Numbers | 5 sets + End of Chapter |
| 4 | Exploring Algebraic Identities | 5 sets + End of Chapter |
| 5 | I’m Up and Down, and Round and Round | 6 sets + End of Chapter |
| 6 | Measuring Space: Perimeter and Area | 3 sets + End of Chapter |
| 7 | The Mathematics of Maybe: Introduction to Probability | 4 sets + End of Chapter |
| 8 | Predicting What Comes Next: Exploring Sequences and Progressions | 3 sets + End of Chapter |