Ganita Manjari · Class 9 · Part II · Chapter 9

Class 9 Maths Ganita Manjari Chapter 9 Exercise 9.1 Solutions

All 17 questions, worked step by step in the CBSE pattern. Each answer frames the converse, tests both statements, and uses diagrams where they help.

Last updated 6 October 2026

17Questions
5Diagrams
Ch 9Propositions and their Converses

How to answer every question

PropositionIf P, then Q

ConverseIf Q, then P

Frame the converse by swapping the “if” part and the “then” part.

Frame the converse Swap the “if” part and the “then” part. Keep other conditions, like “n is a positive integer”, unchanged.
2 Test P
- True: give a reason.
- False: give one counterexample.
Test the converse Q Same way as step 2.
Conclude One line: “P is true but its converse is false.”

Proposition: a statement that is either true or false.

Converse of “if P then Q”: “if Q then P”.

Counterexample: one example that satisfies the “if” part but breaks the “then” part. One is enough to prove a statement false.

Geometry

Question 1: Parallel lines and corresponding angles

P trueQ true

Proposition PIf two lines are parallel, then the corresponding angles formed by a transversal are equal.

Step 1 · Converse QIf the corresponding angles formed by a transversal are equal, then the two lines are parallel.

Q1 · transversal t cutting parallel lines l and m

Step 2 · Is P true?True

In the figure l ∥ m and t is a transversal, so ∠1 = ∠5 by the corresponding angles axiom (accepted in earlier classes).

Step 3 · Is Q true?True

Suppose ∠1 = ∠5 but l and m were not parallel. Then they would meet at a point and form a triangle with the transversal. In that triangle one of the two corresponding angles is an exterior angle and the other is its interior opposite angle, so they cannot be equal. This contradicts ∠1 = ∠5. Hence l ∥ m.

P is true and its converse Q is also true.

Question 2: Square and equal angles

P trueQ false

Proposition PIf a quadrilateral is a square, then all its angles are equal.

Step 1 · Converse QIf all angles of a quadrilateral are equal, then it is a square.

Step 2 · Is P true?True

Every angle of a square is 90°, so all four angles are equal.

Step 3 · Is Q true?False

If four angles of a quadrilateral are equal, each is 360° ÷ 4 = 90°. A rectangle of length 6 cm and breadth 4 cm has all angles equal to 90°, but its adjacent sides are 6 cm and 4 cm, which are unequal. So it is not a square.

Q2 · square and the 6 cm by 4 cm rectangle

P is true but Q is false. Counterexample: a 6 cm by 4 cm rectangle.

Question 3: Incentre and the angle bisectors

StarredP trueQ false

Setup: In triangle ABC the bisectors of ∠B and ∠C meet at the incentre I. Extended, they meet the opposite sides at E (on AC) and F (on AB). This is the starred (challenge) question.

Proposition PIf AB = AC, then IE = IF.

Step 1 · Converse QIf IE = IF, then AB = AC.

Step 2 · Is P true? ProofTrue

Q3 · isosceles triangle with incentre I

AB = AC, so ∠B = ∠C (angles opposite equal sides are equal; Statement 1 of the chapter).
BI and CI bisect equal angles, so ∠IBC = ∠ICB. Call each x. Then ∠B = ∠C = 2x.
In triangle IBC, ∠IBC = ∠ICB, so IB = IC (sides opposite equal angles are equal; this is Statement 2, the converse proved at the start of the chapter).
Compare triangles FBC and ECB. ∠FBC = ∠ECB = 2x, BC = CB (common), ∠FCB = ∠EBC = x. So △FBC ≅ △ECB (ASA), which gives CF = BE.
IF = CF − IC and IE = BE − IB. Since CF = BE and IC = IB, we get IF = IE.

Step 3 · Is Q true? CounterexampleFalse

Take a triangle with ∠A = 60°, ∠B = 80°, ∠C = 40°.

∠IBC = 40° and ∠ICB = 20°, so ∠BIC = 180° − 60° = 120°.
∠FIE = ∠BIC = 120° (vertically opposite angles).
∠A + ∠FIE = 60° + 120° = 180°, so AFIE is a cyclic quadrilateral (opposite angles add up to 180°).
AI bisects ∠A, so ∠FAI = ∠EAI = 30°. Equal angles at A in the same circle give equal chords, so IF = IE.
But AB is opposite ∠C = 40° and AC is opposite ∠B = 80°, so AB ≠ AC.

Q3 · counterexample with A = 60°, B = 80°, C = 40°

Check by measurement: for BC = 10 cm this triangle has AB ≈ 7.4 cm and AC ≈ 11.4 cm, while IE = IF ≈ 2.6 cm. In fact IE = IF whenever ∠A = 60°, isosceles or not.

P is true but its converse Q is false.

Algebra

Question 4: Adding the same number to equal numbers

P trueQ true

Proposition PIf x = y, then a + x = a + y, where x, y and a are any three numbers.

Step 1 · Converse QIf a + x = a + y, then x = y.

Step 2 · Is P true?True

Adding the same number a to two equal numbers gives equal results.

Step 3 · Is Q true?True

Subtract a from both sides: (a + x) − a = (a + y) − a, so x = y.

Both P and Q are true. Together they say we may add the same number to both sides of an equation, or cancel the same number from both sides, and the equation stays true.

Question 5: Product of perfect squares

P trueQ false

Proposition PIf a and b are perfect squares, then ab is a perfect square.

Step 1 · Converse QIf ab is a perfect square, then a and b are perfect squares.

Step 2 · Is P true?True

Let a = m² and b = n² for whole numbers m and n. Then ab = m² × n² = (mn)², which is a perfect square.

Step 3 · Is Q true?False

Take a = 2 and b = 8. Then ab = 16 = 4² is a perfect square, but neither 2 nor 8 is a perfect square. (Another counterexample: a = 3, b = 12 gives ab = 36.)

P is true but Q is false.

Question 6: Squaring equal numbers

P trueQ false

Proposition PIf x = y, then x² = y² (x and y are real numbers).

Step 1 · Converse QIf x² = y², then x = y.

Step 2 · Is P true?True

If x = y, replace y by x: y² = y × y = x × x = x².

Step 3 · Is Q true?False

Take x = 3 and y = −3. Then x² = 9 = y², but x ≠ y.

What is true is shown by factorising:

x² = y² ⇒ x² − y² = (x − y)(x + y) = 0 ⇒ x = y or x = −y

P is true but Q is false. (Q becomes true if x and y are both non-negative.)

Question 7: Cubing equal numbers

P trueQ true

Proposition PIf x = y, then x³ = y³ (x and y are real numbers).

Step 1 · Converse QIf x³ = y³, then x = y.

Step 2 · Is P true?True

Replace y by x: y³ = x × x × x = x³.

Step 3 · Is Q true?True

Yes, unlike Q6. Suppose x³ = y³. Then x³ − y³ = 0, and

x³ − y³ = (x − y)(x² + xy + y²) = 0

So either x − y = 0, or x² + xy + y² = 0. Now x² + xy + y² = ½[x² + y² + (x + y)²], a sum of squares, which is zero only when x = 0, y = 0 and x + y = 0, that is x = y = 0. In every case x = y.

Both P and Q are true.

Divisibility and factors

Question 8: Divisible by 24, and by both 4 and 6

P trueQ false

In Questions 8 to 12, n is a positive integer.

Proposition PIf n is divisible by 24, then it is divisible by both 4 and 6.

Step 1 · Converse QIf n is divisible by both 4 and 6, then it is divisible by 24.

Step 2 · Is P true?True

Let n = 24k. Then n = 4 × (6k) and n = 6 × (4k), so n is divisible by 4 and by 6.

Step 3 · Is Q true?False

Take n = 12. Then 12 ÷ 4 = 3 and 12 ÷ 6 = 2, but 12 ÷ 24 is not a whole number. (36 and 60 also work as counterexamples.)

P is true but Q is false. Being divisible by both 4 and 6 only means n is a multiple of their LCM, which is 12, not 24.

Question 9: Divisible by 60, and by both 5 and 12

P trueQ true

Proposition PIf n is divisible by 60, then it is divisible by both 5 and 12.

Step 1 · Converse QIf n is divisible by both 5 and 12, then it is divisible by 60.

Step 2 · Is P true?True

Let n = 60k. Then n = 5 × (12k) and n = 12 × (5k).

Step 3 · Is Q true?True

Let n be divisible by 12, so n = 12m. Since 5 divides 12m, and 5 is a prime that does not divide 12, 5 must divide m. Write m = 5t. Then n = 12 × 5t = 60t, so 60 divides n. (Quick check: 120, 180, 240 are all multiples of 5 and 12, and all are multiples of 60.)

Both P and Q are true.

The difference between Q8 and Q9 comes from the common factors of the two divisors:

Pair	Common factor (other than 1)	LCM	Product	Is the converse true?
4 and 6	2	12	24	No, LCM ≠ product
5 and 12	none	60	60	Yes, LCM = product

Question 10: Square of a prime and exactly 3 factors

P trueQ true

Proposition PIf n is the square of a prime number, then it has exactly 3 factors.

Step 1 · Converse QIf n has exactly 3 factors, then it is the square of a prime number.

Step 2 · Is P true?True

Let n = p² with p prime. Its factors are 1, p and p², and no others because p has no factor except 1 and p. Example: 49 has factors 1, 7, 49.

Step 3 · Is Q true?True

An odd number of factors means n is a perfect square (proved in this chapter), so n = m² and n ≠ 1. The factors 1, m, m² are already 3. If m were composite, say m = d × e with 1 < d < m, then d would be a fourth factor of n. That gives more than 3 factors, which is not allowed. So m has no factor except 1 and m, which means m is prime.

Both P and Q are true.

Question 11: Product of two unequal primes and 4 divisors

P trueQ false

Proposition PIf n is a product of two unequal prime numbers, then it has exactly 4 divisors.

Step 1 · Converse QIf n has exactly 4 divisors, then it is a product of two unequal prime numbers.

Step 2 · Is P true?True

Let n = p × q with p and q different primes. By unique prime factorisation its divisors are 1, p, q and pq. Example: 15 has divisors 1, 3, 5, 15.

Step 3 · Is Q true?False

Take n = 8. Its divisors are 1, 2, 4, 8, exactly 4, but 8 = 2 × 2 × 2 is not a product of two unequal primes. (27 also works: 1, 3, 9, 27.)

P is true but Q is false. Numbers with exactly 4 divisors are of two kinds: p × q, or p³.

Question 12: No common factor and multiples of 3

P trueQ true

Proposition PIf n and n + 3 have no factors in common, then n is not a multiple of 3.

Step 1 · Converse QIf n is not a multiple of 3, then n and n + 3 have no factors in common.

Step 2 · Is P true?True

Suppose n is a multiple of 3, say n = 3k. Then n + 3 = 3(k + 1) is also a multiple of 3. So 3 is a common factor of n and n + 3, which contradicts the “if” part. Hence n is not a multiple of 3.

Step 3 · Is Q true?True

Let d be a common factor of n and n + 3. Then d divides their difference (n + 3) − n = 3, so d = 1 or d = 3. If d = 3, then 3 divides n, which is not allowed. So d = 1, and the only common factor is 1. Check: (4, 7), (5, 8), (7, 10) all have no common factor.

Both P and Q are true.

Counterexamples to claims about primes

Question 13: Claims that every number of a form is prime

All three claims false

There is no converse to frame here. Each claim says “always prime”, so one value that gives a composite number is enough to disprove it. This is the same idea as Euler disproving Fermat's claim at the start of the chapter.

Method: put n = 1, 2, 3, … one by one, and stop at the first composite value. Then write its factors.

(i) 4n² + 1

n	4n² + 1	Prime or composite?
1	5	prime
2	17	prime
3	37	prime
4	65	composite, 65 = 5 × 13

Counterexample: n = 4. The claim is false.

(ii) n² + n + 11

For n = 1 to 9 the values are 13, 17, 23, 31, 41, 53, 67, 83, 101, all prime. For n = 10: 10² + 10 + 11 = 121 = 11 × 11, which is composite.

Counterexample: n = 10. (n = 11 also fails: 121 + 11 + 11 = 143 = 11 × 13.)

(iii) 4ⁿ + 3

n	4ⁿ + 3	Prime or composite?
1	7	prime
2	19	prime
3	67	prime
4	259	composite, 259 = 7 × 37

Counterexample: n = 4. (If n = 0 is allowed, 4⁰ + 3 = 4 is also composite.)

Question 14: Statements about 2ⁿ − 1 and 2ⁿ + 1

Both statements false

(i) If n is a prime number, then 2ⁿ − 1 is a prime number.

For n = 2, 3, 5, 7 we get 3, 7, 31, 127, all prime. For n = 11 (which is prime): 2¹¹ − 1 = 2048 − 1 = 2047 = 23 × 89, which is composite.

Counterexample: n = 11.

(ii) If n is an even number, then 2ⁿ + 1 is a prime number.

For n = 2 and n = 4 we get 5 and 17, both prime. For n = 6 (which is even): 2⁶ + 1 = 64 + 1 = 65 = 5 × 13, which is composite.

Counterexample: n = 6.

Divisibility shortcuts

Question 15: Is checking 2 and 4 enough to test divisibility by 8?

P trueQ false

Statement: If a number is divisible by 8, then it is divisible by both 2 and 4.

(i) Justify the statementTrue

Let the number be n = 8k. Then n = 2 × (4k), so n is divisible by 2. Also n = 4 × (2k), so n is divisible by 4.

(ii) Is it enough to check 2 and 4?False

No. The converse is false: a number divisible by 2 and 4 need not be divisible by 8.

Counterexample: 12. It is divisible by 2 (12 ÷ 2 = 6) and by 4 (12 ÷ 4 = 3), but 12 ÷ 8 = 1.5, so it is not divisible by 8.

Why this happens: 2 and 4 share a common factor, and every multiple of 4 is already a multiple of 2. So “divisible by 2 and 4” just means “divisible by 4”, which is weaker than “divisible by 8”.

The correct shortcut for 8 is to check the number made by the last three digits:

Number	Last three digits	Divisible by 8?
5128	128 = 8 × 16	Yes
3124	124 ÷ 8 = 15.5	No, although 3124 is divisible by 2 and 4

Question 16: Divisibility by 3 and the sum of digits

P trueQ true

Task: Express the relationship between “a number is divisible by 3” and “sum of the digits is a multiple of 3” using if-then sentences.

Sentence 1 · PTrue

If a number is divisible by 3, then the sum of its digits is a multiple of 3.

Sentence 2 · the converse QTrue

If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.

Both sentences are true. Together: a number is divisible by 3 if and only if the sum of its digits is a multiple of 3.

Why it is true

Take a 3-digit number with digits a, b, c:

100a + 10b + c = (99a + 9b) + (a + b + c) = 3(33a + 3b) + (a + b + c)

The first part, 3(33a + 3b), is always a multiple of 3. So the whole number is a multiple of 3 exactly when a + b + c is. The same works for any number of digits, because 10, 100, 1000, … are each one more than a multiple of 9.

Examples: 4521 has digit sum 4 + 5 + 2 + 1 = 12, a multiple of 3, and 4521 = 3 × 1507. For 4523 the digit sum is 14, not a multiple of 3, and 4523 is not divisible by 3.

Quadrilaterals from two sticks

Question 17: Building quadrilaterals from two sticks

Necessary vs sufficient

Key idea: the two properties are a statement and its converse. “If Q then equal diagonals” says equal diagonals are needed for type Q. “If equal diagonals then Q” says equal diagonals are enough for type Q.

Q17 · equal sticks placed in two different ways

The figure shows the same pair of equal sticks placed in two ways. Both quadrilaterals have equal diagonals, but only the left one is a rectangle.

Part (i): If a quadrilateral is of type Q, then it has equal-length diagonals

(a) Should the two sticks be of equal length?

Yes. Every quadrilateral of type Q has equal diagonals. If the sticks were unequal, the diagonals would be unequal, so the quadrilateral could not be of type Q.

(b) Will it matter how the two sticks are put together?

Yes, it may matter. The property tells us only that equal diagonals are necessary. It does not say that equal diagonals are enough to make type Q. So where and at what angle the sticks cross can still decide whether we get type Q. Example: if Q were “rectangle”, the equal sticks must also cross at their midpoints. Crossing off-centre gives an isosceles trapezium, as in the figure.

Part (ii): If a quadrilateral has equal diagonals, then it is of type Q

(a) Should the two sticks be of equal length?

Yes, use equal sticks. They make the diagonals equal, and then the property guarantees the quadrilateral is of type Q. (This property does not tell us whether unequal sticks could also give type Q, so equal sticks are the safe choice.)

(b) Will it matter how the two sticks are put together?

No. Whatever way the equal sticks cross, the quadrilateral has equal diagonals, so by the property it is of type Q. Both quadrilaterals in the figure would be of type Q.

Property	Equal sticks needed?	Does the way they cross matter?
(i) Type Q ⇒ equal diagonals	Yes, necessary	Yes, it may
(ii) Equal diagonals ⇒ type Q	Yes, equal sticks guarantee type Q	No

Answers at a glance

Seven converses are false (Q2, Q3, Q5, Q6, Q8, Q11, Q15), seven are true (Q1, Q4, Q7, Q9, Q10, Q12, Q16), and Q13, Q14 and Q17 ask for a different task.

Question	Proposition P	Converse Q	Deciding reason
Q1	True	True	Corresponding angles axiom and its converse
Q2	True	False	6 cm by 4 cm rectangle
Q3	True	False	Triangle with A = 60°, B = 80°, C = 40°
Q4	True	True	Add or subtract the same number on both sides
Q5	True	False	a = 2, b = 8
Q6	True	False	x = 3, y = −3
Q7	True	True	x³ − y³ = (x − y)(x² + xy + y²)
Q8	True	False	n = 12
Q9	True	True	5 and 12 have no common factor
Q10	True	True	Only the square of a prime has exactly 3 factors
Q11	True	False	n = 8
Q12	True	True	A common factor of n and n + 3 must divide 3
Q13	Claims false	Not asked	Counterexamples: n = 4, n = 10, n = 4
Q14	Claims false	Not asked	Counterexamples: n = 11, n = 6
Q15	True	False	12 is divisible by 2 and 4, but not by 8
Q16	True	True	Number and its digit sum differ by a multiple of 3
Q17	Equal sticks needed; placement may matter	Equal sticks enough; placement does not matter	Necessary versus sufficient

Source of the questions: NCERT, Ganita Manjari, Grade 9, Part II, Chapter 9, Exercise 9.1. The solutions, explanations and diagrams on this page are our own working.

How to answer every question

Geometry

Algebra

Divisibility and factors

Counterexamples to claims about primes

(i) 4n2 + 1

(ii) n2 + n + 11

(iii) 4n + 3

(i) If n is a prime number, then 2n − 1 is a prime number.

(ii) If n is an even number, then 2n + 1 is a prime number.

Divisibility shortcuts

Quadrilaterals from two sticks

Part (i): If a quadrilateral is of type Q, then it has equal-length diagonals

Part (ii): If a quadrilateral has equal diagonals, then it is of type Q

Answers at a glance

(i) 4n² + 1

(ii) n² + n + 11

(iii) 4ⁿ + 3

(i) If n is a prime number, then 2ⁿ − 1 is a prime number.

(ii) If n is an even number, then 2ⁿ + 1 is a prime number.