Class 9 Maths Ganita Manjari Chapter 3 Solutions (NCERT 2026–27) – The World of Numbers

These Class 9 Maths Ganita Manjari Chapter 3 solutions cover The World of Numbers from the new NCF-2023 textbook (2026–27). Every exercise is solved step by step so you can understand each concept and revise the whole chapter quickly.

Class: 9 Subject: Mathematics Book: Ganita Manjari (Part 1) Chapter: 3 Exercises: 3.1–3.5, End-of-Chapter Session: 2026–27
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Chapter 3 Overview

Chapter 3 of Ganita Manjari, The World of Numbers, tells the story of how the number system grew — from natural numbers and the invention of zero (Śhūnya) by Brahmagupta, to integers, rational numbers (p/q form, operations, density and decimals), and finally irrational numbers such as √2 and π that complete the real number line. The Class 9 Maths Ganita Manjari Chapter 3 solutions below work through every exercise step by step.

Key Concepts & Definitions

Natural numbers (ℕ): 1, 2, 3, … ; Whole numbers: 0, 1, 2, … ; Integers (Z): …, −2, −1, 0, 1, 2, …

Rational number (ℚ): any number of the form p/q with p, q integers and q ≠ 0 (includes all integers, terminating and repeating decimals).

Irrational number: a number that cannot be written as p/q; its decimal is non-terminating and non-repeating (e.g. √2, π).

Real numbers (R): all rational and irrational numbers together — the complete number line.

Absolute value |x|: the distance of x from 0; always ≥ 0. Distance between a and b is |a − b|.

Density: between any two rational numbers there are infinitely many rationals; one of them is their average (a + b)/2.

Key Facts & Formulas

Equality of fractions: a/b = c/d if and only if ad = bc.

Add/subtract: make denominators equal, then add/subtract numerators. Multiply: (a/b)×(c/d) = ac/bd. Divide: (a/b)÷(c/d) = (a/b)×(d/c).

A number between a and b: their average (a + b)/2.

Terminating decimal test: a fraction in lowest terms terminates ⇔ its denominator has only 2 and 5 as prime factors. Number of decimal places = the larger of the powers of 2 and 5.

Irrationality: √2, √3, √5 … (of non-perfect-squares) are irrational; proved by contradiction.

Exercise Set 3.1

1. A merchant in Lothal exchanges bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices, how many copper ingots will he leave with?

SOLUTION 15 ingots per 2 bags → ingots = (15/2) × 12 = 90 copper ingots.

2. Look at the sequence on the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.

SOLUTION They are all prime numbers. The next three primes are 23, 29 and 31.

3. Natural numbers are closed under addition. Are they closed under subtraction? Give a couple of examples to justify your answer.

SOLUTION No. For example, 3 − 5 = −2 and 2 − 7 = −5 are not natural numbers. So natural numbers are not closed under subtraction.

*4. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to ancient base-12 counting systems?

SOLUTION Four fingers × 3 joints = 12 counted with the thumb on one hand. This is the basis of base-12 (duodecimal) counting systems.
Think and Reflect Why does a negative times a negative equal a positive? Answer. If a negative number is a debt, multiplying by a negative means removing that debt. Removing four debts of ₹3 each leaves you ₹12 richer, so (−3) × (−4) = +12.

Exercise Set 3.2

1. The temperature in Ladakh is 4 °C at noon. By midnight it drops by 15 °C. What is the midnight temperature?

SOLUTION4 − 15 = −11 °C.

2. A trader takes a loan (debt) of ₹850. The next day he makes a profit of ₹1,200. The following week he incurs a loss of ₹450. Write this as an equation using integers and find his final financial standing.

SOLUTION (−850) + 1200 + (−450) = 350 − 450 = −₹100 (a net debt of ₹100).

3. Calculate using Brahmagupta’s laws: (i) (–12) × 5   (ii) (–8) × (–7)   (iii) 0 – (–14)   (iv) (–20) ÷ 4

SOLUTION (i) −60; (ii) 56; (iii) 0 − (−14) = 14; (iv) −5.

4. Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., 10 – (–5) = 15).

SOLUTION Subtracting a −5 means removing a debt of ₹5. If you have ₹10 and a ₹5 debt is cancelled, you are ₹5 richer, so 10 − (−5) = 10 + 5 = 15.

Exercise Set 3.3

1. Prove that the following rational numbers are equal: (i) 2/3 and 4/6   (ii) 5/4 and 10/8   (iii) –3/5 and –6/10   (iv) 9/3 and 3

SOLUTION Use a/b = c/d ⇔ ad = bc. (i) 2×6 = 12 = 3×4 ✓. (ii) 5×8 = 40 = 4×10 ✓. (iii) (−3)×10 = −30 = 5×(−6) ✓. (iv) 9/3 = 3 ✓.

2. Find the sum: (i) 2/5 + 3/10   (ii) 7/12 + 5/8   (iii) –4/7 + 3/14

SOLUTION (i) 4/10 + 3/10 = 7/10. (ii) 14/24 + 15/24 = 29/24. (iii) −8/14 + 3/14 = −5/14.

3. Find the difference: (i) 5/6 – 1/4   (ii) 11/8 – 3/4   (iii) –7/9 – (–2/3)

SOLUTION (i) 10/12 − 3/12 = 7/12. (ii) 11/8 − 6/8 = 5/8. (iii) −7/9 + 6/9 = −1/9.

4. Find the product: (i) 2/3 × 3/10   (ii) 7/11 × 5/8   (iii) –4/7 × 5/14

SOLUTION (i) 6/30 = 1/5. (ii) 35/88. (iii) −20/98 = −10/49.

5. Find the quotient: (i) 2/3 ÷ 3/10   (ii) 7/11 ÷ 5/8   (iii) –4/7 ÷ 5/14

SOLUTION (i) 2/3 × 10/3 = 20/9. (ii) 7/11 × 8/5 = 56/55. (iii) −4/7 × 14/5 = −56/35 = −8/5.

6. Show that: (1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3.

SOLUTION LHS = (5/4) × 8/3 = 40/12 = 10/3. RHS = 8/6 + 24/12 = 4/3 + 2 = 10/3. Hence LHS = RHS (distributive law). ✓

7. Simplify using the distributive property: 7/9 (6/7 – 3/4).

SOLUTION 7/9 × 6/7 − 7/9 × 3/4 = 6/9 − 21/36 = 2/3 − 7/12 = 8/12 − 7/12 = 1/12.

8. Find the rational number x such that: 5/6 (x + 3/5) = 5/6 x + 1/2.

SOLUTION LHS = 5/6 x + (5/6)(3/5) = 5/6 x + 1/2 = RHS. The equation is an identity — it is true for every rational number x.
Think and Reflect Why do we need q ≠ 0 in the definition of a rational number p/q? Answer. Division by 0 is undefined — p/0 has no meaning — so the denominator q must be non-zero.

Exercise Set 3.4

1. Represent the rational numbers 2/3, –5/4 and 1½ on a single number line.

SOLUTION Divide each unit interval into the needed equal parts: 2/3 lies two-thirds of the way from 0 to 1; −5/4 = −1¼ lies between −1 and −2; 1½ lies halfway between 1 and 2. Mark these three points on one number line.

2. Find three distinct rational numbers that lie strictly between –1/2 and 1/4.

SOLUTION For example −1/4, 0 and 1/8 all lie between −1/2 and 1/4. (Other correct answers are possible.)

3. Simplify the expression: (–1/4) + (5/12).

SOLUTION −3/12 + 5/12 = 2/12 = 1/6.

4. A tailor has 15¾ metres of fine silk. If making one kurta requires 2¼ metres of silk, exactly how many kurtas can he make?

SOLUTION 15¾ ÷ 2¼ = (63/4) ÷ (9/4) = 63/4 × 4/9 = 63/9 = 7 kurtas.

5. Find three rational numbers between 3.1415 and 3.1416.

SOLUTION For example 3.14151, 3.14155 and 3.14159.

*6. Can you think of other way(s) to find a rational number between any two rational numbers?

SOLUTION Yes — besides the average (a + b)/2, you can write both numbers with a common denominator and pick a fraction with a numerator in between, or convert them to decimals and choose a decimal that lies between them.
Think and Reflect Can √2 be written as a rational number p/q? Answer. No. √2 is irrational — assuming √2 = p/q in lowest terms leads to p and q both being even, a contradiction. So √2 cannot be written as a fraction.

Exercise Set 3.5

1. Without performing long division, determine which of 7/20, 4/15 and 13/250 have terminating decimals and which are repeating. Then check by long division.

SOLUTION 7/20: 20 = 22×5 (only 2s and 5s) → terminating; 7/20 = 0.35. 4/15: 15 = 3×5 (has a 3) → repeating; 4/15 = 0.26666… = 0.2(6). 13/250: 250 = 2×53 (only 2s and 5s) → terminating; 13/250 = 0.052.

2. Perform the long division for 1/13. Identify the repeating block. Does it show cyclic properties for 2/13, 3/13, …? What do you notice?

SOLUTION 1/13 = 0.(076923) — a 6-digit repeating block. 2/13 = 0.(153846), 3/13 = 0.(230769), … All have the same 6-digit period, and the blocks are cyclic rearrangements of the same digits — 1/13 is a cyclic number.

3. Classify the following as rational or irrational (and find the fraction if rational): (i) √81   (ii) √12   (iii) 0.33333…   (iv) 0.123451234512345…   (v) 1.01001000100001…   (vi) 23.560185612239874790120

SOLUTION (i) √81 = 9 → rational (9/1). (ii) √12 = 2√3 → irrational. (iii) 0.(3) = 1/3 → rational. (iv) 0.(12345) = 12345/99999 = 4115/33333 → rational. (v) 1.01001000100001… has no fixed repeating block (zeros keep increasing) → irrational. (vi) 23.560185612239874790120 is a terminating decimal → rational (the digits over the matching power of 10).

4. The number 0.(9) (which means 0.99999…) is a rational number. Using algebra (let x = 0.9̅, multiply by 10, subtract), explain why 0.(9) = 1.

SOLUTION Let x = 0.9999… Then 10x = 9.9999… Subtracting: 10x − x = 9 ⇒ 9x = 9 ⇒ x = 1. So 0.(9) = 1.

*5. The repeating block of 1/7 is a cyclic number. Try to find more numbers n whose reciprocals 1/n produce cyclic repeating blocks.

SOLUTION Such numbers are the full-reptend primes — e.g. 1/7, 1/17, 1/19, 1/23, 1/29 — whose decimal periods are cyclic rearrangements of one block.

Class 9 Maths Ganita Manjari Chapter 3 Solutions — End-of-Chapter Exercises

1. Convert to a terminating or non-terminating repeating decimal by long division: (i) 3/50   (ii) 2/9

SOLUTION (i) 50 = 2×52 → terminating: 3/50 = 0.06. (ii) 9 = 32 → repeating: 2/9 = 0.222… = 0.(2).

2. Prove that √5 is an irrational number.

SOLUTION Assume √5 = p/q in lowest terms (q ≠ 0, p and q co-prime). Then 5q2 = p2, so p2 is divisible by 5, hence p is divisible by 5. Let p = 5k. Then 5q2 = 25k2 ⇒ q2 = 5k2, so q is also divisible by 5. But then p and q share a factor 5 — contradicting “lowest terms”. Hence √5 is irrational.

3. Convert the following decimals to the form p/q: (i) 12.6   (ii) 0.0120   (iii) 3.0(52)   (iv) 1.2(35)   (v) 0.(23)   (vi) 2.0(5)   (vii) 2.12(5)   (viii) 3.12(5)   (ix) 2.(1625) (The bracketed digits repeat.)

SOLUTION (i) 12.6 = 126/10 = 63/5. (ii) 0.0120 = 120/10000 = 3/250. (iii) 3.0(52) = (3052 − 30)/990 = 3022/990 = 1511/495. (iv) 1.2(35) = (1235 − 12)/990 = 1223/990. (v) 0.(23) = 23/99 = 23/99. (vi) 2.0(5) = (205 − 20)/90 = 185/90 = 37/18. (vii) 2.12(5) = (2125 − 212)/900 = 1913/900. (viii) 3.12(5) = (3125 − 312)/900 = 2813/900. (ix) 2.(1625) = (21625 − 2)/9999 = 21623/9999.

4. Locate the following rational numbers on the number line: (i) 0.532   (ii) 1.15

SOLUTION (i) 0.532 lies between 0 and 1 (just past the 0.5 mark): use successive magnification — 0.5 to 0.6, then 0.53 to 0.54, then mark 0.532. (ii) 1.15 lies between 1.1 and 1.2: divide 1 to 2 into tenths, then 1.1 to 1.2 into tenths, and mark 1.15 (midway).

5. Find 6 rational numbers between 3 and 4.

SOLUTION For example 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 (i.e. 31/10, 32/10, …, 36/10).

6. Find 5 rational numbers between 2/5 and 3/5.

SOLUTION Write 2/5 = 20/50 and 3/5 = 30/50. Five between them: 21/50, 22/50, 23/50, 24/50, 25/50 (= 1/2).

7. Find 5 rational numbers between 1/6 and 2/5.

SOLUTION With denominator 30: 1/6 = 5/30 and 2/5 = 12/30. Five between them: 6/30, 7/30, 8/30, 9/30, 10/30 (= 1/5, 7/30, 4/15, 3/10, 1/3).

8. If x/3 + x/5 = 16/15, find the rational number x.

SOLUTION x(1/3 + 1/5) = x(8/15) = 16/15 ⇒ x = (16/15) × (15/8) = 2.

9. Let a and b be two non-zero rational numbers such that a + 1/b = 0. Without assigning values, determine whether ab is positive or negative. Justify.

SOLUTION a + 1/b = 0 ⇒ a = −1/b ⇒ ab = −1. So ab is negative (in fact ab = −1).

10. A rational number has a terminating decimal whose last non-zero digit is in the 4th decimal place. Show it can be written as p/104 with p not divisible by 10. Is the denominator (in lowest form) necessarily divisible by 24 or 54? Give reasons.

SOLUTION A 4-place decimal d = (d × 104)/104 = p/104; since the last non-zero digit is the 4th place, p is not divisible by 10. Not necessarily. 104 = 24×54, but after reducing, the denominator only needs the prime factors that survive. e.g. 0.0625 = 1/16 has denominator 24 (no 5), while 0.0002 = 1/5000 has denominator 54×23. So the lowest-form denominator need not contain both 24 and 54.

11. Without dividing, determine whether the decimal expansion of 18/125 is terminating or not. If it terminates, state the number of decimal places.

SOLUTION 125 = 53 (only 5s) → terminating. Decimal places = max(power of 2, power of 5) = 3. (18/125 = 144/1000 = 0.144.)

12. A rational number in its lowest form has denominator 23 × 5. How many decimal places will its decimal expansion have?

SOLUTION Number of places = larger power among 2 and 5 = max(3, 1) = 3 decimal places (the fraction terminates).

*13. Let a = 7/12 and b = 5/6. Express both as k1/m and k2/m with k2 – k1 > 6, and write five distinct rationals with integer numerators between a and b. Explain why k2 – k1 > n + 1 is needed for n such numbers.

SOLUTION a = 7/12, b = 5/6 = 10/12; here k2 − k1 = 3, too small. Scale by 3: m = 36, a = 21/36, b = 30/36 (k2 − k1 = 9 > 6). Five rationals between: 22/36, 23/36, 24/36, 25/36, 26/36. Between k1 and k2 there are (k2 − k1 − 1) integers; to fit n of them strictly between, we need k2 − k1 − 1 ≥ n, i.e. k2 − k1 > n + 1 guarantees enough integer numerators.

*14. Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that x, y, z must all be zero.

SOLUTION (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx). Substituting 0 and 0: 0 = x2 + y2 + z2. A sum of squares is 0 only when each is 0, so x = y = z = 0.

*15. Show that the rational number (a + b)/2 lies between the rational numbers a and b.

SOLUTION Take a < b. Then a = (a + a)/2 < (a + b)/2 < (b + b)/2 = b. So a < (a + b)/2 < b — the average lies between a and b.

16. Find the lengths of the hypotenuses of all the right triangles in the square root spiral (Fig. 3.14).

SOLUTION Each triangle has one leg from the previous hypotenuse and the other leg 1. Starting with legs 1 and 1: hypotenuses are √2, √3, √4 (= 2), √5, √6, √7, … — in general the nth hypotenuse is √(n + 1).

Common Mistakes to Avoid

Watch out for these

  • Forgetting that a rational number needs q ≠ 0.
  • Sign errors with negative numbers, e.g. (−3)×(−4) = +12 and 10 − (−5) = 15.
  • Not taking the LCM before adding/subtracting fractions.
  • To divide fractions, multiply by the reciprocal — do not “cross-divide” carelessly.
  • Judging terminating vs repeating from the lowest-form denominator (only 2s and 5s ⇒ terminating).
  • Calling √(perfect square) irrational — √81 = 9 is rational; √12 is irrational.

Practice MCQs & Assertion–Reason

1. A number that can be written as p/q (q ≠ 0) is called a:

(a) natural number    (b) rational number    (c) irrational number    (d) whole number

2. Which of these is irrational?

(a) √81    (b) 0.25    (c) √12    (d) 3/7

3. The decimal 0.(3) (0.333…) equals:

(a) 1/3    (b) 3/10    (c) 1/30    (d) 33/100

4. 7/20 has a decimal expansion that is:

(a) non-terminating non-repeating    (b) terminating    (c) non-terminating repeating    (d) not a decimal

5. The value of (−8) × (−7) is:

(a) −56    (b) 56    (c) −15    (d) 15

6. A rational number between 1/4 and 1/2 is:

(a) 3/8    (b) 5/8    (c) 1/8    (d) 3/4

7. The set of integers is denoted by:

(a) N    (b) Q    (c) Z    (d) R

8. |−7/3| equals:

(a) −7/3    (b) 7/3    (c) 3/7    (d) 0

9. √2 is:

(a) a natural number    (b) rational    (c) irrational    (d) an integer

10. 2/9 as a decimal is:

(a) 0.2    (b) 0.(2)    (c) 0.29    (d) 0.92

Answer key: 1-(b), 2-(c), 3-(a), 4-(b), 5-(b), 6-(a), 7-(c), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: √5 is an irrational number.

Reason: The square root of a number that is not a perfect square is irrational.

A-R 2. Assertion: 7/20 has a terminating decimal expansion.

Reason: A fraction terminates when its lowest-form denominator has only 2 and 5 as prime factors.

A-R 3. Assertion: 0.(9) = 1.

Reason: Some terminating decimals also have a repeating-9 form.

A-R 4. Assertion: √81 is an irrational number.

Reason: 81 is a perfect square.

A-R 5. Assertion: Between any two rational numbers there are infinitely many rational numbers.

Reason: The average of two rationals is a rational number lying between them.

Answer key: 1-(A), 2-(A), 3-(B), 4-(D), 5-(A).

Quick Revision Summary

  • Numbers grew in stages: natural → whole (with zero) → integers → rational → real (rational + irrational).
  • A rational number is p/q with q ≠ 0; it has a terminating or repeating decimal.
  • Irrational numbers (√2, π, …) have non-terminating, non-repeating decimals.
  • Terminating ⇔ lowest-form denominator has only 2s and 5s; number of places = larger of the two powers.
  • Between any two rationals lie infinitely many rationals (e.g. their average).
  • √2, √3, √5 … (non-perfect-squares) are proved irrational by contradiction.

How to score full marks in this chapter

Always reduce fractions to lowest terms, take the LCM carefully when adding/subtracting, and for “p/q from a decimal” use the standard rule (subtract the non-repeating part, divide by the right string of 9s and 0s). For irrationality, write the proof-by-contradiction steps clearly.

Frequently Asked Questions

What is Class 9 Maths Ganita Manjari Chapter 3 about?

It covers the growth of the number system — natural numbers, zero, integers, rational numbers, irrational numbers and the real number line — with operations on rationals and decimal expansions.

What is a rational number?

A number that can be written as p/q where p and q are integers and q ≠ 0.

How do you tell if a fraction has a terminating decimal?

In lowest terms, if the denominator has only 2s and 5s as prime factors the decimal terminates; otherwise it is non-terminating and repeating.

Are these Class 9 Maths Ganita Manjari Chapter 3 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Manjari textbook for 2026–27.

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