Class 9 Maths Ganita Manjari Chapter 3 Solutions — 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). Checked against NCERT Ganita Manjari, Grade 9, Part I, first edition April 2026, pages 41–67. This guide follows Exercise Sets 3.1–3.5 and all 16 end-of-chapter questions, with repeating blocks written in brackets to keep their meaning clear.

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 zero (Śhūnya) and Brahmagupta’s arithmetic rules, to integers, rational numbers (pq\frac{p}{q} form, operations, density and decimals), and finally irrational numbers such as 2\sqrt{2} and π that complete the real number line. The explanations distinguish exact values from approximations and include number-line constructions.

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 pq\frac{p}{q} with p, q integers and q≠0q\ne 0 (includes all integers, terminating and repeating decimals).

Irrational number: a number that cannot be written as pq\frac{p}{q}; its decimal is non-terminating and non-repeating (e.g. 2\sqrt{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 distinct rational numbers there are infinitely many rationals; one of them is their average a+b2\frac{a+b}{2}.

Key Facts & Formulas

Equality of fractions: ab=cd\frac{a}{b}=\frac{c}{d} if and only if ad=bcad=bc.

Add/subtract: make denominators equal, then add/subtract numerators. Multiply: (ab)×(cd)=acbd\left(\frac{a}{b}\right)\times \left(\frac{c}{d}\right)=\frac{ac}{bd}. Divide: (ab)÷(cd)=(ab)×(dc)\left(\frac{a}{b}\right)\div \left(\frac{c}{d}\right)=\left(\frac{a}{b}\right)\times \left(\frac{d}{c}\right), with b≠0b\ne 0, d≠0d\ne 0 and c≠0c\ne 0. A divisor cannot be zero.

A number between a and b: their average a+b2\frac{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\sqrt{2}, 3\sqrt{3}, 5…\sqrt{5}\ldots (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 = (152)×12=90\left(\frac{15}{2}\right)\times 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=−23-5=-2 and 2−7=−52-7=-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\times 3 joints = 12 counted with the thumb on one hand. This gives a practical way to count in groups of 12, illustrating base-12 (duodecimal) counting. It is an illustration of the method, not proof of a unique historical origin.
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\left(-3\right)\times \left(-4\right)=+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∘C4-15=-11{}^{\circ}\mathrm{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\left(-850\right)+1200+\left(-450\right)=350-450 = −₹100 (a net debt of ₹100).

3. Calculate using Brahmagupta’s laws: (i) (−12)×5\left(-12\right)\times 5   (ii) (−8)×(−7)\left(-8\right)\times \left(-7\right)   (iii) 0 – (–14)   (iv) (−20)÷4\left(-20\right)\div 4

SOLUTION (i) −60; (ii) 56; (iii) 0−(−14)=140-\left(-14\right)=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)=1510-\left(-5\right)=15).

SOLUTION Subtracting a −5 means removing a debt of ₹5. If your net financial position is +₹10 (for example ₹15 cash minus ₹5 debt) and the ₹5 debt is cancelled, your net position becomes +₹15. Thus 10−(−5)=10+5=1510-\left(-5\right)=10+5=15.

Think and Reflect — common denominators and distributivity

Use a common multiple, preferably the LCM, of the non-zero denominators. For 12+13\frac{1}{2}+\frac{1}{3}, write 36+26=56\frac{3}{6}+\frac{2}{6}=\frac{5}{6}.

Sample verification: (12)(13+16)=(12)(12)=14;\begin{aligned}\left(\frac{1}{2}\right)\left(\frac{1}{3}+\frac{1}{6}\right)&=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\\&=\frac{1}{4}\text{;}\end{aligned} separately (12)(13)+(12)(16)=16+112=14.\begin{aligned}\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{6}\right)&=\frac{1}{6}+\frac{1}{12}\\&=\frac{1}{4}\text{.}\end{aligned} In general ab[cd+ef]=a(cf+de)bdf=acbd+aebf,\begin{aligned}\frac ab\left[\frac cd+\frac ef\right]&=\frac{a(cf+de)}{bdf}\\&=\frac{ac}{bd}+\frac{ae}{bf}\text{,}\end{aligned} with b, d, f non-zero.

Exercise Set 3.3

1. Prove that the following rational numbers are equal: (i) 23\frac{2}{3} and 46\frac{4}{6}   (ii) 54\frac{5}{4} and 108\frac{10}{8}   (iii) −35-\frac{3}{5} and −610-\frac{6}{10}   (iv) 93\frac{9}{3} and 3

SOLUTION Use ab=cd\frac{a}{b}=\frac{c}{d} ⇔ ad=bcad=bc. (i) 2×6=12=3×42\times 6=12=3\times 4 ✓. (ii) 5×8=40=4×105\times 8=40=4\times 10 ✓. (iii) (−3)×10=−30=5×(−6)\left(-3\right)\times 10=-30=5\times \left(-6\right) ✓. (iv) 93=3\frac{9}{3}=3 ✓.

2. Find the sum: (i) 25+310\frac{2}{5}+\frac{3}{10}   (ii) 712+58\frac{7}{12}+\frac{5}{8}   (iii) −47+314-\frac{4}{7}+\frac{3}{14}

SOLUTION (i) 410+310=710\frac{4}{10}+\frac{3}{10}=\frac{7}{10}. (ii) 1424+1524=2924\frac{14}{24}+\frac{15}{24}=\frac{29}{24}. (iii) −814+314=−514-\frac{8}{14}+\frac{3}{14}=-\frac{5}{14}.

3. Find the difference: (i) 56−14\frac{5}{6}-\frac{1}{4}   (ii) 118−34\frac{11}{8}-\frac{3}{4}   (iii) −79−(−23)-\frac{7}{9}-\left(-\frac{2}{3}\right)

SOLUTION (i) 1012−312=712\frac{10}{12}-\frac{3}{12}=\frac{7}{12}. (ii) 118−68=58\frac{11}{8}-\frac{6}{8}=\frac{5}{8}. (iii) −79+69=−19-\frac{7}{9}+\frac{6}{9}=-\frac{1}{9}.

4. Find the product: (i) 23×310\frac{2}{3}\times \frac{3}{10}   (ii) 711×58\frac{7}{11}\times \frac{5}{8}   (iii) −47×514-\frac{4}{7}\times \frac{5}{14}

SOLUTION (i) 630=15\frac{6}{30}=\frac{1}{5}. (ii) 3588\frac{35}{88}. (iii) −2098=−1049-\frac{20}{98}=-\frac{10}{49}.

5. Find the quotient: (i) 23÷310\frac{2}{3}\div \frac{3}{10}   (ii) 711÷58\frac{7}{11}\div \frac{5}{8}   (iii) −47÷514-\frac{4}{7}\div \frac{5}{14}

SOLUTION (i) 23×103=209\frac{2}{3}\times \frac{10}{3}=\frac{20}{9}. (ii) 711×85=5655\frac{7}{11}\times \frac{8}{5}=\frac{56}{55}. (iii) −47×145=−5635=−85-\frac{4}{7}\times \frac{14}{5}=-\frac{56}{35}=-\frac{8}{5}.

6. Show that: (12+34)×83=12×83+34×83.\left(\frac{1}{2}+\frac{3}{4}\right)\times \frac{8}{3}=\frac{1}{2}\times \frac{8}{3}+\frac{3}{4}\times \frac{8}{3}\text{.}

SOLUTION LHS=(54)×83=4012=103.\begin{aligned}\mathrm{LHS}&=\left(\frac{5}{4}\right)\times \frac{8}{3}\\&=\frac{40}{12}\\&=\frac{10}{3}\text{.}\end{aligned} RHS=86+2412=43+2=103.\begin{aligned}\mathrm{RHS}&=\frac{8}{6}+\frac{24}{12}\\&=\frac{4}{3}+2\\&=\frac{10}{3}\text{.}\end{aligned} Hence LHS=RHS\mathrm{LHS}=\mathrm{RHS} (distributive law). ✓

7. Simplify using the distributive property: 79(67−34)\frac{7}{9}\left(\frac{6}{7}-\frac{3}{4}\right).

SOLUTION 79×67−79×34=69−2136=23−712=812−712=112.\begin{aligned}\frac{7}{9}\times \frac{6}{7}-\frac{7}{9}\times \frac{3}{4}&=\frac{6}{9}-\frac{21}{36}\\&=\frac{2}{3}-\frac{7}{12}\\&=\frac{8}{12}-\frac{7}{12}\\&=\frac{1}{12}\text{.}\end{aligned}

8. Find the rational number x such that: 56(x+35)=56x+12\frac{5}{6}\left(x+\frac{3}{5}\right)=\frac{5}{6}x+\frac{1}{2}.

SOLUTION LHS=56x+(56)(35)=56x+12=RHS.\begin{aligned}\mathrm{LHS}&=\frac{5}{6}x+\left(\frac{5}{6}\right)\left(\frac{3}{5}\right)\\&=\frac{5}{6}x+\frac{1}{2}\\&=\mathrm{RHS}\text{.}\end{aligned} The equation is an identity — it is true for every rational number x.
Think and Reflect Why do we need q≠0q\ne 0 in the definition of a rational number pq\frac{p}{q}? Answer. Division by 0 is undefined — p0\frac{p}{0} is not a defined rational number: for p≠0p\ne 0, 0×r=p0\times r=p has no solution, while for p=0p=0 it does not select a unique quotient — so the denominator q must be non-zero.

Number-line activity (page 51)

85=1.6\frac{8}{5}=1.6 lies three fifths beyond 1; −74=−1.75-\frac{7}{4}=-1.75 lies three quarters left of −1.

Think and Reflect: 8/5 and −7/4-2-1.5-1-0.500.511.52−7/48/5
Think and Reflect: 85\frac{8}{5} and −74-\frac{7}{4}

Exercise Set 3.4

1. Represent the rational numbers 23\frac{2}{3}, −54-\frac{5}{4} and 1½ on a single number line.

SOLUTION Divide each unit interval into the needed equal parts: 23\frac{2}{3} lies two-thirds of the way from 0 to 1; −54=−114-\frac{5}{4}=-1\frac{1}{4} lies between −1 and −2; 1½ lies halfway between 1 and 2. Mark these three points on one number line.
Exercise 3.4 Q1: −5/4, 2/3 and 3/2 on one number line-2-1.5-1-0.500.511.52−5/42/33/2
Exercise 3.4 Q1: −54-\frac{5}{4}, 23\frac{2}{3} and 32\frac{3}{2} on one number line

2. Find three distinct rational numbers that lie strictly between −12-\frac{1}{2} and 14\frac{1}{4}.

SOLUTION For example −14-\frac{1}{4}, 0 and 18\frac{1}{8} all lie between −12-\frac{1}{2} and 14\frac{1}{4}. (Other correct answers are possible.)

3. Simplify the expression: (−14)+(512)\left(-\frac{1}{4}\right)+\left(\frac{5}{12}\right).

SOLUTION −312+512=212=16-\frac{3}{12}+\frac{5}{12}=\frac{2}{12}=\frac{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 1534÷214=(634)÷(94)=634×49=639=7\begin{aligned}15\frac{3}{4}\div 2\frac{1}{4}&=\left(\frac{63}{4}\right)\div \left(\frac{9}{4}\right)\\&=\frac{63}{4}\times \frac{4}{9}\\&=\frac{63}{9}\\&=7\end{aligned} 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 distinct rational numbers?

SOLUTION Yes — besides the average a+b2\frac{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.

For a<ba<b, a different weighted average is 2a+b3\frac{2a+b}{3}. Its difference from a is b−a3>0\frac{b-a}{3}>0, and b minus it is 2(b−a)3>0\frac{2\left(b-a\right)}{3}>0, so it lies strictly between. For the common-denominator method, multiply both fractions further if no integer numerator is yet available between them.

Think and Reflect Can 2\sqrt{2} be written as a rational number pq\frac{p}{q}? Answer. No. 2\sqrt{2} is irrational — assuming 2=pq\sqrt{2}=\frac{p}{q} in lowest terms leads to p and q both being even, a contradiction. So 2\sqrt{2} cannot be written as a fraction.

Think and Reflect — irrationality and decimals

For 3\sqrt{3}, assume 3=pq\sqrt{3}=\frac{p}{q} in lowest terms. Then p2=3q2p^{2}=3q^{2}, so 3 divides p; writing p=3kp=3k gives q2=3k2q^{2}=3k^{2}, so 3 also divides q, a contradiction. The same prime-divisibility proof works for 5\sqrt{5} and 7\sqrt{7}. For 10\sqrt{10}, p2=10q2p^{2}=10q^{2} implies both 2 and 5 divide p, hence p=10kp=10k and q2=10k2q^{2}=10k^{2}; both also divide q, contradicting lowest terms.

103=3.3‾\frac{10}{3}=3.\overline{3}; 1112=0.916‾\frac{11}{12}=0.91\overline{6}. In the second decimal, the non-repeating prefix is 91 and only 6 repeats.

If a reduced denominator is 2ᵃ5ᵇ, multiply by the missing factors to make 10max⁡(a,b)10^{\max \left(a,b\right)}. Conversely, a terminating decimal has a power of 10 as denominator before reduction, so no other prime factor can remain.

No real number squares to −1, since every real square is non-negative. The later extension introduces i with i2=−1i^{2}=-1; it is not a point on the real number line.

Constructing 2\sqrt{2}, 3\sqrt{3}, 5\sqrt{5} and n\sqrt{n} (pages 55–56)

Draw OA=1\mathrm{OA}=1 unit. Erect AB=1\mathrm{AB}=1 perpendicular to OA; then OB=2\mathrm{OB}=\sqrt{2}. With compass centre O and radius OB, mark the positive number line at 2\sqrt{2}. To obtain 3\sqrt{3}, erect a new unit perpendicular to OB at B; its endpoint is 3\sqrt{3} from O. For 5\sqrt{5}, use perpendicular legs 2 and 1, or continue the spiral. In general, a perpendicular unit step added to a radius k\sqrt{k} gives radius k+1\sqrt{k+1}. Copy the resulting radius to the number line with a compass.

Exercise Set 3.5

1. Without performing long division, determine which of 720\frac{7}{20}, 415\frac{4}{15} and 13250\frac{13}{250} have terminating decimals and which are repeating. Then check by long division.

SOLUTION 720\frac{7}{20}: 20=22×520=2^{2}\times 5 (only 2s and 5s) → terminating; 720=0.35\frac{7}{20}=0.35. 415\frac{4}{15}: 15=3×515=3\times 5 (has a 3) → repeating; 415=0.26666…=0.26‾\frac{4}{15}=0.26666\ldots =0.2\overline{6}. 13250\frac{13}{250}: 250=2×53250=2\times 5^{3} (only 2s and 5s) → terminating; 13250=0.052\frac{13}{250}=0.052.

Long-division checks: 720\frac{7}{20}: 70=3×20+1070=3\times 20+10, then 100=5×20+0100=5\times 20+0, giving 0.35. For 415\frac{4}{15}: 40=2×15+1040=2\times 15+10, then 100=6×15+10100=6\times 15+10; remainder 10 recurs, so the digit 6 repeats. For 13250\frac{13}{250}: 130=0×250+130130=0\times 250+130, 1300=5×250+501300=5\times 250+50, then 500=2×250500=2\times 250, giving 0.052.

2. Perform the long division for 113\frac{1}{13}. Identify the repeating block. Does it show cyclic properties for 213\frac{2}{13}, 313\frac{3}{13}, …? What do you notice?

SOLUTION

Long division of 1 by 13 gives successive (digit, remainder) pairs (0,10), (7,9), (6,12), (9,3), (2,4), (3,1). Remainder 1 returns, so 113=0.076923‾\frac{1}{13}=0.\overline{076923}. The leading zero is part of the six-digit block.

FractionRepeating decimal
113\frac{1}{13}0.(076923)
213\frac{2}{13}0.(153846)
313\frac{3}{13}0.(230769)
413\frac{4}{13}0.(307692)
513\frac{5}{13}0.(384615)
613\frac{6}{13}0.(461538)
713\frac{7}{13}0.(538461)
813\frac{8}{13}0.(615384)
913\frac{9}{13}0.(692307)
1013\frac{10}{13}0.(769230)
1113\frac{11}{13}0.(846153)
1213\frac{12}{13}0.(923076)

The blocks form two separate families of cyclic rotations: 076923 for numerators 1, 3, 4, 9, 10, 12, and 153846 for numerators 2, 5, 6, 7, 8, 11. In particular, 153846 is not a rotation of 076923. Thus 113\frac{1}{13} does not have the single full cyclic-block property of 17\frac{1}{7}; its period is 6 rather than 12.

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

SOLUTION (i) 81=9\sqrt{81}=9 → rational (91)\left(\frac{9}{1}\right). (ii) 12=23\sqrt{12}=2\sqrt{3} → irrational. (iii) 0.3‾=130.\overline{3}=\frac{1}{3} → rational. (iv) 0.12345‾=1234599999=411533333\begin{aligned}0.\overline{12345}&=\frac{12345}{99999}\\&=\frac{4115}{33333}\end{aligned} → rational. (v) 1.01001000100001… has no fixed repeating block (zeros keep increasing) → irrational. (vi) The terminating decimal is rational: 235601856122398747901201021=58900464030599686975325000000000000000000.\frac{23560185612239874790120}{10^{21}}=\frac{589004640305996869753}{25000000000000000000}\text{.} The final zero does not change its value.

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

SOLUTION Let x=0.9999…x=0.9999\ldots Then 10x=9.9999…10x=9.9999\ldots Subtracting: 10x−x=9⇒9x=9⇒x=1.\begin{aligned}10x-x&=9\\\Rightarrow 9x&=9\\\Rightarrow x&=1\text{.}\end{aligned} So 0.9‾=10.\overline{9}=1.

*5. The repeating block of 17\frac{1}{7} is a cyclic number. Try to find more numbers n whose reciprocals 1n\frac{1}{n} produce cyclic repeating blocks.

SOLUTION

Sample discoveries: n=17n=17 and n=19n=19. We have 117=0.0588235294117647‾\frac{1}{17}=0.\overline{0588235294117647} and 119=0.052631578947368421‾.\frac{1}{19}=0.\overline{052631578947368421}\text{.} Keep the leading zeros when comparing rotations.

For example 217=0.1176470588235294‾\frac{2}{17}=0.\overline{1176470588235294}, a rotation of the 16-digit block for 117\frac{1}{17}. Likewise 219=0.105263157894736842‾,\frac{2}{19}=0.\overline{105263157894736842}\text{,} a rotation of the 18-digit block. The period of 117\frac{1}{17} is 16 and of 119\frac{1}{19} is 18; their multiples through numerators 16 and 18 respectively rotate the same blocks. These examples are called full-reptend primes; that term is optional enrichment.

End-of-Chapter Exercises — Questions 1–16

1. Convert to a terminating or non-terminating repeating decimal by long division: (i) 350\frac{3}{50}   (ii) 29\frac{2}{9}

SOLUTION (i) 50=2×5250=2\times 5^{2} → terminating: 350=0.06\frac{3}{50}=0.06. (ii) 9=329=3^{2} → repeating: 29=0.222…=0.2‾\frac{2}{9}=0.222\ldots =0.\overline{2}.

Long division: for 350\frac{3}{50}, 30<5030<50 gives the first decimal digit 0; bring down another 0: 300÷50=6300\div 50=6 with remainder 0, hence 0.06. For 29\frac{2}{9}, 20=2×9+220=2\times 9+2; the same remainder 2 recurs at every step, hence 0.(2).

2. Prove that 5\sqrt{5} is an irrational number.

SOLUTION Assume 5=pq\sqrt{5}=\frac{p}{q} in lowest terms (q≠0,pq\ne 0,p and q co-prime). Then 5q2=p25q^{2}=p^{2}, so p2p^{2} is divisible by 5, hence p is divisible by 5. Let p=5kp=5k. Then 5q2=25k2⇒q2=5k2,\begin{aligned}5q^{2}&=25k^{2}\\\Rightarrow q^{2}&=5k^{2}\text{,}\end{aligned} so q is also divisible by 5. But then p and q share a factor 5 — contradicting “lowest terms”. Hence 5\sqrt{5} is irrational.

3. Convert the following decimals to the form pq\frac{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=12610=63512.6=\frac{126}{10}=\frac{63}{5}. (ii) 0.0120=12010000=32500.0120=\frac{120}{10000}=\frac{3}{250}. (iii) 3.052‾=3052−30990=3022990=1511495.\begin{aligned}3.0\overline{52}&=\frac{3052-30}{990}\\&=\frac{3022}{990}\\&=\frac{1511}{495}\text{.}\end{aligned} (iv) 1.235‾=1235−12990=1223990.\begin{aligned}1.2\overline{35}&=\frac{1235-12}{990}\\&=\frac{1223}{990}\text{.}\end{aligned} (v) 0.23‾=2399=23990.\overline{23}=\frac{23}{99}=\frac{23}{99}. (vi) 2.05‾=205−2090=18590=3718.\begin{aligned}2.0\overline{5}&=\frac{205-20}{90}\\&=\frac{185}{90}\\&=\frac{37}{18}\text{.}\end{aligned} (vii) 2.125‾=2125−212900=1913900.\begin{aligned}2.12\overline{5}&=\frac{2125-212}{900}\\&=\frac{1913}{900}\text{.}\end{aligned} (viii) 3.125‾=3125−312900=2813900.\begin{aligned}3.12\overline{5}&=\frac{3125-312}{900}\\&=\frac{2813}{900}\text{.}\end{aligned} (ix) 2.1625‾=21625−29999=216239999.\begin{aligned}2.\overline{1625}&=\frac{21625-2}{9999}\\&=\frac{21623}{9999}\text{.}\end{aligned}

4. Locate the following rational numbers on the number line: (i) 0.532   (ii) 1.1(5), meaning 1.15555…

SOLUTION

(i) 0.532=1332500.532=\frac{133}{250}. Magnify [0,1] to [0.5,0.6], then [0.53,0.54]. Divide the last interval into ten equal parts; the second tick after 0.53 is 0.532.

Q4(i): magnified interval from 0.530 to 0.540; tick spacing 0.0010.530.5310.5320.5330.5340.5350.5360.5370.5380.5390.540.532
Q4(i): magnified interval from 0.530 to 0.540; tick spacing 0.001

(ii) The textbook writes 1.15‾=1.15555…1.1\overline{5}=1.15555\ldots , not the terminating decimal 1.15. If x=1.15555…x=1.15555\ldots , then 100x−10x=115.555…−11.555…=104,\begin{aligned}100x-10x&=115.555\ldots -11.555\ldots \\&=104\text{,}\end{aligned} so x=5245x=\frac{52}{45}.

Since 5245=1+745\frac{52}{45}=1+\frac{7}{45}, divide [1,2] into 45 equal parts using the standard parallel-line construction and take the seventh point after 1. An equivalent local construction divides [1.1,1.2] into 9 equal parts and takes the fifth: 1.1+5(0.19)=52451.1+5\left(\frac{0.1}{9}\right)=\frac{52}{45}. It lies strictly between 1.155 and 1.156.

Q4(ii): magnified location of 1.1(5), slightly right of 1.1551.151.1511.1521.1531.1541.1551.1561.1571.1581.1591.161.1(5) = 52/45
Q4(ii): magnified location of 1.1(5), slightly right of 1.155

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. 3110\frac{31}{10}, 3210\frac{32}{10}, …, 3610\frac{36}{10}).

6. Find 5 rational numbers between 25\frac{2}{5} and 35\frac{3}{5}.

SOLUTION Write 25=2050\frac{2}{5}=\frac{20}{50} and 35=3050\frac{3}{5}=\frac{30}{50}. Five between them: 2150\frac{21}{50}, 2250\frac{22}{50}, 2350\frac{23}{50}, 2450\frac{24}{50}, 2550\frac{25}{50} (=12)\left(=\frac{1}{2}\right).

7. Find 5 rational numbers between 16\frac{1}{6} and 25\frac{2}{5}.

SOLUTION With denominator 30: 16=530\frac{1}{6}=\frac{5}{30} and 25=1230\frac{2}{5}=\frac{12}{30}. Five between them: 630\frac{6}{30}, 730\frac{7}{30}, 830\frac{8}{30}, 930\frac{9}{30}, 1030\frac{10}{30} (=15,730,415,310,13).\left(=\frac{1}{5},\frac{7}{30},\frac{4}{15},\frac{3}{10},\frac{1}{3}\right)\text{.}

8. If x3+x5=1615\frac{x}{3}+\frac{x}{5}=\frac{16}{15}, find the rational number x.

SOLUTION x(13+15)=x(815)=1615⇒x=(1615)×(158)=2.\begin{aligned}x\left(\frac{1}{3}+\frac{1}{5}\right)&=x\left(\frac{8}{15}\right)=\frac{16}{15}\\\Rightarrow x&=\left(\frac{16}{15}\right)\times \left(\frac{15}{8}\right)=2\text{.}\end{aligned}

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

SOLUTION a+1b=0⇒a=−1b⇒ab=−1.\begin{aligned}a+\frac{1}{b}&=0\\\Rightarrow a&=-\frac{1}{b}\\\Rightarrow ab&=-1\text{.}\end{aligned} So ab is negative (in fact ab=−1ab=-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 p104\frac{p}{10^{4}} with p not divisible by 10. Is the denominator (in lowest form) necessarily divisible by 242^{4} or 545^{4}? Give reasons.

SOLUTION

Multiplying a decimal whose last non-zero digit is in the fourth decimal place by 10410^{4} gives an integer p. Its units digit is non-zero, so 10 does not divide p. The number is p104=p24×54\frac{p}{10^{4}}=\frac{p}{2^{4}\times 5^{4}}.

Yes: the reduced denominator must be divisible by at least one of 242^{4} or 545^{4}. Since p is not divisible by 10, it cannot contain both a factor 2 and a factor 5. If p is odd, none of the four factors 2 can cancel. If p is not divisible by 5, none of the four factors 5 can cancel. Thus at least one full fourth power remains.

It need not contain both: 0.0625=1160.0625=\frac{1}{16} retains 242^{4} only; 0.0016=16250.0016=\frac{1}{625} retains 545^{4} only. The question says “or”, not “and”.

11. Without dividing, determine whether the decimal expansion of 18125\frac{18}{125} is terminating or not. If it terminates, state the number of decimal places.

SOLUTION 125=53125=5^{3} (only 5s) → terminating. Decimal places = max(power of 2, power of 5) = 3. (18125=1441000=0.144\frac{18}{125}=\frac{144}{1000}=0.144.)

12. A rational number in its lowest form has denominator 23×52^{3}\times 5. How many decimal places will its decimal expansion have?

SOLUTION Number of places = larger power among 2 and 5=max⁡(3,1)=35=\max \left(3,1\right)=3 decimal places (the fraction terminates).

*13. Let a=712a=\frac{7}{12} and b=56b=\frac{5}{6}. Express both as k1m\frac{k_{1}}{m} and k2m\frac{k_{2}}{m} with k2−k1>6k_{2}-k_{1}>6, and write five distinct rationals with integer numerators between a and b. Explain why k2−k1>n+1k_{2}-k_{1}>n+1 is needed for n such numbers.

SOLUTION

Use m=36m=36: a=2136a=\frac{21}{36} and b=3036b=\frac{30}{36}. The numerator gap is 9>69>6, as requested. One sample set of five distinct rationals is 2236\frac{22}{36}, 2336\frac{23}{36}, 2436\frac{24}{36}, 2536\frac{25}{36} and 2636\frac{26}{36}.

Textbook wording issue: between integer numerators k1k_{1} and k2k_{2} there are exactly k2−k1−1k_{2}-k_{1}-1 integers. To choose n of them, the necessary and sufficient condition is k2−k1≥n+1k_{2}-k_{1}\ge n+1. The printed strict condition k2−k1>n+1k_{2}-k_{1}>n+1 is sufficient, but it is not necessary. For example, a gap of 6 already gives five interior numerators.

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

SOLUTION (x+y+z)2=x2+y2+z2+2(xy+yz+zx).\left(x+y+z\right)^{2}=x^{2}+y^{2}+z^{2}+2\left(xy+yz+zx\right)\text{.} Substituting 0 and 0: 0=x2+y2+z20=x^{2}+y^{2}+z^{2}. A sum of squares is 0 only when each is 0, so x=y=z=0x=y=z=0.

*15. Show that the rational number a+b2\frac{a+b}{2} lies between the rational numbers a and b.

SOLUTION Take a<ba<b. Then a=a+a2<a+b2<b+b2=b.\begin{aligned}a&=\frac{a+a}{2}<\frac{a+b}{2}<\frac{b+b}{2}\\&=b\text{.}\end{aligned} So a<a+b2<ba<\frac{a+b}{2}<b — the average lies between a and b.

This strict inequality assumes a≠ba\ne b. If b<ab<a, reverse the labels; if a=ba=b, the average equals both endpoints and is not strictly between them.

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

SOLUTION

The ten triangles in Fig. 3.14 have successive hypotenuse lengths 2\sqrt{2}, 3\sqrt{3}, 2, 5\sqrt{5}, 6\sqrt{6}, 7\sqrt{7}, 222\sqrt{2}, 3, 10\sqrt{10} and 11\sqrt{11} units.

Starting with a segment of length 1, attach a perpendicular unit segment at each new endpoint. If the previous hypotenuse is k\sqrt{k}, the next squared length is (k)2+12=k+1\left(\sqrt{k}\right)^{2}+1^{2}=k+1. Hence the n-th triangle has hypotenuse n+1\sqrt{n+1}.

The exact construction is drawn below. Each outer edge is 1 unit and perpendicular to the radius ending at its starting point.

Ten right triangles forming a square root spiral√2√32√5√6√72√23√10√11OEvery short outer edge = 1 unit
Successive hypotenuses from 2\sqrt{2} to 11\sqrt{11}. Each right-angle marker shows where the new unit segment is erected.

Common Mistakes to Avoid

Watch out for these

  • Forgetting that a rational number needs q≠0q\ne 0.
  • Sign errors with negative numbers, e.g. (−3)×(−4)=+12\left(-3\right)\times \left(-4\right)=+12 and 10−(−5)=1510-\left(-5\right)=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\sqrt{\text{perfect square}} irrational — 81=9\sqrt{81}=9 is rational; 12\sqrt{12} is irrational.

Practice MCQs & Assertion–Reason

Additional ClearStudy practice, separate from the NCERT exercises.

1. A number that can be written as pq\frac{p}{q} (q≠0)\left(q\ne 0\right) is called a:

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

2. Which of these is irrational?

(a) 81\sqrt{81}    (b) 0.25    (c) 12\sqrt{12}    (d) 37\frac{3}{7}

3. The decimal 0.(3) (0.333…) equals:

(a) 13\frac{1}{3}    (b) 310\frac{3}{10}    (c) 130\frac{1}{30}    (d) 33100\frac{33}{100}

4. 720\frac{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)\left(-8\right)\times \left(-7\right) is:

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

6. A rational number between 14\frac{1}{4} and 12\frac{1}{2} is:

(a) 38\frac{3}{8}    (b) 58\frac{5}{8}    (c) 18\frac{1}{8}    (d) 34\frac{3}{4}

7. The set of integers is denoted by:

(a) N    (b) Q    (c) Z    (d) R

8. ∣−73∣|-\frac{7}{3}| equals:

(a) −73-\frac{7}{3}    (b) 73\frac{7}{3}    (c) 37\frac{3}{7}    (d) 0

9. 2\sqrt{2} is:

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

10. 29\frac{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\sqrt{5} is an irrational number.

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

A-R 2. Assertion: 720\frac{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‾=10.\overline{9}=1.

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

A-R 4. Assertion: 81\sqrt{81} is an irrational number.

Reason: 81 is a perfect square.

A-R 5. Assertion: Between any two distinct 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 pq\frac{p}{q} with q≠0q\ne 0; it has a terminating or repeating decimal.
  • Irrational numbers (2,π,…)\left(\sqrt{2},\pi ,\ldots \right) 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 distinct rationals lie infinitely many rationals (e.g. their average).
  • 2\sqrt{2}, 3\sqrt{3}, 5…\sqrt{5}\ldots (non-perfect-squares) are proved irrational by contradiction.

Presenting clear mathematical working

Always reduce fractions to lowest terms, take the LCM carefully when adding/subtracting, and for “pq\frac{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 pq\frac{p}{q} where p and q are integers and q≠0q\ne 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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