---
title: "NCERT Solutions for Class 9 Maths Chapter 3 The World of Numbers"
url: https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions
dateModified: 2026-10-07T15:46:03+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 3 The World of Numbers

This chapter's questions cover foundational number systems, including natural numbers, integers, rational numbers, irrational numbers, and real numbers along with their arithmetic operations and properties.

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## Exercise Set 3.1

### Question 1

*3 marks · Short answer*

A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?

**Solution**

1. Given that for every 2 bags of spices, the merchant receives 15 copper ingots.
2. The number of pairs of 2 bags in 12 bags of spices is $\frac{12}{2} = 6$.
3. Therefore, the total number of copper ingots he will receive is $6 \times 15 = 90$ ingots.

**Answer:** 90 copper ingots

> Common mistake: Multiplying 15 by 12 directly instead of finding the number of batches of 2 bags.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. The given numbers are $11, 13, 17, 19$.
2. These numbers are prime numbers, specifically the prime numbers between 10 and 20.
3. The next three prime numbers following 19 are 23, 29, and 31.

**Answer:** They are prime numbers between 10 and 20. The next three numbers are 23, 29, and 31.

> Common mistake: Including composite odd numbers like 21, 25, or 27 in the sequence.

### Question 3

*3 marks · Short answer*

We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.

**Solution**

1. Natural numbers are not closed under subtraction.
2. For example, taking two natural numbers 3 and 5, their difference $3 - 5 = -2$, which is an integer but not a natural number.
3. Similarly, $5 - 5 = 0$, and 0 is also not a natural number.

**Answer:** No, natural numbers are not closed under subtraction, as shown by $3 - 5 = -2$.

> Common mistake: Confusing natural numbers with integers.

### Question 4

*3 marks · Short answer*

Ancient Indians used the joints of their fingers to count, a practice still seen today. 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 the ancient base-12 counting systems?

**Solution**

1. Each of the four fingers (excluding the thumb) has 3 phalanges or joints.
2. Using the thumb of the same hand to touch and count each joint gives $4 \times 3 = 12$ distinct counts on one hand.
3. This correspondence of counting 12 units on a single hand naturally provided the historical basis for base-12 (duodecimal) counting systems.

**Answer:** You can count 12 on one hand, which directly relates to the historical base-12 counting systems.

> Common mistake: Including the thumb itself as a finger to be counted instead of using it as the pointer.

## Exercise Set 3.2

### Question 1

*3 marks · Short answer*

The temperature in the high-altitude desert of Ladakh is recorded as $4\text{ }^\circ\text{C}$ at noon. By midnight, it drops by $15\text{ }^\circ\text{C}$. What is the midnight temperature?

**Solution**

1. Given: Initial temperature at noon = $4^\circ\text{C}$, drop in temperature by midnight = $15^\circ\text{C}$.
2. The idea used is subtraction of integers where a drop represents a negative change.
3. Working: $\text{Midnight temperature} = 4 - 15 = -11$.
4. Result: The midnight temperature is $-11^\circ\text{C}$.

**Answer:** $-11^\circ\text{C}$

> Common mistake: Subtracting $4$ from $15$ to get $11^\circ\text{C}$ instead of accounting for the direction below zero.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. Loan (debt) of $₹850$ is represented as $-850$.
2. Profit (fortune) of $₹1,200$ is represented as $+1200$, and loss of $₹450$ as $-450$.
3. Equation: $\text{Financial Standing} = -850 + 1200 - 450 = -100$, meaning a net debt of $₹100$.

**Answer:** $-850 + 1200 - 450 = -100$ (Net debt of $₹100$)

> Common mistake: Treating the loan as a positive number.

### Question 3

*3 marks · Short answer*

Calculate the following using Brahmagupta's laws:
(i) $(-12) \times 5$
(ii) $(-8) \times (-7)$
(iii) $0 - (-14)$
(iv) $(-20) \div 4$

**Part (i)**

1. The product of a debt and a fortune is a debt: $(-12) \times 5 = -60$.

Answer (i): $-60$

**Part (ii)**

1. The product of two debts is a fortune: $(-8) \times (-7) = 56$.

Answer (ii): $56$

**Part (iii)**

1. Subtracting a negative number is equivalent to adding its positive counterpart: $0 - (-14) = 0 + 14 = 14$.

Answer (iii): $14$

**Part (iv)**

1. Dividing a negative debt by a positive number results in a debt: $(-20) \div 4 = -5$.

Answer (iv): $-5$

**Answer:** (i) $-60$, (ii) $56$, (iii) $14$, (iv) $-5$

> Common mistake: Mixing up signs in multiplication and subtraction.

### Question 4

*3 marks · Short answer*

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**

1. Let having a debt of $₹5$ be represented as $-5$.
2. If you have $₹10$ and someone removes (subtracts) your debt of $₹5$ ($\text{i.e., } -(-5)$), your financial standing increases because you no longer owe that money.
3. Thus, removing a debt is equivalent to gaining money, so $10 - (-5) = 10 + 5 = 15$.

**Answer:** Removing a debt of $₹5$ from your account increases your wealth by $₹5$, making $10 - (-5) = 15$.

> Common mistake: Thinking subtraction of a negative always results in a negative number.

## Exercise Set 3.3

### Question 1

*3 marks · Proof*

Prove that the following rational numbers are equal:
(i) $\frac{2}{3}$ and $\frac{4}{6}$
(ii) $\frac{5}{4}$ and $\frac{10}{8}$
(iii) $-\frac{3}{5}$ and $\frac{-6}{10}$
(iv) $\frac{9}{3}$ and $3$

**Part (i)**

1. Given the rational numbers $\frac{2}{3}$ and $\frac{4}{6}$.
2. Cross-multiply to check equality: $2 \times 6 = 12$ and $3 \times 4 = 12$.
3. Since $ad = bc$ ($12 = 12$), the rational numbers are equal. Hence proved.

Answer (i): $\frac{2}{3} = \frac{4}{6}$

**Part (ii)**

1. Given the rational numbers $\frac{5}{4}$ and $\frac{10}{8}$.
2. Cross-multiply to check equality: $5 \times 8 = 40$ and $4 \times 10 = 40$.
3. Since $ad = bc$ ($40 = 40$), the rational numbers are equal. Hence proved.

Answer (ii): $\frac{5}{4} = \frac{10}{8}$

**Part (iii)**

1. Given the rational numbers $-\frac{3}{5}$ and $\frac{-6}{10}$.
2. Cross-multiply to check equality: $(-3) \times 10 = -30$ and $5 \times (-6) = -30$.
3. Since $ad = bc$ ($-30 = -30$), the rational numbers are equal. Hence proved.

Answer (iii): $-\frac{3}{5} = \frac{-6}{10}$

**Part (iv)**

1. Given the rational numbers $\frac{9}{3}$ and $3$ (which can be written as $\frac{3}{1}$).
2. Cross-multiply to check equality: $9 \times 1 = 9$ and $3 \times 3 = 9$.
3. Since $ad = bc$ ($9 = 9$), the rational numbers are equal. Hence proved.

Answer (iv): $\frac{9}{3} = 3$

**Answer:** All given pairs of rational numbers are equal by the cross-multiplication rule.

> Common mistake: Forgetting to cross-multiply or misinterpreting negative signs during cross-multiplication.

### Question 2

*3 marks · Short answer*

Find the sum:
(i) $\frac{2}{5} + \frac{3}{10}$
(ii) $\frac{7}{12} + \frac{5}{8}$
(iii) $-\frac{4}{7} + \frac{3}{14}$

**Part (i)**

1. Given expression: $\frac{2}{5} + \frac{3}{10}$.
2. The LCM of the denominators $5$ and $10$ is $10$. Express with a common denominator: $\frac{4}{10} + \frac{3}{10}$.
3. Add the numerators: $\frac{4 + 3}{10} = \frac{7}{10}$.

Answer (i): $\frac{7}{10}$

**Part (ii)**

1. Given expression: $\frac{7}{12} + \frac{5}{8}$.
2. The LCM of $12$ and $8$ is $24$. Express with a common denominator: $\frac{14}{24} + \frac{15}{24}$.
3. Add the numerators: $\frac{14 + 15}{24} = \frac{29}{24}$.

Answer (ii): $\frac{29}{24}$

**Part (iii)**

1. Given expression: $-\frac{4}{7} + \frac{3}{14}$.
2. The LCM of $7$ and $14$ is $14$. Express with a common denominator: $-\frac{8}{14} + \frac{3}{14}$.
3. Add the numerators: $\frac{-8 + 3}{14} = -\frac{5}{14}$.

Answer (iii): $-\frac{5}{14}$

**Answer:** Calculated the sum for all parts.

> Common mistake: Adding denominators directly instead of finding a common denominator using LCM.

### Question 3

*3 marks · Short answer*

Find the difference:
(i) $\frac{5}{6} - \frac{1}{4}$
(ii) $\frac{11}{8} - \frac{3}{4}$
(iii) $-\frac{7}{9} - \left(-\frac{2}{3}\right)$

**Part (i)**

1. Given expression: $\frac{5}{6} - \frac{1}{4}$.
2. The LCM of $6$ and $4$ is $12$. Express with a common denominator: $\frac{10}{12} - \frac{3}{12}$.
3. Subtract the numerators: $\frac{10 - 3}{12} = \frac{7}{12}$.

Answer (i): $\frac{7}{12}$

**Part (ii)**

1. Given expression: $\frac{11}{8} - \frac{3}{4}$.
2. The LCM of $8$ and $4$ is $8$. Express with a common denominator: $\frac{11}{8} - \frac{6}{8}$.
3. Subtract the numerators: $\frac{11 - 6}{8} = \frac{5}{8}$.

Answer (ii): $\frac{5}{8}$

**Part (iii)**

1. Given expression: $-\frac{7}{9} - \left(-\frac{2}{3}\right)$.
2. Rewrite subtraction of a negative as addition: $-\frac{7}{9} + \frac{2}{3}$.
3. The LCM of $9$ and $3$ is $9$. With a common denominator: $-\frac{7}{9} + \frac{6}{9} = -\frac{1}{9}$.

Answer (iii): $-\frac{1}{9}$

**Answer:** Calculated the difference for all parts.

> Common mistake: Mistakes in handling signs when subtracting a negative rational number.

### Question 4

*3 marks · Short answer*

Find the product:
(i) $\frac{2}{3} \times \frac{3}{10}$
(ii) $\frac{7}{11} \times \frac{5}{8}$
(iii) $-\frac{4}{7} \times \frac{5}{14}$

**Part (i)**

1. Given expression: $\frac{2}{3} \times \frac{3}{10}$.
2. Multiply numerators and denominators: $\frac{2 \times 3}{3 \times 10} = \frac{6}{30}$.
3. Simplify by dividing numerator and denominator by $6$: $\frac{1}{5}$.

Answer (i): $\frac{1}{5}$

**Part (ii)**

1. Given expression: $\frac{7}{11} \times \frac{5}{8}$.
2. Multiply numerators and denominators: $\frac{7 \times 5}{11 \times 8} = \frac{35}{88}$.

Answer (ii): $\frac{35}{88}$

**Part (iii)**

1. Given expression: $-\frac{4}{7} \times \frac{5}{14}$.
2. Multiply numerators and denominators: $\frac{-4 \times 5}{7 \times 14} = \frac{-20}{98}$.
3. Simplify by dividing numerator and denominator by $2$: $-\frac{10}{49}$.

Answer (iii): $-\frac{10}{49}$

**Answer:** Calculated the product for all parts.

> Common mistake: Failing to reduce the resulting fraction to its lowest terms.

### Question 5

*3 marks · Short answer*

Find the quotient:
(i) $\frac{2}{3} \div \frac{3}{10}$
(ii) $\frac{7}{11} \div \frac{5}{8}$
(iii) $-\frac{4}{7} \div \frac{5}{14}$

**Part (i)**

1. Given expression: $\frac{2}{3} \div \frac{3}{10}$.
2. Multiply by the reciprocal of the divisor: $\frac{2}{3} \times \frac{10}{3}$.
3. Multiply numerators and denominators: $\frac{2 \times 10}{3 \times 3} = \frac{20}{9}$.

Answer (i): $\frac{20}{9}$

**Part (ii)**

1. Given expression: $\frac{7}{11} \div \frac{5}{8}$.
2. Multiply by the reciprocal of the divisor: $\frac{7}{11} \times \frac{8}{5}$.
3. Multiply numerators and denominators: $\frac{7 \times 8}{11 \times 5} = \frac{56}{55}$.

Answer (ii): $\frac{56}{55}$

**Part (iii)**

1. Given expression: $-\frac{4}{7} \div \frac{5}{14}$.
2. Multiply by the reciprocal of the divisor: $-\frac{4}{7} \times \frac{14}{5}$.
3. Cancel out common factors: $-\frac{4 \times 2}{1 \times 5} = -\frac{8}{5}$.

Answer (iii): $-\frac{8}{5}$

**Answer:** Calculated the quotient for all parts.

> Common mistake: Inverting the first rational number instead of the divisor.

### Question 6

*4 marks · Proof*

Show that: $\left(\frac{1}{2} + \frac{3}{4}\right) \times \frac{8}{3} = \frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3}$.

**Solution**

1. LHS $= \left(\frac{1}{2} + \frac{3}{4}\right) \times \frac{8}{3}$
2. $= \left(\frac{2}{4} + \frac{3}{4}\right) \times \frac{8}{3} = \frac{5}{4} \times \frac{8}{3}$
3. $= \frac{40}{12} = \frac{10}{3}$
4. RHS $= \left(\frac{1}{2} \times \frac{8}{3}\right) + \left(\frac{3}{4} \times \frac{8}{3}\right) = \frac{4}{3} + \frac{2}{1} = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}$
5. Since LHS $=$ RHS, the given statement is verified. Hence proved.

**Answer:** Hence proved.

> Common mistake: Adding fractions without taking a common denominator.

### Question 7

*3 marks · Short answer*

Simplify the following using the distributive property:
$\frac{7}{9} \left(\frac{6}{7} - \frac{3}{4}\right)$.

**Solution**

1. Given expression: $\frac{7}{9} \left(\frac{6}{7} - \frac{3}{4}\right)$
2. Using the distributive property $p(q + r) = pq + pr$, we get: $\left(\frac{7}{9} \times \frac{6}{7}\right) - \left(\frac{7}{9} \times \frac{3}{4}\right)$
3. $= \frac{6}{9} - \frac{7}{12} = \frac{2}{3} - \frac{7}{12}$
4. $= \frac{8}{12} - \frac{7}{12} = \frac{1}{12}$

**Answer:** $\frac{1}{12}$

> Common mistake: Multiplying incorrectly before simplifying the fractions.

### Question 8

*3 marks · Short answer*

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

**Solution**

1. Given equation is $\frac{5}{6}x + \left(-\frac{3}{5}\right) = \frac{5}{6}x + \frac{1}{2}$.
2. Subtract $\frac{5}{6}x$ from both sides of the equation.
3. We get $-\frac{3}{5} = \frac{1}{2}$.
4. Since $-\frac{3}{5} \neq \frac{1}{2}$, the statement is a contradiction.
5. Therefore, there is no value of $x$ that satisfies the given equation.

**Answer:** No solution

> Common mistake: Conflating identical variable terms on both sides as an identity instead of recognizing the contradiction in constant terms.

## Exercise Set 3.4

### Question 1

*3 marks · Short answer*

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

**Solution**

1. Draw a number line with integers marked at equal intervals.
2. To represent $\frac{2}{3}$, divide the unit interval between 0 and 1 into 3 equal parts and mark the 2nd part to the right of 0.
3. To represent $-\frac{5}{4} = -1\frac{1}{4}$, divide the unit interval between $-1$ and $-2$ into 4 equal parts and move 1 part to the left of $-1$.
4. To represent $1\frac{1}{2} = \frac{3}{2}$, mark the point exactly halfway between 1 and 2.
5. The rational numbers $\frac{2}{3}$, $-\frac{5}{4}$, and $1\frac{1}{2}$ are marked on the single number line.

**Answer:** The rational numbers $\frac{2}{3}$, $-\frac{5}{4}$, and $1\frac{1}{2}$ represented on the number line.

> Common mistake: Dividing the wrong unit interval or miscounting the subdivisions.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. First, express the given rational numbers $-\frac{1}{2}$ and $\frac{1}{4}$ with a common denominator.
2. The LCM of 2 and 4 is 4, so $-\frac{1}{2} = -\frac{2}{4}$ and $\frac{1}{4} = \frac{1}{4}$.
3. To easily find three distinct rational numbers between $-\frac{2}{4}$ and $\frac{1}{4}$, multiply both the numerator and denominator of both fractions by a suitable factor, say 4, to get $-\frac{8}{16}$ and $\frac{4}{16}$.
4. Three distinct rational numbers between $-\frac{8}{16}$ and $\frac{4}{16}$ are $-\frac{3}{16}, -\frac{2}{16}, \frac{1}{16}$ (or any other valid choices such as $-\frac{1}{4}, 0, \frac{1}{8}$).
5. Hence, three rational numbers strictly between $-\frac{1}{2}$ and $\frac{1}{4}$ are $-\frac{1}{4}, 0, \frac{1}{8}$.

**Answer:** $-\frac{1}{4}, 0, \frac{1}{8}$

> Common mistake: Failing to make the denominators equal before comparing or finding numbers between them.

### Question 3

*3 marks · Short answer*

Simplify the expression: $-\left(-\frac{1}{4}\right) + \left(\frac{5}{12}\right)$.

**Solution**

1. Given the expression: $-\left(-\frac{1}{4}\right) + \left(\frac{5}{12}\right)$.
2. Using the rule that the product of two negatives is positive, $-\left(-\frac{1}{4}\right) = \frac{1}{4}$.
3. Now add $\frac{1}{4}$ and $\frac{5}{12}$ by finding a common denominator.
4. The least common multiple of $4$ and $12$ is $12$, so convert $\frac{1}{4}$ to an equivalent fraction with denominator $12$: $\frac{1 \times 3}{4 \times 3} = \frac{3}{12}$.
5. Add the numerators: $\frac{3}{12} + \frac{5}{12} = \frac{3 + 5}{12} = \frac{8}{12}$.
6. Simplify the fraction by dividing the numerator and denominator by their highest common factor, $4$, to get $\frac{2}{3}$.

**Answer:** $\frac{2}{3}$

> Common mistake: Failing to change the sign of $-\left(-\frac{1}{4}\right)$ to positive.

### Question 4

*3 marks · Short answer*

A tailor has $15\frac{3}{4}$ metres of fine silk. If making one kurta requires $2\frac{1}{4}$ metres of silk, exactly how many kurtas can he make?

**Solution**

1. Total length of silk available = $15\frac{3}{4} = \frac{63}{4}$ metres.
2. Silk required for one kurta = $2\frac{1}{4} = \frac{9}{4}$ metres.
3. To find the number of kurtas, divide the total length by the length required for one kurta: $\frac{63}{4} \div \frac{9}{4}$.
4. Multiply by the reciprocal of the divisor: $\frac{63}{4} \times \frac{4}{9} = \frac{63}{9}$.
5. Calculate the final result: $\frac{63}{9} = 7$.

**Answer:** $7$ kurtas

> Common mistake: Multiplying the fractions instead of dividing.

### Question 5

*3 marks · Short answer*

Find three rational numbers between $3.1415$ and $3.1416$.

**Solution**

1. Given decimal numbers are $3.1415$ and $3.1416$.
2. Pad the decimals with trailing zeros to make them comparable: $3.14150$ and $3.14160$.
3. Choose numbers whose decimal parts lie strictly between $150$ and $160$ at the fifth decimal place.
4. Three such rational numbers are $3.14151$, $3.14152$, and $3.14155$.

**Answer:** $3.14151$, $3.14152$, $3.14155$

> Common mistake: Failing to equalize the number of decimal places before finding intermediate values.

### Question 6

*3 marks · Short answer*

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

**Solution**

1. One common method is converting the two rational numbers to equivalent fractions with a common denominator and choosing numerators between them.
2. Another method is using decimal representations of the rational numbers and finding terminating or repeating decimals strictly between them.
3. A third method is repeatedly applying the mean (average) formula $\frac{a + b}{2}$ to find successive midpoints between the numbers.

**Answer:** Using equivalent fractions with common denominators, decimal representations, or repeated averaging.

> Common mistake: Stating only one method when multiple ways are asked.

## Exercise Set 3.5

### Question 1

*3 marks · Short answer*

Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: $\frac{7}{20}$, $\frac{4}{15}$ and $\frac{13}{250}$. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

**Solution**

1. Find the prime factorisation of the denominator for each rational number in its lowest form.
2. For $\frac{7}{20}$, the denominator $20 = 2^2 \times 5$, so its decimal expansion terminates ($0.35$).
3. For $\frac{4}{15}$, the denominator $15 = 3 \times 5$, so its decimal expansion is repeating ($0.2\overline{6}$).
4. For $\frac{13}{250}$, the denominator $250 = 2 \times 5^3$, so its decimal expansion terminates ($0.052$).

**Answer:** $\frac{7}{20}$ and $\frac{13}{250}$ are terminating decimals, while $\frac{4}{15}$ is repeating.

> Common mistake: Failing to check if the rational number is in lowest terms before factoring the denominator.

### Question 2

*3 marks · Short answer*

Perform the long division for $\frac{1}{13}$. Identify the repeating block of digits. Does it show cyclic properties if you evaluate $\frac{2}{13}$? Now compute $\frac{3}{13}$, $\frac{4}{13}$, etc. What do you notice?

**Solution**

1. Perform long division to find $\frac{1}{13} = 0.\overline{076923}$, where the repeating block is $076923$.
2. Evaluate $\frac{2}{13} = 0.\overline{153846}$, which is a cyclic shift of the same digits.
3. Compute other fractions like $\frac{3}{13} = 0.\overline{230769}$ and observe that all reciprocals with denominator $13$ produce cyclic permutations of the same six digits.

**Answer:** The repeating block for $\frac{1}{13}$ is $076923$, and all multiples $\frac{k}{13}$ exhibit cyclic properties.

> Common mistake: Arithmetic errors during long division with 13.

### Question 3

*3 marks · Short answer*

Classify the following numbers as rational or irrational:
(i) $\sqrt{81}$
(ii) $\sqrt{12}$
(iii) $0.33333 \dots$
(iv) $0.123451234512345 \dots$
(v) $1.01001000100001 \dots$ (Notice the pattern: Is it repeating a single block?)
(vi) $23.560185612239874790120$
Find the explicit fractions in case they are rational.

**Part (i)**

1. $\sqrt{81} = 9$, which can be written as $\frac{9}{1}$.

Answer (i): Rational, $\frac{9}{1}$

**Part (ii)**

1. $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$, which cannot be expressed as a ratio of integers.

Answer (ii): Irrational

**Part (iii)**

1. $0.33333 \dots = 0.\overline{3} = \frac{1}{3}$, which is a repeating decimal.

Answer (iii): Rational, $\frac{1}{3}$

**Part (iv)**

1. $0.123451234512345 \dots = 0.\overline{12345} = \frac{12345}{99999}$.

Answer (iv): Rational, $\frac{4115}{33333}$

**Part (v)**

1. The decimal $1.01001000100001 \dots$ is non-terminating and non-repeating because the number of zeros between the ones increases by one each time.

Answer (v): Irrational

**Answer:** Classified all six numbers as rational or irrational with their fractions where applicable.

> Common mistake: Confusing non-repeating patterns with repeating decimal blocks.

### Question 4

*3 marks · Proof*

The number $0.\bar{9}$ (which means $0.99999 \dots$) is a rational number. Using algebra (let $x = 0.\bar{9}$, multiply by $10$, and subtract), explain why $0.\bar{9}$ is exactly equal to $1$.

**Solution**

1. Let $x = 0.\overline{9} = 0.99999 \dots$.
2. Multiply both sides by $10$ to get $10x = 9.99999 \dots$.
3. Subtract the first equation from the second: $10x - x = 9.99999 \dots - 0.99999 \dots$.
4. Simplify to get $9x = 9$, which gives $x = 1$. Hence proved.

**Answer:** $0.\overline{9} = 1$

> Common mistake: Assuming $0.\overline{9}$ is strictly less than 1 due to the infinite series representation.

### Question 5

*3 marks · Short answer*

We have seen that the repeating block of $\frac{1}{7}$ is a cyclic number. Try to find more numbers ($n$) whose reciprocals ($\frac{1}{n}$) produce decimals with repeating blocks that are cyclic.

**Solution**

1. Find the decimal expansions of reciprocals of prime numbers like $7$, $17$, $19$, and $23$.
2. Observe that $\frac{1}{17} = 0.\overline{0588235294117647}$ and $\frac{1}{19} = 0.\overline{052631578947368421}$ produce repeating blocks that are cyclic.

**Answer:** Examples of such numbers whose reciprocals produce cyclic decimals are $7$, $17$, $19$, and $23$.

> Common mistake: Not checking enough decimal places to see the full repeating block.

## End-of-Chapter Exercises

### Question 1

*3 marks · Short answer*

Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:
(i) $\frac{3}{50}$
(ii) $\frac{2}{9}$

**Part (i)**

1. Perform long division by dividing $3$ by $50$.
2. The remainder becomes $0$ after two decimal places, giving $0.06$.

Answer (i): $0.06$

**Part (ii)**

1. Perform long division by dividing $2$ by $9$.
2. The remainder never becomes $0$ and the digit $2$ repeats infinitely, giving $0.\bar{2}$.

Answer (ii): $0.\bar{2}$

**Answer:** (i) $0.06$, (ii) $0.\bar{2}$

> Common mistake: Stopping the long division too early for non-terminating repeating decimals.

### Question 2

*4 marks · Proof*

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

**Solution**

1. Let us assume, to the contrary, that $\sqrt{5}$ is a rational number.
2. Therefore, $\sqrt{5} = \frac{p}{q}$, where $p$ and $q$ are integers sharing no common factor other than $1$ ($q \neq 0$).
3. Squaring both sides gives $5 = \frac{p^2}{q^2}$, which means $p^2 = 5q^2$.
4. This implies that $p^2$ is a multiple of $5$, and therefore $p$ is a multiple of $5$. Let $p = 5k$ for some integer $k$.
5. Substituting $p = 5k$ gives $5q^2 = (5k)^2 = 25k^2$, which simplifies to $q^2 = 5k^2$.
6. This implies that $q^2$ is a multiple of $5$, and therefore $q$ is also a multiple of $5$.
7. Thus, both $p$ and $q$ share a common factor of $5$, which contradicts our initial assumption that they are co-prime.
8. Hence proved that $\sqrt{5}$ is irrational.

**Answer:** $\sqrt{5}$ is irrational.

> Common mistake: Forgetting to mention that $p$ and $q$ are co-prime in the initial assumption.

### Question 3

*3 marks · Short answer*

Convert the following decimal numbers in the form of $\frac{p}{q}$.
(i) $12.\bar{6}$
(ii) $0.012\bar{0}$
(iii) $3.05\bar{2}$
(iv) $1.2\bar{35}$
(v) $0.\bar{23}$
(vi) $2.0\bar{5}$
(vii) $2.1\bar{25}$
(viii) $3.1\bar{25}$
(ix) $2.16\bar{25}$

**Part (i)**

1. Let $x = 12.\bar{6}$
2. $10x = 126.\bar{6}$
3. $10x - x = 126.\bar{6} - 12.\bar{6}$
4. $9x = 114 \implies x = \frac{114}{9} = \frac{38}{3}$

Answer (i): $\frac{38}{3}$

**Part (ii)**

1. Let $x = 0.012\bar{0}$
2. $1000x = 12.\bar{0}$
3. $10000x = 120.\bar{0}$
4. $9000x = 108 \implies x = \frac{108}{9000} = \frac{3}{250}$

Answer (ii): $\frac{3}{250}$

**Part (iii)**

1. Let $x = 3.05\bar{2}$
2. $100x = 305.\bar{2}$
3. $1000x = 3052.\bar{2}$
4. $900x = 2747 \implies x = \frac{2747}{900}$

Answer (iii): $\frac{2747}{900}$

**Part (iv)**

1. Let $x = 1.2\bar{35}$
2. $10x = 12.\bar{35}$
3. $1000x = 1235.\bar{35}$
4. $990x = 1223 \implies x = \frac{1223}{990}$

Answer (iv): $\frac{1223}{990}$

**Part (v)**

1. Let $x = 0.\bar{23}$
2. $100x = 23.\bar{23}$
3. $99x = 23 \implies x = \frac{23}{99}$

Answer (v): $\frac{23}{99}$

**Part (vi)**

1. Let $x = 2.0\bar{5}$
2. $10x = 20.\bar{5}$
3. $100x = 205.\bar{5}$
4. $90x = 185 \implies x = \frac{185}{90} = \frac{37}{18}$

Answer (vi): $\frac{37}{18}$

**Part (vii)**

1. Let $x = 2.1\bar{25}$
2. $10x = 21.\bar{25}$
3. $1000x = 2125.\bar{25}$
4. $990x = 2104 \implies x = \frac{2104}{990} = \frac{1052}{495}$

Answer (vii): $\frac{1052}{495}$

**Part (viii)**

1. Let $x = 3.1\bar{25}$
2. $10x = 31.\bar{25}$
3. $1000x = 3125.\bar{25}$
4. $990x = 3094 \implies x = \frac{3094}{990} = \frac{1547}{495}$

Answer (viii): $\frac{1547}{495}$

**Part (ix)**

1. Let $x = 2.16\bar{25}$
2. $100x = 216.\bar{25}$
3. $10000x = 21625.\bar{25}$
4. $9900x = 21409 \implies x = \frac{21409}{9900}$

Answer (ix): $\frac{21409}{9900}$

**Answer:** Converted all given decimals into $\frac{p}{q}$ form.

> Common mistake: Multiplying by incorrect powers of 10 for numbers with non-repeating decimal parts.

### Question 4

*3 marks · Short answer*

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

**Part (i)**

1. To represent $0.532$, locate the interval between $0.5$ and $0.6$ on the number line.
2. Magnify the interval into $10$ equal parts and locate $0.53$ then further subdivide to mark $0.532$.

Answer (i): Point representing $0.532$ between $0.5$ and $0.6$

**Part (ii)**

1. To represent $1.15$, observe that it lies exactly halfway between $1.1$ and $1.2$.
2. Mark the point between $1.1$ and $1.2$ corresponding to $1.15$.

Answer (ii): Point representing $1.15$ between $1.1$ and $1.2$

**Answer:** (i) Between $0.5$ and $0.6$, (ii) Between $1.1$ and $1.2$

> Common mistake: Incorrectly identifying the unit interval for decimal placement.

### Question 5

*3 marks · Short answer*

Find 6 rational numbers between $3$ and $4$.

**Solution**

1. Multiply the numerator and denominator of both numbers by $7$ to make the denominator $7$.
2. Write $3$ as $\frac{21}{7}$ and $4$ as $\frac{28}{7}$.
3. Choose six rational numbers between $\frac{21}{7}$ and $\frac{28}{7}$, which are $\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}$.

**Answer:** $\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}$

> Common mistake: Multiplying by too small a factor, yielding insufficient numbers.

### Question 6

*3 marks · Short answer*

Find 5 rational numbers between $\frac{2}{5}$ and $\frac{3}{5}$.

**Solution**

1. Multiply the numerator and denominator of $\frac{2}{5}$ and $\frac{3}{5}$ by $6$ to get denominators of $30$.
2. The numbers become $\frac{12}{30}$ and $\frac{18}{30}$.
3. Select five rational numbers between them: $\frac{13}{30}, \frac{14}{30}, \frac{15}{30}, \frac{16}{30}, \frac{17}{30}$.

**Answer:** $\frac{13}{30}, \frac{14}{30}, \frac{15}{30}, \frac{16}{30}, \frac{17}{30}$

> Common mistake: Failing to convert to a common denominator before finding intermediate numbers.

### Question 7

*3 marks · Short answer*

Find 5 rational numbers between $\frac{1}{6}$ and $\frac{2}{5}$.

**Solution**

1. Express the given rational numbers $\frac{1}{6}$ and $\frac{2}{5}$ with a common denominator.
2. The least common multiple of $6$ and $5$ is $30$, so $\frac{1}{6} = \frac{5}{30}$ and $\frac{2}{5} = \frac{12}{30}$.
3. To find 5 rational numbers, multiply the numerator and denominator of both fractions by $5 + 1 = 6$ (or a larger factor like $10$) to create enough space.
4. Multiplying by $10$: $\frac{5}{30} = \frac{50}{300}$ and $\frac{12}{30} = \frac{120}{300}$.
5. Choose any 5 distinct rational numbers between $\frac{50}{300}$ and $\frac{120}{300}$, such as $\frac{51}{300}, \frac{52}{300}, \frac{53}{300}, \frac{54}{300}, \frac{55}{300}$.

**Answer:** $\frac{51}{300}, \frac{52}{300}, \frac{53}{300}, \frac{54}{300}, \frac{55}{300}$ (or equivalent fractions like $\frac{17}{90}, \frac{7}{35}$, etc.)

> Common mistake: Failing to make denominators equal before comparing or finding intermediate numbers.

### Question 8

*3 marks · Short answer*

If $\frac{x}{3} + \frac{x}{5} = \frac{16}{15}$, find the rational number $x$.

**Solution**

1. Given the equation $\frac{x}{3} + \frac{x}{5} = \frac{16}{15}$.
2. Take the LCM of the denominators on the left-hand side, which is $15$, to get $\frac{5x + 3x}{15} = \frac{16}{15}$ or $\frac{8x}{15} = \frac{16}{15}$.
3. Multiply both sides by $15$ to get $8x = 16$, and divide by $8$ to find $x = 2$.

**Answer:** $x = 2$

> Common mistake: Adding denominators directly instead of taking the LCM.

### Question 9

*3 marks · Proof*

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

**Solution**

1. Given that $a$ and $b$ are non-zero rational numbers and $a + \frac{1}{b} = 0$.
2. From the given equation, we can transpose $\frac{1}{b}$ to get $a = -\frac{1}{b}$, which implies $ab = -1$.
3. Since the product $ab$ is equal to $-1$, which is strictly less than zero, $ab$ must be negative.

**Answer:** $ab$ is negative.

> Common mistake: Assuming $a$ and $b$ must be positive or negative individually without finding their relation.

### Question 10

*4 marks · Proof*

A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form $\frac{p}{10^4}$, where $p$ is an integer not divisible by $10$. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by $2^4$ or $5^4$? Give reasons.

**Solution**

1. Given that a rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place, let the number be represented with $4$ decimal places as $0.d_1 d_2 d_3 d_4$.
2. This can be written as $\frac{\text{integer}}{10^4}$, where the integer numerator $p$ is not divisible by $10$ (otherwise the decimal would terminate earlier).
3. When written in the lowest form, the denominator is obtained by canceling common factors of $10^4 = 2^4 \times 5^4$ with $p$.
4. Therefore, the denominator in the lowest form is a divisor of $2^4$ or $5^4$ (or both), meaning it is not strictly necessary for the reduced denominator to be a multiple of both $2^4$ and $5^4$, but its prime factors can only be $2$ and $5$ with powers at most $4$.

**Answer:** The number can be written as $\frac{p}{10^4}$; it is not necessary for the reduced denominator to be divisible by both $2^4$ or $5^4$ as common factors may cancel out.

> Common mistake: Confusing the power of $10$ with the exact highest power of $2$ or $5$ in the simplified denominator.

### Question 11

*3 marks · Short answer*

Without performing division, determine whether the decimal expansion of $\frac{18}{125}$ is terminating or non-terminating. If it terminates, state the number of decimal places.

**Solution**

1. Given the rational number $\frac{18}{125}$, find the prime factorization of the denominator $125 = 5^3$.
2. Since the prime factorization of the denominator contains only powers of $5$ ($5^3$), the decimal expansion is terminating.
3. To find the number of decimal places, multiply the numerator and denominator by $2^3$ to make the denominator $10^3$: $\frac{18 \times 8}{5^3 \times 2^3} = \frac{144}{1000} = 0.144$.
4. Thus, the decimal expansion terminates after 3 decimal places.

**Answer:** Terminating, with 3 decimal places.

> Common mistake: Counting the numerator's digits instead of the power of $10$ in the denominator.

### Question 12

*3 marks · Short answer*

A rational number in its lowest form has denominator $2^3 \times 5$. How many decimal places will its decimal expansion have? Explain your answer.

**Solution**

1. Given a rational number in its lowest form with denominator $2^3 \times 5$ (which can be written as $2^3 \times 5^1$).
2. To convert this into a power of $10$, we need equal powers of $2$ and $5$ in the denominator.
3. Multiply the numerator and denominator by $5^2$ to get the denominator as $2^3 \times 5^3 = (2 \times 5)^3 = 10^3$.
4. Since the denominator is a power of $10$ with exponent $3$, the decimal expansion will have 3 decimal places.

**Answer:** 3 decimal places

> Common mistake: Taking the exponent of $2$ or $5$ individually rather than the maximum exponent between them.

### Question 13

*5 marks · Proof*

Let $a = \frac{7}{12}$ and $b = \frac{5}{6}$. Express both $a$ and $b$ in the form $\frac{k_1}{m}$ and $\frac{k_2}{m}$ where $k_1, k_2$ and $m$ are integers and $k_2 - k_1 > 6$. Using the same denominator $m$, write exactly five distinct rational numbers lying between $a$ and $b$ keeping an integer numerator. Explain why the condition $k_2 - k_1 > n + 1$ is necessary to find $n$ such rational numbers between the two rational numbers $a$ and $b$ using this method.

**Solution**

1. Given numbers are $a = \frac{7}{12}$ and $b = \frac{5}{6}$.
2. To find a common denominator, take the LCM of $12$ and $6$, which is $12$.
3. Express $a$ and $b$ with denominator $m = 12$: $a = \frac{7}{12}$ and $b = \frac{5}{6} = \frac{10}{12}$.
4. Here, $k_1 = 7$ and $k_2 = 10$, so $k_2 - k_1 = 10 - 7 = 3$, which is not greater than $n + 1 = 5 + 1 = 6$.
5. To satisfy $k_2 - k_1 > 6$, multiply the numerator and denominator of both fractions by a suitable integer, say $2$: $a = \frac{14}{24}$ and $b = \frac{20}{24}$.
6. Now, $k_1 = 14$ and $k_2 = 20$, so $k_2 - k_1 = 20 - 14 = 6$ (still need strictly greater, so multiply by a larger factor or use $m = 36$).
7. Let us use $m = 36$: $a = \frac{7 \times 3}{12 \times 3} = \frac{21}{36}$ and $b = \frac{5 \times 6}{6 \times 6} = \frac{30}{36}$.
8. Here, $k_1 = 21$ and $k_2 = 30$, giving $k_2 - k_1 = 30 - 21 = 9$, which is greater than $6$.
9. Five distinct rational numbers between $a$ and $b$ with denominator $36$ and integer numerators are $\frac{22}{36}, \frac{23}{36}, \frac{24}{36}, \frac{25}{36}, \frac{26}{36}$.
10. The condition $k_2 - k_1 > n + 1$ is necessary because the number of integers strictly between two integers $k_1$ and $k_2$ is $(k_2 - k_1 - 1)$.
11. To find $n$ such numbers, we must have $k_2 - k_1 - 1 \ge n$, which simplifies to $k_2 - k_1 \ge n + 1$. Hence proved.

**Answer:** Five rational numbers are $\frac{22}{36}, \frac{23}{36}, \frac{24}{36}, \frac{25}{36}, \frac{26}{36}$.

> Common mistake: Forgetting to scale up the fractions so that the difference between the numerators allows enough integer steps.

### Question 14

*4 marks · Proof*

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

**Solution**

1. Given: $x + y + z = 0$ and $xy + yz + zx = 0$.
2. Consider the algebraic identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$.
3. Substitute the given values into the identity: $0^2 = x^2 + y^2 + z^2 + 2(0)$.
4. This gives $x^2 + y^2 + z^2 = 0$.
5. Since $x, y, z$ are rational numbers, their squares $x^2, y^2, z^2$ are non-negative real numbers.
6. The sum of non-negative numbers can be zero only if each number is individually zero.
7. Therefore, $x^2 = 0, y^2 = 0, z^2 = 0$, which implies $x = 0, y = 0, z = 0$. Hence proved.

**Answer:** All the rational numbers $x, y, z$ must be simultaneously zero.

> Common mistake: Assuming that the sum of squares can be zero with non-zero values.

### Question 15

*3 marks · Proof*

Show that the rational number $\frac{(a + b)}{2}$ lies between the rational numbers $a$ and $b$.

**Solution**

1. Let $a$ and $b$ be two rational numbers such that $a < b$.
2. Add $a$ to both sides: $a + a < a + b$, which gives $2a < a + b$, so $a < \frac{a + b}{2}$.
3. Add $b$ to both sides of $a < b$: $a + b < b + b$, which gives $a + b < 2b$, so $\frac{a + b}{2} < b$.
4. Combining both inequalities, we get $a < \frac{a + b}{2} < b$. Hence proved.

**Answer:** The number $\frac{a + b}{2}$ lies between $a$ and $b$.

> Common mistake: Failing to state the initial assumption $a < b$.

### Question 16

*3 marks · Short answer*

Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.

**Solution**

1. By the Baudhāyana-Pythagoras Theorem, the length of the hypotenuse of each right-angled triangle in the spiral is calculated successively.
2. The first triangle has base $1$ and perpendicular $1$, so its hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$.
3. Each subsequent triangle uses the previous hypotenuse as one side and a perpendicular of length $1$ as the other side.
4. Therefore, the lengths of the hypotenuses form the sequence: $\sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \dots$ or simply $1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \dots$

**Answer:** The lengths of the hypotenuses are $\sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \dots$ (continuing for each successive triangle in the spiral).

> Common mistake: Not recognizing the recursive pattern of the hypotenuse lengths.

## Frequently asked questions

### How many exercises and questions are in NCERT Solutions for Class 9 Maths Chapter 3: The World of Numbers?

This chapter for the new NCERT book under the NCF 2023 for the 2026-27 session contains five main exercise sets and an end-of-chapter exercise. Specifically, Exercise Set 3.1 has 4 questions, Exercise Set 3.2 has 4 questions, Exercise Set 3.3 has 8 questions, Exercise Set 3.4 has 6 questions, Exercise Set 3.5 has 5 questions, and the End-of-Chapter Exercises have 16 questions. You can find SwaVid's free PDF and step-by-step solutions for all these questions on this page only.

### What mathematical topics and concepts do the questions cover in this chapter?

The exercises cover a wide range of topics starting from natural numbers and base-12 systems in Exercise Set 3.1 to Brahmagupta's laws of integers and temperature problems in Exercise Set 3.2. Further sets delve into rational numbers, distributive properties, linear equations, cyclic numbers, and algebraic proofs like $0.\bar{9} = 1$. All these core concepts are thoroughly addressed in SwaVid's free PDF and step-by-step solutions available on this page only.

### What are the hardest question types in this chapter and how should I approach them?

The most challenging questions involve proofs and algebraic manipulations, such as the algebraic proof that $0.\bar{9}$ equals 1 and converting repeating decimals to rational form using prime factorization of denominators. To approach them, you should carefully apply algebraic identities, understand decimal expansions, and practice step-by-step derivations. SwaVid's free PDF and step-by-step solutions on this page only provide detailed breakdowns for these difficult proofs.

### How can I write answers for full marks in Class 9 Maths Chapter 3 examinations?

To secure full marks in short answer and proof-based questions, you must clearly state the underlying property or law being used, such as the distributive property of rational numbers or Brahmagupta's rules for integers. Always show intermediate calculation steps clearly rather than just writing the final answer. Following the structured formats in SwaVid's free PDF and step-by-step solutions available on this page only will help you understand how examiners award marks.

### Is the free PDF for Chapter 3: The World of Numbers available for download?

Yes, comprehensive solutions aligned with the new NCERT book for the 2026-27 session are completely accessible for students. You do not need to look elsewhere for study material as SwaVid's free PDF and step-by-step solutions are on this page only.

## Related pages

- [Exercise 3.1 solutions](https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions/exercise-3-1)
- [Exercise 3.2 solutions](https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions/exercise-3-2)
- [Exercise 3.3 solutions](https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions/exercise-3-3)
- [Exercise 3.4 solutions](https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions/exercise-3-4)
- [Exercise 3.5 solutions](https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions/exercise-3-5)
- [Class 9 Maths chapters](https://www.swavid.com/maths/class/9)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student