---
title: "NCERT Solutions for Class 9 Maths Chapter 3 Exercise 3.3"
url: https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions/exercise-3-3
dateModified: 2026-10-07T15:46:03+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 3 Exercise 3.3

Chapter 3: The World of Numbers. Every question from Exercise 3.3, with full working and the final answer.

Free PDF (22 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-9/swavid-ncert-solutions-class-9-maths-chapter-3-the-world-of-numbers-521b6f38aa.pdf

## 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.

## Related pages

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

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