NCERT Solutions for Class 9 Maths Chapter 3 — Exercise Set 3.3#
A rational number is a number of the form $\dfrac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. To add or subtract rational numbers, first make the denominators equal using the LCM. To multiply, multiply the numerators and the denominators. To divide, multiply by the reciprocal.
Two rational numbers $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are equal when $ad = bc$ (cross-multiplication).
1. Prove that the following rational numbers are equal: #
(i) $\dfrac{2}{3}$ and $\dfrac{4}{6}$
We cross-multiply:
$$2 \times 6 = 12$$$$3 \times 4 = 12$$$$\Rightarrow 2 \times 6 = 3 \times 4$$
∴ $\dfrac{2}{3} = \dfrac{4}{6}$.
(Also, dividing the numerator and denominator of $\dfrac{4}{6}$ by 2 gives $\dfrac{2}{3}$.)
(ii) $\dfrac{5}{4}$ and $\dfrac{10}{8}$
We cross-multiply:
$$5 \times 8 = 40$$$$4 \times 10 = 40$$$$\Rightarrow 5 \times 8 = 4 \times 10$$
∴ $\dfrac{5}{4} = \dfrac{10}{8}$.
(iii) $-\dfrac{3}{5}$ and $-\dfrac{6}{10}$
We cross-multiply, writing the numbers as $\dfrac{-3}{5}$ and $\dfrac{-6}{10}$:
$$(-3) \times 10 = -30$$$$5 \times (-6) = -30$$$$\Rightarrow (-3) \times 10 = 5 \times (-6)$$
∴ $-\dfrac{3}{5} = -\dfrac{6}{10}$.
(iv) $\dfrac{9}{3}$ and $3$
We write 3 as $\dfrac{3}{1}$ and cross-multiply:
$$9 \times 1 = 9$$$$3 \times 3 = 9$$$$\Rightarrow 9 \times 1 = 3 \times 3$$
∴ $\dfrac{9}{3} = 3$.
2. Find the sum: #
(i) $\dfrac{2}{5} + \dfrac{3}{10}$
The LCM of 5 and 10 is 10. We find the sum $\dfrac{2}{5} + \dfrac{3}{10}$:
$$= \dfrac{4}{10} + \dfrac{3}{10}$$$$= \dfrac{7}{10}$$
(ii) $\dfrac{7}{12} + \dfrac{5}{8}$
The LCM of 12 and 8 is 24. We find the sum $\dfrac{7}{12} + \dfrac{5}{8}$:
$$= \dfrac{14}{24} + \dfrac{15}{24}$$$$= \dfrac{29}{24}$$
(iii) $-\dfrac{4}{7} + \dfrac{3}{14}$
The LCM of 7 and 14 is 14. We find the sum $-\dfrac{4}{7} + \dfrac{3}{14}$:
$$= -\dfrac{8}{14} + \dfrac{3}{14}$$$$= \dfrac{-8 + 3}{14}$$
$$= -\dfrac{5}{14}$$
3. Find the difference: #
(i) $\dfrac{5}{6} - \dfrac{1}{4}$
The LCM of 6 and 4 is 12. We find the difference $\dfrac{5}{6} - \dfrac{1}{4}$:
$$= \dfrac{10}{12} - \dfrac{3}{12}$$$$= \dfrac{7}{12}$$
(ii) $\dfrac{11}{8} - \dfrac{3}{4}$
The LCM of 8 and 4 is 8. We find the difference $\dfrac{11}{8} - \dfrac{3}{4}$:
$$= \dfrac{11}{8} - \dfrac{6}{8}$$$$= \dfrac{5}{8}$$
(iii) $-\dfrac{7}{9} - \left(-\dfrac{2}{3}\right)$
Subtracting a negative number is the same as adding the positive number. The LCM of 9 and 3 is 9. We find the difference $-\dfrac{7}{9} - \left(-\dfrac{2}{3}\right)$:
$$= -\dfrac{7}{9} + \dfrac{2}{3}$$$$= -\dfrac{7}{9} + \dfrac{6}{9}$$
$$= -\dfrac{1}{9}$$
4. Find the product: #
(i) $\dfrac{2}{3} \times \dfrac{3}{10}$
We find the product $\dfrac{2}{3} \times \dfrac{3}{10}$:
$$= \dfrac{2 \times 3}{3 \times 10}$$$$= \dfrac{6}{30}$$
$$= \dfrac{1}{5}$$
(ii) $\dfrac{7}{11} \times \dfrac{5}{8}$
We find the product $\dfrac{7}{11} \times \dfrac{5}{8}$:
$$= \dfrac{7 \times 5}{11 \times 8}$$$$= \dfrac{35}{88}$$
(iii) $-\dfrac{4}{7} \times \dfrac{5}{14}$
We find the product $-\dfrac{4}{7} \times \dfrac{5}{14}$:
$$= -\dfrac{4 \times 5}{7 \times 14}$$$$= -\dfrac{20}{98}$$
$$= -\dfrac{10}{49}$$
5. Find the quotient: #
(i) $\dfrac{2}{3} \div \dfrac{3}{10}$
Dividing by $\dfrac{3}{10}$ is the same as multiplying by its reciprocal $\dfrac{10}{3}$. We find the quotient $\dfrac{2}{3} \div \dfrac{3}{10}$:
$$= \dfrac{2}{3} \times \dfrac{10}{3}$$$$= \dfrac{20}{9}$$
(ii) $\dfrac{7}{11} \div \dfrac{5}{8}$
We multiply by the reciprocal of $\dfrac{5}{8}$. We find the quotient $\dfrac{7}{11} \div \dfrac{5}{8}$:
$$= \dfrac{7}{11} \times \dfrac{8}{5}$$$$= \dfrac{56}{55}$$
(iii) $-\dfrac{4}{7} \div \dfrac{5}{14}$
We multiply by the reciprocal of $\dfrac{5}{14}$. We find the quotient $-\dfrac{4}{7} \div \dfrac{5}{14}$:
$$= -\dfrac{4}{7} \times \dfrac{14}{5}$$$$= -\dfrac{56}{35}$$
$$= -\dfrac{8}{5}$$
6. Show that: $\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$. #
Solution
We find the left-hand side, $\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3}$:
$$= \left(\dfrac{2}{4} + \dfrac{3}{4}\right) \times \dfrac{8}{3}$$$$= \dfrac{5}{4} \times \dfrac{8}{3}$$
$$= \dfrac{40}{12}$$
$$= \dfrac{10}{3}$$
We find the right-hand side, $\dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$:
$$= \dfrac{8}{6} + \dfrac{24}{12}$$$$= \dfrac{4}{3} + 2$$
$$= \dfrac{4}{3} + \dfrac{6}{3}$$
$$= \dfrac{10}{3}$$$$\Rightarrow \text{LHS} = \text{RHS} = \dfrac{10}{3}$$
∴ $\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$. This is the distributive property of multiplication over addition.
7. Simplify the following using the distributive property: $\dfrac{7}{9}\left(\dfrac{6}{7} - \dfrac{3}{4}\right)$. #
Solution
By the distributive property, $a(b - c) = ab - ac$. We simplify $\dfrac{7}{9}\left(\dfrac{6}{7} - \dfrac{3}{4}\right)$:
$$= \dfrac{7}{9} \times \dfrac{6}{7} - \dfrac{7}{9} \times \dfrac{3}{4}$$$$= \dfrac{6}{9} - \dfrac{21}{36}$$
$$= \dfrac{2}{3} - \dfrac{7}{12}$$
$$= \dfrac{8}{12} - \dfrac{7}{12}$$
$$= \dfrac{1}{12}$$
8. Find the rational number x such that: $\dfrac{5}{6}\left(x + \dfrac{3}{5}\right) = \dfrac{5}{6}x + \dfrac{1}{2}$. #
Solution
We expand the left-hand side using the distributive property:
$$\dfrac{5}{6}\left(x + \dfrac{3}{5}\right) = \dfrac{5}{6}x + \dfrac{5}{6} \times \dfrac{3}{5}$$$$\dfrac{5}{6}\left(x + \dfrac{3}{5}\right) = \dfrac{5}{6}x + \dfrac{1}{2}$$
So the equation becomes:
$$\dfrac{5}{6}x + \dfrac{1}{2} = \dfrac{5}{6}x + \dfrac{1}{2}$$Subtracting $\dfrac{5}{6}x$ from both sides:
$$\dfrac{1}{2} = \dfrac{1}{2}$$This is always true, whatever the value of $x$.
∴ The equation is an identity: it holds for every rational number $x$. For example, $x = 0$ gives $\dfrac{5}{6} \times \dfrac{3}{5} = \dfrac{1}{2}$ on both sides, and $x = 1$ gives $\dfrac{4}{3}$ on both sides. This is just the distributive property again.