NCERT Solutions for Class 9 Maths Chapter 3 — Exercise Set 3.4#
To place $\dfrac{p}{q}$ on the number line, divide each unit into $q$ equal parts and move $p$ parts from 0 — to the right if the number is positive, to the left if it is negative. Between any two rational numbers there are infinitely many rational numbers; one of them is their average $\dfrac{a + b}{2}$.
1. Represent the rational numbers $\dfrac{2}{3}$, $-\dfrac{5}{4}$ and $1\dfrac{1}{2}$ on a single number line. #
Solution
We first write each number as a fraction and see between which integers it lies:
- $\dfrac{2}{3}$ lies between 0 and 1. Divide the unit from 0 to 1 into 3 equal parts and take 2 parts to the right of 0.
- $-\dfrac{5}{4} = -1\dfrac{1}{4}$ lies between −2 and −1. Divide the unit from −1 to −2 into 4 equal parts and take 1 part to the left of −1.
- $1\dfrac{1}{2} = \dfrac{3}{2}$ lies between 1 and 2. Divide the unit from 1 to 2 into 2 equal parts and take 1 part to the right of 1.
2. Find three distinct rational numbers that lie strictly between $-\dfrac{1}{2}$ and $\dfrac{1}{4}$. #
Solution
We write both numbers with the same denominator. The LCM of 2 and 4 is 4, but that leaves too few numerators in between, so we use 8:
$$-\dfrac{1}{2} = -\dfrac{4}{8}$$$$\dfrac{1}{4} = \dfrac{2}{8}$$
The numerators between −4 and 2 are −3, −2, −1, 0 and 1, so the rational numbers $-\dfrac{3}{8}$, $-\dfrac{2}{8}$, $-\dfrac{1}{8}$, $0$ and $\dfrac{1}{8}$ all lie between them.
∴ Three such rational numbers are $-\dfrac{1}{4}$, $0$ and $\dfrac{1}{8}$ (any three from the list above are correct).
3. Simplify the expression: $\left(-\dfrac{1}{4}\right) + \left(\dfrac{5}{12}\right)$. #
Solution
The LCM of 4 and 12 is 12. We simplify $\left(-\dfrac{1}{4}\right) + \left(\dfrac{5}{12}\right)$:
$$= -\dfrac{3}{12} + \dfrac{5}{12}$$$$= \dfrac{2}{12}$$
$$= \dfrac{1}{6}$$
4. A tailor has $15\dfrac{3}{4}$ metres of fine silk. If making one kurta requires $2\dfrac{1}{4}$ metres of silk, exactly how many kurtas can he make? #
Solution
We change both mixed numbers into fractions:
$$15\dfrac{3}{4} = \dfrac{63}{4}$$$$2\dfrac{1}{4} = \dfrac{9}{4}$$
The number of kurtas is the total silk divided by the silk for one kurta:
Number of kurtas $= \dfrac{63}{4} \div \dfrac{9}{4}$
Number of kurtas $= \dfrac{63}{4} \times \dfrac{4}{9}$
Number of kurtas $= \dfrac{63}{9}$
Number of kurtas $= 7$
∴ The tailor can make exactly 7 kurtas, with no silk left over.
5. Find three rational numbers between 3.1415 and 3.1416. #
Solution
We add one more decimal place to both numbers:
$$3.1415 = 3.14150$$$$3.1416 = 3.14160$$
Any decimal from 3.14151 to 3.14159 lies between them.
∴ Three rational numbers are 3.14152, 3.14155 and 3.14158. Each is a terminating decimal, so each is rational; for example, $3.14155 = \dfrac{314155}{100000}$.
*6. Can you think of other way(s) to find a rational number between any two rational numbers? #
Solution
Yes. Some ways, with $a = \dfrac{1}{3}$ and $b = \dfrac{1}{2}$ as an example:
1. The average. $\dfrac{a + b}{2}$ always lies between $a$ and $b$:
$$\dfrac{1}{2}\left(\dfrac{1}{3} + \dfrac{1}{2}\right) = \dfrac{1}{2} \times \dfrac{5}{6} = \dfrac{5}{12}$$2. Equal denominators. Write both numbers with a large common denominator and pick the numerators in between:
$$\dfrac{1}{3} = \dfrac{4}{12}$$$$\dfrac{1}{2} = \dfrac{6}{12}$$
So $\dfrac{5}{12}$ lies between them. Using $\dfrac{8}{24}$ and $\dfrac{12}{24}$ gives three: $\dfrac{9}{24}$, $\dfrac{10}{24}$, $\dfrac{11}{24}$.
3. Adding numerators and denominators (the mediant). For positive denominators, $\dfrac{a_1 + a_2}{b_1 + b_2}$ lies between $\dfrac{a_1}{b_1}$ and $\dfrac{a_2}{b_2}$:
$$\dfrac{1 + 1}{3 + 2} = \dfrac{2}{5}$$and $\dfrac{1}{3} < \dfrac{2}{5} < \dfrac{1}{2}$ because $0.333\ldots < 0.4 < 0.5$.
4. Decimals. Write both as decimals, $0.333\ldots$ and $0.5$, and choose a decimal in between, such as $0.4 = \dfrac{2}{5}$ or $0.45 = \dfrac{9}{20}$.