Exercise 3.1
Question 1
A rational number is called a positive rational number if its numerator and denominator are either both positive integer or negative integers
$\frac{5}{8}, \frac{0}{5}, 7, \frac{-3}{-13}, \frac{-17}{-6}$ are positive integer rational numbers
Question 2
A rational number is called a negative rational number if its numerator and denominator are such that one of then is a positive integers and other is a negative integers
$\frac{-5}{7}, \frac{4}{-3},-6,-\frac{28}{5}$ are negative rational numbers
Question 3
i) we have
$\frac{+3}{-7}=\frac{3 x^{2}}{-7 \times 2}=\frac{3 \times 3}{-7 \times 3}=\frac{3 \times 4}{-7 \times 4}=\frac{3 \times 5}{-7 \times 5}$
$\frac{3}{-7}=\frac{6}{-14}=\frac{9}{-21}=\frac{12}{-28}=\frac{15}{-35}$
Thus , four rational numbers equivalent to $\frac{3}{-7}$are $\frac{6}{-14}, \frac{9}{-21}$,$\frac{12}{-28}, \frac{15}{-35}$
ii) $\frac{-5}{-9}=\frac{(-5) \times(-1)}{(-9) \times(-1)}=\frac{(-5) \times(-2)}{(-9) \times(-2)}=\frac{(-5) \times(-3)}{(-9) \times(-3)}=\frac{(-5) \times(-4)}{(-9) \times(-4)}$
$\frac{-5}{9}=\frac{5}{9}=\frac{10}{18}=\frac{15}{27}=\frac{20}{36} .$
Thes, four rational number equivalent $=\frac{5}{-9}$ are $\frac{5}{9}, \frac{10}{18}, \frac{15}{27}, \frac{20}{36}$
Question 4
Multiply the numerator and the denominator of each rational number by - 1
i) $\frac{4}{-9}=\frac{4 \times(-1)}{(-9) \times(-1)}=-\frac{4}{9} \Rightarrow \frac{4}{-9}=-\frac{4}{9}$
ii) $\frac{17}{-33}=\frac{17 \times(-1)}{(-33) \times(-1)}=\frac{17}{33} \Rightarrow \frac{17}{-33}=-\frac{17}{33}$
iii) $\frac{-15}{-38}=\frac{(-15) \times(-1)}{(-38) \times(-1)}=\frac{15}{38} \Rightarrow \frac{-15}{-38}=\frac{15}{38}$
Question 5
The next four number in the given pattern are
(i) $\frac{-5}{20}, \frac{-6}{24}, \frac{-7}{28}, \frac{-8}{32}$
(ii) $\frac{2 x(-1)}{-3} \times(-1)=\frac{-2}{3}$
$-\frac{10}{15}, \frac{-12}{18},\frac{-14}{21}, \frac{-16}{24}$
Question 6
i)Given rational number are $\frac{-3}{-7}$ and $\frac{15}{35}$
We have $(-3) \times(-5)=15$
and $(-7) \times(-5)=35$
$\frac{(-3) \times(-5)}{(-7) \times(-5)}=\frac{15}{35}$
∴ $\frac{-3}{-7}=\frac{15}{35}$
ii) $-\frac{6}{8}$ and $\frac{10}{-15}$
we have $(-6) \times(-15)=90$
and $\quad 8 \times 10=80 .$
As $90 \neq 80,-6 x-15 \neq 8 \times 10$
$\therefore \frac{-6}{8} \neq \frac{10}{-15}$
iii) $\frac{6}{-10}$ and $-\frac{12}{20}$
we have $6 \times 20=120$ and $-10 x-12=120$
As $\quad 20=120, \quad 6 \times 20=-10 x-12$
$\therefore \frac{6}{-10}=\frac{-12}{20}$
Question 7
i)Given rational numbers are
$\frac{-7}{21}, \frac{3}{9}$
We have $(-7) \times 9=-63,21 \times 3=63$
$\text { As. } \begin{aligned}-63 \neq 63 ;(-7 \times 9) \neq 21 \times 3 \\\therefore \frac{-7}{21} \neq \frac{3}{9} .\end{aligned}$
ii) $-\frac{16}{20}, \frac{20}{-25}$
We have $-16 \times-25=400 ; 20 \times 20=400$.
As $400=400,-16 \times -25=20 \times 20$
iii)$\frac{-3}{5}, \frac{-12}{20}$
We have $-3 \times 20=-60 \quad ; 5 \times-12=-60$
As -60=-60 , $-3 \times 20=-12 \times 5$
$\therefore\frac{-3}{5}=\frac{-12}{20}$
iv)$\frac{8}{-5}, \frac{-24}{15}$
We have $8 \times 15=120 ;-24 \times-5=120$
Both are equal i.e $8 \times 15=-24 \times-5$
$\therefore \frac{8}{-5}=\frac{-24}{15}$ are equal.
Question 8
i) $\frac{5}{4}=\frac{9}{16}=\frac{25}{?}=\frac{-15}{}$
$\frac{5}{4}=\frac{x}{16} \quad \frac{5}{4}=\frac{25}{y} \quad \frac{5}{4}=\frac{-15}{z}$
$16 \times 5=4 \times x \quad 5 x y=25 x 4 \quad 5 x z=-15 \times 4$
$\begin{array}{lll}x=\frac{16 \times 5}{4} & y=\frac{25 \times 4}{5} & z=\frac{-15 \times 4}{5} \\ x=20 & y=20 & z=-12\end{array}$
ii)$\frac{-3}{7}=\frac{9}{14}=\frac{9}{?}=\frac{-6}{?}$
$\frac{-3}{7}=\frac{x}{14} \quad \frac{-3}{7}=\frac{9}{y} \quad \frac{-3}{7}=-\frac{6}{z}$
$7 \times x=14 x-3 \quad-3 x y=9 \times 7 \quad-3 \times z=-6 \times 3$
$\begin{array}{rl}7 x=-42 & y=\frac{9 \times 3}{-3} \\\end{array} \quad z=\frac{-6 \times 7}{-3}$
x=-6 y=-21 z=14
Question 9
i) The given rational number is $\frac{-45}{30}$
Its denominator is positive
HCF of 45, 30 is 15
So divide its numerator and denominator by 9
$\therefore \frac{-45}{30}=\frac{(-45) \div 15}{30 \div 15}=\frac{-3}{2}$
Thus $-\frac{45}{30}=\frac{-3}{2}$ ,which is in standard form
ii) $\frac{16}{-36}$
Convert rational with positive denominator
i.e $\frac{16 \times(-1)}{(-36) \times(-1)}=\frac{-16}{36}$
Now denominator positive
HCF of 16 and 36 is 4
So divide its numerator and denominator by 4
$\therefore \frac{-16}{36}=\frac{-16 \div 4}{36 \div 4}=\frac{-4}{9}$
Thus $-\frac{16}{36}=\frac{-4}{9}$, Which is in standard form
iii) $\frac{-3}{-15}$
Convert above rational number with positive denominator
i.e $\frac{-3}{-15}=\frac{-3 x-1}{-15 x-1}=\frac{3}{15}$
HCF of 3, 15 is 3
So divide its numerator and denominator by 3
i.e $ \frac{3}{15}=\frac{3 \div 3}{15 \div 3}=\frac{1}{5}$
$\therefore \frac{3}{15}=\frac{1}{5}$ Which is in standard form
iv) $\frac{68}{-119}$
Convert above rational number with positive denominator
i.e $\frac{68}{-119}=\frac{68 \times(-1)}{-119 \times(-1)}=\frac{-68^{\circ}}{119}$
HCF of 68, 119 is 17
So divide its numerator and denominator by 17
i.e $\frac{-68}{119}=\frac{-68 \div 17}{119 \div 17} \cdot \frac{-4}{7}$
Thus $\frac{-68}{119}=\frac{-4}{7}$ Which is in standard form
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