Exercise 3.1
Question 1
A rational number is called a postive 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
Exercise 3.2
Question 1
Draw a number line and divide each unit length into 8 equal parts
(daigram to be added)
Question 2
$ \quad A P=P Q=Q B \quad ; T R=R S=U S$
$\begin{array}{ll}P=\frac{7}{3} & R=-\frac{4}{3} \\ Q=\frac{8}{3} & S=\frac{-5}{3}\end{array}$
Question 3
i) true (daigram to be added)
ii) True (daigram to be added)
iii) false (daigram to be added)
Question 4
i) $0>\frac{-4}{7}$, since $\frac{-y}{7}$ lies left to zero in number line
ii) $\frac{5}{-9}=\frac{5 x-1}{-9 x-1}=\frac{-5}{9}, \frac{3}{7}$
$\frac{3}{7}>\frac{-5}{9}$ because in number line $\frac{-5}{9}$ lies left to zero and $\frac{3}{7}$ lies right to zero
iii) $\frac{-9}{-5}=\frac{-9 x-1}{-5 x-1}=\frac{9}{5} ; 0$
$\frac{9}{5}>0$ since $\frac{9}{5}$ lies right to zeso in numler line
$\frac{21}{23}>\frac{-7}{5}$ Since negative number is always less than positive number
Question 5
i) $-4 \times 2=-28 ;-5 \times 5=-25$
As $-25>-28$
So $-5 \times 5>-4 \times 7$
i.e $-\frac{4}{5}<\frac{-5}{7} .$
(ii)
$\begin{aligned}-8 \times 4=-32 & ;-7 \times 5=-35 \\ \text { As }-32>-35 \\ \text { i.e }-8 \times 4>-7 \times 5 \\ \therefore-\frac{8}{5}>\frac{-7}{4} \end{aligned}$
iii)
$\begin{aligned}-7 x-48=& 336 ; 42 \times 8=336 \\ \text { As } 336=336 \\ \therefore-\frac{7}{8} &=\frac{42}{-48} \end{aligned}$
iv)
$\begin{aligned} \frac{1}{-3} ;-\frac{1}{4} & \\ 1 \times 4=4 & ;-3 \times-1=3 \\ \text { As } & 4>3 \\ & i.e \frac{1}{-3}>\frac{-1}{4} \end{aligned}$
$\begin{aligned}v)-\frac{3}{8} ; & \frac{-2}{7} \\-3 \times 7 &=-21 ;-2 \times 8=-16 \\ & As-16>-21 \\ & i \cdot e-\frac{3}{8}<-\frac{2}{7} . \end{aligned}$
vi)
$\begin{aligned}\frac{-4}{3} ; &-\frac{-3}{2} \\-4 \times 2 &=-8 \quad ;-3 \times 3=-9 \\ & \text { As }-8>-9 \\ & \therefore\frac{-4}{3}>\frac{-3}{2} \end{aligned}$
Question 6
i) Writing each number with positive denominator already there are positive
So the given rational number are $-\frac{3}{7},-\frac{3}{2},-\frac{3}{4}$
LCM of their denominator i.e 2,4,7 =28
To write the rational numbers with this LCM 28 as their denominator, we have;
$-\frac{3}{7}=\frac{-3 \times 4}{2 \times 4}=\frac{-12}{28}$
$-\frac{3}{2}=\frac{-3 \times 14}{2 \times 14}=\frac{-42}{28}$
$\frac{-3}{4}=\frac{-3 \times 7}{4 \times 7}=\frac{-21}{28}$
$-42<-21<-12$
Hence the given rational number in ascending order are $\frac{-3}{2}, \frac{-3}{4}, \frac{-3}{7} .$
ii) Write each number with positive denominator
$\frac{5}{-12}=\frac{5 \times(-1)}{-12 \times-1}=\frac{-5}{12}$
$\frac{9}{-24}=\frac{9 \times(-1)}{-24 \times-1}=\frac{-9}{24}$
LCM of their denominator i.e 4,12,16,24 is 48
To write the rational numbers with this LCM i.e 48 as their denominator , we have
$\frac{-3}{4}=\frac{-3 \times 12}{4 \times 12}=\frac{-36}{48}$
$\frac{-5}{12}=\frac{-5 \times 4}{12 \times 4}=\frac{=20}{48}$
$\frac{-9}{24}=\frac{-9 \times 2}{24 \times 2}=\frac{-18}{48}$
$\frac{-7}{16}=\frac{-7 \times 3}{16 \times 3}=\frac{-21}{48}$
As $-36<-21<-20<-18$
∴ Ascending ordes is $-\frac{3}{4}<-\frac{7}{16}<\frac{-5}{12}<\frac{-9}{24}$.
Question 7
i) Write each number with positive denominator
$\frac{17}{-30}=\frac{17 \times(-1)}{-30 \times-1}=\frac{-17}{30}$
LCM of their denominators 10, 20,15,30 is 60
To write rational number with LCM i.e 60 as their denominator
$\frac{-3}{10}=\frac{-3 \times 6}{10 \times 6}=\frac{-18}{60}$
$\frac{-11}{20}=\frac{-11 \times 3}{20 \times 3}=\frac{-33}{60}$
$\frac{-7}{15}=\frac{-7 \times 4}{15 \times 4}=\frac{-28}{60}$
$\frac{-17}{30}=\frac{-17 \times 2}{30 \times 2}=\frac{-34}{60}$
As $-18>-28>-33>-3 y$
Descending ordes is $\frac{-3}{10}>\frac{-7}{15}>\frac{-11}{20}>\frac{-17}{30}$.
ii) Write each number with positive denominator
i.e $\frac{2}{-5}=\frac{2 \times-1}{5 \times-1}=\frac{-2}{5} ; \frac{19}{-30}=\frac{19 \times-1}{-30 \times-1}=\frac{-19}{30}$
LCM of their denominator 5,10 ,15,30 is 30
To write rational number with LCM i.e 30 as their denominators
$\frac{-7}{10}=\frac{-3 \times 3}{10 \times 3}=\frac{-21}{30}$
ii) Given rational number $\frac{-2}{3}$ and $\frac{-1}{3}$ have same denominator
To insert 5 rational numer , multiply both numerator and denominator of each number by (5+1) i.e 6
We have $\frac{-2}{3}=\frac{-2 \times 6}{3 \times 6}=\frac{-12}{18} ; \frac{-1}{3}=\frac{-1 \times 6}{3 \times 6}=\frac{-6}{18}$.
$\because-12<-11<-10<-9<-8<-7<-6$
$\frac{-12}{18}<\frac{-11}{18}<\frac{-10}{18}<\frac{-9}{18}<\frac{-8}{18}<\frac{-7}{18}<\frac{-6}{18}$
$\therefore 5$ rational numbers between $\frac{-2}{3}$ and $\frac{-1}{3}$ are $\frac{-11}{18},-\frac{16}{18}, \frac{-9}{18},-\frac{8}{18},\frac{-7}{18} \quad $ i.e
$\frac{-11}{18}, \frac{-5}{9}, \frac{-1}{2}, \frac{-4}{9}, \frac{-7}{18}$
Question 8
Given rational number $\frac{-4}{5}$ and $-\frac{2}{3}$ have different denominator.
LCM of denominator 5 and 3 is 15
To convert the rational number with same denominator
We have
$\frac{-4}{5} = \frac{-4 \times 3}{5 \times 3}=\frac{-12}{15} ; \frac{-2}{3}=\frac{-2 \times 5}{3 \times 5 .}=\frac{-10}{15}$
We have only one integer between -12 and -10 i.e -11. Thus writting the rational numbers with denominator 15 is not sufficient
To insert 5 rational numbers, multiply both numerator and denominator by (5+1) i.e 6
$\frac{-12}{15}=\frac{-12 \times 6}{15 \times 6}=\frac{-72}{96}$ and $\frac{-10}{15}=\frac{-10 \times 6}{15 \times 6}=\frac{-60}{90}$.
$\begin{aligned} \because&-72<-71<-70<-69<-68<-67<-66<-65<-64 \\ &<-63<-62<-61<-66 \end{aligned}$
We can choose any 5 rational number from these
i.e $-\frac{71}{90},-\frac{70}{90},-\frac{68}{90},-\frac{67}{90},-\frac{65}{90} .$
ii) Given rational number $\frac{1}{2}$ and $\frac{2}{2}$ have different denominator
LCM of denominator 2 and 3 = 6
To convert these rational numbers with same denominator
We have
$-\frac{1}{2}=\frac{-1 \times 3}{2 \times 3}=\frac{-3}{6}$ and $\frac{2}{3}=\frac{2\times 2 }{3\times 2}=\frac{4}{6}$
As $-3<-2<-1<0<1<2<3<4$
We can choose any 5 rational number
i.e $-\frac{2}{6}, \frac{-1}{6}, \frac{1}{6}, \frac{2}{6}, \frac{3}{6} .$
i.e $-\frac{1}{3},-\frac{1}{6}, \frac{1}{6}, \frac{1}{3}, \frac{1}{2}$ are 5 rational number between $-\frac{1}{2}$ and $\frac{2}{3}$
Exercise 3.3
Question 1
i)First expree $\frac{-5}{11}$ as a rational numbers with positive denominator
It is already with positive denominator
Sum = $\frac{3}{11}+\frac{-5}{11}=\frac{3+(-5)}{11}=\frac{-2}{11}$
ii) We have $\frac{5}{-9}=\frac{5 \times(-1)}{-9 \times(-1)}=\frac{-5}{9}$.
Sum $=\frac{4}{9}+\frac{-5}{9}=\frac{4+(-5)}{9}=\frac{4-5}{9}=-\frac{1}{9} .$
iii) We have $\frac{5}{-7}=\frac{5 \times(-1)}{-7 x-1}=\frac{-5}{7}$
$\frac{-2}{-7}=\frac{-2 \times-1}{-7 \times-1}=\frac{2}{7}$
$\Sum=\frac{-5}{7}+\frac{2}{7}=\frac{-5+2}{7}=\frac{-3}{7}$
iv) $\frac{-2}{5}, \frac{3}{4}$
Given rational number have different denominators
LCM of their denominator 5 and 4 is 20
To write the rational number with this LCM i.e 20as their denominator , we have
$\frac{-2}{5}=\frac{-2 \times 4}{5 \times 4}=-\frac{8}{20}$ and $\frac{3}{4}=\frac{3 \times 5}{4 \times 5}=\frac{15}{20}$.
∴ Sum =$\frac{-8}{20}+\frac{15}{20}=\frac{-8+15}{20}=\frac{7}{20}$
Question 2
i)Given rational numbers have different denominators
LCM of denominators 4 and 8 is 8
In order to have their denominators as LCM i.e 8 we have
$+\frac{27}{-4}=\frac{27 \times(-1)}{-4 \times(-1)}=-\frac{27}{4}$
$\frac{-27}{4}+\frac{-15}{8}=\frac{-27 \times 2}{4 \times 2}+\frac{-15}{8}$
$=\frac{-54}{8}+\frac{-15}{8}$
$=\frac{-54-15}{8}=\frac{-69}{8}$
ii) LCM of denominators 8 and 18 is 72
In order to have their denominators as LCM i.e 72 we have
$\begin{aligned} \frac{-1}{18}+\frac{-3}{8} &=\frac{-1 \times 4}{18 \times 4}+\frac{-3 \times 9}{8 \times 9} \\ &=\frac{-4}{72}+\frac{-27}{72} \end{aligned}$
$=\frac{-4+(-27)}{72}$
$=\frac{-31}{72}$
iii)
$\begin{aligned} &-3 \frac{1}{6}+2 \frac{3}{8} \\ &-3 \frac{1}{6}=\frac{-19}{6} ; 2 \frac{3}{8}=\frac{19}{8} \end{aligned}$
LCM of 6 and 8 is 24
In order to have their denominator is their LCM i.e 24
We have
$-3 \frac{1}{6}+2 \frac{3}{8}=\frac{-19}{6}+\frac{19}{8}=\frac{-19 \times 4}{6 \times 4}+\frac{19 \times 3}{8 \times 3}$
$=\frac{-76}{24}+\frac{57}{24}$
$=\frac{-76+57}{24}$
$=\frac{-19}{24} .$
iv) $-2 \frac{4}{5}+4 \frac{3}{10}$
$-2 \frac{4}{5}=\frac{-14}{5}$ and $4 \frac{3}{10}=\frac{43}{10}$
LCM of 5 and 10 is 10
In order to have their denominators is their LCM i.e 10
We have
$-2 \frac{4}{5}+4 \frac{3}{10}=\frac{-14}{5}+\frac{43}{10}=\frac{-14 \times 2}{5 \times 2}+\frac{43 \times 1}{10 \times 1}$
$=\frac{-28}{10}+\frac{43}{10}$
$=\frac{-28+43}{10}$
$=\frac{15}{10}=\frac{3}{2} .$
Question 3
i) $\frac{4}{13}-\frac{-6}{13}=\frac{4}{13}+$ additive inverse of $\left(\frac{-6}{13}\right)$
$=\frac{4}{13}+\frac{6}{13}=\frac{4+6}{13}=\frac{10}{13} .$
ii) $\frac{-2}{3}-\frac{-1}{2}=\frac{2}{3}+$ additive invesse of $\left(\frac{-1}{2}\right)$
$=\frac{-2}{3}+\frac{1}{2} \quad(\mathrm{LCM}$ of 3 and 2 is 6 $)$
$=\frac{-2 \times 2+1 \times 3}{6}$
$=\frac{-4+3}{6}=\frac{-1}{6}$
iii) $\frac{-2}{3}-\frac{5}{9}=\frac{-2}{3}+$ additive invesse of $\left(\frac{+5}{9}\right)$
$=\frac{-2}{3}+\frac{-5}{9}$ (LCM of 3,9 is 9)
$=\frac{-2 \times 3+(-5 \times 1)}{9}$
$=\frac{-6-5}{9}=\frac{-11}{9} .$
Question 4
i) $\frac{5}{63}-\left(\frac{-6}{21}\right)$
$=\frac{5}{63}+\frac{6}{21}$
LCM of 21,63 is 63
$=\frac{5 \times 1+6 \times 3}{63}=\frac{5+18}{63}=\frac{23}{63}$
ii) $\frac{-6}{3}-\left(\frac{-7}{15}\right)$
$=\frac{-6}{3}+\frac{7}{15}$
LCM of 3,15 is 15
$=\frac{-6 \times 5+7 \times 1}{15}=\frac{-30+7}{15}=\frac{-23}{15}$
iii) $3 \frac{1}{8}-\left(-1 \frac{5}{6}\right)$
$\begin{aligned}=\frac{25}{8}+\frac{11}{6} \\ & \mathrm{LCM} \text { of } 8,6 \text { is } 24 \end{aligned}$
$=\frac{25 \times 3+1.1 \times 4}{24}$
$=\frac{75+44}{24}$
$=\frac{119}{24}$
Question 5
Let the other rational number be 'x'
Given Sum = $\frac{2}{5}$
1.e $x+\frac{-4}{9}=\frac{2}{5}$
$x=\frac{2}{5}-\left(-\frac{4}{9}\right)$
$x=\frac{2}{5}+\frac{4}{9}$
LCM of 5,9 is 45
$x=\frac{2 \times 9+4 \times 5}{45}$
$x=\frac{18+20}{45}$
$x=\frac{38}{45}$
∴ The other rational number is $\frac{38}{45}$
Question 6
Let the rational added to $\frac{-5}{2}$ is " $x^{\prime}$
Given the result of sum is $-frac{-7}{8} .$
i.e $x+\left(\frac{-5}{12}\right)=\frac{-7}{8}$
$x=\frac{-7}{8}-\left(\frac{-5}{12}\right)$
$x=\frac{-7}{8}+\frac{5}{12}$
$\mathrm{Lcm}$ of 8,12 is 24
$x=\frac{-7 \times 3+5 \times 2}{24}$
$x=\frac{-21+10}{24}$
$x=\frac{-11}{24}$
∴ Therefore $\frac{-11}{24}$ is to be added to $\frac{-5}{12}$ to get $\frac{-7}{8}$
Question 7
Let the rational number to be substracted from $\frac{-2}{3}$ be $x^{\prime}$
The result is $\frac{-5}{6} .$
i.e $x-\left(\frac{-2}{3}\right)=-\frac{5}{6}$
$x=\frac{=5}{6}+\left(-\frac{2}{3}\right) .$
LCM of 3 and 6 is 6
$x=\frac{-5 \times 1+-2 \times 2}{6}$
$x=\frac{-5+(-4)}{6}$
$x=\frac{-5-4}{6}=\frac{-9}{6}$
$x=-\frac{3}{2}$
∴ $\frac{-3}{2}$ should be substracted from $\frac{-2}{3}$ to get $\frac{-5}{6}$.
Question 8
i) $\frac{2}{3} \times-\frac{7}{8}=\frac{2 x-7}{3 \times 8}=\frac{-7}{12}$
ii) $\frac{-6}{7} \times \frac{5}{7}=\frac{-6 \times 5}{7 \times 7}=-\frac{30}{49}$
iii) $\frac{-2}{9} \times(-5)=\frac{-2 \times-5}{9}=\frac{10}{9}$.
iv) $\frac{-5}{11} \times \frac{11}{-5}=+\frac{8 \times 11}{4 \times+5}=1$
v) $\frac{8}{35} \times \frac{21}{-32}=\frac{18 \times 21}{35 \times-32}=\frac{3}{-20}$
Vi) $\frac{-105}{128} \times\left(-1 \frac{29}{35}\right)=-\frac{105}{128} \times\frac{-64}{35}$
$=\frac{3}{2}$
Question 9
i) $(-6) \div \frac{2}{5}$
$=\frac{(-6)}{(2 / 5)}=\frac{-6 \times 5}{21}=-15$
ii) $\frac{-1}{10} \div \frac{-8}{5}$
$=\frac{(-1 / 10)}{(-8 / 5)}=+\frac{1}{10} \times \frac{5}{+8}$
$=\frac{1}{2 \times 8}=\frac{1}{16}$
iii) $\frac{-65}{14} \div \frac{13}{-7}$
$\begin{aligned}=\frac{-65 / 14}{13 /-7} &=\frac{-65}{14} \times \frac{-7}{13} \\ &=\frac{5}{2} \end{aligned}$
iv)
$\begin{aligned} &(-6) \div 3 \frac{3}{5} \\ &=(-6) \div \frac{18}{5} \end{aligned}$
$=\frac{-6}{(18 / 5)}=\frac{-6 \times 5}{18}=\frac{-5}{3} .$
V) $\frac{-48}{49} \div \frac{72}{-35}$
$=-\frac{48 / 49}{72 /-35}$
$=\frac{-48}{49} \times \frac{-35}{72}$ = $\frac{10}{21}$
vi) $3 \frac{1}{7} \div\left(\frac{-33}{34}\right)$
$=\frac{22}{7} \div\left(\frac{-33}{34}\right)$
$=\frac{22 / 7}{-33 / 34}=\frac{22}{7} \times \frac{34}{-33}$
$=\frac{68}{-21}$
Question 10
Let the other rational number be 'x '
product of them is $\frac{18}{35}$
i.e $\frac{-2}{5} \times x=\frac{18}{35}$
x=$\frac{18 \times 5}{35 \times -2}$
$x=\frac{9}{-7}=\frac{-9}{7} .$
∴ The other rational number is $\frac{-9}{7}$
Question 11
i) $\left(\frac{13}{21} \div \frac{39}{42}\right) \times\left(\frac{-3}{5}\right)$
$=\frac{13 / 21}{39 / 42} \times \frac{-3}{5}$
$=\frac{13}{21} \times \frac{42}{39} \times \frac{-3}{5}=\frac{-2}{5}$
(ii) $\left(-5 \frac{5}{21}\right) \div\left(\frac{7}{11} \times \frac{5}{12}\right)$
$=\frac{-110}{21} \div \frac{35}{132}$
$=\frac{-110 / 21}{35 / 132}$
$\frac{-110}{21} \times \frac{132}{35}$
$=\frac{-22 \times 44}{7 \times 7}=\frac{-768}{49}$
Question 12
i) $\frac{3}{13} \div \frac{-4}{65}$
$=\frac{(3 / 13)}{(-4 / 65)}$ = $\frac{3}{13} \times \frac{65}{-4}$
$=\frac{15}{-4}=\frac{-15}{4}$
∴ The reciprocal is $\frac{-4}{15} .$
ii)$\left(-8 \times \frac{12}{15}\right)-\left(-3 \times \frac{2}{9}\right)$
$=(-4)-\left(-frac{-2}{3}\right)$
$\begin{aligned}=-4+\frac{2}{3} &=\frac{-12+2}{3} \\ &=\frac{-10}{3} . \end{aligned}$
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