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 $\frac{2}{3}$
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