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Chapter 9: Rational Numbers

Exercise 9.1

Q1:

List 9 natural numbers between:

(i) -1 and 0

(ii) -2 and -1

(iii) \(-\frac{4}{5}\; and \;-\frac{2}{3}\)

(iv) \(-\frac{1}{2}\; and \; \frac{2}{3}\)

Sol:

(i) -1 and 0

Let us write -1 and 0 as rational number with denominator 10.   (as we need 9 number between the  numbers given so 9+1=10).

\(\Rightarrow -1=-\frac{10}{10}\) and \( 0=\frac{0}{10}\)

 

\(\ therefore \frac{-10}{10}<\frac{-9}{10}<\frac{-8}{10}<\frac{-7}{10}<\frac{-6}{10}<\frac{-5}{10}<\frac{-4}{10}<\frac{-3}{10}<\frac{-2}{10}<\frac{-1}{10}<0\)

 

\(\Rightarrow -1<\frac{-9}{10}<\frac{-4}{5}<\frac{-7}{10}<\frac{-3}{5}<\frac{-1}{2}<\frac{-2}{5}<\frac{-3}{10}<\frac{-1}{5}<\frac{-1}{10}<0\)

 

Therefore 9 rational numbers between -1 and 0 would be

\(\frac{-9}{10},\frac{-4}{5},\frac{-7}{10},\frac{-3}{5},\frac{-1}{2},\frac{-2}{5},\frac{-3}{10},\frac{-1}{5},\frac{-1}{10}\)

 

(ii) -2 and -1

Let us write -2 and -1 as rational number with denominator 10.   (as we need 9 number between the  numbers given so 9+1=10).

\(\Rightarrow -2=\frac{-20}{10}\) and \( -1=\frac{-10}{10}\)

 

\(\ therefore \frac{-20}{10}<\frac{-19}{10}<\frac{-18}{10}<\frac{-17}{10}<\frac{-16}{10}<\frac{-15}{10}<\frac{-14}{10}<\frac{-13}{10}<\frac{-12}{10}<\frac{-11}{10}<\frac{-10}{10}\)

 

\(\Rightarrow -2<\frac{-19}{10}<\frac{-9}{5}<\frac{-17}{10}<\frac{-8}{5}<\frac{-3}{2}<\frac{-7}{5}<\frac{-13}{10}<\frac{-6}{5}<\frac{-11}{10}<-1\)

 

Therefor 9 rational numbers between -2 and -1 would be,

\(\frac{-19}{10},\frac{-9}{5},\frac{-17}{10},\frac{-8}{5},\frac{-3}{2},\frac{-7}{5},\frac{-13}{10},\frac{-6}{5},\frac{-11}{10}\) .

 

(iii) \(-\frac{4}{5}\; and \;-\frac{2}{3}\)

Let us write \(-\frac{4}{5}\; and \;-\frac{2}{3}\) as rational number with the same denominator.

\(\Rightarrow \frac{-4}{5} = \frac{-72}{90}\; and \;\frac{-2}{3}=\frac{-60}{90}\).

 

\(\ therefore \frac{-72}{90}<\frac{-71}{90}<\frac{-70}{90}<\frac{-69}{90}<\frac{-68}{90}<\frac{-67}{90}<\frac{-66}{90}<\frac{-65}{90}<\frac{-64}{90}<\frac{-63}{90}<\frac{-60}{90}\)

 

\(\Rightarrow \frac{-4}{5}<\frac{-71}{90}<\frac{-7}{9}<\frac{-23}{30}<\frac{-34}{45}<\frac{-67}{90}<\frac{-11}{15}<\frac{-13}{18}<\frac{-32}{45}<\frac{-7}{10}<\frac{-2}{3}\)

 

Therefor 9 rational numbers between  \(-\frac{4}{5}\; and \;-\frac{2}{3}\) are

\(\frac{-71}{90},\frac{-7}{9},\frac{-23}{30},\frac{-34}{45},\frac{-67}{90},\frac{-11}{15},\frac{-13}{18},\frac{-32}{45},\frac{-7}{10}\)

 

(iv) \(\frac{-1}{2}\; and \; \frac{2}{3}\)

Let us write \(\frac{-1}{2}\; and \; \frac{2}{3}\) as rational numbers having same denominator.

\(\frac{-1}{2}=\frac{-6}{12}\; and \; \frac{2}{3}=\frac{8}{12}\)

 

\(\ therefore \frac{-6}{12}< \frac{-5}{12}< \frac{-4}{12}<\frac{-3}{12}<\frac{-2}{12}<\frac{-1}{12}<\frac{0}{12}<\frac{1}{12}<\frac{2}{12}<\frac{3}{12}<\frac{8}{12}\)

 

\(\Rightarrow \frac{-1}{2}< \frac{-5}{12}< \frac{-1}{3}<\frac{-1}{4}<\frac{-1}{6}<\frac{-1}{12}<0<\frac{1}{12}<\frac{1}{6}<\frac{1}{4}<\frac{2}{3}\)

 

\(\frac{-5}{12}, \frac{-1}{3},\frac{-1}{4},\frac{-1}{6},\frac{-1}{12},0,\frac{1}{12},\frac{1}{6},\frac{1}{4}\)

 

Q2: With the pattern given below, write down four more rational number, which comes in continuation:

(i) \(\frac{-3}{5},\frac{-6}{10},\frac{-9}{15},\frac{-12}{20}, ……..\)

(ii) \(\frac{-1}{4},\frac{-2}{8},\frac{-3}{12}, ……..\)

(iii) \(\frac{-1}{6},\frac{-2}{12},\frac{-3}{18},\frac{-4}{24}, ……..\)

(iv) \(\frac{-2}{3},\frac{-4}{6},\frac{-6}{9}, ……..\)

 

Sol:

(i) \(\frac{-3}{5},\frac{-6}{10},\frac{-9}{15},\frac{-12}{20}, ……..\)

\(\Rightarrow \frac{-3}{5},\frac{-3\times 2}{5\times 2},\frac{-3\times 3}{5\times 3},\frac{-3\times 4}{5\times 4},………\)

Therefore, the next four rational numbers following the pattern would be

\(\frac{-3\times 5}{5\times 5},\frac{-3\times 6}{5\times 6},\frac{-3\times 7}{5\times 7},\frac{-3\times 8}{5\times 8}=\frac{-15}{25},\frac{-18}{30},\frac{-21}{35},\frac{-24}{40}\)

 

(ii) \(\frac{-1}{4},\frac{-2}{8},\frac{-3}{12}, ……..\)

\( \Rightarrow \frac{-1\times 1}{4\times 1},\frac{-1\times 2}{4\times 2},\frac{-1\times 3}{4\times 3}, ……..\)

Therefore, the next four rational numbers following the pattern would be

\(\frac{-1\times 4}{4\times 4},\frac{-1\times 5}{4\times 5},\frac{-1\times 6}{4\times 6},\frac{-1\times 7}{4\times 7}\) \(\frac{-4}{16},\frac{-5}{20},\frac{- 6}{24},\frac{- 7}{28}\)

 

(iii) \(\frac{-1}{6},\frac{-2}{12},\frac{-3}{18},\frac{-4}{24}, ……..\)

\(\Rightarrow \frac{-1\times 1}{6\times 1},\frac{-1\times 2}{6\times 2},\frac{-1\times 3}{6\times 3},\frac{-1\times 4}{6\times 4}, ……..\)

Therefore, the next four rational numbers following the pattern would be

\(\frac{-1\times 5}{6\times 5},\frac{-1\times 6}{6\times 6},\frac{-1\times 7}{6\times 7},\frac{-1\times 8}{6\times 8}=\frac{- 5}{30},\frac{- 6}{36},\frac{-7}{42},\frac{-8}{48}\)

 

(iv) \(\frac{-2}{3},\frac{-4}{6},\frac{-6}{9}, ……..\)

\(\Rightarrow \frac{-2\times 1}{3 \times 1},\frac{-2\times 2}{3 \times 2},\frac{-2\times 3}{3 \times 3}, ……..\)

Therefore, the next four rational numbers following the pattern would be

\(\frac{-2\times 4}{3 \times 4},\frac{-2\times 5}{3 \times 5},\frac{-2\times 6}{3 \times 6}, \frac{-2\times 7}{3 \times 7}=\frac{-8}{12},\frac{-10}{15},\frac{-12}{18}, \frac{-14}{21}\)

 

Q3: Write five rational number equivalent to the given fraction:

(i) \(\frac{-2}{7}\)

(ii) \(\frac{-5}{3}\)

(iii) \(\frac{4}{9}\)

Sol:

(i) \(\frac{-2}{7}\)

\(\frac{-2\times 2}{7\times 2}=\frac{-4}{14},\frac{-2\times 3}{7\times 3}=\frac{-6}{21} ,\frac{-2\times 4}{7\times 4}=\frac{-8}{28} ,\frac{-2\times 5}{7\times 5}=\frac{-10}{35} ,\frac{-2\times 6}{7\times 6}=\frac{-12}{42}\)

Thus five rational number equivalent to \(\frac{-2}{7}\) are

\(\frac{-4}{14},\frac{-6}{21}, \frac{-8}{28}, \frac{-10}{35} ,\frac{-12}{42}\)

 

(ii) \(\frac{-5}{3}\)

\(\frac{-5\times 2}{3\times 2}=\frac{-10}{6},\frac{-5\times 3}{3\times 3}=\frac{-15}{9},\frac{-5\times 4}{3\times 4}=\frac{-20}{12},\frac{-5\times 5}{3\times 5}=\frac{-25}{15},\frac{-5\times 6}{3\times 6}=\frac{-30}{18}\)

Thus five rational number equivalent to \(\frac{-5}{3}\) are

\(\frac{-10}{6},\frac{-15}{9},\frac{-20}{12},\frac{-25}{15},\frac{-30}{18},\)

 

(iii) \(\frac{4}{9}\)

\(\frac{4\times 2}{9\times 2}=\frac{8}{18},\frac{4\times 3}{9\times 3}=\frac{12}{27},\frac{4\times 4}{9\times 4}=\frac{16}{36},\frac{4\times 5}{9\times 5}=\frac{20}{45},\frac{4\times 6}{9\times 6}=\frac{24}{54}\)

Thus five rational number equivalent to \(\frac{-5}{3}\) are

\(\frac{8}{18},\frac{12}{27},\frac{16}{36},\frac{20}{45},\frac{24}{54}\)

 

Q4: Draw the number line and represent the following rational numbers on it:

(i) \(\frac{3}{4}\)

(ii) \(\frac{3}{4}\)

(iii) \(\frac{3}{4}\)

(iv) \(\frac{3}{4}\)

 

Sol:

(i) \(\frac{3}{4}\)

(ii)  \(\frac{3}{4}\)

(iii) \(\frac{3}{4}\)

(iv) \(\frac{3}{4}\)

 

Q5: The point P,Q,R,S,T,U,A and B on the number line are such that \(TR=RS=SU\) and \(AP=PQ=QB\). Name the rational numbers represented by P,Q,R and S.

Sol:

Each part which is between the two numbers is divided into 3 parts.

Therefore   \(A=\frac{6}{3},P=\frac{7}{3},Q=\frac{8}{3}\;and\; B=\frac{9}{3}\)

Similarly  \(T=\frac{-3}{3}, R=\frac{-4}{3},S=\frac{-5}{3}\; and\; U=\frac{-6}{3}\)

Thus, the rational numbers represented \(P,Q,R\; and\; S\; are\; \frac{7}{3},\frac{8}{3},\frac{-4}{3},\frac{-5}{3}\) respectively.

 

Q6: Which of the following pairs represent the same rational numbers:

(i) \(\frac{-7}{21}\; and \;\frac{3}{9}\)

(ii) \(\frac{-16}{20}\; and \;\frac{20}{-25}\)

(iii) \(\frac{-2}{-3}\; and \;\frac{2}{3}\)

(iv) \(\frac{-3}{5}\; and \;\frac{-12}{20}\)

(v) \(\frac{-8}{5}\; and \;\frac{-24}{15}\)

(vi) \(\frac{1}{3}\; and \;\frac{-1}{9}\)

(vii) \(\frac{-5}{-9}\; and \;\frac{5}{-9}\)

Sol:

(i)  \(\frac{-7}{21}\; and \;\frac{3}{9}\)

\(\Rightarrow \frac{-7}{21}=\frac{-1}{3}\; and \; \frac{3}{9}=\frac{1}{3}\) \(\ because  \frac{-1}{3}\neq \frac{1}{3}\) \(∴ \frac{-7}{21}\neq \frac{3}{9}\)

 

(ii) \(\frac{-16}{20}\; and \;\frac{20}{-25}\)

\(\Rightarrow \frac{-16}{20}=\frac{-4}{5}\; and \; \frac{20}{-15}=\frac{-4}{5}\) \(\ because \frac{-4}{5}= \frac{-4}{5}\) \(∴ \frac{-16}{20}= \frac{20}{-25}\)

(iii) \(\frac{-2}{-3}\; and \;\frac{2}{3}\)

\(\Rightarrow \frac{-2}{-3}=\frac{2}{3}\; and \;\frac{2}{3}\) \(\ because \frac{2}{3}=\frac{2}{3}\) \(∴ \frac{-2}{-3}=\frac{2}{3}\)

 

(iv) \(\frac{-3}{5}\; and \;\frac{-12}{20}\)

\(\Rightarrow \frac{-3}{5}\; and \;\frac{-12}{20}=\frac{-3}{5}\) \(\ because \frac{-3}{5}= \frac{-3}{5}\) \(∴ \frac{-3}{5}= \frac{-12}{20}\)

 

(v) \(\frac{-8}{5}\; and \;\frac{-24}{15}\)

\(\Rightarrow \frac{-8}{5}\; and \;\frac{-24}{15}=\frac{-8}{5}\) \(\ because \frac{-8}{5}=\frac{-8}{5}\) \(∴ \frac{-8}{5}=\frac{-24}{15}\)

 

(vi) \(\frac{1}{3}\; and \;\frac{-1}{9}\)

\(\Rightarrow \frac{1}{3}\; and \;\frac{-1}{9}\) \(\ because \frac{1}{3} \neq \frac{-1}{9}\) \(∴ \frac{1}{3} \neq \frac{-1}{9}\)

 

(vii) \(\frac{-5}{-9}\; and \;\frac{5}{-9}\)

\(\Rightarrow \frac{-5}{-9}= \frac{5}{9} \; and \;\frac{5}{-9}\) \(\ because \frac{5}{9} \neq \frac{-5}{9}\) \(∴ \frac{-5}{-9} \neq \frac{5}{-9}\)

 

Q7: Rewrite the following rational numbers in the simplest form:

(i) \(\frac{-8}{6}\)

(ii) \(\frac{25}{45}\)

(iii) \(\frac{-44}{72}\)

(iv) \(\frac{-8}{10}\)

Sol:

(i) \(\frac{-8}{6}\) = \(\frac{-8}{6}=\frac{8\div 2}{6\div 2}=\frac{-4}{3}\)  (as 8 and 6 has H.C.F. of 2)

 

(ii) \(\frac{25}{45}\)= \(\frac{25}{45}=\frac{25\div 5}{45\div 5}=\frac{5}{9}\)  (as 25 and 45 has H.C.F. of 5).

 

(iii) \(\frac{-44}{72}\) = \(\frac{-44}{72} =\frac{-44\div 4}{72\div 4}=\frac{-11}{18}\)  (as 44 and 72 has H.C.F. of 4)

 

(iv) \(\frac{-8}{10}\) = \(\frac{-8}{10}=\frac{-8\div 2}{10\div 2}=\frac{-4}{5}\)  (as 8 and 10 has H.C.F. of 2).

 

Q8: Fill in the blanks with the correct symbols out of <,>,and = .

(i) \(\frac{-5}{7}\) ___ \(\frac{2}{3}\)

(ii) \(\frac{-4}{5}\) ___ \(\frac{-5}{7}\)

(iii) \(\frac{-7}{8}\) ___ \(\frac{14}{16}\)

(iv) \(\frac{-8}{5}\) ___ \(\frac{-7}{4}\)

(v) \(\frac{-1}{3}\) ___ \(\frac{-1}{4}\)

(vi) \(\frac{5}{-11}\) ___ \(\frac{-5}{11}\)

(vii) \(0\) ___ \(\frac{-7}{6}\)

Sol:

(i) \(\frac{-5}{7}\) _<_ \(\frac{2}{3}\)

(ii) \(\frac{-4}{5}\) _<_ \(\frac{-5}{7}\)

(iii) \(\frac{-7}{8}\) _=_ \(\frac{14}{16}\)

(iv) \(\frac{-8}{5}\) _>_ \(\frac{-7}{4}\)

(v) \(\frac{-1}{3}\) _<_ \(\frac{-1}{4}\)

(vi) \(\frac{5}{-11}\) _=_ \(\frac{-5}{11}\)

(vii) \(0\) _>_ \(\frac{-7}{6}\)

 

Q9: Write the following rational number in ascending order:

(i) \(\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}\)

(ii) \(\frac{1}{3},\frac{-2}{9},\frac{-4}{3}\)

(iii) \(\frac{-3}{7},\frac{-3}{2},\frac{-3}{4}\)

Sol:

(i) \(\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}\)

\(\Rightarrow \frac{-3}{5}<\frac{-2}{5}<\frac{-1}{5}\)

 

(ii)  \(\frac{1}{3},\frac{-2}{9},\frac{-4}{3}\)

\(\frac{3}{9},\frac{-2}{9},\frac{-12}{9}\)       (converting to same denominator)

\(\frac{-12}{9}<\frac{-2}{9}<\frac{3}{9}\) \(\Rightarrow \frac{-4}{3}<\frac{-2}{9}<\frac{1}{3}\)

 

(iii) \(\frac{-3}{7},\frac{-3}{2},\frac{-3}{4}\)

\(\frac{-12}{28},\frac{-42}{28},\frac{-21}{28}\) \(\frac{-42}{28}<\frac{-21}{28}<\frac{-12}{28}\) \(\Rightarrow \frac{-3}{2}<\frac{-3}{4}<\frac{-3}{7}\)

 

Exercise-9.2

 

Q.1.(i) \( \frac{5}{4}+(\frac{-11}{4})\)

(ii)\( \frac{5}{3}+\frac{3}{5}\)

(iii)\(e \frac{-9}{10}+\frac{22}{15}\)

(iv)\( \frac{-3}{-11}+\frac{5}{9}\)

(v)\( \frac{-8}{19}+\frac{(-2)}{57}\)

(vi)\( \frac{-2}{3}+0\)

(vii)\( -2\frac{1}{3}+4\frac{3}{5}\)

Solution:

(i)\( \frac{5}{4}+(\frac{-11}{4})\)

\( \frac{5}{4}+(\frac{-11}{4})= \frac{5-11}{4}=\frac{-6}{4}=\frac{-3}{2}\)

(ii)\( \frac{5}{3}+\frac{3}{5}\)

\( \frac{5}{3}+\frac{3}{5}\\ \\ = \frac{5\times 5}{3\times 5}+\frac{3\times 3}{5\times 3}\\ \\ =\frac{25}{15}+\frac{9}{15}\\ \\ =\frac{25+9}{15}=\frac{34}{15}=2\frac{4}{15}\)

(iii)\( \frac{-9}{10}+\frac{22}{15}\)

\( \frac{-9}{10}+\frac{22}{15}\\ \\ = \frac{-9\times 3}{10\times 3}+\frac{22\times 2}{15\times 2}\\ \\ =\frac{-27}{30}+\frac{44}{30}\\ \\ =\frac{-27+44}{30}=\frac{17}{30}\)

(iv)\( \frac{-3}{-11}+\frac{5}{9}\)

\( \frac{-3}{-11}+\frac{5}{9}\\ \\ = \frac{-3\times 9}{-11\times 9}+\frac{5\times 11}{9\times 11}\\ \\ =\frac{27}{99}+\frac{55}{99}\\ \\ =\frac{27+55}{99}=\frac{82}{99}\)

(v)\( \frac{-8}{19}+\frac{(-2)}{57}\)

\( \frac{-8}{19}+\frac{(-2)}{57}\\ \\ = \frac{-8\times 3}{19\times 3}+\frac{(-2)\times 1}{57\times 1}\\ \\ =\frac{-24}{57}+\frac{(-2)}{57}\\ \\ =\frac{-24-2}{57}=\frac{-26}{57}\)

(vi)\( \frac{-2}{3}+0\)

\( \frac{-2}{3}+0=\frac{-2}{3} \)

(vii)\( -2\frac{1}{3}+4\frac{3}{5}\)

\( 2\frac{1}{3}+4\frac{3}{5}\\ \\ =\frac{-7}{3}+\frac{23}{5}\\ \\ = \frac{-7\times 5}{3\times 5}+\frac{23\times 3}{5\times 3}\\ \\ =\frac{-35}{15}+\frac{69}{15}\\ \\ =\frac{-35+69}{15}=\frac{34}{15}=2\frac{4}{15}\)

 

Q.2. Find:

(a)\( \frac{7}{24}-\frac{17}{36}\)

\( \frac{7}{24}-\frac{17}{36}\\ \\ = \frac{7\times 3}{24\times 3}-\frac{17\times 2}{36\times 2}\\ \\ =\frac{21}{72}-\frac{34}{72}\\ \\ =\frac{21-34}{72}=\frac{-13}{72}\)

(b)\( \frac{5}{63}-(\frac{-6}{21})\)

\( \frac{5}{63}-(\frac{-6}{21})\\ \\ = \frac{5\times 1}{63\times 1}-(\frac{-6\times 3}{21\times 3})\\ \\ =\frac{5}{63}-(\frac{-18}{63})\\ \\ =\frac{5+18}{63}=\frac{23}{63}\)

(c)\( \frac{-6}{13}-(\frac{-7}{15})\)

\( \frac{-6}{13}-(\frac{-7}{15})\\ \\ = \frac{-6\times 15}{13\times 15}-(\frac{-7\times 13}{15\times 13})\\ \\ =\frac{-90}{195}-(\frac{-91}{195})\\ \\ =\frac{-90-(-91)}{195}=\frac{-90+91}{195}=\frac{1}{195}\)

(d)\( \frac{-3}{8}-\frac{7}{11}\)

\( \frac{-3}{8}-\frac{7}{11}\\ \\ = \frac{-3\times 11}{8\times 11}-\frac{7\times 8}{11\times 8}\\ \\ =\frac{-33}{88}-\frac{56}{88}\\ \\ =\frac{-33-56}{88}=\frac{-89}{88}=-1\frac{1}{88}\)

(e)\( -2\frac{1}{9}-6\)

\( -2\frac{1}{9}-6\\ \\ =\frac{-19}{9}-\frac{6}{1}\\ \\ = \frac{-19\times 1}{9\times 1}-\frac{6\times 9}{1\times 9}\\ \\ =\frac{-19}{9}-\frac{54}{9}\\ \\ =\frac{-19-54}{9}=\frac{-73}{9}=-8\frac{1}{9}\)

 

Q.3.(i)\( \frac{9}{2}\times(\frac{-7}{4})\)

\( \frac{9}{2}\times(\frac{-7}{4})=\frac{9\times (-7)}{2\times 4}=\frac{-63}{8}=-7\frac{7}{8}\)

(ii)\( \frac{3}{10}\times(-9)\)

\( \frac{3}{10}\times(-9)=\frac{3\times (-9)}{10}=\frac{-27}{10}=-2\frac{7}{10}\)

(iii)\( \frac{-6}{5}\times\frac{9}{11}\)

\( \frac{-6}{5}\times\frac{9}{11}=\frac{(-6)\times 9}{5\times 11}=\frac{-54}{55}\)

(iv)\( \frac{3}{7}\times(\frac{-2}{5})\)

\( \frac{3}{7}\times(\frac{-2}{5})=\frac{3\times (-2)}{7\times 5}=\frac{-6}{35}\)

(v)\( \frac{3}{11}\times\frac{2}{5}\)

\( \frac{3}{11}\times\frac{2}{5}=\frac{3\times 2}{11\times 5}=\frac{6}{55}\)

(vi)\( \frac{3}{-5}\times(\frac{-5}{3})\)

\( \frac{3}{-5}\times(\frac{-5}{3})=\frac{3\times (-5)}{-5\times 3}=1\)

 

Q.4. Find the value of:

(i)\( (-4)\div \frac{2}{3}\)

\( (-4)\div \frac{2}{3}=(-4)\times \frac{3}{2}=(-2)\times 3=-6\)

(ii)\( \frac{-3}{5}\div2\)

\( \frac{-3}{5}\div2=\frac{-3}{5}\times \frac{1}{2}=\frac{(-3)\times 1}{5\times 2}=\frac{-3}{10}\)

(iii)\( \frac{-4}{5}\div(-3)\)

\( \frac{-4}{5}\div(-3)=\frac{(-4)}{5}\times \frac{1}{(-3)}=\frac{(-4)\times 1}{5\times (-3)}=\frac{4}{15}\)

(iv)\( \frac{-1}{8}\div\frac{3}{4} \)

\( \frac{-1}{8}\div\frac{3}{4} =\frac{-1}{8}\times \frac{4}{3}=\frac{(-1)\times 1}{2\times 3}=\frac{-1}{6}\)

(v) \( \frac{-2}{13}\div\frac{1}{7} \)

\( \frac{-2}{13}\div\frac{1}{7} =\frac{-2}{13}\times \frac{7}{1}=\frac{(-2)\times 7}{13\times 1}=\frac{-14}{13}=1\frac{-1}{13}\)

(vi) \( \frac{-7}{12}\div(\frac{-2}{13}) \)

\( \frac{-7}{12}\div(\frac{-2}{13}) =\frac{-7}{12}\times \frac{13}{(-2)}=\frac{(-7)\times 13}{12\times (-2)}=\frac{-91}{24}=3\frac{19}{24}\)

(vii)\( \frac{3}{13}\div(\frac{-4}{65}) \)

\( \frac{3}{13}\div(\frac{-4}{65}) =\frac{3}{13}\times \frac{65}{(-4)}=\frac{3\times (-5)}{1\times 4}=\frac{-15}{4}=-3\frac{3}{4}\)

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