Maths Chapter-09: 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)\( \frac{3}{13}\div(\frac{-4}{65}) \)
\( \frac{3}{13}\div(\frac{-4}{65}) =\frac{3}{13}\times \frac{65}{(-4)}=\frac{3\times (-5)}{1\times 4}=\frac{-15}{4}=-3\frac{3}{4}\)