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

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