(i)  $2-\frac{3}{5}=\frac{2\times 5}{5}-\frac{3}{5}= \frac{10-3}{5}=\frac{7}{5}$

(ii)   $4+\frac{7}{8}=\frac{4*8}{8}+\frac{7}{8} = \frac{(4\times 8)+7}{8} = \frac{39}{8}=4\frac{7}{8}$

(iii)  $\frac{3}{5}+\frac{2}{7}=\frac{3\times 7}{5\times 7}+\frac{2\times 5}{7\times 5}=\frac{21+10}{35}=\frac{31}{35}$

(iv)  $\frac{9}{11}-\frac{4}{15}=\frac{9\times 15}{11\times 15}-\frac{4\times 11}{15\times 11}= \frac{135-44}{165}=\frac{91}{165}$

(v)   $\frac{7}{10}+\frac{2}{5}+\frac{3}{2} = \frac{7}{10}+\frac{2\times 2}{5\times 2}+\frac{7+4+15}{10}=\frac{26}{10}=\frac{13}{5}=2\frac{3}{5}$

(vi)  $2\frac{2}{3}+3\frac{1}{2}= \frac{8}{3}+\frac{7}{2}=\frac{8\times 2}{3\times 2}+\frac{7\times 3}{2\times 3}=\frac{16+21}{6}=\frac{37}{6}=6\frac{1}{6}$

(vii)   $8\frac{1}{2}-3\frac{5}{8}=\frac{17}{2}-\frac{29}{8}=\frac{17\times 4}{2\times 4}-\frac{29}{8}=\frac{68-29}{8}=\frac{39}{8}=4\frac{7}{8}$

(i)  $\frac{2}{9}, \frac{2}{3},\frac{8}{21}$

Changing them to like fractions, we obtain

 $$\begin{gathered} \frac{2}{9}=\frac{2\times 7}{9\times 7}=\frac{14}{63} \\ \frac{2}{3}=\frac{2\times 21}{3\times 21}=\frac{42}{63} \\ \frac{8}{21}=\frac{8\times 3}{21\times 3}=\frac{24}{63} \end{gathered}$$

Since 42>24>14,

 $$\begin{gathered} \frac{2}{3}>\frac{8}{21}>\frac{2}{9} \\ \left(\mathrm{ii}\right) \frac{1}{5},\frac{3}{7}, \frac{7}{10} \end{gathered}$$

Changing them to like fractions, we obtain

$$\begin{gathered} \frac{1}{5}=\frac{1\times 14}{5\times 14}=\frac{14}{70} \\ \frac{3}{7}=\frac{3\times 10}{7\times 10}=\frac{30}{70} \\ \frac{7}{10}=\frac{7\times 7}{10\times 7}=\frac{49}{70} \end{gathered}$$

As 49 > 30>14,

 $\frac{70}{10}>\frac{3}{7}>\frac{1}{5}$

Along the first row, sum =  $\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}$

Along the second row, sum = $\frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{15}{11}$

Along the third row, sum = $\frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{15}{11}$

Along the first column, sum = $\frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{15}{11}$

Along the second column, sum = $\frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{15}{11}$

Along the third column, sum =$\frac{2}{11}+\frac{7}{11}+\frac{6}{11}=\frac{15}{11}$

Along the first diagonal, sum = $\frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{15}{11}$

Along the second diagonal, sum =$\frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{15}{11}$

Since the sum of the numbers in each row, in each column, and along the diagonals is the same, it is a magic square.

Length =$12\frac{1}{2} \mathrm{cm} = \frac{25}{2}\mathrm{cm}$

Breadth = $10\frac{2}{3} \mathrm{cm} = \frac{32}{3} \mathrm{cm}$

Perimeter = 2 × (Length + Breadth)

$$\begin{gathered} =2\times \left[\frac{25}{2}+\frac{32}{3}\right]=2\times \left[\frac{(25\times 3)+(32\times 2)}{6}\right]=2\times \left[\frac{75+64}{6}\right] \\ =2\times \frac{139}{6}=\frac{139}{3}=46\frac{1}{3} \mathrm{cm} \end{gathered}$$