1.  Solve:

     (i)    2 - \frac{3}{5}                      (ii)    4 + \frac{7}{8}                (iii)    \frac{3}{5} + \frac{2}{7}                (iv)    \frac{9}{11} - \frac{4}{15}

     (v)    \frac{7}{10} + \frac{2}{5} + \frac{3}{2}        (vi)    2\frac{2}{3} +3 \frac{1}{2}        (vii)    8\frac{1}{2}-3 \frac{5}{8} 

Answers (1)

As we know we have to make the denominator the same in order to add or subtract the fractions. So,

 

 (i)

    2 - \frac{3}{5}=\frac{2}{1}\times\frac{5}{5}-\frac{3}{5}=\frac{10}{5}-\frac{3}{5}=\frac{10-3}{5}=\frac{7}{5}                      

(ii) 

   4 + \frac{7}{8}=\frac{4}{1}\times\frac{8}{8}+\frac{7}{8}=\frac{32}{8}+\frac{7}{8}=\frac{39}{8}                

(iii) 

   \frac{3}{5} + \frac{2}{7}=\frac{3}{5}\times\frac{7}{7}+\frac{2}{7}\times\frac{5}{5}=\frac{21}{35}+\frac{10}{35}=\frac{31}{35}                

(iv) 

  \frac{9}{11} - \frac{4}{15}=\frac{15\times 9-11\times 4}{11\times15}=\frac{135-44}{165}=\frac{91}{165}

               

(v)  

  \frac{7}{10} + \frac{2}{5} + \frac{3}{2}=\frac{7}{10}+\frac{2}{5}\times\frac{2}{2}+\frac{3}{2}\times\frac{5}{5}=\frac{7}{10}+\frac{4}{10}+\frac{15}{10}=\frac{7+4+15}{10}=\frac{26}{10}       

(vi) 

   2\frac{2}{3} +3 \frac{1}{2}=\frac{8}{3}+\frac{7}{2}=\frac{16}{6}+\frac{21}{6}=\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{68}{8}-\frac{29}{8}=\frac{68-29}{8}=\frac{39}{9}=4\frac{1}{3} 

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