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2. Solve :

             (a)   $\frac{1}{18} +\frac{1}{18}$

             (b)    $\frac{8}{15} +\frac{3}{15}$

             (c) $\frac{7}{7} -\frac{5}{7}$

             (d)   $\frac{1}{22} +\frac{21}{22}$

             (e)   $\frac{12}{15} -\frac{7}{15}$

             (f) $\frac{5}{8} +\frac{3}{8}$

             (g)   $1-\frac{2}{3}\left ( 1=\frac{3}{3} \right )$

            ( h)   $\frac{1}{4}+\frac{0}{4}$

             (i)   $3-\frac{12}{5}$

Answers (1)

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(a) $\frac{1}{18} +\frac{1}{18}\ =\ \frac{1+1}{18}\ =\ \frac{2}{18}\ =\ \frac{1}{9}$

(b) $\frac{8}{15} +\frac{3}{15}\ =\ \frac{8+3}{15}\ =\ \frac{11}{15}$

(c) $\frac{7}{7} -\frac{5}{7}\ =\ \frac{7-5}{7}\ =\ \frac{2}{7}$

(d) $\frac{1}{22} +\frac{21}{22}\ =\ \frac{1+21}{22}\ =\ \frac{22}{22}\ =\ 1$

(e) $\frac{12}{15} -\frac{7}{15}\ =\ \frac{12-7}{15}\ =\ \frac{5}{15}\ =\ \frac{1}{ 3}$

(f) $\frac{5}{8} +\frac{3}{8}\ =\ \frac{5+3}{8}\ =\ \frac{8}{8}\ =\ 1$

(g) $1-\frac{2}{3}\ =\ \frac{3}{3}\ -\ \frac{2}{3}\ =\ \frac{3-2}{3}\ =\ \frac{1}{3}$

( h) $\frac{1}{4}+\frac{0}{4}\ =\ \frac{1+0}{4}\ =\ \frac{1}{4}$

(i) $3-\frac{12}{5}\ =\ \frac{15}{5}\ -\ \frac{12}{5}\ =\ \frac{15-12}{5}\ =\ \frac{3}{5}$

Posted by

Devendra Khairwa

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