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1. Solve

            (g)   $\frac{3}{4}-\frac{1}{3}$                        (h)   $\frac{5}{5}-\frac{1}{3}$                             (i)   $\frac{2}{3}+\frac{3}{4}+\frac{1}{2}$

     (j)   $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$                (k)   $1\frac{1}{3}+3\frac{2}{3}$                       (l)   $4\frac{2}{3}+3\frac{1}{4}$

     (m)   $\frac{16}{5}-\frac{7}{5}$                        (n) $\frac{4}{3}-\frac{1}{2}$

Answers (1)

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(g) $\frac{3}{4}-\frac{1}{3}\ =\ \frac{3\times 3\ -\ 1\times 4}{12}\ =\ \frac{5}{12}$

(h) $\frac{5}{5}-\frac{1}{3}\ =\ \frac{5\times 3\ -\ 1\times 5}{15}\ =\ \frac{10}{15}\ =\ \frac{2}{3}$

(i) $\frac{2}{3}+\frac{3}{4}+\frac{1}{2}\ =\ \frac{2\times 4\ +\ 3\times 3}{12}\ +\ \frac{1}{2}\ =\ \frac{17}{12}\ +\ \frac{1}{2}\ =\ \frac{17\times 2\ +\ 1\times 12}{24}$

$=\ \frac{46}{24}\ =\ \frac{23}{12}$

(j) $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\ =\ \frac{1\times 3\ +\ 1\times 2}{6}\ +\ \frac{1}{6}\ =\ \frac{5}{6}\ +\ \frac{1}{6}\ =\ \frac{6}{6}\ =\ 1$

(k) $1\frac{1}{3}+3\frac{2}{3}\ =\ \frac{4}{3}\ +\ \frac{11}{3}\ =\ \frac{15}{3}\ =\ 5$

(l) $4\frac{2}{3}+3\frac{1}{4}\ =\ \frac{14}{3}\ +\ \frac{13}{4}\ =\ \frac{14\times 4\ +\ 13\times 3}{12}\ =\ \frac{95}{12}$ \

(m) $\frac{16}{5}-\frac{7}{5}\ =\ \frac{16\ -\ 7}{5}\ =\ \frac{9}{5}$

(n) $\frac{4}{3}-\frac{1}{2}\ =\ \frac{4\times2\ -\ 1\times3}{6}\ =\ \frac{5}{6}$

Posted by

Sanket Gandhi

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