Question 102(i)
Using suitable rearrangement, find the sum.
$\frac{4}{7} + (\frac{- 4}{9}) + \frac{3}{7} + (\frac{- 13}{9})$

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Solution

Here,

$\Rightarrow \frac{4}{7} + (\frac{- 4}{9}) + \frac{3}{7} + (\frac{- 13}{9})$

$= \frac{4}{7} + \frac{3}{7} + (\frac{- 4}{9}) + (\frac{- 13}{9})$

$= \frac{7}{7} - \frac{17}{9}$

$= 1 - \frac{17}{9}$

$= \frac{9 - 17}{9}$

$= \frac{- 8}{9}$

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Similar questions

\frac{4}{7} + (\frac{- 4}{9}) + \frac{3}{7} + (\frac{- 13}{9})

\frac{4}{7} + (\frac{- 4}{9}) + \frac{3}{7} + (\frac{- 13}{9})

- 5 + \frac{7}{10} + \frac{3}{7} + (- 3) + \frac{5}{14} + (\frac{- 4}{5})

Re-arrange suitably and find the sum in each of the following :

(i) $\frac{11}{12} + \frac{- 17}{3} + \frac{11}{2} + \frac{- 25}{2}$

(ii) $\frac{- 6}{7} + \frac{- 5}{6} + \frac{- 4}{9} + \frac{- 15}{7}$

(iii) $\frac{3}{5} + \frac{7}{3} + \frac{9}{5} + \frac{- 13}{15} + \frac{- 7}{3}$

(iv) $\frac{4}{13} + \frac{- 5}{8} + \frac{- 8}{13} + \frac{9}{13}$

(v) $\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5}$

(vi) $\frac{1}{8} + \frac{5}{12} + \frac{2}{7} + \frac{7}{12} + \frac{9}{7} + \frac{- 5}{16}$

- 5 + \frac{7}{10} + (- 3) + \frac{3}{7} + (\frac{- 4}{5}) + \frac{5}{14}

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