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

Question:

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

Solution:

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

$=\left(\frac{11-25}{2}\right)+\frac{-17}{3}+\frac{11}{12}$

$=\frac{-14}{2}+\frac{-17}{3}+\frac{11}{12}$

$=-7+\frac{-17}{3}+\frac{11}{12}$

$=\frac{-84-68+11}{12}$

$=\frac{-141}{12}$

$=\frac{-47}{4}$

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

$=\left(\frac{-6-15}{7}\right)+\frac{-5}{6}+\frac{-4}{9}$

$=\frac{-21}{7}+\frac{-5}{6}+\frac{-4}{9}$

$=\frac{-21}{7}+\frac{-5}{6}+\frac{-4}{9}$

$=\frac{-378-105-56}{126}$

$=\frac{-539}{126}$

$=\frac{-77}{18}$

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

$=\left(\frac{3+9}{5}\right)+\left(\frac{-7+7}{3}\right)+\frac{-13}{15}$

$=\frac{12}{5}+\frac{-13}{15}$

$=\frac{36-13}{15}$

$=\frac{36-13}{15}$

$=\frac{23}{15}$

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

$=\left(\frac{4-8+9}{13}\right)+\frac{-5}{8}$

$=\frac{5}{13}+\frac{-5}{8}$

$=\frac{40-65}{104}$

$=\frac{-25}{104}$

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

$=\frac{3}{3}+\frac{-2}{5}$

$=\frac{15-6}{15}$

$=\frac{9}{15}$

$=\frac{3}{5}$

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

$=\left(\frac{2-5}{16}\right)+\left(\frac{5+7}{12}\right)+\left(\frac{2+9}{7}\right)$

$=\frac{-3}{16}+\frac{12}{12}+\frac{11}{7}$

$=\frac{-63+336+528}{336}$

$=\frac{801}{336}$

$=\frac{267}{112}$