1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Using the rearrangement property find the sum: (i) $\frac{4}{3}+\frac{3}{5}+\frac{-2}{3}+\frac{-11}{5}$ (ii) $\frac{-8}{3}+\frac{-1}{4}+\frac{-11}{6}+\frac{3}{8}$ (iii) $\frac{-13}{20}+\frac{11}{14}+\frac{-5}{7}+\frac{7}{10}$ (iv) $\frac{-6}{7}+\frac{-5}{6}+\frac{-4}{9}+\frac{-15}{7}$

Open in App
Solution

(i) $\left(\frac{4}{3}+\frac{-2}{3}\right)+\left(\frac{3}{5}+\frac{-11}{5}\right)$ = $\left(\frac{4-2}{3}\right)+\left(\frac{3-11}{5}\right)$ $=\left(\frac{2}{3}+\frac{-8}{5}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{10}{15}+\frac{-24}{15}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{10-24}{15}\right)\phantom{\rule{0ex}{0ex}}=\frac{-14}{15}$. (ii) $\left(\frac{-8}{3}+\frac{-11}{6}\right)+\left(\frac{-1}{4}+\frac{3}{8}\right)$ =$\left(\frac{-16}{6}+\frac{-11}{6}\right)+\left(\frac{-2}{8}+\frac{3}{8}\right)$ =$\left(\frac{-16-11}{6}\right)+\left(\frac{-2+3}{8}\right)$ $=\left(\frac{-27}{6}+\frac{1}{8}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-108}{24}+\frac{3}{24}\right)\phantom{\rule{0ex}{0ex}}=\frac{-105}{24}$ =$\frac{35}{8}$ (iii) $\left(\frac{-13}{20}+\frac{7}{10}\right)+\left(\frac{11}{14}+\frac{-5}{7}\right)$ =$\left(\frac{-13}{20}+\frac{14}{20}\right)+\left(\frac{11}{14}+\frac{-10}{14}\right)$ =$\left(\frac{-13+14}{20}\right)+\left(\frac{11-10}{14}\right)$ $=\left(\frac{1}{20}+\frac{1}{14}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{7}{140}+\frac{10}{140}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{7+10}{140}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{17}{140}\right)\phantom{\rule{0ex}{0ex}}=\frac{17}{140}$. (iv) $\left(\frac{-6}{7}+\frac{-15}{7}\right)+\left(\frac{-5}{6}+\frac{-4}{9}\right)$ =$\left(\frac{-6}{7}+\frac{-15}{7}\right)+\left(\frac{-15}{18}+\frac{-8}{18}\right)$ =$\left(\frac{-6-15}{7}\right)+\left(\frac{-15-8}{18}\right)$ $=\left(\frac{-21}{7}+\frac{-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-3}{1}+\frac{-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-54}{18}+\frac{-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-54-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\frac{-77}{18}$

Suggest Corrections
8
Join BYJU'S Learning Program
Related Videos
Properties of Multiplication of Integers
MATHEMATICS
Watch in App
Join BYJU'S Learning Program