Question

$\text{Evaluate :}\left(i\right)(\frac{2}{-3}\times \frac{5}{4})+(\frac{5}{9}\times \frac{3}{-10})\left(ii\right)(2\times \frac{1}{4})-(\frac{-18}{7}\times \frac{-7}{15})\left(iii\right)(-5\times \frac{2}{15})-(-6\times \frac{2}{9})\left(iv\right)(\frac{8}{5}\times \frac{-3}{2})+(\frac{-3}{10}\times \frac{9}{16})$

Answer 1

$\left(i\right)(\frac{2}{-3}\times \frac{5}{4})+(\frac{5}{9}\times \frac{3}{-10})=\left(\frac{2\times 5}{(-3)\times 4}\right)+\left(\frac{5\times 3}{9\times (-10)}\right)=\left(\frac{1\times 5}{(-3)\times 2}\right)+\left(\frac{1\times 1}{3\times (-2)}\right)=\frac{-5}{6}+\frac{-1}{6}=\frac{-5-1}{6}=\frac{-6}{6}=-1\left(ii\right)(2\times \frac{1}{4})-(\frac{-18}{7}\times \frac{-7}{15})=\left(\frac{2\times 1}{1\times 4}\right)-\left(\frac{(-18)\times (-7)}{7\times 15}\right)=\left(\frac{2\times 1}{1\times 4}\right)-\left(\frac{(-18)\times (-1)}{1\times 15}\right)=\frac{1}{2}-\frac{18}{15}$

$\therefore \text{LCM of 2 and 15=30}=\frac{1\times 15}{2\times 15}-\frac{18\times 2}{15\times 2}(\because \text{LCM of 2 and 15=30})=\frac{15-36}{30}=\frac{-21}{30}=\frac{-7}{10}\left(iii\right)(-5\times \frac{2}{15})-(-6\times \frac{2}{9})=\left(\frac{(-5)\times 2}{1\times 15}\right)-\left(\frac{(-6)\times 2}{1\times 9}\right)=\left(\frac{(-1)\times 2}{1\times 3}\right)-\left(\frac{(-2)\times 2}{1\times 3}\right)=\frac{-2}{3}-\left(\frac{-4}{3}\right)=\frac{-2+4}{3}=\frac{2}{3}\left(iv\right)(\frac{8}{5}\times \frac{-3}{2})+(\frac{-3}{10}\times \frac{9}{16})=\left(\frac{8\times (-3)}{5\times 2}\right)+\left(\frac{(-3)\times 9}{10\times 16}\right)=\left(\frac{4\times (-3)}{5\times 1}\right)+\left(\frac{(-3)\times 9}{10\times 16}\right)=\frac{-12}{5}+\left(\frac{-27}{160}\right)$

$\therefore \text{LCM of 5 and 160=160}=\frac{(-12)\times 32}{5\times 32}+\frac{(-27)\times 1}{160\times 1}=\frac{-384-27}{160}=\frac{-411}{160}$