Verify whether the given statement is true or false:

Solution:

(i) $\left(\frac{5}{9} \div \frac{1}{3}\right) \div \frac{5}{2}=\frac{5}{9} \div\left(\frac{1}{3} \div \frac{5}{2}\right)$

LHS

$\left(\frac{5}{9} \div \frac{1}{3}\right) \div \frac{5}{2}$

$=\left(\frac{5}{9} \times \frac{3}{1}\right) \times \frac{2}{5}$

$=\frac{5 \times 3 \times 2}{9 \times 1 \times 5}$

$=\frac{30}{45}$

$=\frac{2}{3}$

RHS

$\frac{5}{9} \div\left(\frac{1}{3} \div \frac{5}{2}\right)$

$=\frac{5}{9} \div\left(\frac{1}{3} \times \frac{2}{5}\right)$

$=\frac{5}{9} \div\left(\frac{2}{15}\right)$

$=\frac{5}{9} \times\left(\frac{15}{2}\right)=\frac{75}{18}$

$=\frac{25}{6}$

$\mathrm{LHS} \neq \mathrm{RHS}$

FALSE

(ii) $\left[(-16) \div \frac{6}{5}\right] \div \frac{-9}{10}=(-16) \div\left[\frac{6}{5} \div \frac{-9}{10}\right]$

LHS

$=\left[(-16) \div \frac{6}{5}\right] \div \frac{-9}{10}$

$=\left[(-16) \times \frac{5}{6}\right] \times \frac{10}{-9}$

$=\frac{(-16) \times 5 \times 10}{6 \times(-9)}$

$=\frac{800}{54}$

$=\frac{400}{27}$

RHS

$(-16) \div\left(\frac{6}{5} \div \frac{-9}{10}\right)$

$=(-16) \div\left(\frac{6}{5} \times \frac{10}{-9}\right)$

$=-16 \div\left\{\frac{-60}{45}\right\}$

$=-16 \times\left(\frac{-45}{60}\right)$

$=-16 \times\left(\frac{-3}{4}\right)$

$=\frac{48}{4}$

$=12$

LHS $\neq$ RHS

FALSE

(iii) $\left(\frac{-3}{5} \div \frac{-12}{35}\right) \div \frac{1}{14}=\frac{-3}{5} \div\left(\frac{-12}{35} \div \frac{1}{4}\right)$

LHS

$=\left(\frac{-3}{5} \times \frac{35}{-12}\right) \times 14$

$=\frac{(-3) \times 35 \times 14}{5 \times(-12)}$

$=\frac{1470}{60}$

$=\frac{49}{2}$

RHS

$=\frac{-3}{5} \div\left(\frac{-12}{35} \div \frac{1}{4}\right)$

$=\frac{-3}{5} \div\left(\frac{-12}{35} \times \frac{4}{1}\right)$

$=\frac{-3}{5} \div\left(\frac{-12 \times 4}{35}\right)$

$=\frac{-3}{5} \div\left(\frac{-48}{35}\right)$

$=\frac{-3}{5} \times \frac{35}{-48}$

$=\frac{3 \times 35}{5 \times 48}$

$=\frac{105}{240}$

$\mathrm{LHS} \neq \mathrm{RHS}$

FALSE