Question

Simplify:

(i) $\frac{7}{24}\times -48$
(ii) $\frac{-19}{36}\times 16$
(iii) $\frac{-3}{4}\times \frac{4}{3}$
(iv) $-13\times \frac{17}{26}$
(v) $\frac{-13}{5}\times -10$
(vi) $\frac{-9}{16}\times \frac{-64}{27}$

Answer 1

$\begin{array}{l}\text{(i)}\\ \frac{7}{\sout{2}{\sout{4}}_1}\times (-\sout{4}{\sout{8}}^2)\\ =7\times (-2)\\ =-14\\ \left(ii\right)\\ \frac{-19}{\sout{3}{\sout{6}}_9}\times \sout{1}{\sout{6}}^4\\ =\frac{-19}{9}\times 4\\ =\frac{-76}{9}\\ \\ \left(iii\right)\\ \frac{-{\sout{3}}^1}{{\sout{4}}_1}\times \frac{{\sout{4}}^1}{{\sout{3}}_1}\\ =-1\\ \\ \left(iv\right)\\ -13\times \frac{17}{26}\\ =\frac{-\sout{1}{\sout{3}}^1\times 17}{\sout{2}{\sout{6}}^2}\\ =\frac{-17}{2}\\ \\ \left(v\right)\\ \frac{-13}{{\sout{5}}_1}\times (-\sout{1}{\sout{0}}^2)\\ =26\\ \\ \left(vi\right)\\ \frac{(-{\sout{9}}^1)}{\sout{1}{\sout{6}}_1}\times \frac{(-\sout{6}{\sout{4}}^4)}{\sout{2}{\sout{7}}_3}\\ =\frac{4}{3}\\ \end{array}$