Question

Divide the sum of $\frac{8}{3}$ and$\frac{4}{7}$ by the product of $-\frac{3}{7}$ and $\frac{14}{9}$.

Answer 1

Sum of $\frac{8}{3}$ and $\frac{4}{7}$$=\frac{8}{3}+\frac{4}{7}=\frac{8\times 7+4\times 3}{21}=\frac{56+12}{21}$

$\Rightarrow \frac{68}{21}$

Product of $-\frac{3}{7}$ and $\frac{14}{9}$.$=-\frac{3}{7}\times \frac{14}{9}$

$\Rightarrow -\frac{2}{3}$

$\Rightarrow$$\frac{sum}{product}=\frac{{\displaystyle \raisebox{1ex}{$68$}\!\left/ \!\raisebox{-1ex}{$21$}\right.}}{-{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=-\frac{34}{7}$

Hence, $-\frac{34}{7}$ is the correct answer