Question

Simplify the following :

${\left(\frac{3}{7}\right)}^2\times \frac{35}{27}\times {\left(\frac{-1}{5}\right)}^2$.

Answer 1

Step 1: Find the value of ${\left(\frac{3}{7}\right)}^2$ and ${\left(\frac{-1}{5}\right)}^2$.

Given: ${\left(\frac{3}{7}\right)}^2\times \frac{35}{27}\times {\left(\frac{-1}{5}\right)}^2$

$$\begin{gathered} {\left(\frac{3}{7}\right)}^2=\left(\frac{3}{7}\right)\times \left(\frac{3}{7}\right) \\ =\frac{3\times 3}{7\times 7} \\ =\frac{9}{49} \end{gathered}$$

$$\begin{gathered} {\left(\frac{-1}{5}\right)}^2=\left(\frac{-1}{5}\right)\times \left(\frac{-1}{5}\right) \\ =\frac{1}{25} \end{gathered}$$

Step 2: Find the value of ${\left(\frac{3}{7}\right)}^2\times \frac{35}{27}\times {\left(\frac{-1}{5}\right)}^2$.

$$\begin{gathered} {\left(\frac{3}{7}\right)}^2\times \frac{35}{27}\times {\left(\frac{-1}{5}\right)}^2=\frac{9}{49}\times \frac{35}{27}\times \frac{1}{25} \\ =\frac{\cancel{9}}{{\cancel{49}}^7}\times \frac{{\cancel{35}}^{\cancel{5}}}{{\cancel{27}}^3}\times \frac{1}{{\cancel{25}}^5} \\ =\frac{1}{7}\times \frac{1}{3}\times \frac{1}{5} \\ =\frac{1}{105} \end{gathered}$$

Hence, ${\left(\frac{3}{7}\right)}^2\times \frac{35}{27}\times {\left(\frac{-1}{5}\right)}^2=\frac{1}{105}$.