4 Mark Question

Question 14 Marks

By what number should $\Big(\frac{-2}{3}\Big)^{-3}$ be divided so that the quotient may be $\Big(\frac{4}{27}\Big)^{-2}$?

Answer

Let x be the required number
$\therefore\Big(\frac{-2}{3}\Big)^{-3}\div\text{x}=\Big(\frac{4}{27}\Big)^{-2}$
$\Rightarrow\Big(\frac{-3}{2}\Big)^{3}\times\frac{1}{\text{x}}=\Big(\frac{27}{4}\Big)^{2}$
$\Rightarrow\frac{1}{\text{x}}=\Big(\frac{27}{4}\Big)^{2}\div\Big(\frac{-3}{2}\Big)^3$
$\Rightarrow\frac{1}{\text{x}}=\frac{\Big(\frac{27}{4}\Big)^{2}}{\Big(\frac{-3}{2}\Big)^3}$
$\Rightarrow\Big(\frac{27}{4}\Big)^{2}\times\Big(\frac{2}{-3}\Big)^3$
$\Rightarrow\frac{1}{\text{x}}=\frac{27\times27\times2\times2\times2}{4\times4\times(-3)\times(-3)\times(-3)}$
$\Rightarrow\frac{-27}{2}$
$\therefore\text{x}=\frac{-2}{27}$

View full question & answer→

Question 24 Marks

By what number should $\Big(\frac{-2}{3}\Big)^{-3}$ be divided so that the quotient is $\Big(\frac{4}{9}\Big)^{-2}$?

Answer

Let the number be x$\therefore\Big(\frac{3}{-2}\Big)^{-3}\div\text{x}=\Big(\frac{4}{9}\Big)^{-2}$
$\Rightarrow\Big(\frac{3}{-2}\Big)^{3}\div\text{x}=\Big(\frac{9}{4}\Big)^{2}$
$\Rightarrow\frac{\Big(\frac{3}{-2}\Big)^{3}}{\text{x}}=\Big(\frac{9}{4}\Big)^{2}$
$\Rightarrow\frac{\frac{3^3}{-2^3}}{\text{x}}=\frac{9^2}{4^2}$
$\Rightarrow\text{x}=\frac{\Big(\frac{3^3}{-2^3}\Big)}{\Big(\frac{9^2}{4^2}\Big)}$
$\Rightarrow\text{x}=\frac{\Big(\frac{3^3}{-2^3}\Big)}{\Big(\frac{(3^2)^2}{(2^2)^2}\Big)}$
$\Rightarrow\text{x}=\Big(\frac{3^3}{-2^3}\Big)\times\bigg(\frac{(2^2)^2}{(3^2)^2}\bigg)$
$\Rightarrow\text{x}=\Big(\frac{3^3}{-2^3}\Big)\times\Big(\frac{2^4}{3^4}\Big)$
$\Rightarrow\text{x}=\Big(\frac{3^3}{-2^3}\Big)\times\Big(\frac{2^3}{3^3}\Big)\times\Big(\frac{2^1}{3^1}\Big)$
$\Rightarrow\text{x}=\Big(\frac{1}{-1}\Big)\times\Big(\frac{2^1}{3^1}\Big)$
$\Rightarrow\text{x}=\frac{2}{-3}$
$\Rightarrow\text{x}=\frac{2\times-1}{-3\times-1}$

View full question & answer→

Maths 4 Mark Question

Take a timed test