Question

Which one of the following is not equal to ${\left(\frac{100}{9}\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ ?

• A. a) ${\left(\frac{9}{100}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• B. b) $\frac{3}{10}\times \frac{3}{10}\times \frac{3}{10}$
• C. c) $\frac{1}{{\left(\frac{100}{9}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$
• D. d) $\sqrt{\frac{100}{9}\times \frac{100}{9}\times \frac{100}{9}}$

Answer 1

The correct option is D

d) $\sqrt{\frac{100}{9}\times \frac{100}{9}\times \frac{100}{9}}$

Explanation of correct options:

Option (d): $\sqrt{\frac{100}{9}\times \frac{100}{9}\times \frac{100}{9}}$

Let us proceed by simplifying the expression $\sqrt{\frac{100}{9}\times \frac{100}{9}\times \frac{100}{9}}$.

We get, $\sqrt{{\left(\frac{100}{9}\right)}^3}$

$\Rightarrow {\left(\frac{100}{9}\right)}^{\frac{3}{2}}\ne$${\left(\frac{100}{9}\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Explanation of incorrect options:

Option (a): ${\left(\frac{9}{100}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Here, ${\left(\frac{100}{9}\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$$={\left(\frac{9}{100}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[\because {\left(\frac{a}{b}\right)}^{-\frac{m}{n}}={\left(\frac{b}{a}\right)}^{\frac{m}{n}}\right]$

Option (b): $\frac{3}{10}\times \frac{3}{10}\times \frac{3}{10}$

$\Rightarrow \frac{3}{10}\times \frac{3}{10}\times \frac{3}{10}={\left(\frac{3}{10}\right)}^3$

$={\left({\left({\left(\frac{3}{10}\right)}^2\right)}^3\right)}^{\frac{1}{2}}$

$={\left({\left(\frac{9}{100}\right)}^3\right)}^{\frac{1}{2}}$

$={\left({\left(\frac{100}{9}\right)}^{-3}\right)}^{\frac{1}{2}}\left[\because a^m=\frac{1}{a^{-m}}\right]$

$={\left(\frac{100}{9}\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[\because {\left(a^m\right)}^n=a^{mn}\right]$

Option (c): $\frac{1}{{\left(\frac{100}{9}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

$\Rightarrow \frac{1}{{\left(\frac{100}{9}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}={\left(\frac{100}{9}\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$$\left[\because \frac{1}{a^m}=a^{-m}\right]$

Hence, the correct answer is option (d).