Question

Arrange the numbers in decending order: $\sqrt{81},\sqrt{27},\frac{1}{\sqrt{3}},\sqrt{3}$

• A. $\sqrt{81},\sqrt{27},\frac{1}{\sqrt{3}},\sqrt{3}$
• B. $\frac{1}{\sqrt{3}},\sqrt{3},\sqrt{27},\sqrt{81}$
• C. $\sqrt{81},\sqrt{27},\sqrt{3},\frac{1}{\sqrt{3}}$

Answer 1

The correct option is C $\sqrt{81},\sqrt{27},\sqrt{3},\frac{1}{\sqrt{3}}$

Let's express the given numbers such that they have like terms to compare.

$\sqrt{81}=9$

$\sqrt{27}=\sqrt{3\times 3^2}=3\sqrt{3}$

$\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3}\times \sqrt{3}}=\frac{\sqrt{3}}{3}=\frac{1}{3}\times \sqrt{3}$

$\sqrt{3}=\frac{\sqrt{3}}{1}$

Clearly, $9$ is the largest number, i.e., $\sqrt{81}.$

$3\sqrt{3}>\sqrt{3}\text{as}3>1.$
$\therefore \sqrt{27}>\sqrt{3}$

Also $\frac{1}{3}\times \sqrt{3}<\sqrt{3}\text{as}\frac{1}{3}<1.$
$\therefore \sqrt{3}>\frac{1}{\sqrt{3}}$

Hence, the descending order is as follows:
$\sqrt{81},\sqrt{27},\sqrt{3},\frac{1}{\sqrt{3}}$