Question

In the square root of $31,$ what is the integral part?

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is A $5$

$\underset{–}{\text{Need to find:}}$ Integral part in $\sqrt{31}$

We know that $25$ and $36$ are the two perfect squares closest to $31$.

$\because 25<31<36$

$\Rightarrow \sqrt{25}<\sqrt{31}<\sqrt{36}$

$\Rightarrow \underset{–}{5<\sqrt{31}<6}$
$(\because \sqrt{25}=\sqrt{5^2}=(5^2{)}^{\frac{1}{2}}=5^{2\times \frac{1}{2}}=5$, $\sqrt{36}=\sqrt{6^2}=(6^2{)}^{\frac{1}{2}}=6^{2\times \frac{1}{2}}=6)$

So, the square root of $31$ lies between $5$ and $6$.

We know $5$ is the integral part of any number between $5$ and $6.$ For an example, in the number $5.6,$ the integral part is $5$ and the fractional part is $\mathrm{0.6.}$

Hence, as $\sqrt{31}$ belongs to in-between $5$ and $6$, the integral part in the square root of $31$ is $5.$

$\therefore$ Option (a.) is the correct answer.