Question

Choose the order from the following that shows the numbers are arranged from the smallest to the largest.

• A. $\sqrt{241},15.8,\sqrt{264},16.1$
• B. $15.8,16.1,\sqrt{241},\sqrt{264}$
• C. $\sqrt{241},15.8,16.1,\sqrt{264}$
• D. $\sqrt{241},\sqrt{264},15.8,16.1$

Answer 1

The correct option is C $\sqrt{241},15.8,16.1,\sqrt{264}$

$\text{Detailed step-by-step solution:}$
We know that $225<241<256.$
$\Rightarrow \sqrt{225}<\sqrt{241}<\sqrt{256}$
$\Rightarrow 15<\sqrt{241}<16$

Considering the numbers between $15$ and $16$ whose squares are closer to $241$:
$(15.1{)}^2=15.1\times 15.1=228.01(15.2{)}^2=15.2\times 15.2=231.04(15.3{)}^2=15.3\times 15.3=234.09(15.4{)}^2=15.4\times 15.4=237.16(15.5{)}^2=15.5\times 15.5=240.25(15.6{)}^2=15.6\times 15.6=243.36$
$\sqrt{241}$ is closer to 15.5 as compared to $15.6$.
So, $\sqrt{241}\sim \mathrm{15.5.}$

We know that $256<264<289.$
$\sqrt{256}<\sqrt{264}<\sqrt{289}$
$16<\sqrt{264}<17$

Considering the numbers between 16 and 17 whose squares are closer to 264:
$(16.1{)}^2=16.1\times 16.1=259.21(16.2{)}^2=16.2\times 16.2=262.44(16.3{)}^2=16.3\times 16.3=265.69$
$\sqrt{264}$ is closer to 16.2 as compared to 16.3.
So, $264\sim \mathrm{16.2.}$

So, considering all the 4 numbers, the correct order will be:
$15.5<15.8<16.1<16.2$
$\Rightarrow \sqrt{241},15.8,16.1,\sqrt{264}$

So, option C is the correct answer.