Among the following, which is closest to $\sqrt{0.016}$?

Question

Among the following, which is closest in value to $\sqrt{0.016}$?

A. $0.4$

B. $0.04$

C. $0.2$

D. $0.02$

E. $0.13$

My Approach:

$(\frac{16}{1000})^\frac{1}{2} = (\frac{4}{250})^\frac{1}{2} = \frac{2}{5\cdot\sqrt{10}} = \frac{\sqrt{2}}{5\cdot\sqrt{5}}$

But I don't know how to reach to answer.

Thank You!

Answer 1

Just square:

$0.4^2=0.16$

$0.04^2=0.0016$

$0.2^2=0.04$

$0.02^2=0.0004$

$0.13^2=0.0169$

Can you choose?

Answer 2

Though squaring each and checking works equally as well, here's another way to figure it out: $$ \sqrt{0.016}=\frac{\sqrt{1000}}{\sqrt{1000}}\sqrt{0.016}=\frac{\sqrt{16}}{\sqrt{1000}}=\frac{4}{\sqrt{1000}}\approx\frac{4}{\sqrt{1024}}=\frac{4}{\sqrt{2^{10}}}=\frac{4}{2^{5}}=\frac{4}{32}=\frac{1}{8}=0.125 $$

Answer 3

If you rationalize the denominator of your ratio to get $ \ \frac{ \sqrt{10} }{25} \ $ , you can use the fact that $ \ \sqrt{10} \ $ is a little bigger than 3 to estimate that the number in question is a bit larger than $ \ \frac{3}{25} \ = \ 0.12 \ $ . No other choice but (E) is close to that.

Answer 4

This absolutely is the worst way to solve this problem, but I though it might be useful to you, or at least interesting (I bet ability to evaluate things numerically on paper is something interesting).

From \begin{align}\sqrt{0.016}&=\sqrt{16\cdot10^{-3}}=4\sqrt{10^{-1}10^{-2}}=4\cdot10^{-1}\sqrt{10^{-1}}=\frac4{10}\sqrt{1\over10}=\frac25\frac1{\sqrt{10}}\frac{\sqrt{10}}{\sqrt{10}}=\\&=\frac25\frac{\sqrt{10}}{10}=\frac1{25}\sqrt{10}\end{align} you can always evaluate $\sqrt{10}$ (or any square root, and many other different things) numerically to arbitrary precision with numerical root finding algorithms, which given some function $$f:D\subseteq\mathbb R\mapsto\mathbb R$$ and (for most of them) some initial guess $$x_0\in D$$ will provide $x_n\in D$ such that $$\lim_{n\to\infty}f(x_n)=0,$$ that is, the bigger the $n$ gets (more iterations of the algorithm are computed), the more accurate the result $x_n$ gets relative to the actual root $x:\,f(x)=0$. The simplest algorithm is via Newton and goes like this: you want $\sqrt{10}$, i.e. some $x$ such that $x^2=10$, that is, $x^2-10=0$; let $$f(x)=x^2-10.$$ Then by Newton's method you need to differentiate that to get $$f'(x)=2x,$$ and also set some initial guess $x_0$, for which $$x_0=3$$ seems reasonable. The first step goes like this: $$x_1=x_0-\frac{f(x_0)}{f'(x_0)}=3-\frac{9-10}6=3+\frac16=\frac{19}6;$$ the second one: $$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=\frac{19}6-\frac{361/36-10}{38/6}=\frac{19}6-\frac{1/36}{38/6}=\frac{19}6-\frac1{6\cdot38}=\frac{721}{228};$$ you get the idea. Each step produces bigger and bigger nominators and denominators, which ratio approaches $\sqrt{10}$. Long division yields $$x_2=\frac{721}{228}=3.16\overline{228070175438596491\,\,}\approx3.16228,$$ which with $$x=\sqrt{10}=3.16227766016837933199889\dotsc\approx3.16228$$ is accurate to six figures (and only via two iterations).

Now, dividing $3.16228$ by $25$ (remember to cut off at the sixth significant figure to avoid overprecision — you ain't have no more information!) yields $$0.12649,$$ which shows that $0.13$ is the closet number in the given set to $\sqrt{0.016}\,.$