By considering bounds, work out V to a suitable degree of accuracy I keep getting this question in my GCSE papers, but I have no idea how to solve it, and everywhere I look there doesn't seem to be a simple answer. The general question goes like this:

$$v=\sqrt{\frac{a}{b}}$$

$a = 6.43$ correct to 2 decimal places.

$b = 5.514$ correct to 3 decimal places.

By considering bounds, work out the value to $v$ to a suitable degree of accuracy.

(Sorry about the tagging, not sure what this fitted into)
Since $a=6.43$ to two decimal places, $a$ lies between $6.425$ and $6.435$. Similarly, $b$ lies between $5.5135$ and $5.5145$. The smallest possible value of $a/b$ occurs when $a$ is as small as possible and $b$ as large as possible; the largest possible value of $a/b$ occurs when $a$ is as large as possible and $b$ as small as possible. Thus, $$\frac{6.425}{5.5145}\le \frac{a}b\le \frac{6.435}{5.5135},$$ and $$\sqrt{\frac{6.425}{5.5145}}\le v\le \sqrt{\frac{6.435}{5.5135}}\;.$$ These bounds on $v$ are approximately $1.07940$ and $1.08034$, so we know that $v$ is between $1.075$ and $1.085$ and hence that $v=1.08$ is correct to two decimal places. Can we go one place further? The best approximation of $v$ to three decimal places is clearly $1.080$, but it isn’t correct to three places, because $v$ isn’t guaranteed to be between $1.0795$ and $1.0805$: $v$ could be just a hair under $1.0795$.
Let us denote $a&#39; = 6.43,b&#39; = 5.514$ and $a = a&#39;+\delta_1$ and $b = b&#39;+\delta_2$ with $\delta_1\leq 0.01$ and $\delta_2\leq 0.001$. We know that 
$$
v = \sqrt{\frac ab}.
$$

Let us denote $v&#39; = \sqrt{\frac{a&#39;}{b&#39;}}$ and put $\delta = v-v&#39;$. Note that $v,v&#39;\geq 1$ then
$$
|\delta| = |v-v&#39;| = \frac{|v^2-v&#39;^2|}{v+v&#39;} \leq\frac12|v^2-v&#39;^2| = \frac 12\left|\frac{a&#39;+\delta_1}{b&#39;+\delta_2}-\frac{a&#39;}{b&#39;}\right|
$$
$$
=\frac 12\left|\frac{(a&#39;+\delta_1)b&#39; - a&#39;(b&#39;+\delta_2)}{b&#39;(b&#39;+\delta_2)}\right|\leq \frac12\cdot\frac{\delta_1}{5}\leq 0.001
$$
so you have $v\approx \sqrt{\frac{a&#39;}{b&#39;}}= 1.07987...$ with an error smaller than $0.001$ or up to 3 decimal places.