For some reason I thought of this simple algebraic problem while going to sleep last night. Here it is:
A fully grown adult person has a body mass index of 20.2. After gaining 3 kilograms in weight, the same person now has a body mass index of 21.2. How tall is the person to the nearest centimetre?
Let's say the person's initial weight is \(W\) and height is \(h\). This gives us two equations:$$ \begin{align} \frac{W}{h^2} &= 20.2\\ \\ \frac{W+3}{h^2}&=21.2 \end{align} $$ The first equation yields:$$W=20.2 \,h^2$$ and substituting this into the second yields:$$\begin{align} 20.2 \, h^2+3 &= 21.2 \, h^2\\h^2 &= 3\\h &= \pm \sqrt{3}\\ &\approx \pm 1.73 \end{align} $$Thus the person is 1.73 metres tall to the nearest centimetre. I obviously chose easy numbers to work with but the calculation is straightforward for any numbers.
Clearly the problem is of lower secondary high school level of difficulty but for some reason it popped into my head, an echo of my long ago days as a high school teacher of Mathematics. The key to the problem is realising that the height of the person doesn't change because we are dealing with an adult and not a child or adolescent. Units for weight are kilograms and, for height, metres. The problem could be made more difficult by using pounds and inches, forcing conversion between the metric and non-metric systems of measurement.
The issue of accuracy in this problem should be addressed because it must be noted that the BMI is accurate to only one decimal place and yet the height is being reported to two decimal places. BMI is commonly expressed to one decimal place so it's reasonable to suppose that a rounding to the nearest one decimal place has occurred. This means that the first BMI (let's call it BMI_1) and the second BMI (let's call it BMI_2) lie within the following bounds: $$20.15 \leq \text{ BMI}_1 <20.25\\21.15 \leq \text{BMI}_2 <21.25$$Considering the minimum and maximum of these ranges, we get a maximum and minimum difference as follows:$$ \begin{align} \text{minimum}&=21.15-20.25\\ &=0.90\\ \text{maximum} &=21.25-20.15\\ &=1.10 \end{align}$$Going back to our original equation and designating the minimum and maximum heights as \(h_{ \text{min}} \) and \(h_{ \text{max}} \) respectively, we see that:$$\begin{align} 1.1 \, h_{\text{min}}^2 &= 3 \\ h_{ \text{min}} &= \pm \sqrt{3 \div 1.1}\\ & \approx \pm 1.65 \\0.9 \, h_{\text{max}}^2 &= 3 \\ h_{ \text{max}} &=\pm \sqrt{3 \div 0.9}\\ & \approx \pm 1.83 \end{align}$$Thus the height could have ranged between 1.65 metres and 1.83 metres if we record the BMI to the nearest one decimal place. The assumption here is that the weight is 3.00 kilograms which is unlikely of course but I didn't want to complicate matters.
Having realised this, let's rephrase the problem using BMI's accurate to two decimal places as well as a weight accurate to two decimal places as well:
A fully grown adult person has a body mass index of 20.23. After gaining 3.11 kilograms in weight, the same person now has a body mass index of 21.19. How tall is the person to the nearest centimetre?
Using these figures, we see that:$$ \begin{align} 0.96 \, h^2 &=3.11\\ h&= \pm \sqrt{3.11 \div 0.96}\\& \approx \pm 1.80 \end{align}$$Thus we have a height of 1.80 metres to the nearest two decimal places. The moral of this problem is that it's meaningless to have one quantity accurate to one decimal place (BMI) and another accurate to the nearest whole number (weight) and to then ask for a quantity that is dependent on these first two quantities to be expressed with an accuracy of two decimal places. This is simply false accuracy and it can creep into calculations unless we are vigilant, which I wasn't when I originally framed the problem. The more we multiply and divide approximations, the more the margin of error grows.
This post marks my 100th of the year and equals my record for 2022. My record of 105 posts is from 2021, a record I seem likely to break given that three months of the year remain.
