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I am not sure if the question suits in this stack board, but I give it a try.

Let's assume there is a country with a population of 327 million people and a surface of 9.6 million km², resulting in 34 inhabitants per km².

Now, let's say we have a fictive war scenario, in which the country loses and has to give up land to the victors. The population sunk to 100 million people and 10 inhabitants per km² (still assuming that surface was given to victors, thus the country now has a smaller surface!!). How would I calculate the new surface with these two (now changed) variables or would I need more data?

My question now is:

Is it still possible to determine the new surface of the country when both population and population density changed?

I tried many ways to accomplish that, but I really have no idea how to approach it. Maybe it's possible with a system of linear equations? I have no idea.

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  • $\begingroup$ was americas land area smaller before the europeans discovered it,it is related to the logic of your question. $\endgroup$ – trond hansen Feb 8 '19 at 10:56
  • $\begingroup$ Sorry but this has noting to with Earth Science. It's an elemental math question. If X/Y=Z, and I know X and Z, can I calculate Y? Of course you can. $\endgroup$ – Jan Doggen Feb 10 '19 at 14:03
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    $\begingroup$ I'm voting to close this question as off-topic because it just a mathematical question that has nothing to do with Earth Science $\endgroup$ – Jan Doggen Feb 11 '19 at 8:51
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$\text{Poulation density} = \frac{\text{Population}}{\text{Surface}}$

Re-arranging

$\text{Surface} = \frac{\text{Population}}{\text{Poulation density}}$

In your case

$\text{Surface} = \frac{100,000,000\, \text{inhabitants}}{10\, \text{inhabitants}\, \text{km}^{-2}}=10,000,000\, \text{km}^2$

So something doesn't make sense, because those numbers suggests that the surface of the country is now larger than before.

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    $\begingroup$ Oh Camilo, where's your imagination? There could be an off shore volcano creating land. :-). All joking aside, I agree with you, the numbers in the question don't make sense. I think the population densities are wrong. $\endgroup$ – Fred Feb 7 '19 at 8:24

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