Find how many years must elapse before the proportions of red kangaroos and grey kangaroos are reversed, assuming the same rates continue to apply.

Question

I have this question (sorry I'm not able to embed it):

Q.7. There are approximately ten times as many red kangaroos as grey kangaroos in a certain area. If the population of grey kangaroos increases at $11\%$ per annum whereas that of the red kangaroos decreases at $5\%$ per annum, find how many years must elapse before the proportions are reversed, assuming the same rates continue to apply.

Image.

I keep on getting a negative answer for $n$ (number of years) and I can't work out why. Here is what I have done:

1. I wrote out the population formulas for both types of kangaroos:
   Note that $G$ is the initial population of Grey Kangaroos, $R$ is the initial population of red kangaroos, $P$ is the population and $n$ is the number of years.
   Grey: $P=G\cdot 1.11^n$
   Red: $P=R\cdot 0.95^n$

2. The question states that there are $10$ times as many red kangaroos and grey kangaroos. Therefore, I formed the following statement: $$R = 10G.$$

3. The question asks what the value of $n$ will be when this relationship is reversed, or: $$G = 10R.$$

4. Therefore, I solved for $n$ by subbing in $10R$ into the Grey kangaroo formula. $$10R\cdot 1.11^n=R\cdot 0.95^n$$

I won't show all my working, but I arrived at $n=-14.79$ years. This is obviously not correct as it is negative. Can anyone tell me what I have done wrong?

Answer 1

The question asks you for $n$ such that $$ (10G)\times 0.95^n=\frac{1}{10}\times[G\times1.11^n] $$ so you solve the wrong equation. In particular, in order to have reversal, the initially larger population needs to decrease while the initially smaller population needs to increase. You reverse this in your setup.

In any case, the answer is $\approx\frac{\log(100)}{\log(1.11/0.95)}$ which of course is positive.