What exactly it is meant by "average increase in percentage"? I stumble upon this term in while solving this problem:

The average monthly income and expenditure of a person in the year $1995$ is $ \$ 14,000 $ and $ \$ 11,000 $ respectively and that of the year $2000$ is $ \$ 21,000 $ and $ \$ 17,600 $ respectively.Find the average percentage increase in expenditure of the person between $1995$ and $2000$.

ADDED: The solution given in my module goes like this:

Let the cumulative increase in expenditure be $r\%$ then,

$$17,600 = 11,000 \times (1+\frac{r}{100})^5 \Rightarrow 1.6 = (1+\frac{r}{100})^5 \Rightarrow r = 10 $$

But I don't really understand,also how can we possible solve $r$ (by using hand)?For this problem they have utilized the four options given but it's a bit tedious I suppose so I am more interested in any alternative procedure if exists.

  • $\begingroup$ The question is from my data-interpretation module,it includes a line graph which i avoided here by giving the values directly. $\endgroup$ – Quixotic Jan 6 '11 at 13:08
  • $\begingroup$ @Jasper Loy: It is increase and I think it's pretty much clear and $r\% \Rightarrow \frac{r}{100} $ this is also trivial,I am not looking for nitpicking instead I will appreciate well explained answer. $\endgroup$ – Quixotic Jan 6 '11 at 13:22

It is possible to get a good approximate solution to your equation

$$ \left( 1 + \frac{r}{100} \right)^5 = 1.6 $$

by hand if we expand the fifth power and ignore terms of order higher than $(r/100)^2,$ since we know that $r/100$ is small. Thus we have

$$ 1 + \frac{5r}{100} + \frac{10r^2}{10000} = 1.6,$$

from which we get

$$r^2 + 50r - 600 = 0 \quad \text{or} \quad (r-10)(r+60)=0,$$

and so $r=10.$


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