Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The question is from my data-interpretation module,it includes a line graph which i avoided here by giving the values directly. – Quixotic Jan 6 '11 at 13:08
@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. – 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 $r=10.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.