1. ## Increasing by percent

Number of students at university increased from 3000 to 12000 during 2 years. By how much percent was it increasing annually?

How can I figure out this without testing possible answers?

P.S. I have translated this problem from other language and I could translated it a bit wrongly.

2. ## Re: Increasing by percent

Use the "compound interest" formula $\displaystyle A = P (1+i)^n$

A=12000, P = 3000, n=2 years, i = "interest" rate per year (as a decimal)

Solve for i. You should get i = 1 = 100%

3. ## Re: Increasing by percent

I didn't know this formula. Thanks! I'm going to search derivation of it.

But I don't think that I should solve this problem with compound interest formula. There's no word about it in the textbook. Do you know other, more intuitive way solve it?

4. ## Re: Increasing by percent

Well, if you don't want to use cpd. int. formula, then:
Code:
YEAR   100%    POPULATION
0               3000
1    3000       6000
2    6000      12000
2[2(3000)] = 12000

5. ## Re: Increasing by percent

Originally Posted by Jotaro
I didn't know this formula. Thanks! I'm going to search derivation of it.

But I don't think that I should solve this problem with compound interest formula. There's no word about it in the textbook. Do you know other, more intuitive way solve it?
Well assume it is increasing by x% annually.

After one year the population is $\displaystyle 3000*(1+\frac{x}{100})$

After 2 years the population is $\displaystyle 3000 * (1 +\frac{x}{100})(1+\frac{x}{100}) = 12000$. Solve for x.