pepo wrote:

Bunuel wrote:

pepo wrote:

Great Post and thanks Bunuel for sharing it!

I have one question based on the following question:

Example: If $20,000 is invested at 12% annual interest, compounded quarterly, what is the balance after 2 year?

Solution: Balance=20,000∗(1+0.12/4)^2∗4=Balance=20,000∗(1+0.12/4)^2∗4=

=20,000∗(1.03)^8=25,335.4

20,000∗(1.03)^8: How am I supposed to perform this calculation in less than 2 minutes? there must be a shortcut in order to avoid perform the exponents.

Does someone knows something about?

Thanks in advance!

Please read the whole thread:

math-number-theory-percents-91708-40.html#p1202382Thank you.

Already read it, I but don't get the easyness

You are considering this question in complete vacuum and as such not helping yourself in understanding the nuances involved.

You can also look at it this way: for any period >1 , compound interest > simple interest (or CI is JUST greater than SI) for the same rate.

Rate of interest for quarterly compounding = 12/3 = 3%

Number of quarters in 2 years = 8

Thus SI for this= 20000*3*8/100 = 4800 ---> balance after 2 years = 20000+4800=24800 and hence the final answer will be just greater than 24800. Now in GMAT PS, if the options given to you are spread far apart, then this approximation will give you the correct answer.

But if not, then use the binomial theorem to calculate \((1.03)^8\): \((1.03)^8 = (1+0.03)^8 = 1^8*(0.03)^0+8*1^7*(0.03)^1+28*1^6*(0.03)^2\).... (this last term and the following terms will have 0.03 in 0.03^3 and higher powers making these terms very small compared with others, so neglect them).

Thus, \((1.03)^8 = (1+0.03)^8 = 1^8*(0.03)^0+8*1^7*(0.03)^1 = 1+0.24 \approx\) 1.24

Hope this helps.

P.S.: Binomial theorem expansion for \((a+b)^n = \sum_{\substack{0\leq k\leq n}} \binom{n}{k} a^n*b^k\), where \(\binom{n}{k} = {C^n_k}\)