email2vm wrote:
How much more interest will maria receive if she invests 1000$ for one year at x % annual interest, compounded semianually, than if she invest 1000$ for one year at x percent annual interest, compounded annually?
a> 5x
b> 10x
c> x^2/20
d> x^2/40
e>10x+ (x^2/40)
Here is how i solved only to get it wrong:
let x=100%
annually :
SI = 1000*100/100*1 = 1000
total =1000+1000=2000
semiannually:
Si (for 1st 6 months) = 1000
total after 6 months =2000
Si(for last 6 months) = 2000
total after 6 months= 4000
more interest maria will receive = 2000
====looking for the answer choices
100^2/40 != 2000!!!
where am I wrong? Today it seems I am back to square 1
Ravi
Hi,
the correct way is--
whenever amount is compounded other than annually, reduce the interest that many time and increase the time period that many times..lets see in this Q..
it is semi annually that is twice in a year...
so our time becomes 2n and rate of interest = r/2..
1) when annually Amount =\(1000( 1+\frac{r}{100})^1 = 1000(1+\frac{r}{100})\)
2) when semi annually\(1000( 1+\frac{r}{2*100})^2 = 1000(1+\frac{r}{200}^2)+r/100\)
subtract 1 from 2..
\(1000( 1+\frac{r}{2*100})^2 = 1000(1+\frac{r}{200}^2+\frac{r}{100} - 1000( 1+\frac{r}{100})^1\)..
\(1000(1+\frac{r}{200}^2+\frac{r}{100}-(1+\frac{r}{100})\)..
\(1000(\frac{r}{200}^2)\) = \(r^2/40\)
D
Could you please tell me how did you deduce the second condition where it says semi annually.. It seems you have seperated r/100. I couldn't figure out how ?