How to find effective rate of interest?

Question

A man charges at the rate of $10$ paise per rupee per month, payable in advance. What effective rate of interest does he charges per annum?

Here rate is $10\%$ per month. So to change it in per annum, I divided it by $12$ and set $n$ to $1$.

$$\text{effective rate}=\left(1+ \frac{r}{100}\right)^n -1$$

After applying the formula, I got

$$\left(1+ \frac {10}{1200}\right) -1$$

But I am getting the wrong answer and the correct answer is $254.5\%$. Please help me understand this.

Answer 1

My answer is 213.84%

$(1+0.1)^{12}-1=2.1384$

The formula is $(1+\frac{r}{n})^{n}-1$ where r stands for annual interest rate and n stands for number of compounding periods per year. Since 10% is the monthly rate, $\frac {r}{n}$=0.1, n=12

Answer 2

Since the man charges interest at the rate of $10$ paise per rupee per month, payable in advance, this implies that $10$ paise can be treated as interest on $90$ paise.

For example, if you borrow $100$ paise and pay $10\%$ interest in advance, then you have only borrowed $90$ paise. Therefore the monthly interest rate $j$ is calculated as follows $$j=\frac{10}{100-10}=\frac{10}{90}=\frac19$$ Since we already know the monthly interest rate, we can use the following to find the effective interest rate $$r=\left(1+j\right)^{12}-1$$ If you're curious about the annual nominal interest rate, simply multiply the monthly interest rate by $12$. $$i=12j=\frac{12}{9}=\frac{4}{3}$$ Which would allow you to use the following $$r=\left(1+\frac{i}{12}\right)^{12}-1$$ Both approaches eventually show that $r\approx 2.545$, after rounding. $$r=\left(1+\frac19\right)^{12}-1$$ $$\log_{10}\left(r+1\right)=12\log_{10}\left(\frac{10}{9}\right)$$ $$r=10^{12\left(1-\log_{10}9\right)}-1$$ At this point, the author used $$\log_{10}9\approx 0.9542$$ Which leads to the following $$r\approx 10^{0.5496}-1$$ Surely you can take it from here.

Answer 3

You pay $10$ in advance to have $100$, that is you obtain $90$ leaving $10$ as interest; so you have a monthly interest rate of $$ i_m=\frac{10}{90}=11.11\% $$ So you have that the yearly interet rate is $i$, given by $$ (1+i_m)^{12}=1+i\qquad\Longrightarrow\qquad i=(1+i_m)^{12}-1\approx 254.1\% $$