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$10\%$ compounded quarterly, what is the equivalent rate of interest with monthly compounding?

Equivalent rate of interest$= (1+\frac{0.1}{3})^3 -1 =(1+0.033333)^3 -1 =0.1(nearly) =10\%$

Is this solution correct? I am not sure.

Irfan Ahmad
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  • You're looking to solve $1.1=(1+x)^3$? – PC1 Jul 15 '22 at 04:12
  • Interest at rate $r$ compounded $n$ times a year normally means that you have $(1 + \frac{r}{n})^n$ times your initial deposit out the end of the year. So technically, you want to solve $(1 + \frac{0.1}{4})^4 = (1 + \frac{r}{12})^{12}$ for $r$. Unless I'm misunderstanding the question. – Dan Jul 15 '22 at 04:23

1 Answers1

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The way compound interest rates are normally defined, an interest rate of $r$ compounded $n$ times per year means that after one year, you have $(1 + \frac{r}{n})^n$ times your original principal. Thus,

$$(1 + \frac{0.1}{4})^4 = (1 + \frac{r}{12})^{12}$$ $$4 \log 1.025 = 12 \log (1 + \frac{r}{12})$$ $$\log (1 + \frac{r}{12}) = \frac{\log 1.025}{3}$$ $$1 + \frac{r}{12} \approx 1.0082648376090522$$ $$\frac{r}{12} \approx 0.0082648376090522$$ $$r \approx 0.099178051308626$$

So 10% interest compounded quarterly equates to approximately 9.92% interest compounded monthly.

Dan
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