Question

What is the effective rate of interest if $1,000 compounds to $1,331 in 3 years? The effective rate of interest, if $1,000 compounds to $1,331 in 3 years is b

Answer 1

Answer:

The answer is "10".

Explanation:

Given:

T= 3 year

P= $ 1000

A= $ 1331

R=?

Formula:

$\bold{A= P(1+r)^t}$

$\Rightarrow 1331= 1000(1+\frac{r}{100})^3\\\\\Rightarrow (1+\frac{r}{100})^3=\frac{1331}{1000}\\\\\Rightarrow (1+\frac{r}{100})^3=(\frac{11}{10})^3\\\\\Rightarrow (1+\frac{r}{100})=(\frac{11}{10})\\\\\Rightarrow \frac{r}{100}=\frac{11}{10}-1\\\\\Rightarrow \frac{r}{100}=\frac{11-10}{10}\\\\\Rightarrow \frac{r}{100}=\frac{1}{10}\\\\\Rightarrow r=\frac{1}{10}\times 100\\\\\Rightarrow r=10$