Question

The compound interest on $\$1000$ in 2 years compounded yearly is $\$440.$ Find the rate of interest

• A. $20\%$
• B. $10\%$
• C. $15\%$
• D. $12\%$

Answer 1

The correct option is A $20\%$

$\text{Given, P = Principal}=\$1000$
$\text{t = Time period = 2 years}$
$n=1,\text{as interest is compounded 1 times per year}$

$\text{Amount = Principal + Compound Interest}$
$=1000+440$
$=1440$

$\text{Let, rate of interest (in \%)}=r$
$\because \text{Amount}=P{(1+\frac{r}{100\times n})}^{n\times t}$
$\Rightarrow 1440=1000{(1+\frac{r}{100\times 1})}^{1\times 2}$
$\Rightarrow 1440=1000{(1+\frac{r}{100})}^2$
$\Rightarrow \frac{1440}{1000}={(1+\frac{r}{100})}^2$
$\Rightarrow \frac{144}{100}={(1+\frac{r}{100})}^2$

Taking square root both the sides, we get
${\left(\frac{144}{100}\right)}^{\frac{1}{2}}={(1+\frac{r}{100})}^{2\times \frac{1}{2}}$
$\Rightarrow {\left(\frac{12\times 12}{10\times 10}\right)}^{\frac{1}{2}}=1+\frac{r}{100}$
$\Rightarrow {\left(\frac{{12}^2}{{10}^2}\right)}^{\frac{1}{2}}=1+\frac{r}{100}$
$\Rightarrow {\left(\frac{12}{10}\right)}^{2\times \frac{1}{2}}=1+\frac{r}{100}$
$\Rightarrow \frac{12}{10}=1+\frac{r}{100}$
$\Rightarrow \frac{12}{10}-1=\frac{r}{100}$
$\Rightarrow \frac{12}{10}-\frac{10}{10}=\frac{r}{100}$
$\Rightarrow \frac{12-10}{10}=\frac{r}{100}$
$\Rightarrow \frac{2}{10}=\frac{r}{100}$

Cross multiplying, we get
$2\times 100=r\times 10$
$\Rightarrow 200=r\times 10$

Dividing both the sides by $10,$ we get
$\frac{200}{10}=r$
$\Rightarrow r=20$

Hence, required rate of interest is $20\%.$