Question

Help me with the compound interest Problem.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Ranisha took a loan of Rs 12, 00, 000 at the rate of 15 % compounded half yearly for two years. But due to lockdown, the rate of interest reduced by 5 %.

It means,

Principal, P = Rs 12, 00, 000

Rate of interest, r = 10 % per annum compounded half yearly.

Time, n = 2 years

We know

Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded half - yearly for n years is given by

$\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: \: }}$

So, on substituting the values, we get

$\rm \: Amount = 1200000 {\bigg[1 + \dfrac{10}{100} \bigg]}^{2 \times 2}$

$\rm \: Amount = 12000000 {\bigg[1 + \dfrac{1}{10} \bigg]}^{4}$

$\rm \: Amount = 1200000 {\bigg[\dfrac{10 + 1}{10} \bigg]}^{4}$

$\rm \: Amount = 12000000 {\bigg[\dfrac{11}{10} \bigg]}^{4}$

$\rm \: Amount = 1200000 \times \dfrac{14641}{10000}$

$\rm \: Amount = 120 \times 14641$

$\bf\implies \:Amount \: = \: Rs \: 17,56,920$

Now, We know,

$\rm \: Compound\:interest \: = \: Amount \: - \: P$

$\rm \: Compound\:interest \: = \: 1756920 \: - \: 1200000$

$\bf\implies \:Compound\:interest = Rs \: 5,56,920$

$\rule{190pt}{2pt}$

Additional Information :-

1. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by

$\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: \: }}$

2. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi - annually for n years is given by

$\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: \: }}$

3. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is given by

$\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: \: }}$

4. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded monthly for n years is given by

$\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n} \: \: }}$