Question

Kiran borrows 187500 from a bank at 4% per annum compounded annually how much will she have to pay back after 2 years and 4 months

Answer 1

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

Principal, P = 187500

Rate of interest, r = 4 % per annum compounded annually.

Time period, n = 2 years 4 months = 2 1/3 years

We know,

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

$\boxed{\sf{ \: \: Amount \: = \: P {\bigg[1 + \dfrac{r}{100} \bigg]}^{n}\bigg[1 + \dfrac{r}{100} \times \frac{q}{s} \bigg] \: }}$

So, on substituting the values, we get

$\rm \: Amount = 187500 {\bigg[1 + \dfrac{4}{100} \bigg]}^{2}\bigg[1 + \dfrac{4}{100} \times \dfrac{1}{3} \bigg]$

$\rm \: Amount = 187500 {\bigg[1 + \dfrac{1}{25} \bigg]}^{2}\bigg[1 + \dfrac{1}{25} \times \dfrac{1}{3} \bigg]$

$\rm \: Amount = 187500 {\bigg[\dfrac{25 + 1}{25} \bigg]}^{2}\bigg[1 + \dfrac{1}{75}\bigg]$

$\rm \: Amount = 187500 {\bigg[\dfrac{26}{25} \bigg]}^{2}\bigg[\dfrac{75 + 1}{75}\bigg]$

$\rm \: Amount = 187500 {\bigg[\dfrac{26}{25} \bigg]}^{2}\bigg[\dfrac{76}{75}\bigg]$

$\bf\implies \:Amount \: = \: 205504$

So, Kiran have to pay 205504 at the end of 2 years and 4 months.

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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

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

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

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

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

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

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

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