Question

The difference between the compound interest and the simple interest on a certain sum of money for
2 years at 5% per annum is 100. Find the sum.
​

Answer 1

Answer:

$\boxed{\sf \: Sum\:invested\:is\:40000 \: }$

Step-by-step explanation:

Given that,

Rate of interest, r = 5 % per annum

Time period, n = 2 years

Difference between CI and SI = Rs 100

Let assume that sum invested be Rs P.

Now, it is given that

$\sf \: CI - SI = 100$

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

On substituting the values, we get

$\sf \: P {\bigg[1 + \dfrac{5}{100} \bigg]}^{2}  - P - \dfrac{P \times 5 \times 2}{100}  = 100$

$\sf \: P {\bigg[1 + \dfrac{1}{20} \bigg]}^{2}  - P - \dfrac{P}{10}  = 100$

$\sf \: P {\bigg[\dfrac{20 + 1}{20} \bigg]}^{2}  - P - \dfrac{P}{10}  = 100$

$\sf \: P {\bigg[\dfrac{21}{20} \bigg]}^{2}  - P - \dfrac{P}{10}  = 100$

$\sf \: \dfrac{441P}{400}   - P - \dfrac{P}{10}  = 100$

$\sf \: \dfrac{441P - 400P - 40P }{400}  = 100$

$\sf \: \dfrac{41P - 40P }{400}  = 100$

$\sf \: \dfrac{P }{400}  = 100$

$\sf \: P = 400 \times 100$

$\sf\implies \sf \: P = 40000$

Hence,

$\sf\implies \sf \: Sum\:invested\:is\:40000$

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

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

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