Question

A sum of money placed at C.I. triple itself in 9 years. It will amount to nine times itself in? Years

Answer 1

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

Given that,

A sum of money placed at C.I. triple itself in 9 years.

Let assume that

Sum invested = Rs P

So, Amount = 3P

Time, n = 9 years

Let further assume that rate of interest = r % per annum compounded annually.

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 years is given by

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

$\rm \: 3P = P {\bigg[1 + \dfrac{r}{100} \bigg]}^{9}$

$\rm\implies \:\rm \: 3= {\bigg[1 + \dfrac{r}{100} \bigg]}^{9} - - - (1)$

Now, According to statement,

Sum invested = Rs P

Amount = 9P

Rate of interest = r % per annum

Let assume that the required time be n years.

We know,

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

So, on substituting the values, we get

$\rm \: 9P = P {\bigg[1 + \dfrac{r}{100} \bigg]}^{n}$

$\rm \: {3}^{2} = {\bigg[1 + \dfrac{r}{100} \bigg]}^{n}$

$\rm \: {\bigg( {\bigg[1 + \dfrac{r}{100} \bigg]}^{9} \bigg) }^{2} = {\bigg[1 + \dfrac{r}{100} \bigg]}^{n}$

$\rm \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{18} = {\bigg[1 + \dfrac{r}{100} \bigg]}^{n}$

$\rm\implies \:n \: = \: 18 \: years$

Hence,

Sum of money placed at C.I. triple itself in 9 years. It will amount to nine times itself in 18 years.

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

Additional Information :-

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

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

2. 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

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

3. 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

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