Question

Q 10. In how many months will $12,000 yield $1,230 as compound interest at 10% per annum compounded semi-annually?
Ops: A. 18
B. 8
C. 16
D12​

Answer 1

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

Given that,

A sum of $ 12, 000 yield $ 1230 at the rate of 10 % per annum compounded semi - annually.

So,

Principal, p = $ 12, 000

Compound interest, CI = $ 1230

Rate of interest, r = 10 % per annum compounded semi annually.

Let assume that the time period be 'n' years.

We know,

Compound interest (CI) on a certain sum of money Rs p invested at the rate of r % per annum compounded semi - annually for n years is

$\red{\boxed{ \rm{ \: CI \: = \: p \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} - p \: \: }}}$

So, on substituting the values, we get

$\rm :\longmapsto\:1230 = 12000 {\bigg[1 + \dfrac{10}{200} \bigg]}^{2n} - 12000$

$\rm :\longmapsto\:1230 + 12000 = 12000 {\bigg[1 + \dfrac{1}{20} \bigg]}^{2n}$

$\rm :\longmapsto\:13230 = 12000 {\bigg[ \dfrac{20 + 1}{20} \bigg]}^{2n}$

$\rm :\longmapsto\: \dfrac{13230}{12000} = {\bigg[ \dfrac{21}{20} \bigg]}^{2n}$

$\rm :\longmapsto\: \dfrac{1323}{1200} = {\bigg[ \dfrac{21}{20} \bigg]}^{2n}$

$\rm :\longmapsto\: \dfrac{441}{400} = {\bigg[ \dfrac{21}{20} \bigg]}^{2n}$

$\rm :\longmapsto\: {\bigg[\dfrac{21}{20} \bigg]}^{2} = {\bigg[ \dfrac{21}{20} \bigg]}^{2n}$

So, on comparing, we get

$\bf\implies \:2n \: = \: 2$

$\bf\implies \:n \: = \: 1 \: year$

$\bf\implies \:n \: = \: 12 \: months$

• Hence, Option D. is correct.

Additional Information :-

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

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

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

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

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

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

Answer 2

Given,

The principal amount$\[ = 12000\]$

The compound interest$\[ = 1230\]$

The rate of interest in percent$\[ = 10\]$

To find,

The time period.

Solution,

Know that the formula to calculate the compound interest semi-annually is $\[CI = P{\left[ {1 + \frac{R}{{200}}} \right]^{2n}} - P\]$

Apply values.

$\[\begin{array}{l} \Rightarrow 1230 = 12000{\left[ {1 + \frac{{10}}{{200}}} \right]^{2n}} - 12000\\ \Rightarrow 1230 + 12000 = 12000{\left[ {1 + \frac{1}{{20}}} \right]^{2n}}\\ \Rightarrow 13230 = 12000{\left[ {\frac{{21}}{{20}}} \right]^{2n}}\end{array}\]$

Solve further,

$\[\begin{array}{l} \Rightarrow \frac{{13230}}{{12000}} = {\left[ {\frac{{21}}{{20}}} \right]^{2n}}\\ \Rightarrow {\left[ {\frac{{21}}{{20}}} \right]^2} = {\left[ {\frac{{21}}{{20}}} \right]^{2n}}\end{array}\]$

Compare.

$\[\begin{array}{l} \Rightarrow n = 1{\rm{year}}\\ \Rightarrow n = 12\,{\rm{months}}\end{array}\]$

Hence, the time period is $\[12\]$ months.