Question

Calculate the compound interest and amount on RS 1,07,520at 12.5%p.a.for 3years if the interest is Compounded annually​

Answer 1

Answer:

$\boxed{\begin{aligned}& \qquad \:\sf \: Amount = Rs \: 153090 \qquad \: \\ \\& \qquad \:\sf \:Compound\:interest = Rs \: 45570\qquad \end{aligned}} \:$

Step-by-step explanation:

Given that,

Principal, P = Rs 107520

Rate of interest, r = 12.5 % per annum compounded annually

Time period, n = 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 years is given by

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

.So, on substituting the values, we get

$\sf \: Amount = 107520 {\left[ 1 + \dfrac{12.5}{100} \right]}^{3}$

$\sf \: Amount = 107520 {\left[ 1 + \dfrac{1}{8} \right]}^{3}$

$\sf \: Amount = 107520 {\left[ \dfrac{8 + 1}{8} \right]}^{3}$

$\sf \: Amount = 107520 {\left[ \dfrac{9}{8} \right]}^{3}$

$\sf \: Amount = 107520 \times \dfrac{729}{512}$

$\sf \: Amount = 210 \times 729$

$\sf\implies \bf \: Amount = Rs \: 153090$

Now,

$\sf \: Compound\:interest = Amount - Principal$

$\sf \: Compound\:interest = 153090 -107520$

$\sf\implies \bf \: Compound\:interest = Rs \: 45570$

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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

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

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

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

Answer 2

GIVEN IN QUESTION – :

$\sf \: Total \: amount (Principal ) = ₹ 107520$

Rate of interest (r) = 12.5 % per annum compounded annually

Time period (n)= 3 years

We have ,

$\sf Amount=P[ 1 \times \frac{ r}{100} ] ^{n}$

So putting the values in above we get ,

$\sf \: Amount \: = 107520[1 + \frac{12.5}{100} ] ^{3}$

$\sf \: Amount \: = \: 107520(1 + \frac{1}{8} ) ^{ 3}$

$\sf \: Amount = 107520( \frac{9}{8} )^{3}$

$\sf \: Amount \: = 107520 \times \frac{729}{512}$

$\sf \: Amount \: =₹ 153,090$

Now we know that ,

$\sf \: Amount \: = Principal \: + \: compound \: interest$

$\sf \: = > \: Compound \: interest \: = 153090 - 107520 \ \sf \\ \\ \sf \: Compound \: interest = ₹45,570 \:$