Question

find the amount and compound interest on 24000 for 3 year at4 p, a, compound annually​

Answer 1

Principal (P)=Rs. 24000

Time (t)=2 years

Rate (r)=10%

Amount= Principal - (1+r2×100)n×2

=Rs. 24000×(1+10200)2×2

=Rs. 24000×(210200)4

=Rs. 24000×2120×2120×2120×2120

=Rs. 29172

 C.I.= Amount-Principal

=Rs. 29712−Rs. 24000=Rs. 5172

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Answer 2

Question :-

Find the amount and compound interest on Rs 24000 for 3 year at 4 % p.a. compounded annually.

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

Principal, P = Rs 24000

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

Time, 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

$\rm \: Amount \: = \: 24000 \: {\bigg[1 + \dfrac{4}{100} \bigg]}^{3} \: \:$

$\rm \: Amount \: = \: 24000 \: {\bigg[1 + \dfrac{1}{25} \bigg]}^{3} \: \:$

$\rm \: Amount \: = \: 24000 \: {\bigg[\dfrac{25 + 1}{25} \bigg]}^{3} \: \:$

$\rm \: Amount \: = \: 24000 \: {\bigg[\dfrac{26}{25} \bigg]}^{3} \: \:$

$\rm\implies \:Amount \: = \: Rs \: 26996.74$

Now,

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

$\rm \: Compound\:interest \: = \: 26996.74 - 24000$

$\rm\implies \:Compound\:interest = Rs \: 2996.74$

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

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

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