Question

solve this compund interest sum​

Answer 1

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Answer 2

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

Given that,

Atul invested ₹ 2000 for 2 years at a certain rate of interest compounded annually and received ₹ 2200 at the end of first year.

So, it implies

Principal, P = ₹ 2000

Time, n = 1 year

Amount, A = ₹ 2200

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

We know,

Amount on a certain sum of money ₹ P invested at the rate of interest r % per annum compounded annually for n years is

$\boxed{ \rm{ \tt{ Amount=P\bigg(1+\dfrac{r}{100}\bigg)^{n}}}}$

So, on substituting the values, we get

$\rm :\longmapsto\:{ \rm{ \tt{2200=2000\bigg(1+\dfrac{r}{100}\bigg)^{1}}}}$

$\rm :\longmapsto\:{ \rm{ \tt{2200=2000\bigg(1+\dfrac{r}{100}\bigg)}}}$

$\rm :\longmapsto\:{ \rm{ \tt{2200=2000+20r}}}$

$\rm :\longmapsto\:{ \rm{ \tt{2200 - 2000 = 20r}}}$

$\rm :\longmapsto\:{ \rm{ \tt{200 = 20r}}}$

$\bf\implies \:r = 10 \: \% \: per \: annum$

Now,

For second year,

Principal, P = ₹ 2200

Time, n = 1 year

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

So,

$\boxed{ \rm{ \tt{ Amount=P\bigg(1+\dfrac{r}{100}\bigg)^{n}}}}$

So, on substituting the values, we get

$\rm :\longmapsto\:{ \rm{ \tt{ Amount=2200\bigg(1+\dfrac{10}{100}\bigg)^{1}}}}$

$\rm :\longmapsto\:{ \rm{ \tt{ Amount=2200\bigg(1+\dfrac{1}{10}\bigg)^{}}}}$

$\rm :\longmapsto\:{ \rm{ \tt{ Amount=2200\bigg(\dfrac{10 + 1}{10}\bigg)^{}}}}$

$\rm :\longmapsto\:{ \rm{ \tt{ Amount=2200\bigg(\dfrac{11}{10}\bigg)^{}}}}$

$\rm :\longmapsto\:{ \rm{ \tt{ Amount=2420}}}$

So, Amount received after 2 years = ₹ 2420

Additional Information :-

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

$\boxed{ \rm{ \tt{ Amount=P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}}}$

2. Amount on a certain sum of money ₹ P invested at the rate of r % per annum compounded quarterly for n years is

$\boxed{ \rm{ \tt{ Amount=P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}}}$

3. Amount on a certain sum of money ₹ P invested at the rate of r % per annum compounded monthly for n years is

$\boxed{ \rm{ \tt{ Amount=P\bigg(1+\dfrac{r}{1200}\bigg)^{12n}}}}$