Question

Solve this compund interest​

Answer 1

Answer:

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Answer 2

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

Given that,

A sum of money ₹ 5120 amounts to ₹ 7290 at the rate of 12.5 % per annum compounded annually.

So, it implies,

Principal, P = ₹ 5120

Amount, A = ₹ 7290

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

Let time period be 'n' years.

We know,

Amount on a certain sum of money ₹ P invested at the rate of 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 of P, r and n, we get

$\rm :\longmapsto\:{ \rm{ \tt{ 7290=5120\bigg(1+\dfrac{12.5}{100}\bigg)^{n}}}}$

$\rm :\longmapsto\:{ \rm{ \tt{ 729=512\bigg(1+\dfrac{125}{1000}\bigg)^{n}}}}$

$\rm :\longmapsto\:{ \rm{ \tt{ \dfrac{729}{512} =\bigg(\dfrac{1000 + 125}{1000}\bigg)^{n}}}}$

$\rm :\longmapsto\:{ \rm{ \tt{ \dfrac{9 \times 9 \times 9}{8 \times 8 \times 8} =\bigg(\dfrac{1125}{1000}\bigg)^{n}}}}$

$\rm :\longmapsto\:{ \rm{ \tt{ {\bigg(\dfrac{9}{8} \bigg) }^{3} =\bigg(\dfrac{9}{8}\bigg)^{n}}}}$

So, on comparing, we get

$\bf\implies \:n \: = \: 3$

• Hence, Time period is 3 years.

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

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