Question

7. A sum of money becomes 26,471 at an interest rate of 3% per annum compounded annually after a period of one year. Find the sum.​

Answer 1

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

Given that,

A sum of money becomes ₹ 26,471 at an interest rate of 3% per annum compounded annually after a period of one year.

So we have

Amount = ₹ 26471

Rate of interest, r = 3 % per annum compounded annually.

Time, n = 1 year

Let assume that sum invested be ₹ P.

We know,

Amount received on a certain sum of money of ₹ 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

$\rm \: 26471 = P {\bigg[1 + \dfrac{3}{100} \bigg]}^{1}$

$\rm \: 26471 = P {\bigg[\dfrac{100 + 3}{100} \bigg]}^{1}$

$\rm \: 26471 = P {\bigg[\dfrac{103}{100} \bigg]}^{}$

$\rm \: P = 26471 \times {\bigg[\dfrac{100}{103} \bigg]}^{}$

$\rm\implies \:P \: = \: 25700$

Hence,

₹ 25700 becomes ₹ 26,471 at an interest rate of 3% per annum compounded annually after a period of one year.

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

Additional Information :-

1. Amount received on a certain sum of money of ₹ 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 ₹ 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 ₹ 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

SOLUTION —>

GIVEN:A sum of money becomes ₹26,471 at an interest rate of 3% per annum compounded annually after a period of 1 year. Find the sum.

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AMOUNT=₹26,471

RATE = 3%(LET'S TAKE RATE AS r)

Time=1 year

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Formula used=>

$p \times (1 + \frac{r}{100} )^{t}$

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LET'S DO

$p(1 + \frac{3}{100} )^{1} = 26471$

$p \times \frac{103}{100} = 26471$

$p = 25700$

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ADDITIONAL INFORMATION

The amount can be calculated half-yearly. When the amount is calculated half-yearly the rate becomes half.

The amount can be calculated quarterly. When the amount is calculated quarterly the rate becomes 1/4 .