Question

⭐Question″›

A sum of money is put on CI for 2 years at 20%. It would fetch Rs 482 more if the interest is payable half yearly than if it were payable yearly. Find the sum.

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Answer 1

|| Concept ||

$\huge\sf\color{red} {Additional\: Information}$

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$The amount of money lent or borrowed is called the principal.

The additional money charged for its use is called Interest

If the interest due is added to the principal,the total money we receive or pay is called amount due at time

Important formulae :

Amount = principal + Interest

Interest = $\frac{Principal×Rate×time}{100}$

Rate = $\frac{Interest×100}{Principal×time}$

Time = $\frac{Interest×100}{Principal×rate}$

$\large\underline\color{green}{\sf{Given\: :-}}$

• A sum of money is put on CI for 2 years at 20%. It would fetch Rs 482 more if the interest is payable half yearly than if it were payable yearly. Find the sum.

$\large\underline\color{green}{\sf{To\: Find\: :-}}$

• The Sum = ?

$\large\underline\color{green}{\sf{Solution\: :-}}$

Let , the sum of money lent out be Rs..x on compound interest

For compound interest :

$\red{ = \geqslant } \: A\: = P\:(1 \dfrac{R}{100^{T}}$

Where, $\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ A : amount after 2 years

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ P : principal invested

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ R : Rate of interest.

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ T : time period

According to question, we have

• P = Rs. x
• R = 20%
• T = 2 years

$\:\:\:\:\:\:\:\:\:\:$$A_{1}$ = x {1 + $\dfrac{20}{100}$²

$\:\:\:\:\:\:\:\:\:\:\huge\sf\implies$ x (1 + $\dfrac{1}{5}$²

$\:\:\:\:\:\:\:\:\:\:\huge\sf\implies$ $\dfrac{6}{5}$²

$\color{Red}\boxed{Rs.\dfrac{36x}{25}}$

_____________________________

$\large\underline\color{magenta}{\sf{Case2\: :-\: Interest\: payable\:half\:yearly}}$

Where,

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ P : principal invested

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ R : Rate of interest.

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ T : time period

According to question, we have

P = Rs. x

R = $\dfrac{20}{2}$ = 10%

T = 2×2 = 4 years

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $A_{2}$ = x {1 + $\dfrac{20}{100}$⁴

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ x (1 + $\dfrac{1}{10}$⁴

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ x $\dfrac{11}{10}$⁴

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$$\color{red}\boxed{\dfrac{14641x}{10000}}$

__________________________

According to the question,

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $A_{2}$ - $A_{1}$ = $\color{red}{482}$

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $\dfrac{14641x}{10000}}$ - $\dfrac{3643}{25}$ = $\color{red}{482}$

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $\dfrac{146414x\: -\: 14400x}{10000}}$ = $\color{red}{482}$

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ 241x = 482000

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ x = $\dfrac{482×1000}{241}$

$\:\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $\color{Red}\boxed{Rs.2000}$

Hence , answer is ₹2000.

Answer 2

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

Let assume that sum invested be Rs x.

We know,

Compound interest (CI) 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

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

and

Compound interest (CI) received on a certain sum of money of Rs P invested at the rate of r % per annum compounded half yearly for n years is given by

$\boxed{ \rm{ \:CI = P {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: - \: P \: }}$

Now, According to statement, it is given that compound interest is Rs 482 more than when Rs x invested at the rate of 20 % per annum for 2 years compounded half yearly than annually.

It means

$\rm \: CI_{(half\:yearly)} - CI_{(yearly)} = 482$

$\rm \: x{\bigg[1 + \dfrac{20}{200} \bigg]}^{2 \times 2} - x - \bigg(x {\bigg[1 + \dfrac{20}{100} \bigg]}^{2} - x \bigg) = 482$

$\rm \: x{\bigg[1 + \dfrac{1}{10} \bigg]}^{4} - x {\bigg[1 + \dfrac{1}{5} \bigg]}^{2}= 482$

$\rm \: x{\bigg[\dfrac{10 + 1}{10} \bigg]}^{4} - x {\bigg[\dfrac{5 + 1}{5} \bigg]}^{2}= 482$

$\rm \: x{\bigg[\dfrac{11}{10} \bigg]}^{4} - x {\bigg[\dfrac{6}{5} \bigg]}^{2}= 482$

$\rm \: \dfrac{14641}{10000}x - \dfrac{36}{25}x = 482$

$\rm \: \dfrac{14641x - 14400x}{10000} = 482$

$\rm \: \dfrac{241x}{10000} = 482$

$\rm \: \dfrac{x}{10000} = 2$

$\bf\implies \:x \: = \: Rs \: 20000$

So, sum invested is Rs 20000

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

Additional information :-

1. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded 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

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