Question

The difference between the compound interest and the simple interest on a certain sum for 2 years at 5% per annum is $ 40. Find the sum​

Answer 1

Step-by-step explanation:

P=?

R=5%

T=2 years

C.I=P(1+R/100)^n-P

=P(1+1/20)^2-P

=P(21/20)^2-P

=441 P/400-P

=41P/400

S.I=(P×R×T)/100

P×5×2/100

P/10

C.I-S.I =40

41P/400-P/10=40

(41P-40P)/400=40

P/400=40

P=16000

Answer 2

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

Given that,

The difference between the compound interest and the simple interest on a certain sum for 2 years at 5% per annum is $ 40.

Let assume that the sum invested be $ x.

Case :- 1 Case of Simple interest

Principal, P = $ x

Rate of interest, r = 5 % per annum

Time period, n = 2 years

We know,

Simple interest received on a certain sum of money of P invested at the rate of r % per annum for n years is given by

$\boxed{\sf{  \: \: \: Simple \: interest \: = \: \frac{P \times r \times n}{100} \: \: \: }}$

So, on substituting the values, we get

$\rm \: Simple \: interest \: = \: \dfrac{x \times 5 \times 2}{100}$

$\rm\implies \:\rm \: Simple \: interest \: = \: \dfrac{x}{10}$

Case :- 2 Case of Compound interest

Principal, P = $ x

Rate of interest, r = 5 % per annum compounded annually

Time, n = 2 years

We know,

Compound interest 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

$\boxed{\sf{  \:Compound \: interest \: = \: P {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: - \: P \: \: \: }}$

So, on substituting the values, we get

$\rm \: Compound \: interest = x {\bigg[1 + \dfrac{5}{100} \bigg]}^{2} - x$

$\rm \: Compound \: interest = x {\bigg[1 + \dfrac{1}{20} \bigg]}^{2} - x$

$\rm \: Compound \: interest = x {\bigg[\dfrac{20 + 1}{20} \bigg]}^{2} - x$

$\rm \: Compound \: interest = x {\bigg[\dfrac{21}{20} \bigg]}^{2} - x$

$\rm \: Compound \: interest = x {\bigg[\dfrac{441}{400} \bigg]} - x$

$\rm \: Compound \: interest = \dfrac{441x - 400x}{400}$

$\rm\implies \:\rm \: Compound \: interest = \dfrac{41x}{400}$

Now, According to statement,

$\rm \: Compound \: interest \: - \: Simple \: interest \: = \: 40$

So, on substituting the values, we get

$\rm \: \dfrac{41x}{400} - \dfrac{x}{10} = 40$

$\rm \: \dfrac{41x - 40x}{400} = 40$

$\rm \: \dfrac{x}{400} = 40$

$\rm\implies \:x \: = \: 16000$

Hence,

The sum invested is $ 16000

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Additional Information :-

1. 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

$\boxed{\sf{  \:Amount \: = \: P {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: \: }}$

2. 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

$\boxed{\sf{  \:Amount \: = \: P {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: \: }}$

3. 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

$\boxed{\sf{  \:Amount \: = \: P {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: \: }}$

4. 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

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