Question

Principal is 10,000 rupess at a compound intreast of rate 10 percent . Find tbe difference between compound intreast and simple intreast at tbe same rate for 2 years.​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Principal is 10,000 rupess at a compound interest of rate 10 percent per annum. Find tbe difference between compound interest and simple interest at the same rate for 2 years.

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

Case :- 1 Simple interest

Principal, P = Rs 10000

Rate of interest, r = 10 % per annum

Time, n = 2 years

We know,

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

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

So, on substituting the values, we get

$\rm \: SI \: = \: \frac{10000 \times 10 \times 2}{100} \:$

$\rm\implies \:\boxed{\sf{ \: \: SI \: = \: 2000 \: }}$

Case :- 2 Compound interest

Principal, P = Rs 10000

Rate of interest, r = 10 % per annum

Time, n = 2 years

We know,

C8 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

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

So, on substituting the values, we get

$\rm \: CI \: = \: 10000 {\bigg[1 + \dfrac{10}{100} \bigg]}^{2} - 10000$

$\rm \: CI \: = \: 10000 {\bigg[1 + \dfrac{1}{10} \bigg]}^{2} - 10000$

$\rm \: CI \: = \: 10000 {\bigg[ \dfrac{10 + 1}{10} \bigg]}^{2} - 10000$

$\rm \: CI \: = \: 10000 {\bigg[ \dfrac{11}{10} \bigg]}^{2} - 10000$

$\rm \: CI \: = \: 10000 \times \frac{121}{100} - 10000$

$\rm \: CI \: = \: 12100 - 10000$

$\rm\implies \:\boxed{\sf{ \: \: CI \: = \: 2100 \: \: }}$

Hence,

$\rm \: CI \: - \: SI$

$\rm \: = \: 2100 - 2000$

$\rm \: = \: 100$

Thus,

$\rm\implies \:\boxed{\sf{ \:CI \: - \: SI \: = \: Rs \: 100 \: }}$

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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

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

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

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

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

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

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

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

Answer 2

Given :

Principle (P) = 10,000 rupees

Rate (R ) = 10 percent

Number of Year = 2 years

To Find :

The difference between compound interest and simple interest at the same rate for 2 years.

Formula used :

$\rm \: C.I= P(1 + \dfrac{R}{100} {)}^{T} - P$

Where ,

C.I = Compound interest

P = initial principal balance

R = interest rate

T = number of time periods elapsed

How to solve ? :

This $\rm \: C.I= P(1 + \dfrac{R}{100} {)}^{T} - P$ formula is used to calculate compound interest . Put given value in this formula in order to calculated the compound interest value for 2 years

Calculation :

$\rm \implies \: 10,000 \times ( 1 + \dfrac{10}{100} {)}^{2} - 10,000$

$\rm \implies \: 10,000 \times ( 1 + \dfrac{1}{10} {)}^{2} - 10,000$

$\rm \implies \: 10,000 \times ( \dfrac{10 + 1}{10} {)}^{2} - 10,000$

$\rm \implies \: 10,000 \times ( \dfrac{11}{10} {)}^{2} - 10,000$

$\rm \implies \: 10,000 \times \dfrac{121}{100} - 10,000$

$\rm \implies \: 100 \times 121 - 10,000$

$\rm \implies \: 12,100 - 10,000$

$\rm \implies \: 2,100 \: Rs$

Therefore,

Compound interest for 2 years

• ➡ 2,100 Rs

Now :

We have to find Simple Interest So ,

Formula used :

$\rm \: S.I= ( \dfrac{P \times R \: \times \: T }{100})$

where,

• P = Principal,

• R = Rate of Interest,

• T = Time period.

Calculation :

$\rm \: S.I= ( \dfrac{10,000 \times 10\: \times \: 2}{100})$

$\rm \: S.I= ( \dfrac{200,000}{100})$

$\rm \: S.I=2000\: Rs$

Therefore ,

Required Simple Interest is

• ➡ 2000 Rs

Now ,

We have to find difference between compound interest and simple interest at the same rate for 2 years.

So,

Difference between C.I and S.I

$\rm \implies \: C. I - S .I$

$\rm \implies \: 2100 \: Rs - 2000 \: Rs$

$\rm \implies \: \boxed{100\: Rs}$

Hence ,

• Final Required answer is

$\rm \implies \: \boxed{100\: Rs}$