Question

On a particular amount, the Compound interest at the end of 1 year is 40 and the 2nd year is 42.how much money was deposited??​

Answer 1

SOLUTION

GIVEN

On a particular amount, the Compound interest at the end of 1 year is 40 and the 2nd year is 42.

TO DETERMINE

The amount deposited

EVALUATION

Let amount deposited = P

Rate of interest = r%

Then Compound interest at the end of 1 year

$\displaystyle \sf{ = P { \bigg(1 + \frac{r}{100} \bigg)}^{1} - P}$

$\displaystyle \sf{ = \frac{ Pr}{100} }$

Again Compound interest at the end of 2nd year

$\displaystyle \sf{ = P { \bigg(1 + \frac{r}{100} \bigg)}^{2} - P { \bigg(1 + \frac{r}{100} \bigg)}^{1}}$

$\displaystyle \sf{ = P { \bigg(1 + \frac{r}{100} \bigg)}^{} { \bigg(1 + \frac{r}{100} - 1 \bigg)}^{}}$

$\displaystyle \sf{ = P { \bigg(1 + \frac{r}{100} \bigg)}^{} \frac{r}{100} }$

By the given condition

$\displaystyle \sf{ \frac{Pr}{100} = 40 \: \: - - - (1)}$

$\displaystyle \sf{ P { \bigg(1 + \frac{r}{100} \bigg)}^{} \frac{r}{100} = 42 \: \: - - - (2)}$

Equation 2 ÷ Equation 1 gives

$\displaystyle \sf{ { \bigg(1 + \frac{r}{100} \bigg)}^{} = \frac{42}{40} }$

$\displaystyle \sf{ \implies \: \frac{r}{100} = \frac{42}{40} - 1}$

$\displaystyle \sf{ \implies \: \frac{r}{100} = \frac{42 - 40}{40} }$

$\displaystyle \sf{ \implies \: \frac{r}{100} = \frac{2}{40} }$

$\displaystyle \sf{ \implies \: r = 5}$

From Equation 1 we get

$\displaystyle \sf{ \frac{5P}{100} = 40 }$

$\displaystyle \sf{ \implies P= 40 \times \frac{100}{5} }$

$\displaystyle \sf{ \implies P= 40 \times 20 }$

$\displaystyle \sf{ \implies P= 800}$

FINAL ANSWER

The amount Deposited = Rs 800