Question

Tricia sold her car and invested her Php350,000 at 8% compounded quarterly. Find the interest if she invests for 3 years.​

Answer 1

Principal = ₹ 350000 denoted by P

Rate of interest = 8% denoted by R

Time = 3 years denoted by N

$\color{lime}\boxed{ \sf{Amount\:=\:P\: \left[\:1+\:\frac{r}{200} \right]^{4n}}}$

Substituting the values, we get,

$Amount\:=\:350000 \left[\:1+\:\frac{8}{200}\right]^{4 \: \times \: 3 }$

$Amount\:=\:35000\left[\:1+\:\frac{1}{25}\right]^{12}$

$Amount\:=\:350000\left[ \frac{1 \times 25 \: + 1}{25} \right]^{12}$

$Amount\:=\:350000\left[\:\frac{26}{25} \: \right]^{12}$

$Amount\:=\:350000\: \times \: {1.04}^{12}$

$\sf{Amount\:=\:₹\:381600\:\:(approx)}$

$Compound\:Intrest\:=\: Amount\:-\: Principal \\ \\ Compound\:Intrest\:=\:381600 \: - \: 350000 \\ \\ Compound\:Intrest\:=\:₹ \: 31600$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\color{aqua} \boxed{ \begin {array} { |c|c|c|} Statement&Formulas \\ \\ \\❑ \: Amount \: on \: a \: certain \: sum \: of \: money \: of \: P \: invested \: at \: the \: rate \: of \: r \% \: per \: \\ annum \: compounded \: annually \: for \: n \: years \: is \: given \: by &❑ \: Amount\:=\:P[1\:+\: \frac{r}{100}]^n \\ \\ ❑ \: Amount \: on \: a \: certain \: sum \: of \: money \: of \: P \: invested \: at \: the \: rate \: of \: r \% \: per \\ \: annum \: compounded \: quarterly \: for n \: years \: is \: given \: by &❑ \:Amount\:=\:P[1\:+\: \frac{r}{200}]^{4n} \\ \\ ❑ \:Amount \: on \: a \: certain \: sum \: of \: money \: of \: P \: invested \: at \: the \: rat e \: of \: r \% \: per \: \\ annum \: compounded \: monthly \: for \: n \: years \: is \: given \: b y&❑\:Amount\:=\:P\:[1\:+\: \frac{r}{1200}^{12n} \end {array}}$

Answer 2

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

Given that,

Tricia sold her car and invested her Php350,000 at 8% compounded quarterly for 3 years.

It means,

Principal, P = Php350000

Rate, r = 8 % per annum compounded quarterly

Time, n = 3 years

We know,

Compound interest (CI) 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

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

So, on substituting the values, we get

$\rm \: CI \: = \: 350000 {\bigg[1 + \dfrac{8}{400} \bigg]}^{12} - 350000$

$\rm \: CI \: = \: 350000 {\bigg[1 + \dfrac{2}{100} \bigg]}^{12} - 350000$

$\rm \: CI \: = \: 350000 {\bigg[ \dfrac{100 + 2}{100} \bigg]}^{12} - 350000$

$\rm \: CI \: = \: 350000 {\bigg[ \dfrac{102}{100} \bigg]}^{12} - 350000$

$\rm \: CI \: = \: 350000 {(1.02)}^{12} - 350000$

$\rm \: CI \: = \: 350000 (1.268) - 350000$

$\rm \: CI \: = \: 443800 - 350000$

$\rm\implies \:CI \: = \: 93800$

Hence,

$\rm\implies \:\boxed{\sf{ \:Compound \: Interesr \: received \: = \: Php93800 \: }}$

Remark :-

How to evaluate

$\rm \: {(1.02)}^{12}$

Let assume that

$\rm \:x = {(1.02)}^{12}$

Taking log on both sides, we get

$\rm \:logx = log{(1.02)}^{12}$

$\rm \:logx = 12 \: log(1.02)$

$\rm \:logx = 12 \: (0.0086)$

$\rm \:logx = 0.1032$

On taking antilog on both sides, we get

$\rm \: x = antilog(0.1032)$

$\rm \: x = {10}^{0} \times 1.268$

$\rm\implies \:x = 1.268$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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

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