
What sum of money will amount to 18150 in 2 years at 10% per annum, compounded annually? What sum of money will amount to 793170 in 3 years at 10% per annum compounded annually?
Answer
586.5k+ views
Hint: The compound interest is calculated by the formula \[A = P{\left( {1 + \dfrac{r}{100}} \right)^{nt}}\] where A is the final amount, P is initial principal balance, r is the interest rate n is the number of times interest applied per time period and t is the number of time period elapsed.
Complete step-by-step answer:
For the first question we are given the final amount as 18150 and the rate of interest as 10% and it is also given that it is getting compounded annually therefore once in a year also the time elapsed is given as 2 years
So we will put all of these in \[A = P{\left( {1 + \dfrac{r}{100}} \right)^{nt}}\] and try to find the value of P.
\[\begin{array}{l}
\begin{array}{*{20}{l}}
{A = P{{\left( {1 + \dfrac{r}{{100}}} \right)}^{nt}}}\\
{ \Rightarrow 18150 = P{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^{1 \times 2}}}
\end{array}\\
\begin{array}{*{20}{l}}
{ \Rightarrow 18150 = P{{(1.1)}^2}}\\
{ \Rightarrow 18150 = P \times 1.21}\\
{ \Rightarrow \dfrac{{18150}}{{1.21}} = P}\\
{ \Rightarrow P = 15000}
\end{array}
\end{array}\]
Therefore the principal amount is 150.
For the next part also we will use the same formula just the values will be changed accordingly as in place of amount we will put 793170, inplace of t it will be 3 and r and n will be 10 and 1 respectively.
\[\begin{array}{*{20}{l}}
{\therefore A = P{{\left( {1 + \dfrac{r}{n}} \right)}^{nt}}}\\
{ \Rightarrow 793170 = P{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^{1 \times 3}}}\\
{ \Rightarrow 793170 = P{{(1.1)}^3}}\\
{ \Rightarrow 793170 = P \times 1.331}\\
{ \Rightarrow \dfrac{{793170}}{{1.331}} = P}\\
{ \Rightarrow P = 595920.3}
\end{array}\]
So from here we are getting the principal value as 595920.3.
Note: n is the number of times interest is compounded over a year many students make the mistake by replacing it with the total number of times the interest gets compounded. For example if an interest is getting compounded half yearly for 5 years then the value of n is 2 not 10.
Complete step-by-step answer:
For the first question we are given the final amount as 18150 and the rate of interest as 10% and it is also given that it is getting compounded annually therefore once in a year also the time elapsed is given as 2 years
So we will put all of these in \[A = P{\left( {1 + \dfrac{r}{100}} \right)^{nt}}\] and try to find the value of P.
\[\begin{array}{l}
\begin{array}{*{20}{l}}
{A = P{{\left( {1 + \dfrac{r}{{100}}} \right)}^{nt}}}\\
{ \Rightarrow 18150 = P{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^{1 \times 2}}}
\end{array}\\
\begin{array}{*{20}{l}}
{ \Rightarrow 18150 = P{{(1.1)}^2}}\\
{ \Rightarrow 18150 = P \times 1.21}\\
{ \Rightarrow \dfrac{{18150}}{{1.21}} = P}\\
{ \Rightarrow P = 15000}
\end{array}
\end{array}\]
Therefore the principal amount is 150.
For the next part also we will use the same formula just the values will be changed accordingly as in place of amount we will put 793170, inplace of t it will be 3 and r and n will be 10 and 1 respectively.
\[\begin{array}{*{20}{l}}
{\therefore A = P{{\left( {1 + \dfrac{r}{n}} \right)}^{nt}}}\\
{ \Rightarrow 793170 = P{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^{1 \times 3}}}\\
{ \Rightarrow 793170 = P{{(1.1)}^3}}\\
{ \Rightarrow 793170 = P \times 1.331}\\
{ \Rightarrow \dfrac{{793170}}{{1.331}} = P}\\
{ \Rightarrow P = 595920.3}
\end{array}\]
So from here we are getting the principal value as 595920.3.
Note: n is the number of times interest is compounded over a year many students make the mistake by replacing it with the total number of times the interest gets compounded. For example if an interest is getting compounded half yearly for 5 years then the value of n is 2 not 10.
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