Question

Maria invested \[\text{Rs }8000\] in a business. She would be paid interest \[5%\] per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the third year.

Answer 1

Hint: Use compound interest formula for the calculation of amount $A$, given by: \[A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}\]. From this calculate amount $A$ after \[t=2\text{ years}\] and \[t=3\text{ years}\] and then take the difference for the calculation of interest obtained in\[3\text{rd year}\].

Complete step-by-step answer:
Compound interest is the addition of interest to the principal sum of a loan or deposit. It is the result of reinvesting interest, rather than paying it out, so the interest in the next period is then earned on the principal sum plus previously accumulated interest.

The total accumulated amount $A$ , on the principal sum \[P\] plus compound interest $I$ is given by the formula \[A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}\].

 Here, \[A\] is the amount obtained, $t$ is the number of years, \[r\] is the rate, $P$ is the principal and \[n\] is the number of times the interest is given in a year.

The total compound interest generated is given by: $I=A-P$.

Now, we have been given that:

$P=\text{Rs 8000}$, $r=5%=\dfrac{5}{100}=0.05$, $n=1$.

(i) The amount credited at the end of the second year i.e. $t=2\text{ years}$ can be calculated as:

\[\begin{align}

  & \therefore {{A}_{2}}=8000{{\left( 1+\dfrac{0.05}{1} \right)}^{1\times 2}} \\

 & \text{ }=8000{{\left( 1+\dfrac{5}{100} \right)}^{2}} \\

 & \text{ }=8000{{\left( \dfrac{100+5}{100} \right)}^{2}} \\

 & \text{ }=8000{{\left( \dfrac{105}{100} \right)}^{2}} \\

 & \text{ }=8000\times \dfrac{105}{100}\times \dfrac{105}{100} \\
 & \therefore {{A}_{2}}=\text{Rs }8820. \\

\end{align}\]

Hence, the credited amount after two years is \[\text{Rs }8820\].

(ii) Now, to calculate the interest for the third year we need to subtract the amount ${{A}_{2}}$ obtained after two years form the amount ${{A}_{3}}$ obtained after three years.
\[\begin{align}

  & \therefore {{A}_{3}}=8000{{\left( 1+\dfrac{0.05}{1} \right)}^{1\times 3}} \\

 & \text{ }=8000{{\left( 1+\dfrac{5}{100} \right)}^{3}} \\

 & \text{ }=8000{{\left( \dfrac{100+5}{100} \right)}^{3}} \\

 & \text{ }=8000{{\left( \dfrac{105}{100} \right)}^{3}} \\

 & \text{ }=8000\times \dfrac{105}{100}\times \dfrac{105}{100}\times \dfrac{105}{100} \\
 & \therefore {{A}_{3}}=\text{Rs 9261}. \\

\end{align}\]

The interest $I$ for the third year is given by:

$\begin{align}

  & I={{A}_{3}}-{{A}_{2}} \\

 & \text{ }=9261-8820 \\

 & \text{ }=441. \\

\end{align}$

Hence, the interest for the third year is \[\text{Rs }441\].

Note: Here, the value of $n$ must be substituted carefully. We have to read the question carefully as it is given that the rate is compounded annually, therefore, $n=1$ is substituted. We must divide the given rate by 100 and then substitute in the equation.