Question

Rs. 18,000 for $2\dfrac{1}{2}$ years at 10% per annum is compounded annually. Calculate the final amount.

Answer 1

Hint: Use compound interest formula for the calculation of amount ‘$A$’, given by: \[A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}\]. Here, ‘$P$’ is the principal amount, ‘$r$’ is the rate per annum, ‘$t$’ is the time in years and ‘$n$’ is the number of times the interest is given in one 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.
We have been given, \[P=18000,\text{ }r=10%,\text{ }t=2\dfrac{1}{2}=\dfrac{5}{2}years\text{ and }n=1\].
Now, substituting the value of $P,r,t\text{ and }n$ in the formula to find amount, we get,
$\begin{align}
  & A=18000{{\left( 1+\dfrac{10}{100\times 1} \right)}^{1\times \dfrac{5}{2}}} \\
 & =18000{{\left( 1+\dfrac{10}{100} \right)}^{\dfrac{5}{2}}} \\
\end{align}$
Taking L.C.M we get,
\[\begin{align}
  & A=18000{{\left( \dfrac{100+10}{100} \right)}^{\dfrac{5}{2}}} \\
 & =18000{{\left( \dfrac{110}{100} \right)}^{\dfrac{5}{2}}} \\
 & =18000\times \dfrac{{{\left( 110 \right)}^{\dfrac{5}{2}}}}{{{\left( 100 \right)}^{\dfrac{5}{2}}}} \\
 & =18000\times \dfrac{{{\left( 110 \right)}^{\dfrac{5}{2}}}}{{{10}^{5}}} \\
 & =\dfrac{18}{100}\times {{\left( 110 \right)}^{\dfrac{5}{2}}} \\
\end{align}\]
This can be written as,
$\begin{align}
  & A=\dfrac{18}{100}\times {{110}^{2}}\times \sqrt{110} \\
 & =18\times 121\times \sqrt{110} \\
\end{align}$
Substituting,$\sqrt{110}=10.49$, we get,
$\begin{align}
  & A=18\times 121\times 10.49 \\
 & \text{ =22847}\text{.22} \\
\end{align}$
Hence, the final amount after $2\dfrac{1}{2}$ years is Rs. 22,847.22.

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.