Question

Calculate the amount and compound interest on Rs.10000 for 3 years at $12\dfrac{1}{2}\%$ per annum compounded annually.

Answer 1

Hint: To solve this question we will use formula to calculate the amount and compound interest which is given as;
\[\text{A=P}{{\left( 1+\dfrac{r}{n} \right)}^{nt}}\]
Where, A = amount, P = principal, r = rate of interest, n = number of times interest is compounded per unit time 't'. Formula to calculate compound interest \[\text{CI=A+P}\]

Complete step-by-step solution:
Given in the question, we have P = principal = 10800
t = time = 3 years
r = rate \[\Rightarrow 12\dfrac{1}{2}\%=\dfrac{25}{2}\times \dfrac{1}{100} =\dfrac{1}{2}\times \dfrac{1}{4} = \dfrac{1}{8} \]
n = 1 (as compounded annually)
We have formula of amount as \[\text{A=P}{{\left( 1+\dfrac{r}{n} \right)}^{nt}}\]
Substituting value of all variables, we get:
\[\text{A=10800}{{\left( 1+\dfrac{1}{8\times 1} \right)}^{1\times 3}}\]
Solving by opening the bracket we get:
\[\begin{align}
  & \text{A=10800}{{\left(\dfrac{8+1}{8} \right)}^{3}} \\
 &\Rightarrow \text{A=10800}\times \dfrac{9\times 9\times 9}{8\times 8\times 8} \\
 &\Rightarrow \text{A=}\dfrac{1350}{8\times 8}\times 9\times 9\times 9 \\
 &\Rightarrow A=21.09375\times 9\times 9\times 9 \\
 &\Rightarrow A=15377.34375 \\
\end{align}\]
Then, the formula of compound interest is \[\text{CI=A+P}\]
Substituting the value we have,
\[\begin{align}
  & \text{CI=}15377.34375\text{+10800} \\
 & \Rightarrow \text{26177}\text{.3437} \\
\end{align}\]
Hence, the amount is $A=Rs.15377.34375$
And compound interest $CI=RS.\text{26177}\text{.3437}$

Note: The possibility of error in this question can be assuming n = 3 years which would be wrong as the value of t = 3 years, n is the number of times interest is compounded. So, here n = 1 and not 3. Hence, be clear and accurate about the value of n and t.