Mohit invests Rs. 8,000 for 3 years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs.9,440. Calculate the amount at the end of the second year.
A. Rs. 15,729.50
B. Rs. 13,079.80
C. Rs. 12,367.50
D. Rs. 11,139.20

Hint: In this problem, first we need to find the rate of interest. Next, we need to find the interest for the second year and hence calculate the amount at the end of the second year.

Complete step-by-step answer:
The formula for compound interest is shown below.
\[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right)\]
Here, \[A\] is amount, \[P\] is principal, \[r\] is rate of interest and \[n\] is time in years.

Substitute 9440 for\[A\], 8000 for \[P\] and 1 for \[n\] in equation (1) to obtain the rate of interest.
\[
  \,\,\,\,\,9440 = 8000{\left( {1 + \dfrac{r}{{100}}} \right)^1} \\
   \Rightarrow 1 + \dfrac{r}{{100}} = \dfrac{{9440}}{{8000}} \\
   \Rightarrow \dfrac{r}{{100}} = \dfrac{{944}}{{800}} - 1 \\
   \Rightarrow \dfrac{r}{{100}} = \dfrac{{144}}{{800}} \\
   \Rightarrow r = 18 \\
\]
Now, in order to obtain an amount at the end of the second year, substitute 8000 for\[P\], 18 for \[r\] and 2 for \[n\].
\[
  \,\,\,\,\,A = 8000{\left( {1 + \dfrac{{18}}{{100}}} \right)^2} \\
   \Rightarrow A = 8000{\left( {1 + 0.18} \right)^2} \\
   \Rightarrow A = 8000{\left( {1.18} \right)^2} \\
   \Rightarrow A = 11,139.2 \\
\]
Thus, the amount at the end of the second year is Rs.11,139.2.


Note: Key point in the solution is when the rate of interest is compounded annually, always use the compounded interest formula.