Question

The compound interest on \[Rs.2800\] for\[1\dfrac{1}{2}\]years at\[10\% p.a\]
A. \[Rs.441.35\]
B. \[Rs.436.75\]
C. \[Rs.434\]
D. \[Rs.420\]

Answer 1

Hint: Here we would be compounding the money for the first year and the next year separately.

Complete step-by-step answer:
Given, the principal amount\[P = Rs.2800\],total time=\[1\dfrac{1}{2}\]years and rate of interest \[R = 10\% p.a\]
We know the formula for the compound interest is given by \[A = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}\]
The compound interest for\[1\dfrac{1}{2}\]years is given by \[{A_1}\]+\[{A_2}\]
Here, \[{A_1} = \] the compound interest for the first year and \[{A_2} = \] compound interest for the half year after taking the new amount
First, let us calculate \[{A_1}\] which is given by
\[
  {A_1} = 2800{\left( {1 + \dfrac{{10}}{{100}}} \right)^1} \\
   \Rightarrow {A_1} = 2800\left( {\dfrac{{11}}{{100}}} \right) \\
   \Rightarrow {A_1} = 3030Rs \\
\]
Now, we calculate \[{A_2}\], which is given by
\[
  {A_2} = {A_1}{\left( {1 + \dfrac{{10}}{{100}}} \right)^{2n}} \\
  {A_2} = 3080{\left( {1 + \dfrac{{10}}{{100}}} \right)^{2 \times \dfrac{1}{2}}} \\
  {A_2} = 3080\left( {\dfrac{{11}}{{100}}} \right) \\
  {A_2} = 3234Rs \\
\](compounded half yearly)
Therefore, the compound interest is given by \[C.I = {A_2} - P\]
\[ \Rightarrow C.I = 3234 - 2800 = 434Rs\]
Therefore, the option C. \[434Rs\]is the required solution.

Note: Observe that the principal money changes after the money is compounded for a year. Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other words, interest on interest.