Question

Calculate compound interest on ₹\[1000\] over a period of \[1\] year at \[10\% \]per annum if interest is compounded half yearly.

Answer 1

Hint: Compounded half yearly means interest is paid at the end of every six months, so we can take time as $\dfrac{1}{2}$year in two steps and interest is the difference between amount and principal sum.


Complete step by step solution:
$S.I = \dfrac{{P \times R \times T}}{{100}}$
Where S.I is interest
R is rate interest
P is Principal sum
T is time period
For first half year;
\[{P_1} = \]₹$1000$
\[{R_1} = 10\% \]
${T_1} = \dfrac{1}{2}years$
Simple interest at the end of $1st$ year
$S.{I_1} = \dfrac{{{P_1} \times {R_2}{T_1}}}{{100}}$
$ = 1000 \times \dfrac{{10}}{{100}} \times \dfrac{1}{2}$
\[S.{I_1} = 50\]
\[Amount = principal + interest\]
\[{A_1} = 1000 + 50\]
$ = $₹\[1050\]
This amount becomes principal for next half year
For \[{2^{nd}}\]half year,
\[{P_2} = \]₹\[1050\],\[{R_2} = 10\% ,\]${T_2} = \dfrac{1}{2}year$
$S.{I_2} = \dfrac{{{P_2} \times {R_2} \times {T_2}}}{{100}}$
$S.{I_2} = 1050 \times \dfrac{{10}}{{100}} \times \dfrac{1}{2}$
$ = 52.5$
Amount = principal + interest
\[{A_2} = 1050 + 52.5{\text{ }} = \] ₹\[1102.50\]
Compound interest \[ = {A_2}-P\]
\[ = 1102.50-1000\]
$ = $₹\[102.50\]
Also, compound interest\[ = S.{I_1} + S.{I_2}\]
\[ = 50 + 52.50\]
$ = $₹\[102.50\]

Note: We can also use this formula to calculate amount and compound interest.
$A = P{\left( {1 + \dfrac{{nR}}{{100}}} \right)^{T/n}}$
Where A – compound amount
P – Principal sum.
R – Rate of interest
T – Time period
N – Compounded period (in years)