Question

Find the amount and compound interest on Rs 10,000 for 10% C.I. for 1 year compounded \[\dfrac{1}{2}\] yearly, and 10% S.I. for 1 year. Find the difference in interest earned.

Answer 1

Hint:In this particular question use the concept that the compound interest is calculated as, \[C.I.=A-P\], where P = Principal amount and A = amount after compound interest. So, A is calculated as \[A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}\], where r is the rate of interest in percentage, n is the time in years. So, use these concepts to reach the solution of the equation.

Complete step-by-step solution:
Given data: Principal amount = 10000 Rs.
Rate of interest = 10%
Time in years = \[\dfrac{1}{2}\] year
Now, we know that the compound interest is calculated by \[C.I.=A-P\], and here \[A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}\].
So,
\[A=P{{\left( 1+\dfrac{r}{200} \right)}^{2n}}\times \left( 1+\dfrac{tr}{2} \right)\]
\[\begin{align}
  & =10000{{\left( 1+\dfrac{10}{200} \right)}^{\left( 2 \right)}}\times \left( 1+\dfrac{\dfrac{1}{2}\times 10}{100} \right) \\
 & =10000\left( 1.1025 \right)\left( 1.05 \right) \\
 & =11576.25 \\
\end{align}\]
Amount = Rs. 11, 576.25
\[C.I.=A-P=1576.25\]
Simple interest can be calculated as,
\[A=P\left( 1+rt \right)=P+S.I.\]
Here, A = Final amount
P = Principal amount
r = annual interest rate
t = time in years
S. I. = Simple interest
\[S.I.=P\times R\times T\]
\[\begin{align}
  & =10000\times 1\times \dfrac{10}{100} \\
 & =1000 \\
\end{align}\]
So, the difference between O.I & S.I
\[=CI-SI\]
\[=1576.25-1000\]
\[=576.25\] Rs.
So, the final difference is 576.25 Rs.

Note: Whenever we face such types of questions the key concept we have to remember is the formula of compound interest which is stated above. So, first change the given compounded monthly to compounded yearly then convert to fraction in integer by multiplying it by 12 and divide the interest by the same number. So, overall no change will occur.