Question

What is the compound interest on Rs. $ 15000 $ , if Mike took loan from a bank for $ 2\dfrac{1}{2} $ years at $ 8% $
Per annum, compounded quarterly?

Answer 1

Hint: In this question we have to find out compound interest on the sum of 1500, for 2.5 years and rate of interest is 8% per annum. Also, it is given that interest compounded quarterly, in means in one-year interest compounded four times.
Here we have to use the formula $ I=A{{\left( 1+\dfrac{r}{100} \right)}^{n}}-A $ where A is amount on which interest will occurs, r is the rate of interest. As per the question interest compounded quarterly so in place of r we write rate for three month so it is 2% that is we divide the rate of interest by 4 and in place of n we have to multiply it by 4 to interest compounded $ 2\dfrac{1}{2}(4)=10 $ times.

Complete step-by-step answer:
In this question the following information are given
$ \begin{align}
  & A=15000 \\
 & r=8\%per\text{ }annum \\
 & n=2\dfrac{1}{2} \\
\end{align} $
As interest compounded quarterly so rate of interest one quarter for a year so we can write
 $ {{r}_{quater}}=\dfrac{1}{4}(8)=2% $
And interest is compounded $ 2\dfrac{1}{2}(4)=10 $ times so, we can write
\[{{n}_{\left( 2\dfrac{1}{2}years \right)}}=10\]
Now using the formula for the compound interest, we can write
 $ I=A{{\left( 1+\dfrac{r}{100} \right)}^{n}}-A $
So putting the values of $ A=15000 $ , $ {{r}_{quater}}=2% $ and\[{{n}_{\left( 2\dfrac{1}{2}years \right)}}=10\]in the above formula we have
 $ \begin{align}
  & I=A{{\left( 1+\dfrac{r}{100} \right)}^{n}}-A \\
 & I=15000{{\left( 1+\dfrac{2}{100} \right)}^{10}}-15000 \\
\end{align} $
Now we can write the expansion as $ {{\left( 1+\dfrac{r}{100} \right)}^{n}}=1+\dfrac{nr}{100} $ when $ n<<100 $
Since 2 is very less as compared to 100 so we can write
 $ \begin{align}
  & I=15000\left( 1+\dfrac{2(10)}{100} \right)-15000 \\
 & I=15000(1.2)-15000 \\
 & I=18000-15000 \\
 & I=3000 \\
\end{align} $
So compound interest occurs is Rs. 3000.


Note: in the above question we write $ {{\left( 1+\dfrac{r}{100} \right)}^{n}}=1+\dfrac{nr}{100} $ when $ n<<100 $ , this is because if we expand $ {{\left( 1+\dfrac{r}{100} \right)}^{n}}=1+\dfrac{nr}{100} $ we can write
 $ {{\left( 1+\dfrac{r}{100} \right)}^{n}}=1+\dfrac{nr}{100}+{}^{n}{{c}_{2}}{{\left( \dfrac{r}{100} \right)}^{2}}+....up\text{ to higher power} $ . As we do the calculation of higher power of $ {{\left( \dfrac{r}{100} \right)}^{2}} $ and more this is very less.
The term after the $ \dfrac{nr}{100} $ is negligible due to higher power as $ r<<100 $ so we take the rest value as zero.