Questions & Answers

Question

Answers

Answer
Verified

If principal=P , rate of compound interest=\[R\% \] per annum, time =n years, then

Amount \[ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}\]

Now let us assume, in n complete years, a sum of money (say P) put at \[20\% \] compound interest will be more than doubled.

Therefore,

\[

P{\left( {1 + \dfrac{{20}}{{100}}} \right)^n} > 2P \\

{\text{On simplifying we get,}} \\

\Rightarrow {\left( {\dfrac{6}{5}} \right)^n} > 2 \\

{\text{Taking log on both sides,we get}} \\

n\log \left( {\dfrac{6}{5}} \right) > \log 2 \\

{\text{As log}}\left( {\dfrac{6}{5}} \right) = 0.079,{\text{ we get,}} \\

\Rightarrow {\text{n x }}0.079 > 0.301 \\

\Rightarrow {\text{n > 3}}{\text{.81}} \\

\Rightarrow {\text{n}} \sim {\text{4}} \\

\]

If principal=P , rate of compound interest=R% per annum, time =n years, then

1.Amount= \[P{\left( {1 + \dfrac{R}{{100}}} \right)^n}\], when interest is compounded annually

2.Amount= \[P{\left( {1 + \dfrac{{\dfrac{R}{2}}}{{100}}} \right)^{2n}}\], when interest is compounded half-yearly

3.Amount= \[P{\left( {1 + \dfrac{{\dfrac{R}{4}}}{{100}}} \right)^{4n}}\], when interest is compounded quarterly