Question

A man invests ${\text{Rs}}.3000$ for three years at compound interest. After one year, the money amounts to ${\text{Rs}}.3240$. Find the rate of interest and the amount (to the nearest Rupee) due at the end of $3$ years.

Answer 1

Hint: We know ${\text{Total amount}} = P + SI$
Here, $SI$represents simple interest which is given by the formula $\dfrac{{P \times R \times T}}{{100}}$.
Here total amount is given ${\text{Rs}}.3240$ and principal amount is given ${\text{Rs}}.3000$. And a time period of $1$ year. From this we can find the rate and then use this rate in the compound interest formula for $3$ years.
Amount$ = P{\left( {1 + \dfrac{R}{{100}}} \right)^T}$

Complete step-by-step answer:
So, according to the question, a man invests ${\text{Rs}}.3000$ for three years at compound interest and after one year, he gets ${\text{Rs}}.3240$.
As we are given,
$
  {\text{Amount}}\left( A \right) = {\text{Rs}}.3240 \\
  {\text{Principal}}\left( P \right) = {\text{Rs}}{\text{.3000}} \\
 $
So, as we know that ${\text{Total amount}} = P + SI$
Where, $SI$represents simple interest
So, here we get
$
  SI = A - P \\
   = 3240 - 3000 \\
   = {\text{Rs}}{\text{.240}} \\
 $
And the formula of $SI = \dfrac{{P \times R \times T}}{{100}}$. So,
$240 = \dfrac{{3000 \times R \times 1}}{{100}}$ (As we are told after one year, so time $T = 1$ year.)
$R = \dfrac{{24000}}{{3000}} = 8\% $
So, we got the rate of interest$ = 8\% $
Now we have to find the amount after $3$ years. So, by compound interest,
Amount $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^T}$ where $T = 3$. So,
$
  A = 3000{\left( {1 + \dfrac{8}{{100}}} \right)^3} \\
   = 3000\left( {1 + \dfrac{{24}}{{100}} + {}^3{C_2}{{\left( {\dfrac{8}{{100}}} \right)}^2} + {{\left( {\dfrac{8}{{100}}} \right)}^3}} \right) \\
   = 3000\left( {1 + 0.24 + 0.0192 + \dfrac{{512}}{{1000000}}} \right) \\
   = 3000\left( {1.259772} \right) \\
  A = {\text{Rs.}} 3779.136 \\
 $
So, the nearest integer is ${\text{Rs}}{\text{.3779}}$.
So, after three years, the amount will be ${\text{Rs}}{\text{.3779}}$.

Note: Here we used binomial theorem that is
${\left( {1 + x} \right)^n} = 1 + nx + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + {\text{ - - - - - + }}{x^n}$
And we use simple interest to find the rate and compound interest to find the amount after $3$ years.
$SI = \dfrac{{P \times R \times T}}{{100}}$ and for compound interest, $A = P{\left( {1 + \dfrac{R}{{100}}} \right)^T}$.