Question

If Mr. Nirav deposits Rs. $250$ at the beginning of each month in an account that pays an interest of $6\%$ per annum compounded monthly, how many months will be required for the deposit to amount to at least Rs. $6,390$?

Answer 1

$A=Rs.250$
$i=\frac{6}{100\times 12}=0.005$
$M\le Rs.6,390$

Using $M=\frac{A}{i}(1+i)\left[\right(1+r{)}^n-1]$
According to Question,
$6,390\le \frac{250}{0.005}\left(1.005\right)[1.005{)}^m-1]$

$\Rightarrow \frac{6,390\times 0.005}{250\times 1.005}\le (1.005{)}^n-1$

$\Rightarrow \frac{31.95}{251.25}\le (1.005{)}^n-1$

$\Rightarrow 0.1272\le (1.005{)}^n-1$

$\Rightarrow 1.1272\le (1.005{)}^n$

$\Rightarrow (1.005{)}^n\ge 1.1272$

Taking $\log$ on both sides, we have
$n\log 1.005\ge log1.1272$
$n\left(0.00217\right)\ge 0.05200$

$\Rightarrow n\ge \frac{0.05200}{0.00217}$

$\Rightarrow n\ge \frac{5200}{217}$

$n\ge 23.96$
i.e. $n=24$