In how many years will Rs 1800 amount to Rs 2178 at 10%

Solution:

Let the required time be $n$ years.

Rate of interest, $R=10 \%$

Principal amount, $P=$ Rs. 1800

Amount with compound interest, $A=$ Rs. 2178

Now, $A=P \times\left(1+\frac{R}{100}\right)^{n}$

$=$ Rs. $1800 \times\left(1+\frac{10}{100}\right)^{n}$

$=$ Rs. $1800 \times\left(\frac{100+10}{100}\right)^{\mathrm{n}}$

$=$ Rs. $1800 \times\left(\frac{110}{100}\right)^{\mathrm{n}}$

$=$ Rs. $1800 \times\left(\frac{11}{10}\right)^{\mathrm{n}}$

However, amount $=$ Rs. 2178

Now, Rs. $2178=$ Rs. $1800 \times\left(\frac{11}{10}\right)^{n}$

$\Rightarrow \frac{2178}{1800}=\left(\frac{11}{10}\right)^{n}$

$\Rightarrow \frac{121}{100}=\left(\frac{11}{10}\right)^{n}$

$\Rightarrow\left(\frac{11}{10}\right)^{2}=\left(\frac{11}{10}\right)^{n}$

$\Rightarrow n=2$

$\therefore$ Time, $n=2$ years