Find the compound interest on Rs. $2,000$ for $3$ years, compounded annually at $12\%$ per annum.

The correct option is B: $810$

Given:

$P=Rs2000$

$T=3$ years

$R=12\%$ p.a.

n $=$ number of times interest compounded $=1$

Amount after 3 years, A $=P(1+\frac{R}{100}{)}^{nT}$

$=2000(1+\frac{12}{100}{)}^{1\times 3}$

$=2000(\frac{100+12}{100}{)}^3$

$=2000\times \frac{112}{100}\times \frac{112}{100}\times \frac{112}{100}$

$=\frac{2\times 112\times 112\times 112}{1000}$

$=Rs.2809.856$

$\therefore$ Amount, A $=Rs.2809.856$

Compound Interest = Amount - Principal

$=2809.856-2000$

$=Rs.809.856=Rs.810$

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Other method to solve the above problem:

Interst for the first year $=\frac{P\times R\times T}{100}$

$=\frac{2000\times 12\times 1}{100}$

$=Rs240$

Amount after the first year $=P+$ Interest for the first year
$=Rs.2000+Rs.240=Rs.2240$

Interst for the second year $=\frac{2240\times 12\times 1}{100}$

$=Rs.268.8$

Hence, the amount after the second year $=Rs.2240+Rs.268.8=Rs.2508.8$

Interst for the third year $=\frac{2508.8\times 12\times 1}{100}$

$=Rs.301.056$

Amount after the third year $=Rs.2508.8+Rs.301.056=Rs.2809.856$

Compound interest $=$ Final amount $-$ original amount

$=Rs.2809.856-Rs.2000$

$=Rs.809.856$

$\approx Rs.810$