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

The correct option is B: $810$

Given:

$P=Rs\ 2000$

$T=3$ years

$R=12\%$ p.a.

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

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

$=2000(1+\frac{12}{100})^{1\times3}$

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

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

$=\dfrac{2\times112\times112\times112}{1000}$

$=Rs.~2809.856$

$\therefore$ 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 $=\cfrac{P\times R\times T}{100}$

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

$=Rs\ 240$

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

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

$=Rs.\ 268.8$

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

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

$=Rs.\ 301.056$

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

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

$=Rs.\ 2809.856-Rs.\ 2000$

$=Rs.\ 809.856$

$\approx Rs.\ 810$