Question

Calculate the amount and compound interest on 10000 for 3 years at 2% per annum
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Answer 1

$\underline{\bf{Question:-}}$

Calculate the amount and CI on rupees 10,000 for 3 years at 2 percentage per annum compounded annually

$\underline{\bf{Given\:that:-}}$

Principle, p = 10, 000

Time period, n = 3 years

Rate of interest, r = 2 % per annum.

$\underline{\bf{Equation\:used:-}}$

$\sf{:\implies{Amount = P(1+\dfrac{r}{n})^{nt}}}$

$\sf{:\implies{Compound\:Interest = P(1+\dfrac{r}{n})^{nt}-1}}$

$\sf{:\implies{Amount = Principle + Compound\:Interest}}$

$\underline{\bf{Solution:-}}$

$\sf{\implies{Compound\:Interest = P(1+\dfrac{r}{n})^{nt}-1}}$

$\sf{\implies{Compound\:Interest = 10, 000(1+\dfrac{2}{100})^{3}-1}}$

$\sf{\implies{Compound\:Interest = 10, 000(1+0.02)^{3}-1}}$

$\sf{\implies{Compound\:Interest = 10, 000[(1.02)^{3}-1]}}$

$\sf{\implies{Compound\:Interest = 10, 000[1.061208-1]}}$

$\sf{\implies{Compound\:Interest = 10, 000[0.06120]}}$

$\sf{\implies{Compound\:Interest = 612.08}}$

$\boxed{\bf{\implies{Compound\:Interest = 612.08}}}$

$\bf{\implies{Amount = Principle + Compound\:Interest}}$

$\sf{\implies{Amount = 10, 000+ 612.08}}$

$\sf{\implies{Amount = 10,612.08}}$

$\boxed{\bf{\implies{Amount = 10, 612.08}}}$

$\underline{\sf{Additional\:information:-}}$

1. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by

$\boxed{\bf{:\implies{Amount = P(1+\dfrac{r}{n})^{nt}}}}$

2.Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi - annually for n years is given by

$\boxed{\bf{:\implies{Amount = P(1+\dfrac{r}{200})^{2n}}}}$

3. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is given by

$\boxed{\bf{:\implies{Amount = P(1+\dfrac{r}{400})^{4n}}}}$

4. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded monthly for n years is given by

$\boxed{\bf{:\implies{Amount = P(1+\dfrac{r}{1200})^{12n}}}}$