Question

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find the time in years at which the sum of 10,000 will amount to 13,310 at 10 percent per annum, compounded annually.
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Answer 1

$\large\underline{\sf{Solution-}}$

Given that

➢ A sum of Rs 10000 amounts to Rs 13310 at 10 % per annum compounded annually.

Let assume that time taken be n years.

So, we have

Principal, p = Rs 10000

Amount, A = Rs 13310

Rate, r = 10 % per annum

Time = n years

We know,

Amount (A) on certain sum of money of Rs p invested at the rate of r % per annum compounded annually in n years is

$\boxed{ \rm \:A = p {\bigg[1 + \dfrac{r}{100} \bigg]}^{n}}$

So, on substituting the values, we get

$\rm :\longmapsto\:\rm \:13310 = 10000 {\bigg[1 + \dfrac{10}{100} \bigg]}^{n}$

$\rm :\longmapsto\:\rm \:1331 = 1000 {\bigg[1 + \dfrac{1}{10} \bigg]}^{n}$

$\rm :\longmapsto\:\rm \:\dfrac{1331}{1000} = {\bigg[\dfrac{10 + 1}{10} \bigg]}^{n}$

$\rm :\longmapsto\:\rm \:\dfrac{11 \times 11 \times 11}{10 \times 10 \times 10} = {\bigg[\dfrac{11}{10} \bigg]}^{n}$

$\rm :\longmapsto\:\rm \: {\bigg[\dfrac{11}{10} \bigg]}^{3} = {\bigg[\dfrac{11}{10} \bigg]}^{n}$

$\bf\implies \:n = 3$

Additional Information :-

1. Amount (A) on certain sum of money of Rs p invested at the rate of r % per annum compounded semi - annually in n years is

$\boxed{ \rm \:A = p {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n}}$

2. Amount (A) on certain sum of money of Rs p invested at the rate of r % per annum compounded quarterly in n years is

$\boxed{ \rm \:A = p {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n}}$

3. Amount (A) on certain sum of money of Rs p invested at the rate of r % per annum compounded monthly in n years is

$\boxed{ \rm \:A = p {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n}}$

4. Amount (A) on certain sum of money of Rs p invested at the rate of r % per annum for n years, x % per annum for m years compounded annually is

$\boxed{ \rm \:A = p {\bigg[1 + \dfrac{r}{100} \bigg]}^{n}{\bigg[1 + \dfrac{x}{100} \bigg]}^{m}}$

5. Amount (A) on certain sum of money of Rs p invested at the rate of r % per annum compounded annually in n m/s years is

$\boxed{ \rm \:A = p {\bigg[1 + \dfrac{r}{100} \bigg]}^{n}{\bigg[1 + \dfrac{r \times \dfrac{m}{s} }{100} \bigg]}^{}}$