Question

How many year will 36000 amount to 46621.044 at 9% p.a compounded annually

Answer 1

If 36000 amount to 46621.044 at 9% p.a compounded annually then the time period is 3 years.

Step-by-step explanation:

Principal, P = Rs. 36000

Amount, A = Rs. 46621.044

Rate of interest, R = 9%

Let the time period be denoted as “n” years.

The formula for the amount in case of compounding the interest annually is given by,

A = P [1+$\frac{R}{100}$]ⁿ

By substituting the given values in the formula, we get

46621.044 = 36000 [1+$\frac{9}{100}$]ⁿ

⇒ 1.295 = [$\frac{109}{100}$]ⁿ

⇒ 1.295 = [1.09]ⁿ

taking log on both sides

⇒ log 1.295 = log [1.09]ⁿ

⇒ 0.11 = n log 1.09 ….. [since log aᵇ = b log a]

⇒ 0.11 = n * 0.0374

⇒ n = 2.94 ≈ 3 years

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

FORMULA FOR CALCULATING COMPOUND INTEREST

$\sf{A = Final \: Amount}$

$\sf{P = Principal \: Amount}$

$\sf{r = \: Interest \: Rate \: ( Decimal) \: }$

$\sf{t = Time \: ( years)}$

$\sf{n = Number \: of \: times \: interest \: is \: compounded \: per \: year}$

Then

$\displaystyle \: \sf{A = P { \bigg( \: 1 + \frac{r}{n} \bigg)}^{nt} }$

TO DETERMINE

The time ( years) in which the amount ₹ 36000 amount will be compounded to ₹ 46621.044 at 9% p.a compounded annually

CALCULATION

$\sf{A = Final \: Amount} = 46621.044$

$\sf{P = Principal \: Amount} = 36000$

$\displaystyle \: \sf{r = \: Interest \: Rate \: \: } = \frac{9}{100}$

$\sf{t = Time \: ( years)} = \: ?$

$\sf{n = Number \: of \: times \: interest \: is \: compounded \: per \: year} = 1$

So we have

$\displaystyle \: \sf{A = P { \bigg( \: 1 + \frac{r}{n} \bigg)}^{nt} }$

Putting the values we get

$\implies \: \displaystyle \: \sf{46621.044 = 36000 \times { \bigg( \: 1 + \frac{9}{100} \bigg)}^{t} }$

$\implies \: \displaystyle \: \sf{ \frac{46621.044}{36000} = { \bigg( \: 1 + \frac{9}{100} \bigg)}^{t} }$

$\implies \: \displaystyle \: \sf{ \frac{46621044}{36000000} = { \bigg( \: 1 + \frac{9}{100} \bigg)}^{t} }$

$\implies \: \displaystyle \: \sf{ \frac{1295029}{1000000} = { \bigg( \: \frac{109}{100} \bigg)}^{t} }$

$\implies \: \displaystyle \: \sf{ { \bigg( \: \frac{109}{100} \bigg)}^{3} = { \bigg( \: \frac{109}{100} \bigg)}^{t} }$

$\implies \: \displaystyle \: \sf{ { \bigg( \: \frac{109}{100} \bigg)}^{t} = { \bigg( \: \frac{109}{100} \bigg)}^{3} }$

$\implies \: \displaystyle \: \sf{t = 3}$

RESULT

$\boxed{ \sf{ \: THE \: REQUIRED \: TIME \: = 3 \: \: YEARS \: }}$