Amount become $9$ times in $12$ years at Compound interest. In what time it will become $27$ times

Question

Problem : Amount become $9$ times in $12$ years at Compound interest . In what time it will become $27$ times

Solution :

Let principal be $x$.

Amount $= 9x$

Using formula

$A = P(1 + \frac {r}{100})^n$

$9= (1 + \frac {r}{100})^{12} $

Suaring both sides

$81= (1 + \frac {r}{100})^{24} $

$81x= x(1 + \frac {r}{100})^{24} $

Amount will become $81$ times in $24$ years

But how can we find for $27 $times

Answer 1

HINT: Amount is increased $3$ times in $6$ years ( check: in another $6$ years it will be increased another $3$ times, that is, in $12$ years a total of $3\times 3 = 9$ times). Hence, waiting another $6$ years ( a total of $18$ years) the total increase will be $9 \times 3 = 27$ times.

The important idea: in the same period of time an amount will be multiplied by the same quantity.

Answer 2

Instead of squaring both sides, after $9=\left(1+\frac{r}{100}\right)^{12}$, you should solve for $r$: $$9=\left(1+\frac{r}{100}\right)^{12} \implies \left(1+\frac{r}{100}\right)=9^{\frac{1}{12}}$$ Such that you can solve for $r$. After that, you can use your principal formula $A=P\left(1+\frac{r}{100}\right)^n$ to calculate in what time it will become $27$ times the original amount.

I hope this is instructive enough for you to solve it. Since you've put in work yourself (and showed us), I think you'll work it out yourself. If not, let us know :)