A certain amount of money becomes thrice in 5 years at compound interest. How many years it will take to become 9 times?

The correct option is 10 years.

Step-by-Step Explanation:

Let the principal amount be P and the rate of compound interest per annum be r%.

The formula for the total compound amount A after t years is given by:

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

Step 1: Express the condition for the money to become 3 times in 5 years.

Given that the principal amount becomes thrice (3 times) in 5 years, we substitute $A=3P$ and $t=5$ into the compound interest formula:

$3P=P{(1+\frac{r}{100})}^5$

Dividing both sides of the equation by P:

${(1+\frac{r}{100})}^5=3$      (Equation 1)

Step 2: Calculate the time required for the money to become 9 times.

Let T be the required time in years for the money to become 9 times its principal, so $A=9P$:

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

Dividing both sides of the equation by P:

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

Since $9=3^2$, we rewrite the equation as:

${(1+\frac{r}{100})}^T=3^2$

From Equation 1, we know that $3={(1+\frac{r}{100})}^5$. Substituting this expression for 3 into the equation:

${(1+\frac{r}{100})}^T={\left[{(1+\frac{r}{100})}^5\right]}^2$

Applying the power rule of exponents, ${\left(x^a\right)}^b=x^{a\times b}$:

${(1+\frac{r}{100})}^T={(1+\frac{r}{100})}^{5\times 2}$

${(1+\frac{r}{100})}^T={(1+\frac{r}{100})}^{10}$

Equating the exponents as the bases on both sides are identical:

$T=10$

Therefore, it will take 10 years for the amount of money to become 9 times at compound interest.