$$\begin{gathered} \text{Q.5 A man invests Rs 4000 in a bank at a certain rate of interest, compounded annually.} \\ \text{If the ratio of the value of the investment after 3 years to the value of the} \end{gathered}$$

\text{investment after 5 years is}\frac{25}{36}\text{, then the minimum number of years required for the value of the investment to exceed Rs 20000 is:}

The correct answer is 9.

Step-by-step Explanation:

Let the principal amount invested be represented by P, the annual rate of compound interest be R%, and the annual multiplier be r, where:
$r=1+\frac{R}{100}$
Here, the initial investment P is Rs 4000.

The value of the investment after t years, denoted as A(t), is given by the formula:
$A\left(t\right)=P{r}^t$

We are given the ratio of the value of the investment after 3 years to the value of the investment after 5 years:
$\frac{A\left(3\right)}{A\left(5\right)}=\frac{25}{36}$

Substituting the formula for compound interest, we get:
$\frac{P{r}^3}{P{r}^5}=\frac{25}{36}$
Simplifying the left-hand side:
$\frac{1}{r^2}=\frac{25}{36}$
Taking the reciprocal on both sides:
$r^2=\frac{36}{25}$
Taking the square root (since r represents a positive multiplication factor):
$r=\frac{6}{5}=1.2$

Now, we want to find the minimum number of years n required for the value of the investment to exceed Rs 20000:
$A\left(n\right)>20000$
Substitute the values of P and r:
$4000{\left(1.2\right)}^n>20000$
Divide both sides by 4000:
${\left(1.2\right)}^n>5$

Let's calculate the powers of 1.2 step-by-step:
For n = 1: ${1.2}^1=1.2$
For n = 2: ${1.2}^2=1.44$
For n = 3: ${1.2}^3=1.728$
For n = 4: ${1.2}^4=2.0736$
For n = 5: ${1.2}^5\approx 2.488$
For n = 6: ${1.2}^6\approx 2.986$
For n = 7: ${1.2}^7\approx 3.583$
For n = 8: ${1.2}^8\approx 4.300$
For n = 9: ${1.2}^9\approx 5.160$

Since ${1.2}^8<5$ and ${1.2}^9>5$, the minimum integer value of n for which the value of the investment exceeds Rs 20000 is 9.