Aman invests Rs $4000$ in a bank at a certain rate of interest, compounded annually. If the ratio of the value of the investment after $3$ years to the value of the investment after $5$ years is $25: 36$, then the minimum number of years required for the value of the investment to exceed Rs $20000$ is

Formula for Compound Interest:

$A = P(1 + r)^n$

where:

$A =$ final amount

$P =$ principal amount (Rs 4000)

$r =$ rate of interest (in decimals) per year (unknown)

$n =$ number of years

---

Amount after 3 years:

$A_3 = 4000\left(1 + r\right)^3$

Amount after 5 years:

$A_5 = 4000\left(1 + r\right)^5$

Given ratio:

$\Rightarrow \dfrac{A_5}{A_3} = \dfrac{36}{25}$

$\Rightarrow \dfrac{4000\left(1 + r\right)^5}{4000\left(1 + r\right)^3} = \dfrac{36}{25}$

$\Rightarrow \dfrac{\left(1 + r\right)^5}{\left(1 + r\right)^3} = \dfrac{36}{25}$

When dividing powers with the same base, we subtract the exponents.

$\Rightarrow \left(1 + r\right)^{5-3} = \frac{36}{25}$

$\Rightarrow \left(1 + r\right)^2 = \left(\frac{6}{5}\right)^2$

Taking the positive square root:

$\begin{aligned} 1 + r &= \frac{6}{5} \\ r &= 1.2 - 1 \\ r &= 0.2 \end{aligned}$

Hence, $r = 20\%$

---

We need: $4000\left(1 + \frac{20}{100}\right)^n > 20000$

$\Rightarrow 4000(1.2)^n > 20000$

$\Rightarrow (1.2)^n > 5$

---

In the on-screen calculator, keep multiplying $1.2$ by itself, in the 9th time, you'll see it cross $5$. Since we want the minimum number of years, we stop as soon as it crosses 5.

$(1.2)^8 \approx 4.30$

$(1.2)^9 \approx 5.16$

Hence, we need at least $\boxed{9}$ years.

---

To check (for more clarity):

After 8 years: $A = 4000 \times (1.2)^8 = 4000 \times 4.30 = \text{Rs } 17200$

After 9 years: $A = 4000 \times (1.2)^9 = 4000 \times 5.16 = \text{Rs } 20640$