Question

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

Answer 1

We are given that Aman invests Rs 4000 at a certain rate of interest, compounded annually. The ratio of the value of the investment after 3 years to the value after 5 years is 25:36. Let the rate of interest be $r$ per annum. The formula for the compound interest is:
$A = P \left(1 + \frac{r}{100}\right)^t$
where: - $A$ is the amount after time $t$. - $P$ is the principal. - $r$ is the annual interest rate, and $t$ is the number of years.
We are given that:
$\frac{A_3}{A_5} = \frac{25}{36}$

Using the compound interest formula for 3 years and 5 years:
$\frac{4000 \left(1 + \frac{r}{100}\right)^3}{4000 \left(1 + \frac{r}{100}\right)^5} = \frac{25}{36}$

Simplifying:
$\frac{\left(1 + \frac{r}{100}\right)^3}{\left(1 + \frac{r}{100}\right)^5} = \frac{25}{36}$

Taking the reciprocal:
$\left(1 + \frac{r}{100}\right)^2 = \frac{36}{25}$

Taking the square root:
$1 + \frac{r}{100} = \frac{6}{5}$

Solving for $r$:
$\frac{r}{100} = \frac{1}{5} \implies r = 20\%$

Thus, the rate of interest is 20%.

Now, to find the minimum number of years for the investment to exceed Rs 20000, we use the formula for compound interest:
$20000 = 4000 \left(1 + \frac{20}{100}\right)^t$
$5 = 1.2^t$

Taking the logarithm of both sides:
$\log(5) = t \log(1.2)$
$t = \frac{\log(5)}{\log(1.2)} \approx \frac{0.69897}{0.07918} \approx 8.83$

Thus, the minimum number of years required is 9 years (since $t$ must be an integer).
Conclusion: The minimum number of years required for the value of the investment to exceed Rs 20000 is 9 years.

Answer 2

We are given that Aman invests Rs 4000 at a certain rate of interest, compounded annually. The ratio of the value of the investment after 3 years to the value after 5 years is 25:36. Let the rate of interest be $r$ per annum. The formula for the compound interest is:
$A = P \left(1 + \frac{r}{100}\right)^t$
where: - $A$ is the amount after time $t$. - $P$ is the principal. - $r$ is the annual interest rate, and $t$ is the number of years.
We are given that:
$\frac{A_3}{A_5} = \frac{25}{36}$

Using the compound interest formula for 3 years and 5 years:
$\frac{4000 \left(1 + \frac{r}{100}\right)^3}{4000 \left(1 + \frac{r}{100}\right)^5} = \frac{25}{36}$

Simplifying:
$\frac{\left(1 + \frac{r}{100}\right)^3}{\left(1 + \frac{r}{100}\right)^5} = \frac{25}{36}$

Taking the reciprocal:
$\left(1 + \frac{r}{100}\right)^2 = \frac{36}{25}$

Taking the square root:
$1 + \frac{r}{100} = \frac{6}{5}$

Solving for $r$:
$\frac{r}{100} = \frac{1}{5} \implies r = 20\%$

Thus, the rate of interest is 20%.

Now, to find the minimum number of years for the investment to exceed Rs 20000, we use the formula for compound interest:
$20000 = 4000 \left(1 + \frac{20}{100}\right)^t$
$5 = 1.2^t$

Taking the logarithm of both sides:
$\log(5) = t \log(1.2)$
$t = \frac{\log(5)}{\log(1.2)} \approx \frac{0.69897}{0.07918} \approx 8.83$

Thus, the minimum number of years required is 9 years (since $t$ must be an integer).
Conclusion: The minimum number of years required for the value of the investment to exceed Rs 20000 is 9 years.