Question

The money invested in a company is compounded continuously. If Rs. 400 invested today becomes Rs. 800 in 6 years, then at the end of 30 years, it will become (in Rs.) ______.

• A. 18101.76
• B. 12800
• C. 9050.88
• D. 12804

Hint:

Shortcut for Doubling Time: If money doubles every 6 years, then in 30 years it will undergo $30 / 6 = 5$ complete doubling cycles.
$400 \times (2)^5 = 400 \times 32 = 12800$. No $e$ or $\ln$ is required!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This problem involves continuous exponential growth. We are given the time it takes for an initial investment to double, and we must find the total value of the investment after a much longer specified time period.

Step 2: Detailed Explanation:
The mathematical formula for continuous compounding is:
$A = P e^{rt}$
where:
$A$ = Final amount
$P$ = Principal (initial) investment
$r$ = Continuous growth rate constant
$t$ = Time
Condition 1: Doubling Time
We are told that $P = 400$ grows to $A = 800$ in $t = 6$ years.
$800 = 400 e^{r(6)}$
Divide both sides by 400:
$2 = e^{6r}$
Take the natural logarithm ($\ln$) of both sides to solve for the rate $r$:
$\ln(2) = \ln(e^{6r})$
$\ln(2) = 6r \Rightarrow r = \frac{\ln(2)}{6}$
Condition 2: Future Value
We want to find the final amount $A$ after $t = 30$ years using the same initial $P = 400$ and rate $r$.
$A = 400 e^{r(30)}$
Substitute the exact expression we found for $r$:
$A = 400 e^{\left( \frac{\ln(2)}{6} \right) \times 30}$
Simplify the exponent:
$\frac{30}{6} = 5$
So, the equation becomes:
$A = 400 e^{5 \ln(2)}$
Use the logarithmic power rule ($n \ln x = \ln(x^n)$) to move the 5 inside:
$A = 400 e^{\ln(2^5)}$
$A = 400 e^{\ln(32)}$
Because the exponential function $e^x$ and the natural logarithm $\ln(x)$ are perfect inverse functions, they cancel each other out ($e^{\ln(X)} = X$):
$A = 400 \times 32$
$A = 12800$

Step 3: Final Answer:
It will become Rs. 12800, matching option (b).