Question

The money invested in a company is compounded continuously. If \textrupee~200 invested today becomes \textrupee~400 in 6 years, then at the end of 33 years it will become

• A. $1600\sqrt{2}$
• B. $3200\sqrt{2}$
• C. $12800\sqrt{2}$
• D. $6400\sqrt{2}$

Hint:

Algebra Tip: Instead of solving for the actual interest rate $R$, treat the entire expression $(1+\frac{R}{100})$ as a single constant variable. This drastically simplifies calculations involving exponential powers.

Answer 1

The Correct Option is D

Concept: Mathematics - Exponential Growth and Compound Interest.

Step 1: Identify the initial given parameters. Let the principal amount be $P = 200$. The final amount after $N = 6$ years is $A = 400$. The standard formula used here for compounding over intervals is $A = P\left(1+\frac{R}{100}\right)^N$.

Step 2: Set up the equation to find the growth factor. Substitute the initial values into the formula: $400 = 200\left(1+\frac{R}{100}\right)^6$.

Step 3: Solve for the constant growth multiplier. Divide both sides by 200 to get: $2 = \left(1+\frac{R}{100}\right)^6$. Taking the 6th root of both sides gives us the single-year growth factor: $1+\frac{R}{100} = 2^{\frac{1}{6}}$.

Step 4: Set up the equation for the new time period. We need to find the final amount $A$ after $N = 33$ years. Using the same principal $P = 200$, the equation is: $A = 200\left(1+\frac{R}{100}\right)^{33}$.

Step 5: Substitute the growth multiplier and evaluate. Substitute $2^{\frac{1}{6}}$ in place of $\left(1+\frac{R}{100}\right)$: $A = 200\left(2^{\frac{1}{6}}\right)^{33}$. Multiply the exponents: $\frac{33}{6} = \frac{11}{2}$. $A = 200(2^{\frac{11}{2}}) = 200(2^5 \cdot 2^{\frac{1}{2}})$. Since $2^5 = 32$ and $2^{\frac{1}{2}} = \sqrt{2}$, the equation becomes: $A = 200(32\sqrt{2}) = 6400\sqrt{2}$. $$ \therefore \text{The amount at the end of 33 years will be } 6400\sqrt{2}. $$