Question

A certain sum placed at compound interest triples itself in 5 years. In how many years will it amount to nine times at the same rate of interest?

• A. 8 years
• B. 10 years
• C. 12 years
• D. 14 years
• E. 16 years

Answer 1

The Correct Option is B

Step 1: Understand the problem.
We are given that a sum of money placed at compound interest triples itself in 5 years. We are asked to find in how many years it will amount to nine times at the same rate of interest.

Step 2: Use the compound interest formula.
The compound interest formula is given by:
\[ A = P \left(1 + \frac{r}{100}\right)^t \] where:
- \( A \) is the amount after time \( t \),
- \( P \) is the principal,
- \( r \) is the rate of interest,
- \( t \) is the time in years.

We are told that the amount triples itself in 5 years, so:
\[ 3P = P \left(1 + \frac{r}{100}\right)^5 \] Dividing both sides by \( P \) (assuming \( P \neq 0 \)):
\[ 3 = \left(1 + \frac{r}{100}\right)^5 \] Taking the 5th root of both sides:
\[ \left(1 + \frac{r}{100}\right) = 3^{1/5} \] We know that \( 3^{1/5} \approx 1.24573 \). Therefore:
\[ 1 + \frac{r}{100} = 1.24573 \] Solving for \( r \):
\[ \frac{r}{100} = 1.24573 - 1 = 0.24573 \] \[ r = 24.573\% \] So, the rate of interest is approximately 24.573%.

Step 3: Find the time when the amount becomes 9 times the principal.
Now, we are asked to find when the amount will become 9 times the principal. Using the compound interest formula again:
\[ 9P = P \left(1 + \frac{r}{100}\right)^t \] Dividing both sides by \( P \):
\[ 9 = \left(1 + \frac{r}{100}\right)^t \] Substituting \( 1 + \frac{r}{100} = 1.24573 \):
\[ 9 = 1.24573^t \] Taking the logarithm of both sides:
\[ \log(9) = t \log(1.24573) \] Using logarithms:
\[ \log(9) \approx 0.9542 \quad \text{and} \quad \log(1.24573) \approx 0.096 \] Solving for \( t \):
\[ t = \frac{0.9542}{0.096} \approx 10 \]

Step 4: Conclusion.
The amount will become nine times the principal in approximately 10 years.

Final Answer:
The correct option is (B): 10 years.