A sum becomes two times in 6 years at a certain rate of interest at simple interest. Find the time in which the same amount will be 12 times at the same rate of interest.

Answer

To solve the problem step by step, we will follow the logic laid out in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We know that a sum of money doubles in 6 years at a certain rate of simple interest. We need to find out how long it will take for the same sum to become 12 times at the same rate.

2. **Identifying the Initial Conditions:**
- Let the principal amount (P) be \(100\) (for simplicity).
- If the amount doubles in 6 years, the final amount (A) after 6 years is \(200\) (which is \(2 \times 100\)).

3. **Calculating the Simple Interest (SI):**
- The simple interest earned in 6 years can be calculated as:
\[ SI = A - P = 200 - 100 = 100 \]

4. **Using the Simple Interest Formula:**
- The formula for simple interest is:
\[ SI = \frac{P \times R \times T}{100} \]
- Substituting the known values (SI = 100, P = 100, T = 6):
\[ 100 = \frac{100 \times R \times 6}{100} \]
- Simplifying this gives:
\[ 100 = R \times 6 \]
- Therefore, we can find the rate \(R\):
\[ R = \frac{100}{6} \approx 16.67\% \]

5. **Finding the Time for the Amount to Become 12 Times:**
- Now, we want to find out how long it will take for the amount to become \(12 \times 100 = 1200\).
- The interest earned when the amount is \(1200\) is:
\[ SI = A - P = 1200 - 100 = 1100 \]

6. **Using the Simple Interest Formula Again:**
- Using the same formula for simple interest:
\[ 1100 = \frac{100 \times R \times T}{100} \]
- Substituting \(R = \frac{100}{6}\):
\[ 1100 = \frac{100 \times \frac{100}{6} \times T}{100} \]
- This simplifies to:
\[ 1100 = \frac{100 \times T}{6} \]
- Rearranging gives:
\[ 1100 \times 6 = 100 \times T \]
- Thus:
\[ T = \frac{1100 \times 6}{100} = 66 \]

7. **Conclusion:**
- Therefore, the time required for the amount to become 12 times is **66 years**.