Find the number of years in which an amount invested at 8% p.a. simple interest doubles itself.

Options

A. 12 years
B. 11 years
C. 12.5 years
D. 13 years

Answer

To find the number of years in which an amount invested at 8% per annum simple interest doubles itself, we can follow these steps:

### Step 1: Understand the formula for Simple Interest
The formula for calculating simple interest (SI) is:
\[ SI = \frac{P \times R \times T}{100} \]
where:
- \(SI\) = Simple Interest
- \(P\) = Principal amount (initial investment)
- \(R\) = Rate of interest per annum
- \(T\) = Time in years

### Step 2: Determine the condition for doubling the amount
If the amount doubles, then the total amount \(A\) becomes:
\[ A = 2P \]
The simple interest earned in this case will be:
\[ SI = A - P = 2P - P = P \]

### Step 3: Set up the equation
From the simple interest formula, we know:
\[ SI = \frac{P \times R \times T}{100} \]
Substituting \(SI = P\) and \(R = 8\%\), we get:
\[ P = \frac{P \times 8 \times T}{100} \]

### Step 4: Simplify the equation
We can cancel \(P\) from both sides (assuming \(P \neq 0\)):
\[ 1 = \frac{8 \times T}{100} \]

### Step 5: Solve for \(T\)
Rearranging the equation gives:
\[ T = \frac{100}{8} \]
Calculating this gives:
\[ T = 12.5 \text{ years} \]

### Conclusion
The number of years in which an amount invested at 8% per annum simple interest doubles itself is **12.5 years**.