Problem Setup: Simple Interest Rate

The question asks for the annual simple interest rate ($R$) at which a principal amount ($P$) doubles itself in 10 years ($T$).

• Time ($T$) = 10 years.
• The final amount ($A$) is twice the principal ($P$), so $A = 2P$.
• The Simple Interest ($SI$) earned is the difference between the final amount and the principal: $SI = A - P$.

Calculating Simple Interest Earned

Since the sum doubles ($A = 2P$), the simple interest earned is:

$SI = A - P = 2P - P = P$

This means the interest earned over 10 years is equal to the principal amount.

Finding the Simple Interest Rate

The formula for simple interest is:

$SI = \frac{P \times R \times T}{100}$

Substitute the known values ($SI = P$ and $T = 10$ years) into the formula:

$P = \frac{P \times R \times 10}{100}$

To solve for $R$, we can simplify the equation. Assuming the principal $P$ is not zero, we can divide both sides by $P$:

$1 = \frac{R \times 10}{100}$

Now, solve for $R$:

$1 \times 100 = R \times 10$

$100 = 10R$

$R = \frac{100}{10}$

$R = 10$

The rate is 10% per annum.