Question

# In how many years will a sum of money double itself with the rate of 10% per annum simple interest?

Hint: To solve the problem, we should know the definition of annual simple interest. We have,
Simple Interest (I) = $\dfrac{P\times R\times t}{100}$
Where, P= principal amount
R = simple interest annual rate
t = time period of the annual simple interest
Here, we have R = 10% and have to calculate t for the sum of the money (that is P) to double.

In this question, we are left with two unknowns, P and t. However, we also have an additional condition. This condition tells that within the required time (which we have to calculate), the sum of money doubles itself. Thus, if originally, we had principal amount as P, finally, this amount would become 2P. Thus, simple interest (I) becomes 2P-P = P. Since, simple interest is basically the amount accumulated over the total principal amount. Further, for simplification, we can write,
$\dfrac{R}{100}=\dfrac{10}{100}=0.1$
Thus, we have,
I=$\dfrac{P\times R\times t}{100}$
Since, I = P (as calculated above), we have,
P = $\dfrac{P\times R\times t}{100}$
We can cancel P from both sides. Thus, we have,
1=$\dfrac{R\times t}{100}$
Plugging in the known values, we have,
1= 0.1$\times$t
Since, $\dfrac{R}{100}$=0.1
Now,
t=10 years
Hence, it will take 10 years for the sum of money to double itself with the rate of 10% per annum simple interest.

Note: While solving questions related to principal interest, it is important to keep in mind that simple interest calculated from the formula, Simple Interest (I) = $\dfrac{P\times R\times t}{100}$ , doesn’t represent the total amount of money. In fact, the total amount is the sum of Principal amount (P) and simple interest. Thus, in this case, when money was doubled, the total amount was 2P and simple interest was P.