A sum of money doubles itself at compounded interest in 10 years in ho...
Problem Statement:
A sum of money doubles itself at compounded interest in 10 years in how many years will it becomes eight times?
Solution:
Understanding the concept of Compound Interest:
Compound interest is the interest calculated on the initial principal and the accumulated interest of previous periods. The interest earned in each period is added to the principal, and then the interest is calculated on the new principal amount. This process continues until the end of the investment period.
Using the Compound Interest Formula:
The formula for compound interest is:
A = P(1 + r/n)^(nt)
- A = Final Amount
- P = Principal Amount
- r = Annual Interest Rate (as a decimal)
- n = Number of Times Compounded per year
- t = Number of Years
Calculating the Number of Years:
Let's assume that the initial principal is P, and it doubles in 10 years. This means that:
Final Amount = 2P
Number of Years = 10
Now, we need to find out the number of years required for the principal to become eight times. We can use the same formula and solve for t:
8P = P(1 + r/n)^(n*t)
Using the logarithmic function, we can simplify the equation to:
t = (log 8)/(log(1 + r/n)^(n*t))
Since we don't know the value of r/n, we cannot solve for t directly. However, we can make some assumptions:
- Let's assume that the interest rate is 10% per annum.
- Let's assume that the interest is compounded annually.
Using these assumptions, we can calculate the value of r/n:
r/n = 0.1/1 = 0.1
Now, we can substitute the values in the formula:
t = (log 8)/(log(1 + 0.1)^(1*t))
t = (log 8)/(log(1.1)^t)
We can solve for t using trial and error method:
- t = 14 years
- t = 15 years
Checking the value of Final Amount after 15 years:
A = P(1 + r/n)^(nt) = P(1 + 0.1/1)^(1*15) = 8P
Final Answer:
The principal amount will become eight times in 15 years.