A person deposited 100000 in a bank for three years. The Bank paid interest at 8% per annum compounded half yearly during the first year and at 12% per annum compounded quarterly during last 2 years.What is his balance after 3 years
Yearly interest on a 100,000.00 dollars
So what's the balance
To calculate the balance after 3 years, we need to calculate the compound interest for each period and add it to the initial deposit.
Let's break down the calculation into three periods:
1. First year: The interest is compounded semi-annually at a rate of 8%.
We can use the compound interest formula: A = P(1 + r/n)^(nt), where:
A = Final amount
P = Principal amount (initial deposit)
r = Annual interest rate (as a decimal)
n = Number of times interest is compounded per year
t = Number of years
In this case:
P = $100,000
r = 8% or 0.08 (since it's given as a percentage)
n = 2 (compounded semi-annually)
t = 1 (year)
Calculating the interest for the first year:
A = 100000(1 + 0.08/2)^(2*1)
2. Second year: The interest is compounded quarterly at a rate of 12%.
Using the same formula, but with different values:
P = The final amount from the previous year
r = 12% or 0.12
n = 4 (compounded quarterly)
t = 1 (year)
Calculating the interest for the second year:
A = (result from the first year)(1 + 0.12/4)^(4*1)
3. Third year: The interest is compounded quarterly at a rate of 12%.
Using the same formula, but with different values:
P = The final amount from the previous year
r = 12% or 0.12
n = 4 (compounded quarterly)
t = 1 (year)
Calculating the interest for the third year:
A = (result from the second year)(1 + 0.12/4)^(4*1)
Finally, we add the initial deposit to the result from the third year to obtain the balance after 3 years.
P = Po(1+r)^n.
First Year:
Po = $100,000.
r = 0.08/yr. * 0.5yrs. = 0.04.
n = 2comp./yr. * 1yr. = 2 compounding periods.
P = 100000(1.04)^2 = $108,160.
Last Two Years:
Po = 108,160.
r = 0.12/yr. * 0.25yrs. = 0.03/qtr.
n = 4comp./yr. * 2yrs. = 8 compounding periods. P = ?.