We will use the compound interest formula to solve these compound interest word problems.
Example #1
A deposit of $3000 earns 2% interest compounded semiannually. How much money is in the bank after for 4 years?
Solution
B = P( 1 + r)^{n}
P = $3000
r = 2% annual interest rate / 2 interest periods = 1% semiannual interest rate
n = number of payment periods = number of interest periods times number of years
n = 2 times 4 = 8
B = 3000( 1 + 1%)^{8} = 3000(1 + 0.01)^{8} = 3000(1.01)^{8 }
B = 3000(1.082856)
B = 3248.57
After four years, there will be 3248.57 dollars in the bank account.
Example #2
A deposit of $2150 earns 6% interest compounded quarterly. How much money is in the bank after for 6 years?
Solution
B = P( 1 + r)^{n}
P = $2150
r = 6% annual interest rate / 4 interest periods = 1.5% quarterly interest rate
n = number of payment periods = number of interest periods times number of years
n = 4 times 6 = 24
B = 2150( 1 + 1.5%)^{24} = 2150(1 + 0.015)^{24} = 2150(1.015)^{24 }
B = 2150(1.4295)
B = 3073.425
After 6 years, there will be 3073.425 dollars in the bank account.
Example #3
A deposit of $495 earns 3% interest compounded annually. How much money is in the bank after for 3 years?
Solution
B = P( 1 + r)^{n}
P = $495
r = 3% annual interest rate / 1 interest period = 3% annual interest rate
n = number of payment periods = number of interest periods times number of years
n = 1 times 3 = 3
B = 495( 1 + 3%)^{3} = 495(1 + 0.03)^{3} = 495(1.03)^{3 }
B = 495(1.092727)
B = 540.89
After 3 years, there will be 540.89 dollars in the bank account.
May 07, 21 02:29 PM
A time-series data shows information about the same subject or element of a sample or population for different periods of time.
Basic math formulas
Algebra word problems