# Compound interest word problems

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.

## Recent Articles 1. ### Find the Multiplicity of a Zero

Oct 20, 21 04:45 AM

Learn how to find the multiplicity of a zero with this easy to follow lesson

## Check out some of our top basic mathematics lessons.

Formula for percentage

Math skills assessment

Compatible numbers

Surface area of ​​a cube 