Question

Purchasing a Car

Now you have to decide how to save enough money to purchase a used car in three years. You have the
$1000 that you saved up and you plan to continue working. According to your estimates, you can save an
additional $60 per month to put towards the car purchase. After conducting some research at the banks,
you have decided on two options (see below). You need to figure out which option will yield the most
money after the three years.
Option #1-CD for 3 years
Interest rate of 3% compounded monthly.
No money can be added to the CD.
However you can save your money on the side.
Option # 2-CD for 1 year
Interest rate of 2% compounded quarterly.
You can add money at the end of each year.
You will renew it each year for 3 years.
Work Shown:

Answer 1

Explanation:

To determine which option will yield the most money after three years, let's calculate the final amount for each option.

Option #1 - CD for 3 years:

Principal (initial investment) = $1000

Interest rate = 3% per year (compounded monthly)

No additional money can be added

To calculate the final amount, we can use the formula for compound interest:

A = P * (1 + r/n)^(n*t)

Where:

A = Final amount

P = Principal (initial investment)

r = Interest rate (as a decimal)

n = Number of times the interest is compounded per year

t = Number of years

For Option #1:

P = $1000

r = 3% = 0.03 (as a decimal)

n = 12 (compounded monthly)

t = 3 years

A = $1000 * (1 + 0.03/12)^(12*3)

Calculating the final amount for Option #1, we get:

A = $1000 * (1 + 0.0025)^(36)

A ≈ $1000 * (1.0025)^(36)

A ≈ $1000 * 1.0916768

A ≈ $1091.68

Option #2 - CD for 1 year:

Principal (initial investment) = $1000

Interest rate = 2% per year (compounded quarterly)

Money can be added at the end of each year

To calculate the final amount, we need to consider the annual additions and compounding at the end of each year.

First Year:

P = $1000

r = 2% = 0.02 (as a decimal)

n = 4 (compounded quarterly)

t = 1 year

A = $1000 * (1 + 0.02/4)^(4*1)

A ≈ $1000 * (1.005)^(4)

A ≈ $1000 * 1.0202

A ≈ $1020.20

At the end of the first year, the total amount is $1020.20.

Second Year:

Now we add an additional $60 to the previous amount:

P = $1020.20 + $60 = $1080.20

r = 2% = 0.02 (as a decimal)

n = 4 (compounded quarterly)

t = 1 year

A = $1080.20 * (1 + 0.02/4)^(4*1)

A ≈ $1080.20 * (1.005)^(4)

A ≈ $1080.20 * 1.0202

A ≈ $1101.59

At the end of the second year, the total amount is $1101.59.

Third Year:

Again, we add $60 to the previous amount:

P = $1101.59 + $60 = $1161.59

r = 2% = 0.02 (as a decimal)

n = 4 (compounded quarterly)

t = 1 year

A = $1161.59 * (1 + 0.02/4)^(4*1)

A ≈ $1161.59 * (1.005)^(4)

A ≈ $1161.59 * 1.0202

A ≈ $1185.39

At the end of the third year, the total amount is $1185.39.

Comparing the final amounts:

Option #1: $1091.68

Option #2: $1185.39

Therefore, Option #2 - CD for 1 year with an interest rate of 2% compounded quarterly and the ability to add money at the end of each year will yield the most money after three years.