To determine how much Ingrid needs to deposit at the end of each quarter to afford the $22,000 car in 8 years, we can use the future value of an ordinary annuity formula. The formula for future value $FV$ of an ordinary annuity is:

$FV = P \times \frac{(1 + r)^n - 1}{r}$

Where:

• $FV$ is the future value (which is $22,000),
• $P$ is the amount deposited at the end of each period (the unknown),
• $r$ is the interest rate per period,
• $n$ is the total number of periods.

Step 1: Identify the known values.

• Future value $FV = 22,000$,
• Interest rate per quarter $r = \frac{5.9\%}{4} = \frac{0.059}{4} = 0.01475$ (since it's compounded quarterly),
• Total number of periods $n = 8 \text{ years} \times 4 \text{ quarters/year} = 32 \text{ quarters}$.

Step 2: Rearrange the formula to solve for $P$ (the quarterly deposit):

$P = \frac{FV}{\frac{(1 + r)^n - 1}{r}}$

Substitute the known values:

$P = \frac{22,000}{\frac{(1 + 0.01475)^{32} - 1}{0.01475}}$

Now let's calculate this step by step.

1. $1 + 0.01475 = 1.01475$,
2. $1.01475^{32} = 1.558307$,
3. $1.558307 - 1 = 0.558307$,
4. $\frac{0.558307}{0.01475} = 37.85$.

Thus,

$P = \frac{22,000}{37.85} = 581.31$

Final Answer:

Ingrid must deposit $581.31 at the end of each quarter in the account to save enough to buy the car.

Would you like further details or have any questions?

Here are 5 related questions:

1. What if the interest rate was compounded monthly instead of quarterly?
2. How much would Ingrid need to deposit if the car costs $25,000 instead?
3. What happens if Ingrid wants to buy the car in 6 years instead of 8?
4. How does the deposit change if the interest rate increases to 6.5%?
5. What would be the future value if Ingrid deposited $600 each quarter?

Tip: Always verify that the compounding frequency (quarterly, monthly, etc.) matches the frequency of the deposits in savings problems.