ICMBA Exam Study Guide
This final set of problems from the textbook provides a comprehensive review of all Time Value of Money (TVM) concepts. The problems are diverse, ranging from calculating loan payments and amortization schedules to solving for unknown variables like interest rates and the number of periods. Mastering these problems is essential for your ICMBA exam.
- Loan Amortization: Understand how to calculate equal, annual payments and how to construct an amortization schedule that breaks down each payment into interest and principal. This is a common and important exam topic.
- Solving for Unknowns: Problems that require you to find an unknown interest rate, a missing cash flow, or the number of periods test your algebraic skills and your ability to apply TVM formulas in reverse.
- Practical Applications: Many of these problems are framed as real-world scenarios, such as retirement planning, loan repayment, and endowment creation. Focus on translating the narrative into the correct TVM framework.
- Ethics in Finance: Pay attention to the ethics-based problems, as they test your understanding of how TVM concepts apply to real-world business practices and ethical dilemmas.
Key Formulas:
Present Value (Ordinary Annuity): \( PV = PMT \times \left[ \frac{1 - \frac{1}{(1+r)^n}}{r} \right] \)
Loan Payment (PMT): \( PMT = \frac{PV}{\left[ \frac{1 - \frac{1}{(1+r)^n}}{r} \right]} \)
Future Value (Continuous Compounding): \( FV = PV \times e^{rt} \)
Present Value (Perpetuity): \( PV = \frac{PMT}{r} \)
Problems and Solutions (P5-41 to P5-62)
Problem P5-41: Annuities and compounding
Problem Statement: Janet Boyle intends to deposit $300 per year in a credit union for the next 10 years, and the credit union pays an annual interest rate of 8%.
- Determine the future value that Janet will have at the end of 10 years, given that end-of-period deposits are made and no interest is withdrawn, if (1) $300 is deposited annually and the credit union pays interest annually. (2) $150 is deposited semiannually and the credit union pays interest semiannually. (3) $75 is deposited quarterly and the credit union pays interest quarterly.
- Use your findings in part a to discuss the effect of more frequent deposits and compounding of interest on the future value of an annuity.
Solution:
Theoretical Foundation: This problem demonstrates how a change in the frequency of both deposits and compounding affects the future value of an annuity. More frequent compounding, even with the same nominal annual rate, leads to a higher future value due to earning "interest on interest" more often.
Part a: Future value in 10 years
Using the ordinary annuity formula \( FV_n = PMT \times \left[ \frac{(1+r)^n - 1}{r} \right] \) with adjustments for compounding frequency:
- (1) Annual deposits:
- \( PMT = \$300 \), \( r = 8\% \), \( n = 10 \)
- \( FV_{10} = \$300 \times \left[ \frac{(1.08)^{10} - 1}{0.08} \right] = \$4,345.97 \)
- (2) Semiannual deposits:
- \( PMT = \$150 \), \( r = \frac{8\%}{2} = 4\% \), \( n = 10 \times 2 = 20 \)
- \( FV_{20} = \$150 \times \left[ \frac{(1.04)^{20} - 1}{0.04} \right] = \$4,466.19 \)
- (3) Quarterly deposits:
- \( PMT = \$75 \), \( r = \frac{8\%}{4} = 2\% \), \( n = 10 \times 4 = 40 \)
- \( FV_{40} = \$75 \times \left[ \frac{(1.02)^{40} - 1}{0.02} \right] = \$4,528.84 \)
Part b: Discussion
As the frequency of deposits and compounding increases (from annual to semiannual to quarterly), the future value of the annuity also increases. This is because interest is being compounded more frequently, and the earlier deposits have more opportunities to earn interest on interest. This demonstrates that more frequent deposits and compounding are more beneficial to an investor.
Tricky Area:
The most important part of this problem is to correctly adjust all three variables for the compounding frequency: the payment amount (PMT), the interest rate (r), and the number of periods (n). Failing to adjust any of these will lead to an incorrect result.
Problem P5-42: Deposits to accumulate future sums
Problem Statement: For each of the cases shown in the following table, determine the amount of the equal, annual, end-of-year deposits necessary to accumulate the given sum at the end of the specified period, assuming the stated annual interest rate.
| Case | Sum to be accumulated | Accumulation period (years) | Interest rate |
|---|---|---|---|
| A | $5,000 | 3 | 12% |
| B | $100,000 | 20 | 7% |
| C | $30,000 | 8 | 10% |
| D | $15,000 | 12 | 8% |
Solution:
Theoretical Foundation: This problem is a reverse application of the future value of an ordinary annuity. You are given the future value and need to solve for the periodic payment (PMT). The formula is \( PMT = \frac{FV_n}{\left[ \frac{(1+r)^n - 1}{r} \right]} \).
- Case A:
- \( PMT = \frac{\$5,000}{\left[ \frac{(1.12)^3 - 1}{0.12} \right]} = \frac{\$5,000}{3.3744} = \$1,481.67 \)
- Case B:
- \( PMT = \frac{\$100,000}{\left[ \frac{(1.07)^{20} - 1}{0.07} \right]} = \frac{\$100,000}{40.9955} = \$2,439.30 \)
- Case C:
- \( PMT = \frac{\$30,000}{\left[ \frac{(1.10)^8 - 1}{0.10} \right]} = \frac{\$30,000}{11.4359} = \$2,623.36 \)
- Case D:
- \( PMT = \frac{\$15,000}{\left[ \frac{(1.08)^{12} - 1}{0.08} \right]} = \frac{\$15,000}{18.9771} = \$790.43 \)
Tricky Area:
The key here is setting up the problem correctly. You are solving for the periodic payment (PMT), which requires you to first calculate the future value interest factor for an annuity in the denominator. A common error is to confuse this with a present value problem.
Problem P5-43: Creating a retirement fund
Problem Statement: To supplement your planned retirement in exactly 42 years, you estimate that you need to accumulate $220,000 by the end of 42 years from today. You plan to make equal, annual, end-of-year deposits into an account paying 8% annual interest.
- How large must the annual deposits be to create the $220,000 fund by the end of 42 years?
- If you can afford to deposit only $600 per year into the account, how much will you have accumulated by the end of the forty-second year?
Solution:
Theoretical Foundation: This problem is a real-world application of both solving for a periodic payment and finding the future value of an ordinary annuity. It's a great example of how TVM concepts are used for long-term financial planning.
Part a: Annual deposits needed
Given: \( FV_{42} = \$220,000 \), \( r = 8\% \), \( n = 42 \).
\( PMT = \frac{\$220,000}{\left[ \frac{(1.08)^{42} - 1}{0.08} \right]} = \frac{\$220,000}{333.1593} = \$660.33 \)
Part b: Future value with $600 deposits
Given: \( PMT = \$600 \), \( r = 8\% \), \( n = 42 \).
Tricky Area:
The problems are straightforward but require careful calculation, especially with a long time horizon like 42 years. A small rounding error can lead to a significant difference in the final answer. Double-check your numbers in the future value interest factor calculation.
Problem P5-44: Finding interest rates
Problem Statement: You want to borrow $24,000 for a tax payment. Your friend offers you a loan, claiming a 10% interest rate. He calculates the interest as $24,000 x 10% = $2,400, deducts this amount, and gives you $21,600. You are to repay $24,000 in one year. Has your friend charged you 10% interest? What is the real interest rate charged on the loan?
Solution:
Theoretical Foundation: This is a crucial ethics-focused problem that tests your understanding of the difference between the stated interest rate and the actual rate of return. The real interest rate is based on the actual amount of money you receive versus the amount you repay.
The face value of the loan is $24,000, and the stated interest rate is 10%. However, you only receive **$21,600** today, and you are repaying **$24,000** in one year. The interest paid is the difference: \( \$24,000 - \$21,600 = \$2,400 \).
To find the real interest rate, we use the formula \( r = \frac{\text{Interest paid}}{\text{Principal received}} \).
No, your friend has not charged you 10% interest. The **real interest rate** is **11.11%**. This is a higher rate because the interest was deducted upfront from the principal you received, effectively increasing the cost of the loan.
Tricky Area:
The main trap is accepting the stated interest rate without calculating the effective rate based on the actual cash flows. The problem highlights the difference between an interest rate based on the face value of a loan and one based on the actual amount disbursed. This is a common practice in payday loans and other non-traditional lending, so understanding this concept is vital.
Problem P5-45: Deposits to create a perpetuity
Problem Statement: You have decided to endow a university with a scholarship. It is expected to cost $6,000 per year to attend the university into perpetuity. You expect to give the university the endowment in 10 years and will accumulate it by making equal annual (end-of-year) deposits into an account. The rate of interest is expected to be 10% for all future time periods.
- How large must the endowment be?
- How much must you deposit at the end of each of the next 10 years to accumulate the required amount?
Solution:
Theoretical Foundation: This is a two-part problem that links the concept of a perpetuity to the future value of an ordinary annuity. First, you calculate the present value of the perpetual payments (the size of the endowment). Second, you treat that value as the future value of your savings plan and solve for the periodic payments.
Part a: Endowment size
The endowment needs to generate $6,000 a year forever. This is a perpetuity. The size of the endowment is its present value.
The endowment must be **$60,000**.
Part b: Annual deposits
The $60,000 found in part a is the future value that you need to accumulate in 10 years. Now we solve for the annual deposit (PMT) of an ordinary annuity.
\( PMT = \frac{\$60,000}{\left[ \frac{(1.10)^{10} - 1}{0.10} \right]} = \frac{\$60,000}{15.9374} = \$3,764.12 \)
Tricky Area:
This problem is tricky because it links two concepts together. A common mistake is to try to do the entire calculation in one step. The correct approach is to break it into two separate problems: first, find the required future value (the endowment size), and second, solve for the periodic payment needed to reach that future value.
Problem P5-46: Inflation, time value, and annual deposits
Problem Statement: John Kelley wants to buy a vacation home that costs $200,000 today when he retires in 25 years. He believes prices will increase at an average inflation rate of 5% per year. He can earn 9% annually on his investments and plans to make equal, end-of-year deposits over the next 25 years to fund the purchase.
- Inflation is expected to average 5% per year for the next 25 years. What will John's dream house cost when he retires?
- How much must John invest at the end of each of the next 25 years to have the cash purchase price of the house when he retires?
- If John invests at the beginning instead of at the end of each of the next 25 years, how much must he invest each year?
Solution:
Theoretical Foundation: This problem combines future value of a single amount (for inflation) with solving for the periodic payment of both an ordinary annuity and an annuity due. It's a comprehensive retirement planning problem that requires careful step-by-step calculations.
Part a: Future cost of the house
This is a future value of a single amount calculation. \( PV = \$200,000 \), \( r = 5\% \), \( n = 25 \).
Part b: Annual deposits (Ordinary Annuity)
The amount from part a ($677,270.99) is the future value we need to accumulate. We solve for PMT with \( r = 9\% \) and \( n = 25 \).
\( PMT = \frac{\$677,270.99}{\left[ \frac{(1.09)^{25} - 1}{0.09} \right]} = \frac{\$677,270.99}{84.7007} = \$8,000.77 \)
Part c: Annual deposits (Annuity Due)
Since the deposits are at the beginning of the year, we use the annuity due factor, which is the ordinary annuity factor multiplied by \( (1+r) \). The PMT will be lower than in part b.
\( PMT_{due} = \frac{\$677,270.99}{84.7007 \times 1.09} = \frac{\$677,270.99}{92.3238} = \$7,335.75 \)
Tricky Area:
This is a two-stage problem. The first step is to correctly calculate the future cost of the house, which serves as the target future value for the second part. The final part is tricky because it requires you to switch from an ordinary annuity to an annuity due, which has a different PMT value even though the total future value is the same. Remember the annuity due multiplier!
Problem P5-47: Loan payment
Problem Statement: Determine the equal, annual, end-of-year payment required each year over the life of the loans shown in the following table to repay them fully during the stated term of the loan.
| Loan | Principal | Interest rate | Term of loan (years) |
|---|---|---|---|
| A | $12,000 | 8% | 3 |
| B | $60,000 | 12% | 10 |
| C | $75,000 | 10% | 30 |
| D | $4,000 | 15% | 5 |
Solution:
Theoretical Foundation: This problem requires you to solve for the periodic payment (PMT) of an ordinary annuity. The principal of the loan is the present value, and the payments are the annuity. The formula is \( PMT = \frac{PV}{\left[ \frac{1 - \frac{1}{(1+r)^n}}{r} \right]} \).
- Loan A:
- \( PMT = \frac{\$12,000}{\left[ \frac{1 - \frac{1}{(1.08)^3}}{0.08} \right]} = \frac{\$12,000}{2.5771} = \$4,656.40 \)
- Loan B:
- \( PMT = \frac{\$60,000}{\left[ \frac{1 - \frac{1}{(1.12)^{10}}}{0.12} \right]} = \frac{\$60,000}{5.6502} = \$10,619.64 \)
- Loan C:
- \( PMT = \frac{\$75,000}{\left[ \frac{1 - \frac{1}{(1.10)^{30}}}{0.10} \right]} = \frac{\$75,000}{9.4269} = \$7,956.14 \)
- Loan D:
- \( PMT = \frac{\$4,000}{\left[ \frac{1 - \frac{1}{(1.15)^5}}{0.15} \right]} = \frac{\$4,000}{3.3522} = \$1,193.26 \)
Tricky Area:
The main challenge is not confusing this with a future value problem. You are given the present value of the loan and need to find the series of equal payments that, when discounted back to the present, equal that loan amount. Make sure to use the correct present value interest factor for an annuity in your calculation.
Problem P5-48: Loan amortization schedule
Problem Statement: Joan Messineo borrowed $15,000 at a 14% annual rate of interest to be repaid over 3 years. The loan is amortized into three equal, annual, end-of-year payments.
- Calculate the annual, end-of-year loan payment.
- Prepare a loan amortization schedule showing the interest and principal breakdown of each of the three loan payments.
- Explain why the interest portion of each payment declines with the passage of time.
Solution:
Theoretical Foundation: This is a classic amortization problem that requires you to calculate the loan payment and then track the principal balance and interest payments over time. The key is to remember that the interest is calculated on the remaining principal balance, which decreases with each payment.
Part a: Annual loan payment
Using the loan payment formula: \( PV = \$15,000 \), \( r = 14\% \), \( n = 3 \).
Part b: Amortization schedule
The schedule tracks the beginning balance, interest, principal, and ending balance for each year.
| Year | Beginning Balance | Payment | Interest (14%) | Principal (Payment - Interest) | Ending Balance |
|---|---|---|---|---|---|
| 1 | $15,000.00 | $6,461.34 | $2,100.00 | $4,361.34 | $10,638.66 |
| 2 | $10,638.66 | $6,461.34 | $1,489.41 | $4,971.93 | $5,666.73 |
| 3 | $5,666.73 | $6,461.34 | $793.34 | $5,668.00* | $0.00 |
*Due to rounding, the final principal payment may be slightly different. In this case, the final payment is the remaining balance plus interest for the year.
Part c: Explanation for declining interest
The interest portion of each payment declines over time because the **outstanding principal balance is decreasing**. Since the loan payment is a fixed amount, as the interest portion decreases, the principal portion of the payment increases. This is a fundamental characteristic of an amortizing loan.
Tricky Area:
The most common mistake is to miscalculate the annual payment. If the payment is incorrect, the entire amortization schedule will be wrong. Remember that interest is always calculated on the **beginning balance** for that period, not the initial loan amount.
Problem P5-49: Loan interest deductions
Problem Statement: Liz Rogers just closed a $10,000 business loan that is to be repaid in three equal, annual, end-of-year payments. The interest rate on the loan is 13%. As part of her firm's detailed financial planning, Liz wishes to determine the annual interest deduction attributable to the loan. (Because it is a business loan, the interest portion of each loan payment is tax-deductible to the business.)
- Determine the firm's annual loan payment.
- Prepare an amortization schedule for the loan.
- How much interest expense will Liz's firm have in each of the next 3 years as a result of this loan?
Solution:
Theoretical Foundation: This problem is very similar to the previous one, but with an added focus on the tax-deductibility of interest payments. It's a great example of how financial calculations are used in real-world tax and financial planning.
Part a: Annual loan payment
Using the loan payment formula: \( PV = \$10,000 \), \( r = 13\% \), \( n = 3 \).
Part b: Amortization schedule
The schedule breaks down each payment into interest and principal. This is the crucial step for determining the tax-deductible interest expense.
| Year | Beginning Balance | Payment | Interest (13%) | Principal (Payment - Interest) | Ending Balance |
|---|---|---|---|---|---|
| 1 | $10,000.00 | $4,235.15 | $1,300.00 | $2,935.15 | $7,064.85 |
| 2 | $7,064.85 | $4,235.15 | $918.43 | $3,316.72 | $3,748.13 |
| 3 | $3,748.13 | $4,235.15 | $487.26 | $3,747.89* | $0.00 |
*Due to rounding, the final principal payment may be slightly different. The total payment must cover the final balance plus interest.
Part c: Annual interest expense
The annual interest expense is simply the interest portion of each payment from the amortization schedule:
- Year 1: $1,300.00
- Year 2: $918.43
- Year 3: $487.26
Tricky Area:
The trickiest part is correctly constructing the amortization table, particularly ensuring that each period's interest is calculated on the correct beginning balance. The total interest deduction over the life of the loan is the sum of these amounts, which is a valuable piece of information for a business owner.
Problem P5-50: Loan rates of interest
Problem Statement: You want to buy a new car that costs $48,000. Dealer A offers a zero-interest 2-year loan with monthly payments. Dealer B requires a $10,000 down payment and offers 24 monthly installments of $1,500. The market interest rate is 6%.
- What is the net cost today of the two options? Which option offers you the cheapest financing?
- Use a financial calculator or spreadsheet to help you calculate what the interest rate would be if the financing cost from Dealer A was equal to that of Dealer B.
Solution:
Theoretical Foundation: This problem compares two different financing options. The best way to compare them is to calculate the present value of all cash flows (outflows) associated with each option. The option with the lowest present value is the cheapest.
Part a: Net cost today (PV)
The market rate is 6% annually, so the monthly rate is \( \frac{6\%}{12} = 0.5\% = 0.005 \). The term is 2 years, or 24 months.
- Dealer A:
- Loan amount = $48,000, 24 monthly payments. Since it's a zero-interest loan, the payments are \( \frac{\$48,000}{24} = \$2,000 \) per month.
- Net Cost = Present value of 24 payments of $2,000 at 0.5% monthly.
- \( PV_A = \$2,000 \times \left[ \frac{1 - \frac{1}{(1.005)^{24}}}{0.005} \right] = \$45,043.68 \)
- Dealer B:
- Net Cost = Down payment + Present value of 24 monthly payments.
- \( PV_B = \$10,000 + \left( \$1,500 \times \left[ \frac{1 - \frac{1}{(1.005)^{24}}}{0.005} \right] \right) = \$10,000 + \$33,782.76 = \$43,782.76 \)
The cheapest financing is from **Dealer B** ($43,782.76) because the present value of its total cash flows is lower than Dealer A's ($45,043.68).
Part b: Interest rate for Dealer A to equal Dealer B's cost
We want to find the interest rate \( r \) where the present value of Dealer A's payments equals the present value of Dealer B's payments, which is $43,782.76. The loan amount for Dealer A is $48,000, which is repaid over 24 months. The monthly payment is $2,000.
By using a financial calculator or a spreadsheet, you can solve for \( r \). This gives a monthly rate of approximately **0.86%**. The annual rate would be \( 0.86\% \times 12 \approx 10.32\% \).
Tricky Area:
The main challenge is setting up the problem correctly for comparison. You must use the market interest rate to discount the future cash flows for both options. Part b is a reverse TVM problem where you solve for the interest rate, which is a key skill to master.
Problem P5-51: Growth rates
Problem Statement: You are given the series of cash flows shown in the following table. a. Calculate the compound annual growth rate between the first and last payment in each stream. b. If year-1 values represent initial deposits in a savings account paying annual interest, what is the annual rate of interest earned on each account? c. Compare and discuss the growth rate and interest rate found in parts a and b, respectively.
| Year | A | B | C |
|---|---|---|---|
| 1 | $500 | $1,500 | $2,500 |
| 2 | 560 | 1,550 | 2,600 |
| 3 | 640 | 1,610 | 2,650 |
| 4 | 720 | 1,680 | 2,650 |
| 5 | 800 | 1,760 | 2,800 |
| 6 | 1,850 | 2,850 | |
| 7 | 1,950 | 2,900 | |
| 8 | 2,060 | ||
| 9 | 2,170 | ||
| 10 | 2,280 |
Solution:
Theoretical Foundation: This problem distinguishes between a compound annual growth rate (CAGR) and a simple interest rate. The CAGR formula is used to find the average growth rate of a series of values over time, while the interest rate is the return on an investment.
Part a: Compound Annual Growth Rate (CAGR)
Using the formula \( CAGR = \left( \frac{FV}{PV} \right)^{\frac{1}{n-1}} - 1 \). Note that \( n-1 \) is used because the number of growth periods is one less than the number of data points.
- **Stream A:** \( CAGR = \left( \frac{\$800}{\$500} \right)^{\frac{1}{4}} - 1 = 12.47\% \)
- **Stream B:** \( CAGR = \left( \frac{\$2,280}{\$1,500} \right)^{\frac{1}{9}} - 1 = 4.77\% \)
- **Stream C:** \( CAGR = \left( \frac{\$2,900}{\$2,500} \right)^{\frac{1}{6}} - 1 = 2.50\% \)
Part b: Annual Rate of Interest
The annual rate of interest is the same as the CAGR when the cash flows are the start and end values of a single deposit. We use the same formula.
- **Account A:** 12.47%
- **Account B:** 4.77%
- **Account C:** 2.50%
Part c: Comparison and discussion
The growth rate and the interest rate are fundamentally the same in this context. They both measure the compound annual return on an initial amount over a period of time. However, the interest rate is a specific return on an investment, while the CAGR is a more general term for the average growth of a variable over a period. In a simple savings account, the growth rate is the interest rate. In a mixed stream of cash flows, the actual return might be different from the CAGR.
Tricky Area:
The primary trick here is to be precise with the number of periods for the CAGR calculation. For a stream with 'n' payments, there are only 'n-1' growth periods. Using 'n' instead of 'n-1' is a common mistake. The comparison in part c also requires you to understand the nuanced relationship between a growth rate and an interest rate.
Problem P5-52: Rate of return
Problem Statement: Rishi Singh has $1,500 to invest. His investment counselor suggests an investment that pays no stated interest but will return $2,000 at the end of 3 years. Rishi is considering another investment, of equal risk, that earns an annual return of 8%. Which investment should he make, and why?
- What annual rate of return will Rishi earn with this investment?
- Rishi is considering another investment, of equal risk, that earns an annual return of 8%. Which investment should he make, and why?
Solution:
Theoretical Foundation: This problem requires you to solve for an unknown interest rate from the future value formula. It then connects this to a real-world investment decision by comparing the calculated rate to an alternative investment's rate of return.
Part a: Annual rate of return
We need to solve for \( r \) in the formula \( FV_n = PV \times (1+r)^n \). Given: \( PV = \$1,500 \), \( FV = \$2,000 \), \( n = 3 \).
\( 1.3333 = (1+r)^3 \)
\( (1.3333)^{1/3} = 1+r \)
\( 1.1006 = 1+r \)
\( r = 0.1006 \text{ or } 10.06\% \)
Part b: Investment choice
Rishi's first investment offers a return of 10.06%. The alternative, which has equal risk, offers a return of 8%. Since the first investment offers a higher return, Rishi should choose the **investment that returns $2,000 in 3 years**. As long as both investments have the same risk, the one with the higher return is the better choice.
Tricky Area:
The problem is straightforward, but it's essential to understand that comparing the returns is the correct way to make a decision when the risk is the same. Simply looking at the final amount is not sufficient if the initial investment amounts or time periods are different.
Problem P5-53: Rate of return and investment choice
Problem Statement: Clare Jaccard has $5,000 to invest. She is considering four equally risky investments, each requiring an initial $5,000 payment and providing a single amount at the end of its life.
| Investment | Single amount | Investment life (years) |
|---|---|---|
| A | $8,400 | 6 |
| B | $15,900 | 15 |
| C | $7,600 | 4 |
| D | $13,000 | 10 |
- Calculate, to the nearest 1%, the rate of return on each of the four investments.
- Which investment would you recommend to Clare, given her goal of maximizing the rate of return?
Solution:
Theoretical Foundation: This problem requires solving for the interest rate (\( r \)) for a series of investments. The objective is to find the investment with the highest return, which is the best choice given that all investments have the same risk.
Part a: Rate of return
Using the formula \( r = \left( \frac{FV}{PV} \right)^{\frac{1}{n}} - 1 \), where \( PV = \$5,000 \):
- **Investment A:** \( r = \left( \frac{\$8,400}{\$5,000} \right)^{\frac{1}{6}} - 1 = 8.99\% \approx 9\% \)
- **Investment B:** \( r = \left( \frac{\$15,900}{\$5,000} \right)^{\frac{1}{15}} - 1 = 8.00\% \approx 8\% \)
- **Investment C:** \( r = \left( \frac{\$7,600}{\$5,000} \right)^{\frac{1}{4}} - 1 = 11.04\% \approx 11\% \)
- **Investment D:** \( r = \left( \frac{\$13,000}{\$5,000} \right)^{\frac{1}{10}} - 1 = 10.02\% \approx 10\% \)
Part b: Recommendation
Clare should choose **Investment C**, as it offers the highest rate of return at approximately 11%.
Tricky Area:
When solving for the rate of return, make sure to use a high level of precision in your intermediate steps to avoid rounding errors that could change your final answer, especially when rounding to the nearest percent. It's also easy to confuse the rate of return with the total nominal return, which is not the correct metric for comparison.
Problem P5-54: Rate of return: Annuity
Problem Statement: What is the rate of return on an investment of $10,606 if the company will receive $2,000 each year for the next 10 years?
Solution:
Theoretical Foundation: This problem is a reverse application of the present value of an ordinary annuity. You are given the present value (the investment cost), the periodic payment, and the number of periods. You need to solve for the unknown interest rate (\( r \)).
Given: \( PV = \$10,606 \), \( PMT = \$2,000 \), \( n = 10 \).
The formula is \( PV = PMT \times \left[ \frac{1 - \frac{1}{(1+r)^n}}{r} \right] \).
This equation cannot be solved algebraically for \( r \). You must use a financial calculator or a spreadsheet's IRR (Internal Rate of Return) function. The result is approximately **12%**.
Tricky Area:
The main trick is knowing that you cannot solve this equation manually. You must use a financial tool. The problem tests your knowledge of the correct formula and your ability to use the right tools for the job, rather than your algebraic skills.
Problem P5-55: Monthly loan payments
Problem Statement: Ricky is considering purchasing an apartment costing $700,000. He will pay a 30% down payment and take out a mortgage for the remainder. He wants the plan with the lowest monthly payment from the following offers.
| Bank | Interest rate | Term (years) |
|---|---|---|
| A | 3.5% | 15 |
| B | 3% | 20 |
| C | 4% | 25 |
| D | 4.5% | 18 |
- What are the monthly payments for plans offered by the four banks?
- Which plan should Ricky choose?
Solution:
Theoretical Foundation: This problem is a direct application of the loan payment formula. It also involves non-annual compounding and the calculation of the loan principal after a down payment. You must find the monthly payment for each option and then compare them.
First, find the loan principal: \( \$700,000 \times (1 - 0.30) = \$490,000 \).
We use the formula \( PMT = \frac{PV}{\left[ \frac{1 - \frac{1}{(1+r)^n}}{r} \right]} \) with monthly adjustments for r and n.
Part a: Monthly payments
- Bank A:
- \( r = \frac{3.5\%}{12} \), \( n = 15 \times 12 = 180 \)
- \( PMT = \frac{\$490,000}{\left[ \frac{1 - \frac{1}{(1 + 0.035/12)^{180}}}{0.035/12} \right]} = \$3,514.88 \)
- Bank B:
- \( r = \frac{3\%}{12} \), \( n = 20 \times 12 = 240 \)
- \( PMT = \frac{\$490,000}{\left[ \frac{1 - \frac{1}{(1 + 0.03/12)^{240}}}{0.03/12} \right]} = \$2,719.98 \)
- Bank C:
- \( r = \frac{4\%}{12} \), \( n = 25 \times 12 = 300 \)
- \( PMT = \frac{\$490,000}{\left[ \frac{1 - \frac{1}{(1 + 0.04/12)^{300}}}{0.04/12} \right]} = \$2,593.42 \)
- Bank D:
- \( r = \frac{4.5\%}{12} \), \( n = 18 \times 12 = 216 \)
- \( PMT = \frac{\$490,000}{\left[ \frac{1 - \frac{1}{(1 + 0.045/12)^{216}}}{0.045/12} \right]} = \$3,184.81 \)
Part b: Decision
Ricky should choose **Bank C** as it has the lowest monthly payment of $2,593.42.
Tricky Area:
The main challenge is consistently using the correct monthly interest rate and number of periods. A common mistake is to use the annual interest rate or the number of years directly in the formula without converting them to monthly values. This problem is a good test of your attention to detail in a multi-part calculation.
Problem P5-56: Interest rate for an annuity
Problem Statement: Anna Waldheim was awarded a judgment of $2,000,000. She is offered a settlement of $156,000 per year for 25 years. Her counteroffer is $255,000 per year for 25 years. Both assume payments at the end of each year and have a present value of $2,000,000.
- What interest rate assumption have the defendants used in their offer (rounded to the nearest whole percent)?
- What interest rate assumption have Anna and her lawyer used in their counteroffer (rounded to the nearest whole percent)?
- Anna is willing to settle for an annuity that carries an interest rate assumption of 9%. What annual payment would be acceptable to her?
Solution:
Theoretical Foundation: This problem is a reverse application of the present value of an ordinary annuity. You are given the PV, PMT, and n and must solve for the interest rate \( r \). Part c is a direct application of the standard PV of an annuity formula.
Part a: Defendants' interest rate
Given: \( PV = \$2,000,000 \), \( PMT = \$156,000 \), \( n = 25 \).
Using a financial calculator or spreadsheet, the interest rate \( r \) is approximately **6%**.
Part b: Anna's interest rate
Given: \( PV = \$2,000,000 \), \( PMT = \$255,000 \), \( n = 25 \).
Using a financial calculator or spreadsheet, the interest rate \( r \) is approximately **12%**.
Part c: Acceptable annual payment
Given: \( PV = \$2,000,000 \), \( r = 9\% \), \( n = 25 \). We solve for PMT.
An annual payment of **$203,612.35** would be acceptable to her.
Tricky Area:
The main challenge is knowing that parts a and b require a financial calculator or spreadsheet to solve for the unknown interest rate, as the formula cannot be solved algebraically. Part c is a straightforward application of the loan payment formula, but you must use the present value of the judgment as the PV, not the annuity payment.
Problem P5-57: Loan rates of interest
Problem Statement: You have a credit card debt of $50,000 with a 32% interest rate, compounded monthly. You find three alternative cards with different interest rates and handling fees. You plan to pay off the debt in one year with equal monthly payments.
| Card | Interest rate | Outstanding amount |
|---|---|---|
| A | 30% | $50,000 |
| B | 29% | $50,500 |
| C | 28% | $51,000 |
- If you pay off the debt in one year with equal monthly amounts, how much in interest payments have you saved by using (1) Card A, (2) Card B, and (3) Card C rather than keeping your existing card?
- Which card should you choose?
Solution:
Theoretical Foundation: This problem requires a multi-step analysis to compare different loan options. You must calculate the monthly payment for each card, determine the total payments, and then find the total interest paid for each. The best option is the one that minimizes total interest payments and fees.
Part a: Interest savings
First, calculate the monthly payment for the existing card (32% annual, 12 months, $50,000 PV).
\( \text{Total Payments}_{\text{old}} = \$4,785.66 \times 12 = \$57,427.92 \)
\( \text{Total Interest}_{\text{old}} = \$57,427.92 - \$50,000 = \$7,427.92 \)
Now, calculate the total interest for each new card.
- Card A (30%, $50,000):
- \( PMT_A = \frac{\$50,000}{\left[ \frac{1 - \frac{1}{(1 + 0.30/12)^{12}}}{0.30/12} \right]} = \$4,727.65 \)
- \( \text{Total Payments}_A = \$4,727.65 \times 12 = \$56,731.80 \)
- \( \text{Total Interest}_A = \$56,731.80 - \$50,000 = \$6,731.80 \)
- \( \text{Savings}_A = \$7,427.92 - \$6,731.80 = \$696.12 \)
- Card B (29%, $50,500):
- \( PMT_B = \frac{\$50,500}{\left[ \frac{1 - \frac{1}{(1 + 0.29/12)^{12}}}{0.29/12} \right]} = \$4,710.02 \)
- \( \text{Total Payments}_B = \$4,710.02 \times 12 = \$56,520.24 \)
- \( \text{Total Interest}_B = \$56,520.24 - \$50,500 = \$6,020.24 \)
- \( \text{Savings}_B = \$7,427.92 - \$6,020.24 = \$1,407.68 \)
- Card C (28%, $51,000):
- \( PMT_C = \frac{\$51,000}{\left[ \frac{1 - \frac{1}{(1 + 0.28/12)^{12}}}{0.28/12} \right]} = \$4,705.51 \)
- \( \text{Total Payments}_C = \$4,705.51 \times 12 = \$56,466.12 \)
- \( \text{Total Interest}_C = \$56,466.12 - \$51,000 = \$5,466.12 \)
- \( \text{Savings}_C = \$7,427.92 - \$5,466.12 = \$1,961.80 \)
Part b: Decision
You should choose **Card C**. Although it has the highest outstanding amount initially due to the handling fee, it results in the greatest total interest savings because of its lowest interest rate. The savings from the lower interest rate more than offset the upfront fee.
Tricky Area:
The main trap is to simply look at the interest rate and assume the lowest is the best without accounting for the fees. Because the fees are added to the principal, they are also subject to interest. You must perform a complete cash flow analysis for each option to make the correct decision, comparing the total interest paid for each.
Problem P5-58: Number of years to equal future amount
Problem Statement: For each of the following cases, determine the number of years it will take for the initial deposit to grow to equal the future amount at the given interest rate.
| Case | Initial deposit | Future amount | Interest rate |
|---|---|---|---|
| A | $300 | $1,000 | 7% |
| B | $12,000 | $15,000 | 5% |
| C | $9,000 | $20,000 | 10% |
| D | $100 | $500 | 9% |
| E | $7,500 | $30,000 | 15% |
Solution:
Theoretical Foundation: This problem requires you to solve for the number of periods (\( n \)) in the future value of a single amount formula. The calculation involves using logarithms to isolate the exponent \( n \).
Using the formula \( FV_n = PV \times (1+r)^n \) and solving for \( n \):
- Case A: \( n = \frac{\ln(1,000/300)}{\ln(1.07)} = 17.78 \text{ years} \)
- Case B: \( n = \frac{\ln(15,000/12,000)}{\ln(1.05)} = 4.57 \text{ years} \)
- Case C: \( n = \frac{\ln(20,000/9,000)}{\ln(1.10)} = 8.35 \text{ years} \)
- Case D: \( n = \frac{\ln(500/100)}{\ln(1.09)} = 18.66 \text{ years} \)
- Case E: \( n = \frac{\ln(30,000/7,500)}{\ln(1.15)} = 9.92 \text{ years} \)
Tricky Area:
The main challenge is the algebraic manipulation to solve for \( n \). Remember to first divide both sides by PV, then take the logarithm of both sides. This is a fundamental skill for solving a wide variety of TVM problems.
Problem P5-59: Time to accumulate a given sum
Problem Statement: Manuel Rios wishes to determine how long it will take an initial deposit of $10,000 to double.
- If Manuel earns 10% annual interest on the deposit, how long will it take for him to double his money?
- How long will it take if he earns only 7% annual interest?
- How long will it take if he can earn 12% annual interest?
- Reviewing your findings in parts a, b, and c, indicate what relationship exists between the interest rate and the amount of time it will take Manuel to double his money.
Solution:
Theoretical Foundation: This problem is a direct application of the future value formula, solving for \( n \). It also introduces the "Rule of 72," a simple approximation for doubling time, and asks you to explain the inverse relationship between the interest rate and the time it takes to double an investment.
Given: \( PV = \$10,000 \), \( FV = \$20,000 \). We need to solve for \( n \).
Part a: 10% interest
Using the formula \( n = \frac{\ln(FV/PV)}{\ln(1+r)} \): \( n = \frac{\ln(20,000/10,000)}{\ln(1.10)} = 7.27 \text{ years} \). By the Rule of 72: \( \frac{72}{10} = 7.2 \text{ years} \).
Part b: 7% interest
Using the formula: \( n = \frac{\ln(2)}{\ln(1.07)} = 10.24 \text{ years} \). By the Rule of 72: \( \frac{72}{7} \approx 10.29 \text{ years} \).
Part c: 12% interest
Using the formula: \( n = \frac{\ln(2)}{\ln(1.12)} = 6.12 \text{ years} \). By the Rule of 72: \( \frac{72}{12} = 6 \text{ years} \).
Part d: Relationship
There is an **inverse relationship** between the interest rate and the time it takes to double your money. As the interest rate increases, the time required to double the investment decreases. This is because higher interest rates generate more interest in each period, which in turn compounds faster, leading to a shorter doubling time.
Tricky Area:
The trick is the "why" in part d. You must explain that the relationship is inverse and that the reason is the power of compounding. The Rule of 72 is a great tool, but the precise calculation reveals the exact relationship.
Problem P5-60: Number of years to provide a given return
Problem Statement: In each of the following cases, determine the number of years that the given annual end-of-year cash flow must continue to provide the given rate of return on the given initial amount.
| Case | Initial amount | Annual cash flow | Rate of return |
|---|---|---|---|
| A | $1,000 | $250 | 11% |
| B | $150,000 | $30,000 | 15% |
| C | $80,000 | $10,000 | 10% |
| D | $600 | $275 | 9% |
| E | $17,000 | $3,500 | 6% |
Solution:
Theoretical Foundation: This problem requires you to solve for the number of periods (\( n \)) for a loan or investment that is structured as an annuity. You are given the present value, the periodic payment, and the interest rate. You must use the present value of an ordinary annuity formula and solve for \( n \).
Using the formula \( PV = PMT \times \left[ \frac{1 - \frac{1}{(1+r)^n}}{r} \right] \) and solving for \( n \):
- Case A: \( n = - \frac{\ln\left(1 - \frac{1,000 \times 0.11}{250}\right)}{\ln(1.11)} = 4.88 \text{ years} \)
- Case B: \( n = - \frac{\ln\left(1 - \frac{150,000 \times 0.15}{30,000}\right)}{\ln(1.15)} = 9.92 \text{ years} \)
- Case C: \( n = - \frac{\ln\left(1 - \frac{80,000 \times 0.10}{10,000}\right)}{\ln(1.10)} = 16.54 \text{ years} \)
- Case D: \( n = - \frac{\ln\left(1 - \frac{600 \times 0.09}{275}\right)}{\ln(1.09)} = 2.45 \text{ years} \)
- Case E: \( n = - \frac{\ln\left(1 - \frac{17,000 \times 0.06}{3,500}\right)}{\ln(1.06)} = 6.22 \text{ years} \)
Tricky Area:
The main challenge is the algebraic manipulation. It's easy to make a sign error with the natural logarithm. The problem is a solid test of your ability to use the TVM formulas in a non-standard way to solve for an unknown variable.
Problem P5-61: Time to repay installment loan
Problem Statement: Mia Salto wishes to determine how long it will take to repay a loan with initial proceeds of $14,000 where annual end-of-year installment payments of $2,450 are required.
- If Mia can borrow at a 12% annual rate of interest, how long will it take for her to repay the loan fully?
- How long will it take if she can borrow at a 9% annual rate?
- How long will it take if she has to pay 15% annual interest?
- Reviewing your answers in parts a, b, and c, describe the general relationship between the interest rate and the amount of time it will take Mia to repay the loan fully.
Solution:
Theoretical Foundation: This problem requires solving for the number of periods (\( n \)) of an annuity. It highlights the inverse relationship between the interest rate and the time it takes to repay a loan, given a constant payment.
Given: \( PV = \$14,000 \), \( PMT = \$2,450 \). We need to solve for \( n \).
Part a: 12% interest
Using the formula \( n = - \frac{\ln\left(1 - \frac{PV \times r}{PMT}\right)}{\ln(1+r)} \):
Part b: 9% interest
Part c: 15% interest
Part d: Relationship
There is a **direct relationship** between the interest rate and the time it takes to repay a loan with a constant payment. As the interest rate increases, a larger portion of each payment goes toward interest, leaving less to pay down the principal. This means it takes longer to fully repay the loan.
Tricky Area:
The main challenge is the final "why." The intuition might be that a higher rate means you pay faster, but the opposite is true for a constant payment. You must explain that a higher interest rate consumes a greater percentage of the fixed payment, leaving less for the principal.
Problem P5-62: ETHICS PROBLEM
Problem Statement: A manager at a "Check Into Cash" business defends his business practice as simply "charging what the market will bear." "After all," says the manager, "we don't force people to come in the door." How would you respond to this ethical defense of the payday-advance business?
Solution:
Theoretical Foundation: This problem moves beyond simple calculation to the ethical implications of financial practice. It highlights the difference between a free market ideal and the real-world conditions of asymmetric information and vulnerability. The response should draw on concepts of economic necessity and exploitation.
A response could challenge the manager's defense on several grounds:
- **Exploiting Vulnerability:** The "market" for payday loans is typically composed of individuals in dire financial straits who have few other options. The high interest rates (which can exceed 300% annually) are not a reflection of a fair market but rather an exploitation of a captive audience.
- **Asymmetric Information:** Payday lenders often do not fully or clearly disclose the true costs of their loans in an easily understandable way. The borrower may not fully grasp the long-term impact of the high interest rates, creating a cycle of debt.
- **Moral and Social Responsibility:** While "not forcing" someone to take a loan is technically true, a company has a moral and social responsibility to not exploit vulnerable populations. The practice can be seen as predatory, pushing people into a cycle of debt from which they cannot escape.
In a truly free and transparent market with perfect information and no coercion, the manager's argument might hold. However, in the context of payday lending, these conditions are not met, and the ethical defense is therefore weak.
Tricky Area:
The trick here is to avoid a simple "it's unethical" response and to instead ground your argument in core economic and ethical principles. You should specifically mention concepts like **asymmetric information** and **vulnerable populations** to provide a robust and well-reasoned answer that shows a deeper understanding of the issue.