Finance Solver: compound interest and depreciation Problem set A One question at a time

Write down N for each of these.

1. (a) 5 years, compounded quarterly

   • N = years × payments per year (no payments, so PpY = CpY).
   • N = 5 × 4 = 20

2. (b) 30 months, compounded monthly

   • Monthly: one period each month, so N is the number of months.
   • N = 30

3. (c) 12 years, compounded half-yearly

   • N = years × payments per year.
   • N = 12 × 2 = 24

4. (d) 3 years, compounded monthly

   • N = years × payments per year.
   • N = 3 × 12 = 36

Find the final value.

1. (a) €4000 at 3.2% per year, compounded yearly, for 6 years

   • N = 6, I(%) = 3.2, PV = −4000, Pmt = 0, FV = ?, PpY = 1, CpY = 1
   • Solve for FV: 4 832.13…
   • The investment is worth €4 832.13.

2. (b) €7500 at 2.4% per year, compounded monthly, for 5 years

   • N = 60, I(%) = 2.4, PV = −7500, Pmt = 0, FV = ?, PpY = 12, CpY = 12
   • Solve for FV: 8 455.21…
   • The investment is worth €8 455.21.

How much must be invested now at 3.6% per year, compounded quarterly, to have $10 000 in 8 years?

• N = 8 × 4 = 32
• N = 32, I(%) = 3.6, PV = ?, Pmt = 0, FV = 10 000, PpY = 4, CpY = 4
• Solve for PV: −7 507.28…
• Invest $7 507.28 now (PV is negative: it is money paid in).

£2000 is invested at 5% per year, compounded yearly. After how many complete years is it first worth more than £3000?

• N = ?, I(%) = 5, PV = −2000, Pmt = 0, FV = 3000, PpY = 1, CpY = 1
• Solve for N: 8.31…
• N = 8.31… is part-way through year 9, so it passes £3000 after 9 complete years.

$5000 grows to $6200 in 4 years, compounded yearly. Find the annual interest rate.

• N = 4, I(%) = ?, PV = −5000, Pmt = 0, FV = 6200, PpY = 1, CpY = 1
• Solve for I(%): 5.53…
• The annual interest rate is 5.53%.

$4000 is invested at 3.6% per year, compounded monthly. How long will it take for the investment to be worth $5000? Give your answer in complete months, and then in years and months.

• N = ?, I(%) = 3.6, PV = −4000, Pmt = 0, FV = 5000, PpY = 12, CpY = 12
• Solve for N: 74.49… (months)
• Interest is added at the end of each month, so 75 months.
• 75 months = 6 years 3 months.

A laptop costs £1200 and loses 22% of its value each year. (a) Find its value after 3 years. (b) After how many complete years is it first worth less than £300?

• (a) N = 3, I(%) = −22, PV = −1200, Pmt = 0, FV = ?, PpY = 1, CpY = 1
• Solve for FV: 569.46…
• The laptop is worth £569.46 (depreciation is a negative I(%)).
• (b) N = ?, I(%) = −22, PV = −1200, Pmt = 0, FV = 300, PpY = 1, CpY = 1
• Solve for N: 5.58…
• After 5 years it is worth £346.46; after 6 years £270.24. So 6 complete years.

€8000 is invested at 4.5% per year, compounded yearly, for 5 years. Inflation is 2% per year. Find the value of the investment in real terms.

• Real rate: \(r = c - i = 4.5 - 2 = 2.5\)
• N = 5, I(%) = 2.5, PV = −8000, Pmt = 0, FV = ?, PpY = 1, CpY = 1
• Solve for FV: 9 051.27…
• The value in real terms is €9 051.27.

Maria has £6000 to save for 10 years. Account X pays 5.2% per year, compounded annually. Account Y pays 5.1% per year, compounded monthly. Which account gives her more at the end of 10 years, and by how much?

• Account X: N = 10, I(%) = 5.2, PV = −6000, Pmt = 0, FV = ?, PpY = CpY = 1
• Solve for FV: 9961.13
• Account Y: N = 120, I(%) = 5.1, PV = −6000, Pmt = 0, FV = ?, PpY = CpY = 12
• Solve for FV: 9980.96
• Account Y gives more, by £19.82.

A tractor is bought for $30 000 and loses 16% of its value each year. (a) Find its value after 3 years. (b) After how many complete years is it first worth less than half of its price?

• (a) N = 3, I(%) = −16, PV = −30000, Pmt = 0, FV = ?, PpY = 1, CpY = 1
• Solve for FV: 17 781.12 (depreciation is a negative I(%))
• (b) N = ?, I(%) = −16, PV = −30000, Pmt = 0, FV = 15000, PpY = 1, CpY = 1
• Solve for N: 3.98…
• After 3 years it is worth $17 781.12; after 4 years $14 936.14. So 4 complete years.

$12 000 is invested at 3.9% per year for 15 years. Find how much more it is worth if the interest is compounded monthly rather than yearly.

• Monthly: N = 15 × 12 = 180
• N = 180, I(%) = 3.9, PV = −12 000, Pmt = 0, FV = ?, PpY = 12, CpY = 12
• Solve for FV: 21 519.47…
• Yearly: N = 15
• N = 15, I(%) = 3.9, PV = −12 000, Pmt = 0, FV = ?, PpY = 1, CpY = 1
• Solve for FV: 21 301.71…
• Difference: 21 519.47 − 21 301.71 = $217.76