Amortisation and annuities Problem set A Worked solutions

A loan of £12 000 charges 5.4% per year, compounded monthly, and is repaid monthly over 4 years. Find (a) the monthly payment, (b) the total repaid, (c) the total interest.

• (a) N = 48, I(%) = 5.4, PV = 12 000, Pmt = ?, FV = 0, PpY = 12, CpY = 12
• Solve for Pmt: −278.53…
• The monthly payment is £278.53 (money paid out, so negative).
• (b) Total repaid = 48 × 278.53… = £13 369.50
• (c) Total interest = 13 369.50 − 12 000 = £1 369.50

Chen can afford to repay $450 a month for 25 years. A bank charges 3.9% per year, compounded monthly. Find the largest loan Chen can take.

• N = 25 × 12 = 300
• N = 300, I(%) = 3.9, PV = ?, Pmt = −450, FV = 0, PpY = 12, CpY = 12
• Solve for PV: 86 152.27…
• The largest loan is $86 152.27.

€150 is saved at the end of every month in an account paying 2.7% per year, compounded monthly. Find the amount after 8 years.

• N = 8 × 12 = 96
• N = 96, I(%) = 2.7, PV = 0, Pmt = −150, FV = ?, PpY = 12, CpY = 12
• Solve for FV: 16 053.42…
• The account holds €16 053.42.

£200 000 is paid into a fund earning 3% per year, compounded yearly. Equal amounts are taken out at the end of each year for 20 years, emptying the fund. Find each amount.

• N = 20, I(%) = 3, PV = −200 000, Pmt = ?, FV = 0, PpY = 1, CpY = 1
• Solve for Pmt: 13 443.14…
• Each withdrawal is £13 443.14 (money received, so positive).

A couple want to have $20 000 in 6 years' time for a house deposit. They pay an equal amount into an account at the end of each month. The account pays 2.4% per year, compounded monthly. (a) How much must they pay in each month? (b) How much of the $20 000 is interest?

• (a) N = 72, I(%) = 2.4, PV = 0, Pmt = ?, FV = 20000, PpY = 12, CpY = 12
• Solve for Pmt: −258.53…
• They must pay in $258.53 each month.
• (b) Total paid in = 72 × 258.5348… = $18 614.51
• Interest = 20000 − 18 614.51 = $1385.49

Sofia borrows €9000 at 6% per year, compounded monthly, and repays €250 a month. How many payments does she make?

• N = ?, I(%) = 6, PV = 9000, Pmt = −250, FV = 0, PpY = 12, CpY = 12
• Solve for N: 39.79…
• N = 39.79…, so 40 payments, the last one smaller than €250.

A retirement account holds $60 000 and earns 2.4% per year, compounded monthly. $800 is taken out at the end of each month. For how many months does the money last?

• N = ?, I(%) = 2.4, PV = −60 000, Pmt = 800, FV = 0, PpY = 12, CpY = 12
• Solve for N: 81.34…
• N = 81.34…. With N = 81 and Pmt = 800 the solver gives FV = 272.19, a balance too small for another full $800 withdrawal.
• The money lasts for 81 full withdrawals.

Tom borrows £9000 to buy a van. Option 1 charges 7.2% per year, compounded monthly, over 3 years. Option 2 charges 6.6% per year, compounded monthly, over 4 years. Both are repaid with equal payments at the end of each month. For each option find (a) the monthly payment and (b) the total repaid. (c) Which option costs less overall, and by how much?

• Option 1: N = 36, I(%) = 7.2, PV = 9000, Pmt = ?, FV = 0, PpY = CpY = 12 → Pmt = −278.72
• Option 2: N = 48, I(%) = 6.6, PV = 9000, Pmt = ?, FV = 0, PpY = CpY = 12 → Pmt = −213.85
• Totals: 36 × 278.7176… = £10 033.83 and 48 × 213.8499… = £10 264.80
• Option 1 costs less overall, by £230.96.

A house loan of £180 000 charges 3.3% per year, compounded monthly, and is repaid monthly over 25 years. (a) Find the monthly payment. (b) Find the amount still owed after 10 years.

• (a) N = 25 × 12 = 300
• N = 300, I(%) = 3.3, PV = 180 000, Pmt = ?, FV = 0, PpY = 12, CpY = 12
• Solve for Pmt: −881.93…
• The monthly payment is £881.93.
• (b) After 10 years, N = 120 payments have been made. Keep Pmt = −881.93…, change N and solve for FV.
• N = 120, I(%) = 3.3, PV = 180 000, Pmt = −881.93…, FV = ?, PpY = 12, CpY = 12
• Solve for FV: −125 078.58…
• The amount still owed is £125 078.58 (FV is negative: it is still to be paid).