AI HL · 12.2
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Amortisation and annuities
12.2.2
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A loan is repaid monthly over 12 years. What is N?
Have a go first. Then open the working one line at a time.
N = 12 × 12 = 144
N is the number of payments.
Tomás borrows €18 500 to buy a car. The loan charges 4.8% per year, compounded monthly, and is repaid with equal payments at the end of each month for 5 years. (a) Find the monthly payment. (b) Find the total interest Tomás pays. (c) Tomás can afford at most €300 a month. Find the largest amount he could borrow on the same terms.
Have a go first. Then open the working one line at a time.
N = 60, I(%) = 4.8, PV = 18 500, FV = 0, PpY = CpY = 12 → Pmt = −347.43
He receives the loan, so PV is positive. FV is 0 because the loan is repaid. €347.43 a month.
60 × 347.425… − 18 500 = €2345.51
Everything he pays, less what he borrowed. €2345.80 from the rounded payment is accepted.
N = 60, I(%) = 4.8, Pmt = −300, FV = 0, PpY = CpY = 12 → PV = 15 974.66
Now the payment is known and the loan is the unknown. The largest loan is €15 974.66.
A family borrows £240 000 to buy a house. The loan charges 4.2% per year, compounded monthly, and is repaid with equal monthly payments over 30 years. (a) Find the monthly payment. (b) Find the total amount repaid.
Have a go first. Then open the working one line at a time.
N = 360, I(%) = 4.2, PV = 240 000, FV = 0, PpY = CpY = 12 → Pmt = −1173.64
30 years of monthly payments is 360. £1173.64 a month.
360 × 1173.641… = £422 510.84
Nearly £182 511 of that is interest.
Priya pays $80 000 into a fund that earns 3.6% per year, compounded annually. She takes out an equal amount at the end of each year for 25 years, which empties the fund. Find the amount she takes out each year.
Have a go first. Then open the working one line at a time.
N = 25, PpY = CpY = 1
One withdrawal a year for 25 years.
PV = −80 000, FV = 0
She pays the money in, so PV is negative. The fund is emptied, so FV is 0.
I(%) = 3.6 → Pmt = 4906.75
$4906.75 a year. It comes back positive because she receives it.
Luca saves €250 at the end of every month in an account paying 3% per year, compounded monthly. Find the amount in the account after 10 years.
Have a go first. Then open the working one line at a time.
N = 120, PpY = CpY = 12
Monthly for 10 years.
PV = 0, Pmt = −250
Nothing is in the account at the start. He pays €250 in each month, so Pmt is negative.
I(%) = 3 → FV = 34 935.35
He paid in €30 000. The other €4935.35 is interest.
Mei wants to have €15 000 in 4 years' time. She will pay an equal amount into an account at the end of each month. The account pays 3% per year, compounded monthly. (a) How much must she pay in each month? (b) How much of the €15 000 is interest?
Have a go first. Then open the working one line at a time.
N = 48, I(%) = 3, PV = 0, Pmt = ?, FV = 15 000, PpY = CpY = 12
Nothing in the account at the start. The goal is the future value.
Pmt = −294.51
Negative: money paid out. She pays €294.51 a month.
48 × 294.5149… = 14 136.72; interest = 15 000 − 14 136.72 = 863.28
She pays in €14 136.72 and the account adds €863.28.
Hana borrows $26 000 at 5.4% per year, compounded monthly. She repays the loan with equal payments at the end of each month for 6 years. (a) Find the monthly payment. (b) Find the amount Hana still owes after 4 years.
Have a go first. Then open the working one line at a time.
N = 72, I(%) = 5.4, PV = 26 000, FV = 0, PpY = CpY = 12 → Pmt = −423.57
$423.57 a month.
N = 4 × 12 = 48
After 4 years she has made 48 payments. Leave Pmt in its box, so the solver keeps its full value.
N = 48, I(%) = 5.4, PV = 26 000, Pmt = −423.569…, PpY = CpY = 12 → FV = −9615.49
FV is what is left of the loan after 48 payments. It is negative because she would have to pay it. She still owes $9615.49.
A loan of €15 000 is repaid with payments of €295 at the end of each month for 5 years. Interest is compounded monthly. Find the annual interest rate, to 3 significant figures.
Have a go first. Then open the working one line at a time.
N = 5 × 12 = 60, PpY = CpY = 12
Monthly payments for 5 years.
PV = 15 000, Pmt = −295, FV = 0
The loan is received, so PV is positive. The payments are paid out, so Pmt is negative. FV is 0 because the loan is repaid.
I(%) = 6.7143…
The annual rate is 6.71%.
Arjun has £120 000 in a fund that earns 4.5% per year, compounded monthly. He takes out £900 at the end of each month. Find the number of full withdrawals of £900 he can make.
Have a go first. Then open the working one line at a time.
PV = −120 000, Pmt = 900, FV = 0
The money is paid into the fund, so PV is negative. He receives the withdrawals, so Pmt is positive. FV is 0 when the fund is empty.
I(%) = 4.5, PpY = CpY = 12 → N = 185.19…
Round down: 185 full withdrawals.
N = 185 → FV = 166.67
After 185 withdrawals £166.67 is still in the fund, which is less than £900. 185 months is 15 years and 5 months.
A van costs $32 000. Kofi pays a deposit of 15% and borrows the rest at 6.6% per year, compounded monthly. He repays $550 at the end of each month, and then one smaller final payment. (a) Find the amount Kofi borrows. (b) Find the number of full payments of $550. (c) Find the final payment.
Have a go first. Then open the working one line at a time.
85% of 32 000 = 27 200
The deposit is 15%, so he borrows the other 85%: $27 200.
I(%) = 6.6, PV = 27 200, Pmt = −550, FV = 0, PpY = CpY = 12 → N = 57.88…
N is not a whole number: 57 full payments, then a smaller one.
N = 58, I(%) = 6.6, PV = 27 200, Pmt = −550, PpY = CpY = 12 → FV = 67.19
FV is positive: paying $550 as payment 58 would overpay by $67.19.
550 − 67.19 = 482.81
The final payment is $482.81. Check: with N = 57 the solver gives FV = −480.16, what he owes after payment 57, and a month's interest on that brings it to $482.81.
Rana needs a loan of £14 000. Lender A charges 6.0% per year, compounded monthly, over 4 years. Lender B charges 5.4% per year, compounded monthly, over 5 years. Both are repaid with equal payments at the end of each month. Find the total Rana repays to each lender, and say which loan costs less overall.
Have a go first. Then open the working one line at a time.
N = 48, I(%) = 6.0, PV = 14 000, Pmt = ?, FV = 0, PpY = CpY = 12 → Pmt = −328.79
Total repaid = 48 × 328.7904… = £15 781.94.
N = 60, I(%) = 5.4, PV = 14 000, Pmt = ?, FV = 0, PpY = CpY = 12 → Pmt = −266.77
Total repaid = 60 × 266.7706… = £16 006.23.
A costs £1781.94 in interest; B costs £2006.23
Lender A costs £224.29 less overall, although its monthly payment is higher (£328.79 against £266.77).