Year 12 · Financial mathematics

F4Investments and loans 0 / 16

YR 11

A1 A2 M1 M2 F1 S1 S2

YR 12

A4 M6 M7 F4 F5 S4 S5 N2 N3

💡 Quick tips — memory joggers

In $FV = PV(1+r)^{n}$, $r$ is the rate per period and $n$ the number of periods — for monthly compounding divide the annual rate by 12 and multiply the years by 12.
Appreciation grows, depreciation shrinks: use $(1+r)^{n}$ when a value rises and $(1-r)^{n}$ for declining-balance depreciation.
On a reducing-balance loan interest is charged on the outstanding balance, so early repayments are mostly interest; total interest $=$ (repayment $\times$ number of repayments) $-$ amount borrowed.
A credit card’s interest-free period only applies if the closing balance is paid in full by the due date — otherwise interest is usually charged from the purchase date.

Level 1 · Fluency

An amount of $8000 is invested at 4% p.a. compounded annually. Which expression gives the value of the investment after 5 years?
A. 8000 × 0.04 × 5 B. 8000(1.04)⁵ C. 8000(0.04)⁵ D. 8000 + 0.04 × 5

Level 2 · Application

$6000 is invested at 5.2% p.a. compounded quarterly for 3 years. Calculate the future value of the investment, correct to the nearest cent.
Level 3 · Further Application
$12 000 is invested at a rate of r% p.a. compounded annually. After 6 years the investment has grown to $16 000.
1. (a) Write an equation, in terms of r, for the value of the investment after 6 years. (1 mark)
2. (b) Hence find the annual interest rate r, correct to two decimal places. (2 marks)
(a) 16 000 = 12 000(1 + r)⁶.
(b) (1 + r)⁶ = 16000/12000 = 1.3333…, so 1 + r = 1.3333^1/6 = 1.04912. r = 4.91% p.a.
$9000 is invested at a rate of r% p.a. compounded annually. After 5 years the investment has grown to $12 000.
1. (a) Write an equation, in terms of r, for the value of the investment after 5 years. (1 mark)
2. (b) Hence find the annual interest rate r, correct to two decimal places. (2 marks)
(a) 12 000 = 9000(1 + r)⁵. (b) r = 5.92% p.a.

(a) 12 000 = 9000(1 + r)⁵.
(b) (1 + r)⁵ = 12000/9000 = 1.33333, so 1 + r = 1.33333^1/5 = 1.05922. r = 5.92% p.a.
$15 000 is invested at 5% p.a. compounded annually. After how many whole years will the investment first exceed $20 000?
Level 1 · Fluency

$5000 earns 6% p.a. simple interest. How much interest is earned in 4 years?

Level 2 · Application

$9000 is invested for 8 years. Account P pays 6.1% p.a. simple interest; Account Q pays 6.1% p.a. compounded annually. Calculate how much more interest Account Q earns than Account P over the 8 years.
Level 3 · Further Application
Bianca and Noah each invest $2200 for 4 years. Bianca earns 6.5% p.a. simple interest. Noah earns 6% p.a. compounded annually.
1. (a) Calculate the value of each investment after 4 years. (2 marks)
2. (b) Determine who has the greater investment after 4 years, and by how much. (1 mark)
(a) Bianca: 2200(1 + 0.065 × 4) = $2,772.00. Noah: 2200(1.06)⁴ = $2,777.45.
(b) Bianca’s is greater by $-5.45.
Ava and Leo each invest $3000 for 5 years. Ava earns 7% p.a. simple interest. Leo earns 6.5% p.a. compounded annually.
1. (a) Calculate the value of each investment after 5 years. (2 marks)
2. (b) Determine who has the greater investment after 5 years, and by how much. (1 mark)
(a) Ava: $4,050.00. Leo: $4,110.26. (b) Leo, by $60.26.

(a) Ava: 3000(1 + 0.07 × 5) = 3000 × 1.35 = $4,050.00. Leo: 3000(1.065)⁵ = $4,110.26.
(b) Leo has the greater investment, by $4,110.26 − $4,050.00 = $60.26.
Ruby invests $5000 at 5% p.a. simple interest. Max invests $5000 at 4.8% p.a. compounded annually.
1. (a) Calculate the value of each investment after 8 years. (2 marks)
2. (b) State who has more after 8 years, and by how much. (1 mark)
(a) Ruby: $7,000.00. Max: $7,275.46. (b) Max, by $275.46.

(a) Ruby: 5000(1 + 0.05 × 8) = 5000 × 1.4 = $7,000.00. Max: 5000(1.048)⁸ = $7,275.46.
(b) Max has more, by $7,275.46 − $7,000.00 = $275.46.
Level 1 · Fluency

$6000 is invested at 5% p.a. Which compounding period gives the greatest future value over one year — annually, quarterly or monthly?

Level 2 · Application

$6000 is invested for 4 years at 5% p.a. Calculate the future value if interest is compounded (i) annually and (ii) monthly, and state the difference.
Level 3 · Further Application
Priya invests $8000 at 4.8% p.a. She may choose annual or monthly compounding, and a term of 5 or 10 years.
1. (a) Find the future value for monthly compounding over 10 years. (2 marks)
2. (b) Explain why increasing the term from 5 to 10 years increases the future value by more than double the 5-year interest. (1 mark)
(a) 8000(1 + 0.048/12)¹²⁰ = $12,916.22.
(b) Interest compounds — in the second 5 years interest is earned on a much larger balance (which already includes the first 5 years’ interest), so the extra interest is greater than the first 5 years’ interest.
Marcus invests $10 000 at 6% p.a. compounded monthly.
1. (a) Find the future value after 8 years. (2 marks)
2. (b) Explain why choosing monthly rather than annual compounding increases the future value. (1 mark)
(a) $16,141.42. (b) See solution.

(a) r = 0.06/12 = 0.005, n = 8 × 12 = 96. FV = 10 000(1.005)⁹⁶ = $16,141.42.
(b) With monthly compounding interest is added 12 times a year, so interest starts earning further interest sooner — the balance grows on a larger amount more often than with annual compounding.
Two accounts each start with $7000 at 5% p.a. compounded annually. Account A runs for 4 years; Account B runs for 8 years.
1. (a) Find the interest earned by each account. (2 marks)
2. (b) Explain why Account B earns more than double the interest of Account A. (1 mark)
(a) A: $1,508.54; B: $3,342.19. (b) See solution.

(a) A: 7000(1.05)⁴ − 7000 = $8,508.54 − $7,000 = $1,508.54. B: 7000(1.05)⁸ − 7000 = $10,342.19 − $7,000 = $3,342.19.
(b) In the second 4 years interest is earned on the larger balance that already includes the first 4 years’ interest, so the growth accelerates — more than doubling the interest.
Level 1 · Fluency

A term deposit guarantees 3.5% p.a. Shares have averaged 8% p.a. but the return varies each year. Which is the lower-risk investment?

Level 2 · Application

$15 000 is invested for 5 years. Option A is a term deposit at 3.8% p.a. compounded annually. Option B is shares returning an average of 8% p.a. Calculate the future value of each and state the difference.
Level 3 · Further Application
Dan has $15 000 to invest for 5 years and compares a term deposit at 3.8% p.a. (compounded annually) with shares averaging 8% p.a.
1. (a) Calculate the future value of each option. (2 marks)
2. (b) Give one reason, other than return, why Dan might still choose the term deposit. (1 mark)
(a) Term deposit $18,074.99; shares (average) $22,039.92.
(b) The term deposit is guaranteed and lower-risk; the share return is an average only and the value could fall, so the term deposit protects the capital.
Sara has $10 000 to invest for 6 years and compares a term deposit at 4% p.a. (compounded annually) with a share fund averaging 9% p.a.
1. (a) Calculate the future value of each option. (2 marks)
2. (b) Give one reason, other than return, why Sara might still choose the term deposit. (1 mark)
(a) Term deposit $12,653.19; share fund (average) $16,771.00. (b) See solution.

(a) Term deposit: 10 000(1.04)⁶ = $12,653.19. Share fund: 10 000(1.09)⁶ = $16,771.00.
(b) The term deposit’s return is guaranteed, whereas the 9% is only an average — the fund could fall in value, so the term deposit better protects Sara’s capital.
Tom compares investing $8000 for 5 years in (i) a term deposit at 3.6% p.a. compounded annually and (ii) shares averaging 10% p.a.
1. (a) Calculate the future value of each option. (2 marks)
2. (b) Explain why comparing only the future values may be misleading. (1 mark)
(a) Term deposit $9,547.49; shares (average) $12,884.08. (b) See solution.

(a) Term deposit: 8000(1.036)⁵ = $9,547.49. Shares: 8000(1.10)⁵ = $12,884.08.
(b) The 10% share return is only an average; actual yearly returns vary and the shares could fall, so the higher future value carries higher risk and is not guaranteed like the term deposit.
Level 1 · Fluency

Prices rise with inflation at 3% p.a. Which formula gives the cost in 5 years of an item costing $200 today?

Level 2 · Application

The inflation rate is 3.1% p.a. A weekly grocery bill is currently $560. Calculate the expected cost of the same weekly groceries in 12 years, correct to the nearest cent.
Level 3 · Further Application
A salary of $72 000 today needs to keep pace with inflation of 3.2% p.a.
1. (a) Calculate the equivalent salary needed in 8 years to have the same buying power. (2 marks)
2. (b) A worker’s salary rises to $88 000 in that time. State whether their buying power has increased or decreased, and justify. (1 mark)
(a) 72000(1.032)⁸ = $92,633.93.
(b) $88 000 is less than $92,633.93, so buying power has decreased.
A salary of $65 000 today needs to keep pace with inflation of 2.9% p.a.
1. (a) Calculate the equivalent salary needed in 10 years to have the same buying power. (2 marks)
2. (b) A worker’s salary rises to $85 000 in that time. State whether their buying power has increased or decreased, and justify. (1 mark)
(a) $86,510.06. (b) Decreased.

(a) 65 000(1.029)¹⁰ = $86,510.06.
(b) $85 000 is less than $86,510.06, so the salary has not kept pace with inflation — buying power has decreased.
A council rate is $2400 today and inflation is 3.5% p.a.
1. (a) Calculate the equivalent amount needed in 6 years to have the same real value. (2 marks)
2. (b) If the rate is actually increased to $3000 in 6 years, state whether it has risen faster or slower than inflation, and justify. (1 mark)
(a) $2,950.21. (b) Faster than inflation.

(a) 2400(1.035)⁶ = $2,950.21.
(b) $3000 is more than $2,950.21, so the rate has risen faster than inflation.
Level 1 · Fluency

An asset worth $10 000 appreciates at 5% p.a. What is its value after 1 year?

Level 2 · Application

An antique vase is bought for $2400 and appreciates at 5.8% p.a. Calculate its value after 14 years, correct to the nearest dollar.
Level 3 · Further Application
Shares are bought for $12 000 and are worth $15 800 after 3 years.
1. (a) Calculate the average annual compound growth rate, correct to two decimal places. (2 marks)
2. (b) Assuming the same rate continues, find the value after a further 2 years. (1 mark)
(a) (15800/12000)^1/3 − 1 = 9.60% p.a.
(b) 15800(1 + 0.0960)² = $18,980.50.
A painting is bought for $8000 and is worth $11 500 after 4 years.
1. (a) Calculate the average annual compound growth rate, correct to two decimal places. (2 marks)
2. (b) Assuming the same rate continues, find the value after a further 3 years. (1 mark)
(a) 9.50% p.a. (b) $15,098.72.

(a) (11500/8000)^1/4 − 1 = 1.4375^1/4 − 1 = 1.09497 − 1 = 9.50% p.a.
(b) 11 500(1.095)³ = $15,098.72.
A block of land is bought for $250 000 and is worth $340 000 after 5 years.
1. (a) Calculate the average annual compound growth rate, correct to two decimal places. (2 marks)
2. (b) Assuming the same rate continues, find the value after a further 4 years. (1 mark)
(a) 6.34% p.a. (b) $434,775.98.

(a) (340000/250000)^1/5 − 1 = 1.36^1/5 − 1 = 1.06343 − 1 = 6.34% p.a.
(b) 340 000(1.0634)⁴ = $434,775.98.
Level 1 · Fluency

On a line graph of a share’s monthly price, what does a steep upward section represent?

Level 2 · Application

A share’s price at the end of five consecutive years was $4.20, $4.85, $4.50, $5.60 and $5.15. Calculate the overall percentage change from Year 1 to Year 5, correct to one decimal place.
Level 3 · Further Application
The table shows a share’s closing price at the end of each year.

Year 1 2 3 4 5
Price $4.20 $4.85 $4.50 $5.60 $5.15
1. (a) Between which two consecutive years was the percentage fall the greatest? Justify with a calculation. (2 marks)
2. (b) Describe what a period of high volatility would look like on the graph. (1 mark)
(a) Falls: Yr2→3 = (4.50−4.85)/4.85 = -7.2%; Yr4→5 = (5.15−5.60)/5.60 = -8.0%. The greatest fall is Year 4 → Year 5.
(b) Large up-and-down swings between consecutive points (a jagged line), rather than a smooth trend.
The table shows a share’s closing price at the end of each year.

Year 1 2 3 4 5
Price $2.50 $2.90 $3.40 $3.10 $3.75
1. (a) Between which two consecutive years was the percentage rise the greatest? Justify with a calculation. (2 marks)
2. (b) Describe what a steady upward trend would look like on the graph. (1 mark)
(a) Rises: Yr1→2 = (2.90−2.50)/2.50 = 16.0%; Yr2→3 = (3.40−2.90)/2.90 = 17.2%; Yr4→5 = (3.75−3.10)/3.10 = 21.0%. The greatest rise is Year 4 → Year 5.
(b) A line that climbs consistently from left to right, each year higher than the last, without large dips.
The table shows a share’s closing price at the end of each year.

Year 1 2 3 4 5
Price $6.40 $6.10 $6.95 $7.20 $7.80
1. (a) Between which two consecutive years was the percentage rise the greatest? Justify with a calculation. (2 marks)
2. (b) Explain why a line graph is more useful than a table for seeing the overall trend. (1 mark)
(a) Rises: Yr2→3 = (6.95−6.10)/6.10 = 13.9%; Yr3→4 = (7.20−6.95)/6.95 = 3.6%; Yr4→5 = (7.80−7.20)/7.20 = 8.3%. The greatest rise is Year 2 → Year 3.
(b) A line graph shows the shape of the change — rises, falls and steepness — at a glance, making the overall trend and volatility easier to see than reading individual numbers.
Level 1 · Fluency

An investor owns 500 shares each paying a dividend of $0.40. What is the total dividend received?

Level 2 · Application

An investor owns 800 shares currently priced at $13.40 each. Each share pays an annual dividend of $0.62. Calculate the dividend yield, correct to two decimal places.
Level 3 · Further Application
Mia owns 1500 shares in a company. The market price is $27.00 per share and the dividend yield is 3%.
1. (a) Calculate the dividend paid per share. (1 mark)
2. (b) Hence calculate Mia’s total annual dividend. (2 marks)
(a) Dividend per share = 3% × $27.00 = $0.81.
(b) Total = 1500 × $0.81 = $1,215.00.
Raj owns 2000 shares in a company. The market price is $16.00 per share and the dividend yield is 5%.
1. (a) Calculate the dividend paid per share. (1 mark)
2. (b) Hence calculate Raj’s total annual dividend. (2 marks)
(a) $0.80 (b) $1,600.00

(a) Dividend per share = 5% × $16.00 = $0.80.
(b) Total = 2000 × $0.80 = $1,600.00.
Ella owns 3500 shares. The market price is $12.00 per share and the dividend yield is 5.5%.
1. (a) Calculate the dividend paid per share. (1 mark)
2. (b) Hence calculate Ella’s total annual dividend. (2 marks)
(a) $0.66 (b) $2,310.00

(a) Dividend per share = 5.5% × $12.00 = $0.66.
(b) Total = 3500 × $0.66 = $2,310.00.
Level 1 · Fluency

Which formula gives the salvage value S of an asset of initial value V₀ depreciating at rate r per period for n periods (declining balance)?
A. S = V₀(1 − r)ⁿ B. S = V₀(1 + r)ⁿ C. S = V₀ − rn D. S = V₀ − Dn

Level 2 · Application

A machine bought for $52 000 depreciates by 16% p.a. using the declining-balance method. Calculate its salvage value after 6 years, correct to the nearest dollar.
Level 3 · Further Application
A $40 000 asset can be depreciated by straight-line at $6000 per year, or by declining-balance at 18% p.a.
1. (a) Find the value after 5 years under each method. (2 marks)
2. (b) State which method gives the higher value after 5 years. (1 mark)
(a) Straight-line: 40000 − 6000 × 5 = $10,000.00. Declining-balance: 40000(0.82)⁵ = $14,829.59.
(b) The declining-balance method gives the higher value.
A $25 000 asset can be depreciated by straight-line at $3500 per year, or by declining-balance at 20% p.a.
1. (a) Find the value after 4 years under each method. (2 marks)
2. (b) State which method gives the higher value after 4 years. (1 mark)
(a) Straight-line $11,000.00; declining-balance $10,240.00. (b) Straight-line.

(a) Straight-line: 25 000 − 3500 × 4 = $11,000.00. Declining-balance: 25 000(0.80)⁴ = 25 000 × 0.4096 = $10,240.00.
(b) The straight-line method gives the higher value after 4 years.
A $60 000 machine can be depreciated by straight-line at $9000 per year, or by declining-balance at 25% p.a.
1. (a) Find the value after 3 years under each method. (2 marks)
2. (b) State which method gives the higher value after 3 years. (1 mark)
(a) Straight-line $33,000.00; declining-balance $25,312.50. (b) Straight-line.

(a) Straight-line: 60 000 − 9000 × 3 = $33,000.00. Declining-balance: 60 000(0.75)³ = 60 000 × 0.421875 = $25,312.50.
(b) The straight-line method gives the higher value after 3 years.
Level 1 · Fluency

A home loan is repaid by equal monthly instalments. As the loan is repaid, what happens to the portion of each repayment that is interest?

Level 2 · Application

Use the table of present-value interest factors to answer.A loan of $300 000 is charged interest at 6% p.a. compounded monthly and repaid over 20 years. Using the table, calculate the monthly repayment, correct to the nearest cent.

N r = 0.004 r = 0.005 r = 0.006
240 154.0933 139.5808 127.0084
300 174.5210 155.2069 138.9683
360 190.5977 166.7916 147.3214
Level 3 · Further Application
A couple borrows $420 000 to buy a home. Interest is 6% p.a. compounded monthly and the loan is repaid in equal monthly instalments over 25 years.

N r = 0.005
300 155.2069
1. (a) State the monthly interest rate and the number of repayments. (1 mark)
2. (b) Using the factor above, calculate the monthly repayment, correct to the nearest cent. (2 marks)
3. (c) Hence calculate the total interest paid over the life of the loan. (2 marks)
(a) r = 0.005 per month; N = 300 repayments.
(b) Repayment = 420000 ÷ 155.2069 = $2,706.07.
(c) Total paid = $2,706.07 × 300 = $811,819.77; interest = $811,819.77 − $420 000 = $391,819.77.
A family borrows $500 000 to buy a home. Interest is 6% p.a. compounded monthly and the loan is repaid in equal monthly instalments over 20 years.

N r = 0.005
240 139.5808
1. (a) State the monthly interest rate and the number of repayments. (1 mark)
2. (b) Using the factor above, calculate the monthly repayment, correct to the nearest cent. (2 marks)
3. (c) Hence calculate the total interest paid over the life of the loan. (2 marks)
(a) r = 0.005 per month; N = 240 repayments.
(b) Repayment = 500 000 ÷ 139.5808 = $3,582.15.
(c) Total paid = $3,582.15 × 240 = $859,716.00; interest = $859,716.00 − $500 000 = $359,716.00.
A borrower takes a $180 000 loan. Interest is 4.8% p.a. compounded monthly and it is repaid in equal monthly instalments over 20 years.

N r = 0.004
240 154.0933
1. (a) State the monthly interest rate and the number of repayments. (1 mark)
2. (b) Using the factor above, calculate the monthly repayment, correct to the nearest cent. (2 marks)
3. (c) Hence calculate the total interest paid over the life of the loan. (2 marks)
(a) r = 0.004 per month; N = 240 repayments.
(b) Repayment = 180 000 ÷ 154.0933 = $1,168.12.
(c) Total paid = $1,168.12 × 240 = $280,348.80; interest = $280,348.80 − $180 000 = $100,348.80.
Level 1 · Fluency

A credit card charges interest that is “compounded daily”. Over 30 days, how many times is interest applied to the balance?

Level 2 · Application

A credit card charges 19.9% p.a. interest, compounded daily. A balance of $3200 is left unpaid for 52 days, with no repayments. Calculate the balance owing after 52 days, correct to the nearest cent.
Level 3 · Further Application
Nina owes $500 on a credit card charging 17% p.a. compounded daily. She makes no purchases but repays $250 exactly 15 days after the statement date.
1. (a) Calculate the amount owing immediately before the $250 repayment. (2 marks)
2. (b) Hence calculate the amount owing immediately after the repayment. (2 marks)
(a) 500(1 + 0.17/365)¹⁵ = $503.50.
(b) $503.50 − $250 = $253.50.
Sam owes $800 on a credit card charging 18% p.a. compounded daily. He makes no purchases but repays $400 exactly 20 days after the statement date.
1. (a) Calculate the amount owing immediately before the $400 repayment. (2 marks)
2. (b) Hence calculate the amount owing immediately after the repayment. (2 marks)
(a) $807.93 (b) $407.93

(a) 800(1 + 0.18/365)²⁰ = $807.93.
(b) $807.93 − $400 = $407.93.
Priya owes $1200 on a credit card charging 20.5% p.a. compounded daily. She makes no purchases but repays $600 exactly 30 days after the statement date.
1. (a) Calculate the amount owing immediately before the $600 repayment. (2 marks)
2. (b) Hence calculate the amount owing immediately after the repayment. (2 marks)
(a) $1,220.38 (b) $620.38

(a) 1200(1 + 0.205/365)³⁰ = $1,220.38.
(b) $1,220.38 − $600 = $620.38.
Level 1 · Fluency

Name one fee, other than interest, that a credit card commonly charges.

Level 2 · Application

A card has a $99 annual fee and charges a $30 late fee. In one year the holder pays the annual fee, is charged 3 late fees, and pays $146 in interest. Calculate the total cost of the card for the year.
Level 3 · Further Application
A cardholder is considering two cards for a year in which they expect to pay about $180 in interest.

Card A Card B
Annual fee $0 $120
Interest rate 21.9% p.a. 12.9% p.a.
Interest expected $305 $180
1. (a) Calculate the total yearly cost of each card (fee + interest). (2 marks)
2. (b) Recommend a card and justify your choice. (1 mark)
(a) Card A: $0 + $305 = $305. Card B: $120 + $180 = $300.
(b) Card B is $5 cheaper for this level of spending; if a balance is regularly carried, the lower interest rate on Card B outweighs its annual fee.
A cardholder who carries a balance is considering two cards for one year.

Card A Card B
Annual fee $0 $90
Interest rate 22.9% p.a. 13.9% p.a.
Interest expected $250 $150
1. (a) Calculate the total yearly cost of each card (fee + interest). (2 marks)
2. (b) Recommend a card and justify your choice. (1 mark)
(a) Card A: $0 + $250 = $250. Card B: $90 + $150 = $240.
(b) Card B is $10 cheaper for someone carrying a balance; its lower interest rate more than offsets the $90 annual fee.
A cardholder who expects to carry a balance for a year compares two cards.

Card A Card B
Annual fee $0 $150
Interest rate 20.9% p.a. 11.9% p.a.
Interest expected $420 $240
1. (a) Calculate the total yearly cost of each card (fee + interest). (2 marks)
2. (b) Recommend a card and justify your choice. (1 mark)
(a) Card A: $0 + $420 = $420. Card B: $150 + $240 = $390.
(b) Card B is $30 cheaper for someone who carries a balance; the lower interest rate more than offsets its annual fee. If the balance were always paid in full (no interest), Card A with the $0 fee would be cheaper.
Level 1 · Fluency

Order these from the typically highest to lowest interest rate: home loan, credit card, personal loan.

Level 2 · Application

A $4000 debt is carried for 2 years. On a credit card it is charged 20% p.a. simple; as a personal loan, 10% p.a. simple. Calculate the interest under each and the saving from the personal loan.
Level 3 · Further Application
A person needs to borrow $6000 for 3 years and compares a credit card (20% p.a.) with a personal loan (11% p.a.), treating both as simple interest for an estimate.
1. (a) Estimate the interest cost of each option over 3 years. (2 marks)
2. (b) Explain why, in practice, the credit-card cost is usually even higher than this estimate. (1 mark)
(a) Card ≈ 6000 × 0.20 × 3 = $3,600.00; loan ≈ 6000 × 0.11 × 3 = $1,980.00.
(b) Credit-card interest compounds (usually daily) on the balance, so the true cost is higher than a flat simple-interest estimate.
A person needs to borrow $8000 for 2 years and compares a credit card (21% p.a.) with a personal loan (10.5% p.a.), treating both as simple interest for an estimate.
1. (a) Estimate the interest cost of each option over 2 years. (2 marks)
2. (b) Explain why, in practice, the credit-card cost is usually even higher than this estimate. (1 mark)
(a) Card ≈ $3,360.00; loan ≈ $1,680.00. (b) See solution.

(a) Card ≈ 8000 × 0.21 × 2 = $3,360.00; loan ≈ 8000 × 0.105 × 2 = $1,680.00.
(b) Credit-card interest actually compounds (usually daily) on the balance, so the true cost exceeds this flat simple-interest estimate.
A person needs to borrow $10 000 for 3 years and compares a credit card (19.9% p.a.) with a personal loan (9.9% p.a.), treating both as simple interest for an estimate.
1. (a) Estimate the interest cost of each option over 3 years. (2 marks)
2. (b) State one non-interest factor that might still make the personal loan the better choice. (1 mark)
(a) Card ≈ $5,970.00; loan ≈ $2,970.00. (b) See solution.

(a) Card ≈ 10 000 × 0.199 × 3 = $5,970.00; loan ≈ 10 000 × 0.099 × 3 = $2,970.00.
(b) A personal loan has fixed scheduled repayments that guarantee the debt is cleared by a set date, whereas a credit card allows minimum-only payments that can leave the debt outstanding for years.
Level 1 · Fluency

A credit-card statement lists a “minimum payment” of 2% of the balance. On a $2500 balance, how much is the minimum payment?

Level 2 · Application

A $2500 balance is charged 19% p.a. interest (compounded monthly ≈ 1.583% per month). If only the $50 minimum is paid this month, calculate the balance at the start of next month (after interest and the payment).
Level 3 · Further Application
A statement shows a $4200 balance, interest 19% p.a., and a minimum payment of 2% ($84).
1. (a) Calculate one month’s interest (use 19% ÷ 12 per month). (1 mark)
2. (b) Explain why paying only the minimum each month means the balance takes years to clear. (2 marks)
(a) 4200 × 0.19/12 = $66.50.
(b) One month’s interest ($66.50) is almost as large as the $84 minimum payment, so very little of each payment reduces the principal. As the balance barely falls, interest stays high and the debt clears extremely slowly.
A statement shows a $3000 balance, interest 22% p.a., and a minimum payment of 2.5% ($75).
1. (a) Calculate one month’s interest (use 22% ÷ 12 per month). (1 mark)
2. (b) Explain why paying only the minimum each month means the balance takes years to clear. (2 marks)
(a) 3000 × 0.22/12 = $55.00.
(b) One month’s interest ($55.00) is a large share of the $75.00 minimum payment, so only about $20 reduces the principal each month. The balance falls very slowly, interest stays high, and the debt takes years to clear.
A statement shows a $6000 balance, interest 20% p.a., and a minimum payment of 2% ($120).
1. (a) Calculate one month’s interest (use 20% ÷ 12 per month). (1 mark)
2. (b) Explain why increasing the monthly payment above the minimum clears the debt much sooner. (2 marks)
(a) 6000 × 0.20/12 = $100.00.
(b) At the minimum, $100.00 of the $120 is interest, so only about $20 reduces the principal. Paying more than the minimum sends a larger part to the principal, so the balance falls faster, less interest is charged the next month, and the debt clears much sooner.
Level 1 · Fluency

What is a credit-card “interest-free period”?

Level 2 · Application

A card offers up to 44 days interest-free if the balance is paid in full each month. A holder makes a $1500 purchase and pays it in full within the interest-free period. If the card’s rate is 20% p.a., roughly how much interest do they avoid over that period (use 44 days)?
Level 3 · Further Application
Explain the interest-free period by describing, for a $1500 purchase on a card with a 44-day interest-free period:
1. (a) what happens if the holder pays the closing balance in full by the due date; and (2 marks)
2. (b) what happens if they pay only part of it. (1 mark)
(a) They pay no interest at all on the purchase — the interest-free period means the full $1500 costs nothing extra if cleared by the due date.
(b) The interest-free benefit is usually lost: interest is charged (often back-dated to the purchase date) on the outstanding amount, so carrying any balance can be costly.
Explain the interest-free period by describing, for a $2500 purchase on a card with a 55-day interest-free period:
1. (a) what happens if the holder pays the closing balance in full by the due date; and (2 marks)
2. (b) why a cash advance is usually treated differently. (1 mark)
(a) No interest at all is charged on the $2500 — paid in full by the due date, the purchase costs nothing extra.
(b) A cash advance usually attracts interest immediately from the day of withdrawal (no interest-free period) and often a cash-advance fee, so it is far more costly than a purchase.
A holder makes a $1800 purchase on a card with a 44-day interest-free period and a 20% p.a. rate.
1. (a) If they pay in full by the due date, state the interest charged. (1 mark)
2. (b) If they instead pay nothing and carry the balance, estimate the interest for the first 44 days and explain what the interest-free offer required. (2 marks)
(a) $0 — paid in full within the interest-free period.
(b) Interest ≈ 1800 × 0.20 × 44/365 = $43.40. The interest-free offer required the closing balance to be paid in full by the due date, so carrying the balance forfeits it and interest (often back-dated to the purchase date) applies.
Level 1 · Fluency

A store account charges 24% p.a. compounded daily. What daily rate is used in the compound-interest formula?

Level 2 · Application

A $1450 purchase on a store account is left unpaid for 60 days and is charged 22.9% p.a. compounded daily. Calculate the interest charged over the 60 days, correct to the nearest cent.
Level 3 · Further Application
A store offers “12 months interest-free” on a $2200 purchase, but if any balance remains after 12 months, interest is back-dated at 24.9% p.a. (treat as compounded monthly, 12 months).
1. (a) A customer still owes $300 at the 12-month mark. Calculate the interest back-dated on the full $2200 for the 12 months. (2 marks)
2. (b) Explain why this makes the “interest-free” deal risky. (1 mark)
(a) Interest = 2200(1 + 0.249/12)¹² − 2200 = $614.85.
(b) Interest is charged on the whole original amount from the purchase date, not just the leftover $300 — so failing to clear the balance in time can cost hundreds of dollars.
A store offers “18 months interest-free” on a $3600 purchase, but if any balance remains after 18 months, interest is back-dated at 22.9% p.a. (treat as compounded monthly, 18 months).
1. (a) A customer still owes $500 at the 18-month mark. Calculate the interest back-dated on the full $3600 for the 18 months. (2 marks)
2. (b) Explain why this makes the “interest-free” deal risky. (1 mark)
(a) Interest = 3600(1 + 0.229/12)¹⁸ − 3600 = $1,459.14.
(b) Interest is charged on the whole original $3600 from the purchase date, not just the $500 still owing, so failing to clear the balance in time costs far more than expected.
A store offers “24 months interest-free” on a $4800 purchase, but if any balance remains after 24 months, interest is back-dated at 25.9% p.a. (treat as compounded monthly, 24 months).
1. (a) A customer still owes $700 at the 24-month mark. Calculate the interest back-dated on the full $4800 for the 24 months. (2 marks)
2. (b) Explain why a customer should aim to clear the balance before the deadline. (1 mark)
(a) Interest = 4800(1 + 0.259/12)²⁴ − 4800 = $3,213.32.
(b) Interest is back-dated on the entire original $4800 from the purchase date (not just the $700 left), so missing the deadline can add thousands of dollars — the balance should be cleared before the interest-free period ends.

N	r = 0.004	r = 0.005	r = 0.006
240	154.0933	139.5808	127.0084
300	174.5210	155.2069	138.9683
360	190.5977	166.7916	147.3214

N	r = 0.004	r = 0.005	r = 0.006
240	154.0933	139.5808	127.0084
300	174.5210	155.2069	138.9683
360	190.5977	166.7916	147.3214

N	r = 0.004	r = 0.005	r = 0.006
240	154.0933	139.5808	127.0084
300	174.5210	155.2069	138.9683
360	190.5977	166.7916	147.3214

	Card A	Card B
Annual fee	$0	$120
Interest rate	21.9% p.a.	12.9% p.a.
Interest expected	$305	$180