← IGCSE / O Level Mathematics

IGCSE / O Level Mathematics · Topic 1 Number · Chapter 8 · pp. 73–82

Money and finance

Money questions use the number skills you already have: multiplying by a rate, taking a percentage, and using a multiplier. This chapter applies them to changing currency, working out pay, profit and loss, and the interest a bank adds to savings or loans.

Learning objectives

  1. Convert between currencies using an exchange rate.
  2. Work out gross pay, net pay, overtime, bonuses and piece-work pay.
  3. Find profit, loss and discount, and give profit or loss as a percentage of the cost price.
  4. Use the simple interest formula, and rearrange it to find the time, rate or principal.
  5. Work out compound interest, growth and depreciation with a multiplier.

Book exercises: 8.1 to 8.6 and Student assessments 1 and 2 (pp. 73–82).

Currency conversions

🔑 Key ideas

  • An exchange rate tells you how much of another currency one unit buys, e.g. €1 = A$1.50.
  • From € to the other currency: multiply by the rate.
  • From the other currency back to €: divide by the rate.
  • Sense check: if €1 buys more than 1 unit, the foreign number should be the bigger one.
€ euros €400 A$ dollars A$600 × 1.50 ÷ 1.50 Exchange rate: €1 = A$1.50 Multiply going out, divide coming back
One rate works both ways: multiply to change euros into dollars, divide to change dollars back into euros.
Country€1 buys
Australia1.50 Australian dollars (A$)
India75 rupees
Zimbabwe412 800 Zimbabwe dollars (ZIM$)
South Africa15 rand
Turkey4.0 Turkish lira (L)
Japan130 yen
Kuwait0.35 dinar
USA1.15 US dollars (US$)

💡 The book's rate table for Exercise 8.1. Rates change every day; these are the 2017 values the book uses.

📘 Example 1 · How many Australian dollars can be bought for €400?

  1. €1 buys A$1.50.
  2. Euros to dollars, so multiply: 400 × 1.50 = 600.
  3. €400 buys A$600.
  4. Check: more dollars than euros, as expected, because €1 buys more than A$1. ✔

📘 Example 2 · How much does it cost to buy A$940?

  1. A$1.50 costs €1.
  2. Dollars to euros, so divide: 940 ÷ 1.50 = 626.666…
  3. Round to the nearest cent: €626.67.
  4. Check: 626.67 × 1.50 = 940.005 ≈ A$940. ✔

📘 Example 3 · €1 = 75 rupees. Change €60 into rupees, then 9000 rupees into euros.

  1. Euros to rupees, multiply: 60 × 75 = 4500 rupees.
  2. Rupees to euros, divide: 9000 ÷ 75 = €120.

✏️ Try it

  1. €1 = US$1.15. Change €80 into US dollars.
  2. €1 = US$1.15. How many euros does US$230 cost?
  3. €1 = 130 yen. How many euros does 6500 yen cost?
Answers

1. 80 × 1.15 = US$92 · 2. 230 ÷ 1.15 = €200 · 3. 6500 ÷ 130 = €50

Next: earnings →

Earnings

WordMeaning
Basic payHours in the normal working week × the hourly rate.
OvertimeExtra hours beyond the basic week, paid at a higher rate: time and a half (1.5 × rate) or double time (2 × rate).
BonusAn extra payment sometimes added to basic pay.
Piece workPaid for the number of articles made, not for the time taken.
Gross payEverything earned before anything is taken off.
DeductionsTax, insurance and pension contributions taken from gross pay.
Net payNet pay = Gross pay − Deductions: what you actually receive.
Gross pay $471.25 Basic $380 Overtime $71.25 Bonus $20 After deductions (20%) Net pay $377 −$94.25 = gross
Example 4 as bars. Basic pay, overtime and bonus add up to gross pay; deductions (dashed) come off to leave net pay.

📘 Example 4 · Ana works 40 hours at $9.50 an hour, plus 5 hours of overtime at time and a half. She gets a $20 bonus and 20% is deducted. Find her gross and net pay.

  1. Basic pay: 40 × 9.50 = $380.
  2. Overtime rate: 1.5 × 9.50 = $14.25 an hour.
  3. Overtime pay: 5 × 14.25 = $71.25.
  4. Gross pay: 380 + 71.25 + 20 = $471.25.
  5. Deductions: 20% of 471.25 = 0.2 × 471.25 = $94.25.
  6. Net pay: 471.25 − 94.25 = $377.00.
  7. Check: 80% of 471.25 = 0.8 × 471.25 = 377. ✔

📘 Example 5 · Piece work: Tom is paid $3 for every 10 bowls he makes. He makes 240 bowls. How much is he paid?

  1. Count the batches of 10: 240 ÷ 10 = 24 batches.
  2. Pay per batch × batches: 24 × 3 = $72.

✏️ Try it

  1. Gross pay is $350 and deductions are $84. What is the net pay?
  2. Lin works 37 hours at $12 an hour plus 4 hours at double time. Find her gross pay.
  3. A machinist is paid $0.80 per shirt and sews 135 shirts. How much is she paid?
Answers

1. 350 − 84 = $266 · 2. 37 × 12 = 444, 4 × 24 = 96, total $540 · 3. 135 × 0.80 = $108

Next: profit and loss →

Profit and loss

🔑 Key ideas

  • Cost price (C.P.): what it costs to make or buy. Selling price (S.P.): what it is sold for.
  • S.P. > C.P.: profit = S.P. − C.P. · S.P. < C.P.: loss = C.P. − S.P.
  • A discount is a reduction in the selling price, often given as a percentage.
  • Percentage profit or loss is always worked out on the cost price.
  • When something is worth less over time, it depreciates.
S.P. − C.P.profit (or loss) ÷ C.P.cost price × 100make a % % profitor % loss C.P. $7500 S.P. $4500 loss $3000 3000 ÷ 7500 × 100 = 40% loss
Top: the flowchart for a percentage profit or loss. Bottom: the car in Example 8 loses $3000, which is 40% of its cost price.

📘 Example 6 · A trader buys oranges in boxes of 12 dozen for $14.40 a box. He buys 3 boxes and sells all the oranges for 12c each. Find his profit or loss.

  1. Cost price: 3 × $14.40 = $43.20.
  2. Number of oranges: 3 boxes × 144 = 432.
  3. Selling price: 432 × 12c = 5184c = $51.84.
  4. S.P. is bigger, so it is a profit: 51.84 − 43.20 = $8.64 profit.
  5. Check another way: each orange costs 1440c ÷ 144 = 10c, so the profit is 2c each, and 432 × 2c = $8.64. ✔

📘 Example 7 · A skirt usually costing $35 is sold at a 15% discount. What is the discount?

  1. 15% as a decimal is 0.15.
  2. 0.15 × 35 = $5.25 discount.
  3. So the sale price is 35 − 5.25 = $29.75.

📘 Example 8 · A woman buys a car for $7500 and sells it two years later for $4500. Find her loss as a percentage of the cost price.

  1. Loss: 7500 − 4500 = $3000.
  2. Divide by the cost price: 3000 ÷ 7500 = 0.4.
  3. × 100: 40% loss.

📘 Example 9 · A bike bought for $240 is sold for $300. Find the percentage profit.

  1. Profit: 300 − 240 = $60.
  2. Divide by the cost price (not the selling price): 60 ÷ 240 = 0.25.
  3. × 100: 25% profit.
  4. Check: 240 × 1.25 = 300. ✔

✏️ Try it

  1. 50 pens cost $15 and are sold at 40c each. Find the profit.
  2. C.P. $120, S.P. $150. Find the percentage profit.
  3. A phone bought for $400 is sold for $340. Find the percentage loss.
Answers

1. 50 × 40c = $20, profit $5 · 2. 30 ÷ 120 × 100 = 25% · 3. 60 ÷ 400 × 100 = 15%

Next: simple interest →

Simple interest

🔑 Key ideas

  • Interest is money a bank adds to a deposit (or charges on a loan).
  • The money deposited is the principal P. The rate is r% per annum (p.a., each year). The time is t years.
  • Simple interest is worked out on the principal only, so it is the same every year.
  • SI = . Total amount = P + SI.
To findUse
Interest SISI =
Time tt =
Rate rr =
Principal PP =

💡 If a question gives the total amount, subtract the principal first to get the interest.

$250 Year 0 $270 Year 1 $290 Year 2 $310 Year 3 $330 Year 4 $350 Year 5 $370 Year 6 P $250 at 8% p.a.: + $20 interest every year
Simple interest grows by the same step each year (darker blocks), because 8% is always taken of the original $250.

📘 Example 10 · Find the simple interest on $250 deposited for 6 years at 8% p.a.

  1. Write down P = 250, t = 6, r = 8.
  2. SI = = .
  3. SI = $120.

📘 Example 11 · How long will $250 at 8% take to earn $80 interest?

  1. Substitute: 80 = .
  2. Simplify the right side: 80 = 20t.
  3. Divide by 20: t = 4, so it takes 4 years.
  4. Check: 250 × 4 × 8 ÷ 100 = 80. ✔

📘 Example 12 · What rate makes $750 earn $180 in 4 years?

  1. Substitute: 180 = .
  2. Simplify: 180 = 30r.
  3. Divide by 30: r = 6, so the rate is 6% per year.

📘 Example 13 · Find the principal that earns $120 in 6 years at 4%.

  1. Substitute: 120 = = .
  2. Multiply both sides by 100: 12 000 = 24P.
  3. Divide by 24: P = $500.
  4. Check: 500 × 6 × 4 ÷ 100 = 120. ✔

✏️ Try it

  1. Find the simple interest on $400 at 5% p.a. for 3 years.
  2. $600 earns $90 simple interest in 5 years. Find the rate.
  3. $1000 at 4% p.a. grows to $1200 in total. How many years did that take?
Answers

1. 400 × 3 × 5 ÷ 100 = $60 · 2. r = 9000 ÷ 3000 = 3% · 3. SI = $200, t = 20 000 ÷ 4000 = 5 years

Next: compound interest →

Compound interest

🔑 Key ideas

  • Compound interest is paid on the principal and on the interest already added.
  • Use a multiplier: an increase of r% is × (1 + ). 10% up is × 1.10.
  • After n years: total = P × (1 + )n, and the interest I = P(1 + )n − P.
  • A decrease (depreciation) uses a multiplier below 1: 15% down is × 0.85.
  • Interest added every 6 months at an APR of 8.5%: use 8.5 ÷ 2 = 4.25% per period, and count the periods.
  • Compound growth is a geometric sequence: exponential growth.
€500 000 Start × 1.10 €550 000 End of year 1 × 1.10 €605 000 End of year 2 × 1.10 €665 500 End of year 3 500 000 × 1.10³ = 665 500 → interest = €165 500
The builder's €500 000 loan at 10% compound interest: each year multiplies the amount owed by 1.10.
Simple Compound 1,100 1,100 Year 1 1,200 1,210 Year 2 1,300 1,331 Year 3 1,400 1,464.10 Year 4 1,500 1,610.51 Year 5
$1000 at 10% a year. Simple interest adds $100 every year; compound interest adds 10% of a growing amount, so the gap widens each year.

📘 Example 14 · A builder borrows €500 000 at 10% compound interest p.a. and repays it after 3 years. How much interest does he pay?

  1. 10% up is a multiplier of 1.10.
  2. Three years: 500 000 × 1.103 = 500 000 × 1.331 = €665 500.
  3. Interest: 665 500 − 500 000 = €165 500.
  4. Check year by year: 550 000 → 605 000 → 665 500. ✔

📘 Example 15 · How long will it take a debt to double at 27% compound interest p.a.?

  1. 27% up is a multiplier of 1.27.
  2. After 1 year: 1.27P. After 2 years: 1.272P = 1.61P.
  3. After 3 years: 1.273P = 2.05P, which is more than 2P.
  4. The debt has doubled after 3 years.

📘 Example 16 · Find the compound interest on a $600 loan for 3 years at an APR of 5%.

  1. 5% up is a multiplier of 1.05.
  2. Total: 600 × 1.053 = 694.575, so $694.58 (2 d.p.).
  3. Interest: 694.58 − 600 = $94.58.

📘 Example 17 · $3000 is invested for 18 months at an APR of 8.5%, with interest added every 6 months. Find the interest.

  1. Rate per 6 months: 8.5 ÷ 2 = 4.25%, a multiplier of 1.0425.
  2. Number of periods: 18 months ÷ 6 = 3.
  3. Total: 3000 × 1.04253 = 3398.99 (2 d.p.).
  4. Interest: 3398.99 − 3000 = $398.99.

📘 Example 18 · Depreciation: a $12 000 car loses 20% of its value each year. What is it worth after 2 years?

  1. 20% down leaves 80%, a multiplier of 0.8.
  2. Two years: 12 000 × 0.82 = 12 000 × 0.64.
  3. Value: $7680.
  4. Check: 12 000 → 9600 → 7680. ✔

🧪 Try it: simple vs compound interest

Simple interest

    Compound interest

      ✏️ Try it

      1. Find the compound interest on $2000 at 3% p.a. for 2 years.
      2. A town of 5000 people grows by 4% a year. How many people after 2 years?
      3. How many years will it take a debt to double at 20% compound interest p.a.?
      Answers

      1. 2000 × 1.032 = 2121.80, interest $121.80 · 2. 5000 × 1.042 = 5408 · 3. 1.23 = 1.728, 1.24 = 2.0736, so 4 years

      Next: summary and quiz →

      Chapter summary

      • Currency: multiply by the rate to change € into the other currency, divide to change back.
      • Gross pay = basic + overtime + bonus. Net pay = gross pay − deductions.
      • Time and a half = 1.5 × rate; double time = 2 × rate. Piece work pays per item made.
      • Profit = S.P. − C.P. Percentage profit or loss = profit or loss ÷ cost price × 100.
      • Simple interest SI = Ptr ÷ 100: the same amount every year. Rearrange it to find t, r or P.
      • Compound interest: total = P × (1 + r/100)n; interest = total − P. Depreciation uses a multiplier below 1.

      📄 Practice questions

      The book's Exercises 8.1 to 8.6 and Student assessments 1 and 2 are in the printable assignment for this chapter.

      Mini-quiz

      1. €1 buys A$1.50. How many Australian dollars do you get for €60?
      2. €1 buys 130 yen. How many euros does 3900 yen cost?
      3. Gross pay is $420 and deductions are $95. What is the net pay?
      4. The basic rate is $8 an hour. What is one hour of overtime at time and a half?
      5. In piece work, an employee is paid according to…
      6. An item costs $50 and sells for $65. What is the percentage profit?
      7. A car bought for $9200 is sold for $6900. What is the percentage loss?
      8. What is the simple interest on $300 at 6% per year for 4 years?
      9. $500 earns $150 simple interest at 6% per year. How many years is that?
      10. Which multiplier adds 4% each year?
      11. $1000 is invested at 10% compound interest for 2 years. What is the total amount?
      12. A car loses 15% of its value each year. What is the multiplier for one year?

      Source: Cambridge IGCSE Mathematics Core and Extended (4th edition, Pimentel and Wall), Chapter 8 (Money and finance). Explanations, the Try it answers, the tool and the quiz are Guru Jazzy's own. This page is not made or endorsed by Cambridge or Hodder Education.