Number for IGCSE Mathematics: Methods and Worked Examples

The arithmetic foundation of the entire syllabus

Number is the largest and most foundational topic area in Cambridge IGCSE Mathematics (0580). Every other branch of the syllabus, from algebra to statistics, depends on fluency with numerical operations, and examiners test this fluency across all four papers. A candidate who masters Number secures marks not only in dedicated Number questions but also in multi-step problems where arithmetic accuracy determines whether the final answer is correct.

The sections below progress from the classification of numbers through operations, indices, ratio, estimation, and the Extended-only topics of surds and exponential growth. Each section states the key definitions, demonstrates the method through worked examples, and flags the errors that most frequently cost marks.

Types of number

The IGCSE syllabus requires candidates to identify and work with several categories of number. A clear grasp of these classifications prevents confusion in later topics, particularly when questions ask whether a result is rational or irrational.

Highest common factor and lowest common multiple

The HCF of two numbers is the largest integer that divides both exactly. The LCM is the smallest positive integer that both numbers divide into exactly. The standard method uses prime factorisation.

Worked example: Find the HCF and LCM of 60 and 84.

1. Prime factorise each number: 60 = 2 x 2 x 3 x 5 = 2^2 x 3 x 5. And 84 = 2 x 2 x 3 x 7 = 2^2 x 3 x 7.
2. HCF: take the lowest power of each shared prime factor. Both share 2^2 and 3^1, so HCF = 2^2 x 3 = 12.
3. LCM: take the highest power of every prime factor present in either number. LCM = 2^2 x 3 x 5 x 7 = 420.

Sets and Venn diagrams

Set theory appears in both Core and Extended tiers, though Extended candidates face more complex problems involving three sets and formal notation.

Venn diagram problems typically require candidates to place given numbers of elements into the correct regions, then read off unions, intersections, or complements. The method is systematic: always fill the intersection region first, then work outward.

Worked example: In a class of 30 students, 18 study French, 14 study Spanish, and 5 study neither. How many study both?

1. Students studying at least one language: 30 - 5 = 25.
2. Using the addition formula: n(F union S) = n(F) + n(S) - n(F intersection S). So 25 = 18 + 14 - n(F intersection S).
3. Solving: n(F intersection S) = 32 - 25 = 7. Seven students study both languages.

Powers, roots, and indices

The laws of indices underpin a wide range of IGCSE questions, from simplifying algebraic expressions to solving equations. Both tiers require the basic laws; Extended candidates must also handle negative and fractional indices fluently.

The index laws

• Multiplication: a^m x a^n = a^(m+n)
• Division: a^m / a^n = a^(m-n)
• Power of a power: (a^m)^n = a^(mn)
• Zero index: a^0 = 1 (for any non-zero a)
• Negative index: a^(-n) = 1/a^n
• Fractional index: a^(1/n) = the nth root of a; a^(m/n) = (the nth root of a)^m

Worked example: Simplify (2^3 x 2^5) / 2^4.

1. Numerator: 2^3 x 2^5 = 2^(3+5) = 2^8.
2. Division: 2^8 / 2^4 = 2^(8-4) = 2^4 = 16.

Worked example (Extended): Evaluate 27^(2/3).

1. Interpret the fractional index: 27^(2/3) = (27^(1/3))^2.
2. The cube root of 27 is 3.
3. 3^2 = 9. So 27^(2/3) = 9.

Fractions, decimals, and percentages

Conversion fluency between these three representations is tested repeatedly. Candidates should be able to move between them without hesitation.

Operations with fractions

Addition and subtraction require a common denominator. Multiplication proceeds numerator-by-numerator and denominator-by-denominator. Division means multiplying by the reciprocal of the divisor.

Worked example: Calculate 2/3 + 3/4.

1. Find the LCM of 3 and 4: LCM = 12.
2. Convert: 2/3 = 8/12; 3/4 = 9/12.
3. Add: 8/12 + 9/12 = 17/12 = 1 and 5/12.

Estimation and limits of accuracy

Estimation questions require candidates to round each value to one significant figure (or as specified) before performing the calculation. The purpose is to produce a quick approximation, and examiners penalise candidates who work with the exact values instead.

Worked example: Estimate the value of (48.7 x 0.523) / 9.86.

1. Round to 1 significant figure: 48.7 becomes 50; 0.523 becomes 0.5; 9.86 becomes 10.
2. Calculate: (50 x 0.5) / 10 = 25 / 10 = 2.5.

Upper and lower bounds

When a measurement is given to a stated degree of accuracy, the true value lies within half a unit either side. For a length recorded as 5.3 cm to one decimal place, the lower bound is 5.25 cm and the upper bound is 5.35 cm.

Candidates must reason carefully about which bound to use in compound calculations. To maximise a quotient, use the upper bound of the numerator and the lower bound of the denominator. To minimise it, reverse the choice.

Ratio and proportion

Ratio questions appear in two main forms: dividing a quantity in a given ratio, and using proportion to find unknown values.

Worked example: Divide 180 in the ratio 2 : 3 : 5.

1. Sum the parts: 2 + 3 + 5 = 10.
2. Value of one part: 180 / 10 = 18.
3. The three shares are 2 x 18 = 36, 3 x 18 = 54, and 5 x 18 = 90.
4. Check: 36 + 54 + 90 = 180.

Direct and inverse proportion

In direct proportion, as one quantity increases the other increases at the same rate: y = kx (or y is proportional to x^n for more complex relationships). In inverse proportion, as one quantity increases the other decreases: y = k/x.

Worked example: y is inversely proportional to x. When x = 4, y = 6. Find y when x = 8.

1. Since y = k/x, substitute: 6 = k/4, so k = 24.
2. When x = 8: y = 24/8 = 3.

Percentages: increases, decreases, and reverse problems

Percentage calculations extend beyond simple "find 15% of 240" into multi-step scenarios involving increase, decrease, reverse percentages, and compound interest.

Percentage change

Percentage change = (change / original) x 100. A 15% increase on a value V gives V x 1.15. A 15% decrease gives V x 0.85. These multipliers simplify repeated-percentage problems considerably.

Worked example: A shirt originally costs 40 pounds. It is reduced by 20%, then the sale price is increased by 10%. What is the final price?

1. After 20% reduction: 40 x 0.80 = 32.
2. After 10% increase on the sale price: 32 x 1.10 = 35.20.
3. The final price is 35.20 pounds. Note this is not the same as a net 10% reduction, which would give 36.

Reverse percentages

If a price after a 20% increase is 96, the original price is not found by subtracting 20% of 96. The correct approach identifies 96 as 120% of the original.

1. 96 represents 120% of the original.
2. 1% of the original = 96 / 120 = 0.80.
3. 100% of the original = 0.80 x 100 = 80.

Compound interest

Compound interest accrues on both the principal and previously accumulated interest. The formula is A = P(1 + r/100)^n, where P is the principal, r the annual rate, and n the number of years.

Worked example: 2000 pounds is invested at 5% compound interest per year for 3 years. Find the total amount.

1. A = 2000 x (1.05)^3.
2. 1.05^3 = 1.157625.
3. A = 2000 x 1.157625 = 2315.25.

Surds (Extended only)

Surds are irrational numbers left in root form to preserve exactness. The IGCSE Extended tier requires candidates to simplify surds and rationalise denominators.

Simplifying surds

The key principle is that the square root of (a x b) equals the square root of a multiplied by the square root of b. To simplify, extract the largest perfect square factor.

Worked example: Simplify the square root of 72.

1. Find the largest perfect square factor of 72: 72 = 36 x 2.
2. The square root of 72 = the square root of 36 multiplied by the square root of 2 = 6 multiplied by the square root of 2.

Rationalising the denominator

A fraction with a surd in the denominator is rationalised by multiplying both numerator and denominator by the surd.

Worked example: Rationalise 5 / (square root of 3).

1. Multiply numerator and denominator by the square root of 3: (5 x square root of 3) / (square root of 3 x square root of 3).
2. The denominator becomes 3.
3. Result: (5 x square root of 3) / 3.

Exponential growth and decay (Extended only)

These problems model situations where a quantity increases or decreases by a fixed percentage over equal time intervals. Population growth, radioactive decay, and depreciation are the classic contexts.

The general formula is y = a x b^t, where a is the initial amount, b is the growth (or decay) factor, and t is the number of time periods. For growth, b is greater than 1. For decay, b is between 0 and 1.

Worked example: A car worth 12,000 pounds depreciates by 15% each year. What is it worth after 4 years?

1. Decay factor: 1 - 0.15 = 0.85.
2. Value after 4 years: 12,000 x (0.85)^4.
3. 0.85^4 = 0.52200625.
4. Value = 12,000 x 0.52200625 = 6264.08 (to the nearest penny).

Common mistakes across the Number topic

Self-check questions

1. List all the prime numbers between 30 and 50.
2. Find the HCF and LCM of 36 and 90 using prime factorisation.
3. In a group of 40 students, 22 play football, 17 play tennis, and 8 play neither. How many play both sports? Draw a Venn diagram to represent this information.
4. Simplify: (3^4 x 3^2) / 3^5.
5. Evaluate 8^(-2/3) without a calculator.
6. Convert the recurring decimal 0.272727... into a fraction in its simplest form.
7. A length is measured as 7.4 cm, correct to one decimal place. Write down the upper and lower bounds of this measurement.
8. Divide 450 in the ratio 2 : 3 : 4.
9. After a 12% increase, a salary is 33,600 pounds. What was the original salary?
10. Simplify the square root of 200, giving your answer in the form a times the square root of b, where a and b are integers and b is as small as possible.
11. A population of 5000 bacteria doubles every 3 hours. Write a formula for the population P after t hours, and find the population after 12 hours.
12. Rationalise the denominator of 8 / (square root of 2) and simplify.

These questions span both Core and Extended content. Questions 5, 10, 11, and 12 are aimed specifically at Extended tier candidates, though Core students may find them useful for building deeper understanding. Working through each question in full, showing all steps, replicates the discipline that IGCSE examiners reward with method marks even when the final answer contains an arithmetic slip.