Cambridge IGCSELive bank

Mathematics — IGCSE

Core and extended routes are both covered. Agree your tier with your teacher before drilling the harder topics.

Syllabus code 0580 · checked against the board on 2026-09-13 · 15 questions available on this subject's topics (27 marks)

Topic map

Practice available now (15 questions)

Simplify (2x³y)² × 3x⁻²y⁴ ÷ 6xy², leaving your answer in terms of x and y.

2 marks · Simplify · Powers, roots and index laws

Show the mark scheme

Answer: 2x³y⁴

  1. Square the bracket: 4x⁶y²
  2. Multiply: 4x⁶y² × 3x⁻²y⁴ = 12x⁴y⁶
  3. Divide: 12x⁴y⁶ ÷ 6xy² = 2x³y⁴

Work in three lines — bracket, multiply, divide. Learners who try to do it in one line lose the index of x more often than any other single mark in the paper.

Work out 8^(2/3).

1 mark · Work out · Powers, roots and index laws

Show the mark scheme

Answer: 4

  1. 8^(2/3) means (cube root of 8)²
  2. Cube root of 8 = 2
  3. 2² = 4

Read a fractional index as "root, then power". Root first is easier arithmetic, and it keeps the numbers small enough to do without a calculator.

A jacket is marked at P1,250. In a sale the price falls by 20%. The sale price is later increased by 20% to become the new normal price. Show that the new normal price is P1,200.

2 marks · Show that · Percentage change, including repeated change over time

Show the mark scheme

Answer: P1,250 × 0.80 = P1,000; P1,000 × 1.20 = P1,200

  1. Sale price: 1250 × 0.80 = 1000
  2. New price: 1000 × 1.20 = 1200

This is the 20%-up-is-not-20%-down trap. Both marks belong to the two calculations; the answer is already printed, so quoting it earns nothing.

Expand and simplify (2x − 3)(x + 5).

1 mark · Expand and simplify · Simplifying, expanding and factorising

Show the mark scheme

Answer: 2x² + 7x − 15

  1. Four products: 2x·x, 2x·5, −3·x, −3·5
  2. 2x² + 10x − 3x − 15
  3. Collect: 2x² + 7x − 15

Write all four products before collecting. Skipping a term is the single most common error in this question type, and it costs the whole mark rather than part of it.

Solve 3(2x − 1) = 2(x + 4) + 5.

3 marks · Solve · Linear and simultaneous equations

Show the mark scheme

Answer: x = 4

  1. Expand both sides: 6x − 3 = 2x + 8 + 5
  2. Simplify the right: 6x − 3 = 2x + 13
  3. Collect: 4x = 16, so x = 4

Expand both sides first, then collect. Collecting before expanding is where sign errors are born.

Find the nth term of the sequence 5, 8, 11, 14, …

1 mark · Find · Sequences and the nth term

Show the mark scheme

Answer: 3n + 2

  1. Common difference = 3, so the rule starts 3n
  2. n = 1 gives 3, but the first term is 5, so add 2

One mark, five seconds of checking: substitute n = 1 and n = 2 into your own rule. It is the cheapest verification in the whole paper.

Each exterior angle of a regular polygon is 24°. How many sides does the polygon have?

1 mark · Find · Angles, polygons and parallel lines

Show the mark scheme

Answer: 15

  1. Exterior angles of a polygon sum to 360°
  2. 360 ÷ 24 = 15 sides

The exterior angle fact does all the work in polygon questions. Interior angles total 180(n − 2), which is slower and easier to get wrong.

A, B, C and D lie on a circle in that order. Angle ABC = 74°. Find angle ADC, giving a reason for your answer.

3 marks · Find, giving a reason · Circle theorems

Show the mark scheme

Answer: 106°, because opposite angles of a cyclic quadrilateral add to 180°.

  1. ABCD is a cyclic quadrilateral
  2. Opposite angles sum to 180°
  3. 180 − 74 = 106°

Circle theorem questions carry a reason mark. A correct number with no theorem named typically loses one third of the marks available.

In a right-angled triangle the hypotenuse is 9.6 m. One of the sides next to an angle is 7.2 m. Work out that angle, giving your answer to one decimal place.

3 marks · Work out · Pythagoras and right-angled trigonometry

Show the mark scheme

Answer: 41.4°

  1. The two sides involved are the adjacent (7.2 m) and the hypotenuse (9.6 m), so use cosine
  2. cos θ = 7.2 ÷ 9.6 = 0.75
  3. θ = 41.4°

Label opposite, adjacent and hypotenuse on the diagram before choosing a ratio. Most marks lost here are lost to using sine because it was written first on the formula sheet.

Find the median of 4, 7, 2, 9, 5, 7, 3.

1 mark · Find · Averages from lists and frequency tables

Show the mark scheme

Answer: 5

  1. Order the list: 2, 3, 4, 5, 7, 7, 9
  2. Seven values, so the median is the 4th
  3. Median = 5

Order first, always. With an even count, take the mean of the two middle values — that is where 5.5 style answers come from.

A bag contains 5 red counters and 3 blue counters. Two counters are taken at random without replacement. Find the probability that both counters are red, giving your answer as a fraction in its lowest terms.

3 marks · Find · Probability, with and without replacement

Show the mark scheme

Answer: 5/14

  1. First counter red: 5/8
  2. Second counter red, given the first was red: 4/7
  3. 5/8 × 4/7 = 20/56 = 5/14

Without replacement changes the second fraction on both the top and the bottom. Write the second fraction separately — it is usually a mark of its own.

A straight line passes through the points (2, 5) and (6, 13). Find the equation of the line in the form y = mx + c.

3 marks · Find · Graphs, gradients and curve sketching

Show the mark scheme

Answer: y = 2x + 1

  1. Gradient m = (13 − 5) ÷ (6 − 2) = 8 ÷ 4 = 2
  2. Substitute (2, 5): 5 = 2(2) + c, so c = 1
  3. Equation: y = 2x + 1

One mark for the gradient, one for substituting a point, one for the finished equation. Show each of the three lines — a bare correct equation can score one mark, not three.

Showing 12 of 15. Open the bank to filter by topic, format or assessment objective.