MK Maths resources · Set 01
GCSE Algebra Diagnostic
Twenty original questions arranged from core algebraic fluency to higher-tier reasoning. Work without notes where possible and show each stage clearly.
Questions
The labels indicate the principal level and skill, although several questions connect more than one curriculum objective.
Simplify 4a + 3a.
Simplify 7x − 2x + 5.
Expand 3(2x + 5).
Find the value of 3x² + 4 when x = −2.
Solve 5x + 7 = 27.
Solve 3(x − 2) = 15.
The first four terms of a sequence are 5, 8, 11, 14. Find an expression for its nth term.
Expand and simplify (x + 4)(x + 2).
Factorise fully 6x + 18.
Factorise x² + 7x + 12.
Solve 2x + 5 = 3x − 4.
Make a the subject of v = u + at.
Simplify (3x²y)(2xy³).
Solve x² − 5x + 6 = 0.
Factorise fully 9x² − 25.
Solve simultaneously: x + y = 9 and 2x − y = 6.
Solve the inequality 3x − 4 < 11.
The first four terms of a sequence are 2, 6, 12, 20. Find an expression for its nth term.
Simplify (x² − 9) / (x² + 2x − 3). State the values of x for which the original expression is undefined.
A rectangle has sides x + 3 cm and x − 1 cm. Its area is 60 cm². Form and solve an equation to find the dimensions of the rectangle.
Answers
- 1. 7a
- 2. 5x + 5
- 3. 6x + 15
- 4. 16
- 5. x = 4
- 6. x = 7
- 7. 3n + 2
- 8. x² + 6x + 8
- 9. 6(x + 3)
- 10. (x + 3)(x + 4)
- 11. x = 9
- 12. a = (v − u) / t
- 13. 6x³y⁴
- 14. x = 2 or x = 3
- 15. (3x − 5)(3x + 5)
- 16. x = 5, y = 4
- 17. x < 5
- 18. n(n + 1)
- 19. (x − 3) / (x − 1), where x ≠ −3 and x ≠ 1
- 20. 10 cm by 6 cm
Selected worked solutions
Question 5 · Linear equation
Start with 5x + 7 = 27. Subtract 7 from both sides to obtain 5x = 20. Divide both sides by 5, giving x = 4.
Question 10 · Factorising a quadratic
We need two numbers with a product of 12 and a sum of 7. These numbers are 3 and 4, so x² + 7x + 12 = (x + 3)(x + 4).
Question 12 · Rearranging a formula
From v = u + at, subtract u from both sides: v − u = at. Divide both sides by t to isolate a: a = (v − u) / t, where t ≠ 0.
Question 16 · Simultaneous equations
Add x + y = 9 and 2x − y = 6. The y terms cancel, leaving 3x = 15, so x = 5. Substitute into x + y = 9 to obtain 5 + y = 9, hence y = 4.
Question 18 · Quadratic sequence
The first differences are 4, 6 and 8, so the second difference is constant. The terms can be recognised as 1 × 2, 2 × 3, 3 × 4 and 4 × 5. Therefore the nth term is n(n + 1).
Question 19 · Algebraic fraction
Factorise the numerator and denominator: x² − 9 = (x − 3)(x + 3), while x² + 2x − 3 = (x + 3)(x − 1). Cancel the common factor to obtain (x − 3) / (x − 1). The original denominator is zero when x = −3 or x = 1, so both values must be excluded.
Question 20 · Forming and solving an equation
The area gives (x + 3)(x − 1) = 60. Expanding and rearranging gives x² + 2x − 63 = 0, which factorises to (x + 9)(x − 7) = 0. The possible values are −9 and 7, but −9 would produce negative side lengths. Therefore x = 7 and the rectangle measures 10 cm by 6 cm.
This is an original MK Maths resource aligned with algebra skills commonly assessed by AQA and Pearson Edexcel GCSE Mathematics and Cambridge IGCSE Mathematics. It is not endorsed by any examination board.