Practice Questions
OxfordAQA International GCSE Mathematics: Linear equations and forming equations from problems (9260) – Practice Questions
Original practice questions with worked answers on linear equations, graphs and forming equations for OxfordAQA International GCSE Mathematics.
- Subject
- Mathematics
- Level
- IGCSE
- Topic
- Linear equations and forming equations from problems
- Author
- Marlbridge Academic Team
- Updated
Aligned to OxfordAQA IGCSE Mathematics (9260), Version 5.1 (for exams May/June 2018 onwards). Official specification .
Syllabus page (what it covers and how it is assessed): OxfordAQA IGCSE Mathematics.
Syllabus points this page covers
9260
- 3.2.3 Solving equations and inequalities
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These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.
These questions cover references A19 and A22 of section 3.2.3 of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). All of the content is Core, so it is also assessable on the Extension papers. Show your working: each answer below is split into steps so you can see where your own working earns or loses credit. Read the study guide and revision notes first if you need to. See also the course hub, the checklist and the free diagnostics.
Questions
1. Solve 9x − 4 = 5x + 20. [2]
2. Solve 5(2x + 3) − 2(x − 4) = 47. [3]
3. Solve (x − 4)/3 + (3x − 1)/4 = 6. [4]
4. This question is about solving 6 − 2x = x + 1 using a graph.
(a) Complete the table of values for y = 6 − 2x. [1]
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| y = 6 − 2x | 6 | −2 | |||
| y = x + 1 | 1 | 2 | 3 | 4 | 5 |
(b) Draw both lines for 0 ≤ x ≤ 4 and read from your graph an estimate, to 1 decimal place, of the value of x where 6 − 2x = x + 1. [2] (c) Solve the equation algebraically and compare with your estimate. [2]
5. A taxi firm charges £3.20 plus £1.45 for each kilometre travelled.
(a) Write a formula for the cost, C pounds, of a journey of n kilometres. [2] (b) Work out the cost of a 12 km journey. [1] (c) A journey costs £24.95. Work out its length. [2]
6. The angles of a quadrilateral are x°, (2x + 15)°, (3x − 25)° and (4x + 10)°. Work out the size of the largest angle. [4]
7. An isosceles triangle has two equal sides of length (3x − 1) cm and a base of length (x + 9) cm. Its perimeter is 49 cm. Work out the length of each side. [4]
8. A mother is 4 times as old as her son. In 6 years’ time she will be 3 times as old as him. Work out the mother’s age now. [4]
9. Printer X charges a £60 set-up fee plus 8p per copy. Printer Y has no set-up fee and charges 14p per copy.
(a) Write an equation and solve it to find the number of copies at which the two printers cost the same. [3] (b) For 750 copies, which printer is cheaper and what is its cost? [2]
10. Five maths books and three science books cost £61.25 in total. Two maths books and six science books cost £54.50 in total. Form two equations and solve them to find the price of one maths book and one science book. [6]
11. A rectangle has length (3x + 2) cm and width (x + 4) cm. A square has side length (x + 7) cm. The rectangle and the square have the same perimeter.
(a) Show that 8x + 12 = 4x + 28. [2] (b) Solve the equation. [2] (c) Which shape has the larger area, and by how much? [3]
12. A triangle has a side of length 3(x − 4) cm, and a square has a side of length (2x − 15) cm. A student claims the two lengths are equal. Form an equation, solve it, and explain what your solution tells you about the student’s claim. [4]
Answers
1. 9x − 5x = 20 + 4, so 4x = 24 [1], giving x = 6 [1] [2] Examiner insight: A line of collecting terms is worth showing. A bare answer that happens to be wrong leaves nothing to reward.
2. Expand: 10x + 15 − 2x + 8 = 47 [1]. Collect: 8x + 23 = 47, so 8x = 24 [1]. x = 3 [1] [3] Examiner insight: The usual slip is −2(x − 4) becoming −2x − 8. Write the expansion on its own line so the sign change is visible.
3. Multiply every term by 12: 4(x − 4) + 3(3x − 1) = 72 [1]. Expand: 4x − 16 + 9x − 3 = 72 [1]. 13x − 19 = 72, so 13x = 91 [1]. x = 7 [1] [4] Examiner insight: Multiplying only the fractions and not the 6 is a common loss. Substituting x = 7 gives 1 + 5 = 6, a quick check worth doing.
4. (a) y = 4, 2, 0 for x = 1, 2, 3 [1] (b) Both lines drawn correctly, crossing between x = 1 and x = 2 [1]. Estimate x ≈ 1.7 [1] (c) 6 − 1 = x + 2x, so 3x = 5 [1]. x = 5/3 ≈ 1.67, which agrees with the graph [1] [5] Examiner insight: The solution is the x-coordinate of the crossing. Writing the y-value, about 2.7, answers a different question.
5. (a) C = 3.20 + 1.45n, with the fixed charge 3.20 [1] and the rate 1.45 multiplied by n [1] (b) 3.20 + 1.45 × 12 = 3.20 + 17.40 = £20.60 [1] (c) 3.20 + 1.45n = 24.95, so 1.45n = 21.75 [1]. n = 15 km [1] [5] Examiner insight: Define your letters. A formula such as C = 3.20 + 1.45n is easy to follow when it follows “C is the cost in pounds, n is the distance in km”.
6. The angle sum of a quadrilateral is 360°, so x + (2x + 15) + (3x − 25) + (4x + 10) = 360 [1]. 10x = 360, so x = 36 [1]. Largest angle is 4x + 10 = 4(36) + 10 = 154° [1]. Check: 36 + 87 + 83 + 154 = 360 [1] [4] Examiner insight: Stopping at x = 36 answers the equation but not the question. The largest angle is the one the question asks for.
7. Perimeter: 2(3x − 1) + (x + 9) = 49 [1]. 6x − 2 + x + 9 = 49, so 7x + 7 = 49 [1]. x = 6 [1]. Sides are 3(6) − 1 = 17 cm, 17 cm and 6 + 9 = 15 cm [1] [4] Examiner insight: Counting the equal side twice is the key step. Adding the three expressions only once is a common slip.
8. Let s be the son’s age now, so the mother is 4s. In 6 years: mother 4s + 6 [1], son s + 6 [1]. Equation: 4s + 6 = 3(s + 6), so 4s + 6 = 3s + 18 and s = 12 [1]. Mother is 4 × 12 = 48 years [1] [4] Examiner insight: Add 6 to both ages before multiplying. Using 4s + 6 = 3s + 6 forgets that the son also gets older.
9. (a) Costs in pence for n copies are 6000 + 8n and 14n [1]. Equate: 6000 = 6n [1]. n = 1000 copies [1] (b) Printer X: 60 + 0.08 × 750 = £120 [1]. Printer Y: 0.14 × 750 = £105, so Printer Y is cheaper at £105 [1] [5] Examiner insight: Keep to one unit. Mixing pounds and pence in the same equation is the main danger here.
10. Let m be the price of a maths book and s of a science book, in pounds. 5m + 3s = 61.25 [1] and 2m + 6s = 54.50 [1]. Multiply the first by 2: 10m + 6s = 122.50. Subtract the second: 8m = 68 [1]. m = 8.50 [1]. 3s = 61.25 − 42.50 = 18.75, so s = 6.25 [1]. Check: 2(8.50) + 6(6.25) = 17 + 37.50 = 54.50. A maths book costs £8.50 and a science book costs £6.25 [1] [6] Examiner insight: Test your pair of prices in the equation that was not used to find the second price. An arithmetic slip often shows up at that point.
11. (a) Rectangle perimeter: 2[(3x + 2) + (x + 4)] = 2(4x + 6) = 8x + 12 [1]. Square perimeter: 4(x + 7) = 4x + 28. Equal perimeters give 8x + 12 = 4x + 28 [1] (b) 4x = 16 [1], so x = 4 [1] (c) Rectangle: 14 × 8 = 112 cm² [1]. Square: 11² = 121 cm² [1]. The square is larger by 9 cm² [1] [7] Examiner insight: A “show that” works best when you write each expansion line. Do not start from the answer; derive it.
12. The student’s claim gives 3(x − 4) = 2x − 15 [1]. 3x − 12 = 2x − 15 [1], so x = −3 [1]. Then each length is 3(−7) = −21 and 2(−3) − 15 = −21. A length cannot be negative, so no such triangle and square exist and the claim is false [1] [4] Examiner insight: Getting x = −3 is not the end. Putting it back in the lengths shows the impossible situation, and a sentence of explanation completes the answer.
Where marks are usually lost
- Dropping a minus sign when expanding a bracket that has a negative multiplier.
- Clearing fractions but leaving a whole-number term unmultiplied.
- Giving the y-coordinate of an intersection when the equation’s solution is the x-coordinate.
- Giving a graph estimate to more accuracy than the graph supports.
- Leaving x as the final answer when the question asks for an angle, a length, an age or a cost.
- Mixing units, such as pounds and pence, in one equation.
- Forgetting to define the letters used.
- Skipping the check that a solution is sensible (negative lengths, parts of people).
Next steps
- Revise the key results: revision notes
- Learn the methods from scratch: study guide
- Course hub: OxfordAQA IGCSE Mathematics
- Printable checklist
- Find your gaps: all free 10-minute diagnostics
- Book a free trial class
Official syllabus
OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.2.3, references A19 and A22.
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