Edexcel International GCSE Mathematics A 4MA1: Expressions, formulae and linear equations – Practice Questions

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 sections 2.3 Expressions and formulae and 2.4 Linear equations of Topic 2, Equations, formulae and identities, in the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on it. Questions and parts labelled (Higher) use the Higher tier only statement on changing the subject (subject appearing twice, or a power or root of the subject). Everything else is for both tiers; question 7 is in the style of the Higher example for 2.4, which uses the same Foundation skill with more fractions. A calculator may be used on every 4MA1 paper, but show your working: most marks are for method.

Learn the content first in the study guide and the revision notes. For index laws, expanding and factorising, use the separate algebraic manipulation practice questions. The course hub is Edexcel IGCSE Mathematics and the printable checklist lists every statement.

Questions

1. Write each of these in its simplest algebraic form.

(a) a × 7 × b [1] (b) n × n × n × 2 [1] (c) Add 2 to x, then divide the result by 3. [1]

2. Calculate the value of each expression. Show your substitution.

(a) 5a − 2b when a = −3 and b = −4.5 [2] (b) (p² − q)/(2p) when p = 1/2 and q = −3/4 [2]

3. A phone contract costs £12 per month, plus £0.05 for each text sent.

(a) Write a formula for the total cost, £C, of m months on the contract with t texts sent in total. [2] (b) Calculate C when m = 6 and t = 340. [2] (c) Over 12 months the total cost was £163. Calculate the number of texts sent. [2]

4. A rectangular lawn is x metres long and y metres wide. A path 1 metre wide runs all the way round the outside of the lawn.

(a) Derive a formula, in its simplest form, for the area A m² of the path. [3] (b) Calculate the area of the path when x = 8.5 and y = 6. [2]

5. Rearrange each formula to make the letter in brackets the subject.

(a) s = (u + v)t/2 [u] [2] (b) y = 5 − 3x/4 [x] [2] (c) V = πr²h [r], where r > 0 [2]

6. Solve these equations. Show clear algebraic working.

(a) 7x − 4 = 3x + 18 [2] (b) 3(2y + 5) = 4 − 2(y − 1) [3] (c) (5w − 2)/3 = 2w + 1 [2]

7. Solve (x + 3)/4 − (2x − 5)/6 = 1/3. Show clear algebraic working. [4]

8. The angles of a quadrilateral are (2a + 15)°, (3a − 10)°, (a + 40)° and (4a − 5)°.

(a) Form an equation in a and solve it. [3] (b) Find the size of the largest angle. [2]

9. (Higher)

(a) Make x the subject of y = (3x + 2)/(x − 4). [4] (b) Use your answer to find x when y = 5. [1]

10. (Higher)

(a) Make a the subject of k = √((a + 2)/(a − 1)). [4] (b) Find a when k = 2. [1]

11. At a cinema an adult ticket costs £a and a child ticket costs £4 less than an adult ticket. Two adult tickets and three child tickets cost £53 altogether.

(a) Show that 5a − 12 = 53. [2] (b) Find the cost of an adult ticket and of a child ticket. [2] (c) Write a formula for the total cost, £C, of x adult tickets and y child tickets, in terms of a, x and y. [2] (d) (Higher) Make a the subject of your formula. [3] (e) (Higher) At another cinema with the same pricing rule, 4 adults and 5 children pay £88. Use your answer to (d) to find the adult price there. [1]

12. Temperatures in degrees Fahrenheit, F, and degrees Celsius, C, are linked by F = 9C/5 + 32.

(a) Calculate F when C = −12.5. [2] (b) Make C the subject of the formula. [2] (c) Find the temperature at which the Celsius and Fahrenheit readings are equal. [3]

Answers

1. (a) 7ab [1] (b) 2n³ [1] (c) (x + 2)/3 [1] Examiner insight: In (c) writing x + 2/3 loses the mark, because without a bracket or a full fraction line only the 2 is divided by 3.

2. (a) 5(−3) − 2(−4.5) [1] = −15 + 9 = −6 [1] (b) Numerator (1/2)² − (−3/4) = 1/4 + 3/4 = 1 and denominator 2(1/2) = 1 [1]; value 1 [1] Examiner insight: The method mark is for a correct substitution shown; a wrong final answer with no substitution written down scores zero.

3. (a) 12m for the monthly charge and 0.05t for the texts [1]; C = 12m + 0.05t [1] (b) C = 12(6) + 0.05(340) = 72 + 17 [1] = 89 [1] (c) 163 = 12(12) + 0.05t, so 0.05t = 19 [1]; t = 19 ÷ 0.05 = 380 texts [1] Examiner insight: A formula must include the “C =” part; writing only 12m + 0.05t is an expression and usually loses the accuracy mark.

4. (a) Lawn plus path is (x + 2) by (y + 2) [1]; A = (x + 2)(y + 2) − xy [1] = 2x + 2y + 4 [1] (b) A = 2(8.5) + 2(6) + 4 [1] = 33 m² [1] Examiner insight: Using x + 1 and y + 1 for the outer rectangle is the usual error; the path adds 1 m on each side, so 2 m to each dimension.

5. (a) 2s = (u + v)t, so u + v = 2s/t [1]; u = 2s/t − v [1] (b) 3x/4 = 5 − y [1]; x = 4(5 − y)/3 (or (20 − 4y)/3) [1] (c) r² = V/(πh) [1]; r = √(V/(πh)) [1] Examiner insight: Any correct equivalent form scores full marks, but the final answer must have the new subject alone on one side with no copy of it on the other.

6. (a) 7x − 3x = 18 + 4, so 4x = 22 [1]; x = 5.5 [1] (b) 6y + 15 = 4 − 2y + 2 [1]; 8y = −9 [1]; y = −9/8 = −1.125 [1] (c) 5w − 2 = 3(2w + 1) = 6w + 3 [1]; w = −5 [1] Examiner insight: “Show clear algebraic working” means an answer found by trial and improvement, even if correct, may score no marks.

7. Multiply every term by 12: 3(x + 3) − 2(2x − 5) = 4 [1]; 3x + 9 − 4x + 10 = 4 [1]; −x = −15 [1]; x = 15 [1] Examiner insight: The second method mark depends on −2 × −5 = +10; writing −10 is the usual slip, and follow-through then applies only to the remaining steps.

8. (a) (2a + 15) + (3a − 10) + (a + 40) + (4a − 5) = 360 [1]; 10a + 40 = 360 [1]; a = 32 [1] (b) Angles 79°, 86°, 72°, 123° [1]; largest 123° [1] Examiner insight: Giving a = 32 as the answer to (b) loses the mark; check the four angles add to 360° before choosing the largest.

9. (a) y(x − 4) = 3x + 2 [1]; xy − 4y = 3x + 2 [1]; xy − 3x = 4y + 2, so x(y − 3) = 4y + 2 [1]; x = (4y + 2)/(y − 3) [1] (b) x = (20 + 2)/(5 − 3) = 11 [1] Examiner insight: The factorising step x(y − 3) is the key mark; an answer that still has x on the right-hand side scores only the early method marks.

10. (a) k² = (a + 2)/(a − 1) [1]; k²a − k² = a + 2 [1]; k²a − a = k² + 2, so a(k² − 1) = k² + 2 [1]; a = (k² + 2)/(k² − 1) [1] (b) a = (4 + 2)/(4 − 1) = 2 [1] Examiner insight: Square both sides before doing anything else; taking the square root of individual terms such as √a earns no credit.

11. (a) Child ticket costs a − 4, so 2a + 3(a − 4) = 53 [1]; 2a + 3a − 12 = 53, giving 5a − 12 = 53 [1] (b) 5a = 65, a = 13 [1]; adult £13, child £9 [1] (c) ax for the adults and (a − 4)y for the children [1]; C = ax + (a − 4)y [1] (d) C = ax + ay − 4y [1]; C + 4y = a(x + y) [1]; a = (C + 4y)/(x + y) [1] (e) a = (88 + 20)/(4 + 5) = 108/9 = £12 [1] Examiner insight: On a “show that” the printed result earns nothing by itself; the line with the brackets, 2a + 3(a − 4), must appear.

12. (a) F = 9(−12.5)/5 + 32 = −22.5 + 32 [1] = 9.5 [1] (b) F − 32 = 9C/5 [1]; C = 5(F − 32)/9 [1] (c) Set F = C: C = 9C/5 + 32 [1]; 5C = 9C + 160, so −4C = 160 [1]; C = −40, so −40 °C = −40 °F [1] Examiner insight: In (c) the first mark is for replacing F by C (or C by F) to form one linear equation in one unknown, so write that equation down before solving.

Where marks are usually lost

• Substituting a negative value without brackets, so −4.5 is subtracted instead of −2 × −4.5 being added.
• Writing an expression such as 12m + 0.05t when a formula “C = …” is asked for.
• Adding the path width once instead of twice to each side of a rectangle.
• Dividing only the last term when rearranging: u = 2s/t − v is right, u = 2s − v/t is not.
• Expanding −2(y − 1) as −2y − 2 instead of −2y + 2.
• Multiplying only one term by the LCM when clearing fractions.
• Higher: forgetting to factorise, so the subject is left on both sides.
• Higher: taking a square root of separate terms instead of squaring both sides first.
• Giving the value of the letter instead of the angles, prices or temperature asked for.
• Rounding an exact answer such as −9/8 when no rounding is requested.

Next steps

• Refresh the methods: revision notes
• Re-learn anything you missed: study guide
• Earlier algebra skills: use of symbols and algebraic manipulation study guide
• Course hub: Edexcel IGCSE Mathematics
• Track coverage: printable checklist
• Check other topics: all free 10-minute diagnostics
• Want help from a teacher? Book a free trial class

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 2, Equations, formulae and identities: section 2.3 Expressions and formulae and section 2.4 Linear equations (Foundation and Higher tier content).