Algebra: Exam Practice

This lesson pulls the whole Algebra strand of OCR GCSE Mathematics (J560) together with exam-style questions and full worked solutions. The questions range across both tiers: Foundation-level items (grades $1$1–$5$5) and Higher-only items (grades $4$4–$9$9, marked [Higher]). For each question the marks and tier are given, and the solution explains where the method and accuracy marks fall. Work each one before reading the solution, then map your performance to the grade-band guidance. The skills assessed span AO1 (technique), AO2 (reasoning and communication) and AO3 (problem solving).

How Marks Are Awarded

It is worth understanding how an OCR mark scheme is built before you attempt the questions. Marks fall into three broad types. Method marks (M) are given for using a correct process — for example, expanding both brackets, or substituting correctly into the quadratic formula — even if a later arithmetic slip spoils the final number. Accuracy marks (A) are given for a correct value, but usually only if the method mark has already been earned. Communication or reasoning marks reward a clearly stated reason, a correct unit, or a properly drawn number line. The single most important consequence is this: show every step. A bare answer with no working risks scoring zero even when it is correct, whereas full working can pick up most of the marks even when the final answer is wrong. As you read each solution below, notice exactly where each mark is described.

Exam-Style Questions

Question 1 — Simplifying (2 marks, Foundation/Higher, grades 2–3)

Simplify $5a + 3b - 2a + 4b$5a+3b−2a+4b.

Solution: Collect like terms: $5a - 2a = 3a$5a−2a=3a and $3b + 4b = 7b$3b+4b=7b. The answer is $3a + 7b$3a+7b. One method mark is for collecting either set of like terms correctly; the accuracy mark is for the complete, correct expression. A frequent error here is to combine the $a$a-terms and $b$b-terms into a single term such as $7ab$7ab — but $a$a and $b$b are different labels and cannot be merged, so the two terms must stay separate.

Question 2 — Expanding and simplifying (3 marks, Foundation/Higher, grades 3–4)

Expand and simplify $(x + 6)(x - 2)$(x+6)(x−2).

Solution: Form all four products: $x^{2} - 2x + 6x - 12$x2−2x+6x−12. Collect the middle terms: $x^{2} + 4x - 12$x2+4x−12. A method mark is awarded for at least three correct terms before simplifying, another for collecting, and the accuracy mark for the final expression.

Question 3 — Solving a linear equation (3 marks, Foundation/Higher, grades 3–4)

Solve $4(x - 1) = 2x + 10$4(x−1)=2x+10.

Solution: Expand: $4x - 4 = 2x + 10$4x−4=2x+10. Subtract $2x$2x: $2x - 4 = 10$2x−4=10. Add $4$4: $2x = 14$2x=14. Divide by $2$2: $x = 7$x=7. Method marks are for expanding and for collecting terms; the accuracy mark is for $x = 7$x=7. Check: $4(6) = 24$4(6)=24 and $2(7) + 10 = 24$2(7)+10=24 ✓.

Question 4 — Forming and solving (4 marks, Foundation/Higher, grades 4–5)

The angles of a quadrilateral are $x$x, $2x$2x, $(x + 30)$(x+30) and $(x + 50)$(x+50) degrees. Work out the largest angle. Give a reason for the equation you write.

Solution: Angles in a quadrilateral sum to $360^{\circ}$360∘ (this is the reason mark), so $x + 2x + (x + 30) + (x + 50) = 360$x+2x+(x+30)+(x+50)=360. Then $5x + 80 = 360$5x+80=360, so $5x = 280$5x=280 and $x = 56$x=56. The four angles are $56^{\circ}$56∘, $112^{\circ}$112∘, $86^{\circ}$86∘ and $106^{\circ}$106∘, so the largest is $112^{\circ}$112∘. Marks: forming the equation, the reason, solving, and identifying the largest angle.

Question 5 — Sequences (3 marks, Foundation/Higher, grades 3–4)

Find the $n$nth term of the sequence $6, 10, 14, 18, \dots$6,10,14,18,… and use it to find the $25$25th term.

Solution: The terms go up by $4$4 each time, so the common difference is $4$4 and the $n$nth term starts $4n$4n. Checking at $n = 1$n=1, $4(1) = 4$4(1)=4, but the first term is actually $6$6, so we add $2$2: the $n$nth term is $4n + 2$4n+2. To find the $25$25th term, substitute $n = 25$n=25: $4(25) + 2 = 102$4(25)+2=102. The method mark is for $4n$4n, the accuracy mark for the complete formula $4n + 2$4n+2, and the final mark for $102$102. Always check the constant by substituting $n = 1$n=1, as it is easy to add the wrong number.

Question 6 — Solving a quadratic by factorising (3 marks, Foundation/Higher, grades 4–5)

Solve $x^{2} + 2x - 15 = 0$x2+2x−15=0.

Solution: Two numbers multiplying to $-15$−15 and adding to $2$2 are $5$5 and $-3$−3, so $(x + 5)(x - 3) = 0$(x+5)(x−3)=0. Therefore $x = -5$x=−5 or $x = 3$x=3. The method mark is for the correct factorisation; the accuracy marks are for both roots. A quick check confirms the answer: $(-5)^{2} + 2(-5) - 15 = 25 - 10 - 15 = 0$(−5)2+2(−5)−15=25−10−15=0 ✓, and $3^{2} + 2(3) - 15 = 9 + 6 - 15 = 0$32+2(3)−15=9+6−15=0 ✓. Losing one root, or reading the signs from the brackets the wrong way round, are the two most common ways to drop marks here.

Question 7 — Simultaneous equations (4 marks, Foundation/Higher, grades 5–6)

Solve $3x + 2y = 16$3x+2y=16 and $5x - 2y = 8$5x−2y=8.

Solution: The $y$y-terms are $+2y$+2y and $-2y$−2y (opposite signs), so add the equations: $(3x + 5x) + (2y - 2y) = 16 + 8$(3x+5x)+(2y−2y)=16+8, which gives $8x = 24$8x=24 and $x = 3$x=3. Substitute into $3x + 2y = 16$3x+2y=16: $9 + 2y = 16$9+2y=16, so $2y = 7$2y=7 and $y = 3.5$y=3.5. Marks: deciding to add (the key decision, based on the opposite signs), finding $x$x, substituting back, and finding $y$y. Check in the second equation: $5(3) - 2(3.5) = 15 - 7 = 8$5(3)−2(3.5)=15−7=8 ✓. Note that the $y$y-terms cancelled because their signs were opposite; had both been $+2y$+2y you would have subtracted instead.

Question 8 — Inequalities (3 marks, Foundation/Higher, grades 4–5)

Solve $5 - 2x \ge -3$5−2x≥−3 and list the positive integer solutions.

Solution: Subtract $5$5 from both sides: $-2x \ge -8$−2x≥−8. Now divide by $-2$−2 and reverse the inequality sign because we are dividing by a negative: $x \le 4$x≤4. The positive integers satisfying $x \le 4$x≤4 are $1, 2, 3, 4$1,2,3,4. The method mark is for reaching $-2x \ge -8$−2x≥−8, the accuracy mark for correctly reversing to $x \le 4$x≤4, and the final mark for the correct integer list. The reversal is the most-tested point here, and the word "positive" means $0$0 is excluded from the list.

Question 9 — Quadratic formula [Higher] (3 marks, Higher, grades 6–7)

[Higher] Solve $2x^{2} + 3x - 4 = 0$2x2+3x−4=0, giving your answers to $2$2 decimal places.

Solution: First write down the coefficients with their signs: $a = 2$a=2, $b = 3$b=3, $c = -4$c=−4. Compute the discriminant: $b^{2} - 4ac = 9 - 4(2)(-4) = 9 + 32 = 41$b2−4ac=9−4(2)(−4)=9+32=41. Substitute into the formula: $x = \dfrac{-3 \pm \sqrt{41}}{4}$x=4−3±41​​. Since $\sqrt{41} \approx 6.403$41​≈6.403, the two roots are $x \approx \dfrac{-3 + 6.403}{4} \approx 0.85$x≈4−3+6.403​≈0.85 and $x \approx \dfrac{-3 - 6.403}{4} \approx -2.35$x≈4−3−6.403​≈−2.35. Marks: correct substitution into the formula, correct value of the discriminant, and both answers rounded to $2$2 decimal places. The instruction "$2$2 decimal places" is a strong hint that the formula, not factorising, is the intended method here.

Question 10 — Composite functions [Higher] (3 marks, Higher, grades 6–7)

[Higher] Given $f(x) = 3x + 1$f(x)=3x+1 and $g(x) = x - 4$g(x)=x−4, work out $fg(x)$fg(x) and hence $fg(6)$fg(6).

Solution: Do $g$g first: $g(x) = x - 4$g(x)=x−4. Then $f(x - 4) = 3(x - 4) + 1 = 3x - 12 + 1 = 3x - 11$f(x−4)=3(x−4)+1=3x−12+1=3x−11. So $fg(x) = 3x - 11$fg(x)=3x−11, and $fg(6) = 3(6) - 11 = 7$fg(6)=3(6)−11=7. Marks: applying $g$g first, simplifying $fg(x)$fg(x), and evaluating at $6$6.

Question 11 — Straight-line graph (4 marks, Foundation/Higher, grades 5–6)

A straight line passes through the points $(1, 4)$(1,4) and $(4, 13)$(4,13). Find the equation of the line in the form $y = mx + c$y=mx+c.