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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 1–5) and Higher-only items (grades 4–9, 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).
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.
Simplify 5a+3b−2a+4b.
Solution: Collect like terms: 5a−2a=3a and 3b+4b=7b. The answer is 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-terms and b-terms into a single term such as 7ab — but a and b are different labels and cannot be merged, so the two terms must stay separate.
Expand and simplify (x+6)(x−2).
Solution: Form all four products: x2−2x+6x−12. Collect the middle terms: 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.
Solve 4(x−1)=2x+10.
Solution: Expand: 4x−4=2x+10. Subtract 2x: 2x−4=10. Add 4: 2x=14. Divide by 2: x=7. Method marks are for expanding and for collecting terms; the accuracy mark is for x=7. Check: 4(6)=24 and 2(7)+10=24 ✓.
The angles of a quadrilateral are x, 2x, (x+30) and (x+50) degrees. Work out the largest angle. Give a reason for the equation you write.
Solution: Angles in a quadrilateral sum to 360∘ (this is the reason mark), so x+2x+(x+30)+(x+50)=360. Then 5x+80=360, so 5x=280 and x=56. The four angles are 56∘, 112∘, 86∘ and 106∘, so the largest is 112∘. Marks: forming the equation, the reason, solving, and identifying the largest angle.
Find the nth term of the sequence 6,10,14,18,… and use it to find the 25th term.
Solution: The terms go up by 4 each time, so the common difference is 4 and the nth term starts 4n. Checking at n=1, 4(1)=4, but the first term is actually 6, so we add 2: the nth term is 4n+2. To find the 25th term, substitute n=25: 4(25)+2=102. The method mark is for 4n, the accuracy mark for the complete formula 4n+2, and the final mark for 102. Always check the constant by substituting n=1, as it is easy to add the wrong number.
Solve x2+2x−15=0.
Solution: Two numbers multiplying to −15 and adding to 2 are 5 and −3, so (x+5)(x−3)=0. Therefore x=−5 or 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 ✓, and 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.
Solve 3x+2y=16 and 5x−2y=8.
Solution: The y-terms are +2y and −2y (opposite signs), so add the equations: (3x+5x)+(2y−2y)=16+8, which gives 8x=24 and x=3. Substitute into 3x+2y=16: 9+2y=16, so 2y=7 and y=3.5. Marks: deciding to add (the key decision, based on the opposite signs), finding x, substituting back, and finding y. Check in the second equation: 5(3)−2(3.5)=15−7=8 ✓. Note that the y-terms cancelled because their signs were opposite; had both been +2y you would have subtracted instead.
Solve 5−2x≥−3 and list the positive integer solutions.
Solution: Subtract 5 from both sides: −2x≥−8. Now divide by −2 and reverse the inequality sign because we are dividing by a negative: x≤4. The positive integers satisfying x≤4 are 1,2,3,4. The method mark is for reaching −2x≥−8, the accuracy mark for correctly reversing to 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 is excluded from the list.
[Higher] Solve 2x2+3x−4=0, giving your answers to 2 decimal places.
Solution: First write down the coefficients with their signs: a=2, b=3, c=−4. Compute the discriminant: b2−4ac=9−4(2)(−4)=9+32=41. Substitute into the formula: x=4−3±41. Since 41≈6.403, the two roots are x≈4−3+6.403≈0.85 and x≈4−3−6.403≈−2.35. Marks: correct substitution into the formula, correct value of the discriminant, and both answers rounded to 2 decimal places. The instruction "2 decimal places" is a strong hint that the formula, not factorising, is the intended method here.
[Higher] Given f(x)=3x+1 and g(x)=x−4, work out fg(x) and hence fg(6).
Solution: Do g first: g(x)=x−4. Then f(x−4)=3(x−4)+1=3x−12+1=3x−11. So fg(x)=3x−11, and fg(6)=3(6)−11=7. Marks: applying g first, simplifying fg(x), and evaluating at 6.
A straight line passes through the points (1,4) and (4,13). Find the equation of the line in the form y=mx+c.
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