OCR GCSE Mathematics: Algebraic formulae, rearranging and kinematics (J560) – Revision Notes

Full explanations and worked examples are in the study guide. These notes condense OCR GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), section 6.02 (Algebraic formulae, 6.02a to 6.02e) for the final weeks. Four formulae are Higher tier only and are marked. The specification says no formula sheet is provided: all formulae you must know are listed under 6.02d.

Use these notes with the practice questions, the OCR GCSE Mathematics hub, the printable checklist, the exam preparation page and the free 10-minute diagnostics.

Formulae you must know

Ref	Formula	Tier
6.02d	Circumference C = 2πr = πd	Both tiers
6.02d	Area of a circle A = πr²	Both tiers
6.02d	Pythagoras a² + b² = c²	Both tiers
6.02d	sin θ = o/h, cos θ = a/h, tan θ = o/a	Both tiers
6.02d	Quadratic formula x = (−b ± √(b² − 4ac))/(2a)	Higher tier only
6.02d	Sine rule a/sin A = b/sin B = c/sin C	Higher tier only
6.02d	Cosine rule a² = b² + c² − 2bc cos A	Higher tier only
6.02d	Area of a triangle = ½ab sin C	Higher tier only
6.02e	v = u + at	Both tiers
6.02e	s = ut + ½at²	Both tiers
6.02e	v² = u² + 2as	Both tiers

For 6.02e, a is constant acceleration, u initial velocity, v final velocity, s displacement from the position at t = 0, and t time taken. The specification says you do not need to know the definition of each letter, so a question will define them.

6.02a Forming formulae

Method
1. Give each quantity a letter.
2. Fixed amount: a number on its own.
3. Rate: number × letter.
4. Add the parts. Brackets if a sum is multiplied.
5. Test with one easy value.

Reminder: a hire firm charges £18 plus £0.45 per mile. For m miles, C = 18 + 0.45m. Test m = 20: 18 + 9 = 27, which is sensible.

Reminder: a frame is x cm wide and (x + 9) cm long. Its perimeter is P = 2(x + x + 9) = 4x + 18.

6.02b Substituting

Method
1. Write the formula, then rewrite it with each value in brackets.
2. Brackets, then powers and roots, then × and ÷, then + and −.
3. Algebraic fraction: work out top and bottom separately.
4. Root sign: finish everything under it first.

Reminder: 3a² − b with a = −2, b = −5.

3(−2)² − (−5) = 3(4) + 5 = 17

Distinction: (−2)² = 4, but −2² = −4. Brackets decide.

6.02c Changing the subject

Method
1. Aim: get the new subject alone.
2. Remove fractions by multiplying both sides.
3. Expand brackets.
4. Collect every term with the new subject on one side.
5. Factorise out the subject (if it appears twice).
6. Divide, or undo the power or root.

Situation	Move
Subject once	Undo operations in reverse order
Subject twice	Collect, factorise, divide
Power of subject	Isolate the power, then take the root
Root of subject	Isolate the root, then square both sides
Reciprocal of subject	Combine into one fraction, then invert

Reminder (power): m = k/n². Then mn² = k, so n² = k/m and n = √(k/m).

Reminder (twice): ab = 3a + 5. Then ab − 3a = 5, a(b − 3) = 5, so a = 5/(b − 3).

Reminder (reciprocal): 1/f = 1/u + 1/v gives v = fu/(u − f).

Reminder (root): x = 3√(y/z) gives y = x²z/9. Square the whole left side, including the 3.

6.02d Standard formulae in use

Circle. For a circle with diameter 10 cm, C = 10π cm and A = 25π cm². Leave π in the answer when asked for exact form.

Right-angled triangles.

Find hypotenuse:   add the squares, then √
Find shorter side: subtract the squares, then √
Find a side with an angle:   choose sin, cos or tan from the sides involved
Find an angle:   use the inverse function

SOH-CAH-TOA: label the sides first (hypotenuse opposite the right angle, opposite the angle, adjacent to it).

Higher tier only: quadratic formula. Write a, b, c with their signs, work out b² − 4ac, then take both ± values. If a negative b is squared, the result is positive.

Higher tier only: choosing a rule for a non-right-angled triangle.

You know	Use
A side and its opposite angle, plus one more side or angle	Sine rule
Two sides and the angle between them	Cosine rule for the third side
All three sides	Cosine rule for an angle
Two sides and the angle between them (area)	½ab sin C

Reminder: for sides 5 and 8 with 40° between them, a² = 25 + 64 − 80 cos 40° = 27.72, so a = 5.26. The area is ½ × 5 × 8 × sin 40° = 12.9.

If the cosine of an angle is negative, the angle is obtuse.

6.02e Kinematics

Quantity wanted	Quantities given	Formula
v	u, a, t	v = u + at
s	u, a, t	s = ut + ½at²
v, u or a	the other two of u, v, a plus s	v² = u² + 2as

Method
1. List u, v, a, s, t. Mark the unknown.
2. Hidden values: "starts from rest" means u = 0; "stops" means v = 0.
3. Choose the formula with exactly one unknown.
4. Keep signs consistent (upwards positive means a = −9.8 for gravity).

Reminder: from rest with a = 2.5 for 6 s, v = 0 + 2.5 × 6 = 15 and s = ½ × 2.5 × 36 = 45.

Higher tier only: if the unknown is t in s = ut + ½at², you get a quadratic in t. Reject a negative time.

Checking your answer

Each topic has a quick test that takes under a minute.

• Formed formula: put in a value you can work out in your head.
• Rearranged formula: substitute a simple pair of numbers into both versions and see that they agree. For v = fu/(u − f), try f = 2 and u = 3: 1/2 = 1/3 + 1/v gives 1/v = 1/6, so v = 6, and 2 × 3/(3 − 2) = 6.
• Right-angled triangle: the hypotenuse must be the longest side.
• Cosine rule or sine rule (Higher tier only): the longest side must be opposite the largest angle.
• Kinematics: if a is negative and the object is slowing, v must be smaller than u.
• Units: lengths in cm or m, areas in cm² or m², time in s, velocity in m/s and acceleration in m/s².

Must-know distinctions

• Expression, formula and equation: an expression has no equals sign, a formula links quantities, an equation is solved for an unknown.
• Substituting (find a value) versus rearranging (change the layout with no numbers).
• Find a hypotenuse (add) versus find a shorter side (subtract).
• Sine rule (pairs of side and opposite angle) versus cosine rule (two sides and the angle between).
• Circumference (length, cm) versus area (cm²).

Quick self-test

1. A taxi costs £4 plus 90p per kilometre. Write a formula for the cost £C of k km and find the cost for 12 km.
2. Find 3a² − b when a = −2 and b = −5.
3. Make x the subject of y = 6x − 11.
4. Make r the subject of V = ⅓πr²h.
5. Make p the subject of 3p + 2 = q(p − 4).
6. A circle has diameter 10 cm. Give the circumference and area in terms of π.
7. A right-angled triangle has shorter sides 8 cm and 15 cm. Find the hypotenuse.
8. In a right-angled triangle the adjacent side is 6 cm and the hypotenuse is 10 cm. Find the angle between them to 1 decimal place.
9. Using v = u + at, find a when u = 8, v = 26 and t = 6.
10. A particle has u = 5, a = 2 and s = 18. Use v² = u² + 2as to find v to 2 decimal places.
11. (Higher tier only) Solve x² + 6x + 2 = 0 to 2 decimal places.

Answers

1. C = 4 + 0.9k, so C = 4 + 10.8 = £14.80.
2. 3(4) + 5 = 17.
3. x = (y + 11)/6.
4. r = √(3V/(πh)).
5. 3p − qp = −4q − 2, so p(3 − q) = −4q − 2 and p = (4q + 2)/(q − 3).
6. C = 10π cm, A = 25π cm².
7. √(64 + 225) = √289 = 17 cm.
8. cos θ = 6/10 = 0.6, so θ = 53.1°.
9. 26 = 8 + 6a, so a = 3.
10. v² = 25 + 72 = 97, so v = 9.85.
11. x = (−6 ± √(36 − 8))/2 = (−6 ± √28)/2 = −0.35 or −5.65.

Where marks are usually lost

• Dropping brackets when substituting a negative number.
• Squaring only part of an expression when undoing a square root.
• Leaving the new subject on both sides after rearranging.
• Dividing only one term in a numerator.
• Adding instead of subtracting when finding a shorter side.
• Rounding a middle value too early, then losing accuracy in the final answer.
• Giving a cosine rule angle without checking whether it should be obtuse.
• Giving only one quadratic root when two are needed.
• Choosing the wrong kinematics formula because a hidden value such as u = 0 was missed.
• Writing a final answer with no units, or the wrong ones (cm instead of cm²).

Official syllabus

OCR GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017, published by OCR. Section 6.02, Algebraic formulae.