A5: Rearranging Formulae
Use standard mathematical formulae; rearrange formulae to change the subject
Use standard mathematical formulae; rearrange formulae to change the subject
Rearrange v = u + at to make t the subject.
Solution:
Step 1: Subtract u from both sides: v - u = at
Step 2: Divide both sides by a: v - u⁄a = t
Answer: t = v - u⁄a
Rearrange A = πr² to make r the subject.
Solution:
Step 1: Divide both sides by π: A⁄π = r²
Step 2: Take the square root: √(A⁄π) = r
Answer: r = √(A⁄π)
Rearrange A = lb to make b the subject.
Solution:
Divide both sides by l: A⁄l = b
Answer: b = A⁄l
Rearrange s = d⁄t to make d the subject.
Solution:
Multiply both sides by t: st = d
Answer: d = st
Rearrange y = mx + c to make x the subject.
Solution:
Step 1: Subtract c from both sides: y - c = mx
Step 2: Divide both sides by m: y - c⁄m = x
Answer: x = y - c⁄m
Rearrange A = ½bh to make h the subject.
Solution:
Step 1: Multiply both sides by 2: 2A = bh
Step 2: Divide both sides by b: 2A⁄b = h
Answer: h = 2A⁄b
Rearrange E = ½mv² to make v the subject.
Solution:
Step 1: Multiply both sides by 2: 2E = mv²
Step 2: Divide both sides by m: 2E⁄m = v²
Step 3: Take square root: v = √(2E⁄m)
Rearrange ax + b = cx + d to make x the subject.
Solution:
Step 1: Subtract cx from both sides: ax - cx + b = d
Step 2: Subtract b from both sides: ax - cx = d - b
Step 3: Factorise left side: x(a - c) = d - b
Step 4: Divide by (a - c): x = d - b⁄a - c
Rearranging = solving an equation where the 'answer' is a letter. Reverse the operations acting on the target letter, one layer at a time.
Rearrange T = 5c + 8 to make c the subject.
Subtract 8: T − 8 = 5c → Divide by 5: c = (T − 8)/5
M1 for subtracting 8 first, A1 for the division. Undo in reverse order: +8 was LAST, so it comes off FIRST.
Rearrange V = πr²h to make r the subject.
÷ πh: V/(πh) = r² → Square root: r = √(V/(πh))
M1 for dividing by πh, A1 for the root. The root undoes the SQUARE — last-in-first-out again.
Exam wording to expect: "Make … the subject", "Rearrange … to give an equation for …".
Q1: Rearrange a = 3b + 2 to make b the subject.
Q2: Rearrange F = ma to make a the subject.
Q3: Rearrange C = 2πr to make r the subject.
Q4: Rearrange v² = u² + 2as to make s the subject.
Q5: Rearrange P = 2(l + w) to make w the subject.
Q6: Rearrange ax - 3 = bx + 5 to make x the subject.
Q7: Rearrange y = 4(x − 3) to make x the subject
Q8: Rearrange A = 2(lw + wh) to make w the subject
The formula for converting Celsius to Fahrenheit is F = 1.8C + 32. Rearrange to make C the subject, then convert 95°F to Celsius.
Subtract 32: F − 32 = 1.8C → Divide by 1.8: C = (F − 32)/1.8
95°F: C = (95 − 32)/1.8 = 63/1.8 = 35°C
M1 for the rearrangement, A1 for C = (F − 32)/1.8, A1 for 35°C. Applied rearrangements (conversion, physics) are the standard exam context.
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. a − 2 = 3b → b = (a − 2)/3
Q2. Divide by m: a = F/m
Q3. Divide by 2π: r = C/(2π)
Q4. Subtract u², divide by 2a: s = (v² − u²)/(2a)
Q5. P/2 = l + w → w = P/2 − l
Q6. ax − bx = 8 → x(a − b) = 8 → x = 8/(a − b) — factorise when x appears twice
Q7. y/4 = x − 3 → x = y/4 + 3
Q8. A/(2l) − wh = lw → hmm, divide first: A/2 = lw + wh = w(l + h) → w = A/(2(l + h))
Rearranging: 2–3 marks on every paper — it is the gateway skill for science formulas, geometry and compound measures. Two-variable targets (Q6 style) mark the grade-5 boundary.
Links to: solving equations (A17), substitution (A2), compound measures (R1).
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
The kinetic energy formula is E = ½mv². (a) Rearrange to make v the subject. (b) Find v when E = 100 and m = 8.
Solution:
(a) Step 1: 2E = mv². Step 2: v² = 2E⁄m. Step 3: v = √(2E⁄m)
(b) v = √(200⁄8) = √25 = 5 m/s
1. Wrong: Rearranging v = u + at to make t: t = v - u/a Correct: t = v - u⁄a (the whole of v - u is divided by a)
2. Wrong: From A = πr², rearranging to r = A⁄π Correct: r² = A⁄π, so r = √(A⁄π) (don't forget the square root)
3. Wrong: Rearranging ax + b = cx + d by moving ax and d: ax - d = cx Correct: ax - cx = d - b, then factorise: x(a - c) = d - b
6 marks: The formula for the surface area of a cylinder (excluding ends) is A = 2πrh. (a) Rearrange to make h the subject. (b) Rearrange to make r the subject. (c) A can has A = 314 cm² and r = 5 cm. Find h. (Use π = 3.14)
(a) h = A⁄2πr
(b) r = A⁄2πh
(c) h = 314⁄2 × 3.14 × 5 = 314⁄31.4 = 10 cm
Mark scheme: (a) 1 mark. (b) 1 mark. (c) 2 marks for substitution, 1 mark for correct answer, 1 mark for units.
The stopping distance formula is d = v²⁄2a where d = distance, v = initial speed, a = deceleration.
(a) Rearrange to make v the subject.
(b) If a car decelerates at 8 m/s² and stops in 25 m, what was its initial speed?
(c) Explain why doubling the speed more than doubles the stopping distance.
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