The short answers
Question 1 · what to plot
Wrong axes cost more than one mark
The graph is the first method-of-analysis mark, and the marks after it depend on it. In the June 2025 /51 report, candidates who chose the wrong axes lost the validity mark and the mark for the constant as well.
Question 2 · significant figures
The sig-fig mark comes after the gradient
No Paper 5 scheme from March 2021 to June 2026 puts a significant-figure condition on the gradient or the y-intercept. The condition comes one step later, on the constant you calculate from them and on the final value.
Choosing what to plot
Question 1 The method-of-analysis marks in the planning question.Question 1 gives you a relationship and asks you to plan an experiment to test it. The method of analysis starts with one decision: what goes on each axis so that the relationship gives a straight line. Everything after that (the validity statement, the constants from the gradient and intercept) is marked against the graph you chose.
Four checks before you write the axes down
- Rearrange into y = mx + c. Both axes must hold quantities you measure. The June 2023 /52 report: plotting (R + k) on an axis “is not possible because k is unknown”.
- Known power: plot that function. If the equation has 1/h or 1/r3 in it, plot against 1/h or 1/r3.
- Unknown power: take logs. If the power is a constant you have to find (aq), plot lg of one quantity against lg of the other. The gradient gives the power.
- Name both axes in words, or draw labelled axes. The examiner reports for every series from March 2021 to November 2025 say credit is not given for writing y = mx + c under the equation. Since June 2026 the schemes add a note: drawn axes take priority. If your sketch and your sentence disagree, the sketch is what gets marked.
Do not accept logarithms.June 2026 /54 mark scheme, Question 1, method of analysis
Logs aren't a safe default. In 20 of the 42 schemes, the graph mark carries a line refusing a log graph, because the relationship was already straight in the right variables.
Predict, then check: the three June 2026 relationships
Pick the graph you'd plan to plot. The answer is the one the June 2026 mark scheme gives.
June 2026 /51 · P and q are constants
Show the scheme's answer
lg T against lg a (ln T against ln a also accepted). The power q is unknown, so only logs make the line straight. The gradient is −q, and P comes from the y-intercept: P = (1/2π)√(M/L) × 10y-intercept.
June 2026 /52 · K is a constant
Show the scheme's answer
I against 1/h. Rearranged, I = KAfE × (1/h), so the gradient is KAfE and K = gradient / (AfE). The line passes through the origin. Here the scheme also accepted lg I against lg h, with a gradient of −1.
June 2026 /54 · α and Z are constants, v is kept constant
Show the scheme's answer
1/c2 against 1/r3. Divide through by v2: 1/c2 = [α / (πρv2)] × (1/r3) + Z/v2. So α = πρv2 × gradient and Z = y-intercept × v2. The powers are known, and the scheme says “Do not accept logarithms.” A log graph also fails here because the + Z term stops lg c against lg r being a straight line.
From the examiner reports
Wrong axes the reports named
| Paper | Plotted instead | What the report says |
|---|---|---|
| June 2021 /51 | t against R | Correct axes were often t against 1/R. |
| June 2022 /51 | V against d, or V against 1/d | Correct axes were often 1/V against d. |
| June 2023 /51 | t2 against 1/sin θ | Named as an incorrect suggestion. |
| June 2023 /52 | E against 1/R; (R + k) on an axis | (R + k) can't be plotted because k is unknown. |
| June 2024 /51 | ln TC against t | “An incorrect rearrangement of the equation.” The scheme wanted ln (TC − TR) against t. |
Significant figures after the graph
Question 2 Where the sig-fig marks sit, and how many figures each one wants.Question 2, step by step
Which steps carry a significant-figure mark
- TableProcessed values: same s.f. as the raw data, or one more.s.f. checked
- GraphPoints, error bars, best-fit and worst lines.no s.f. mark
- Gradient and interceptMarked on read-offs and substitution. Keep an extra figure.no s.f. mark
- ConstantFrom the gradient or intercept: match the data, or one more.s.f. mark
- Final valueAt least 2 s.f., right power of ten.s.f. mark
The gradient and the y-intercept
I checked all 42 schemes from March 2021 to June 2026. The gradient and y-intercept marks are for reading two points off your line, more than half its length apart, and showing the substitution. None of them sets a number of significant figures. Cambridge's own 9702 help page says the same: keeping an extra figure in the gradient and intercept is “not subject to any penalty”, and it cuts rounding errors in the steps that follow.
The constant you calculate from them
This is where the mark is. All four June 2026 schemes give the constant from the gradient its own line: “given to 2 or 3 sf”. The range isn't fixed. It comes from the data. In June 2025 /51 E had two significant figures, and the report says the constants R and Z followed it:
R and Z should be given to two (or three) significant figures.June 2025 /51 examiner report, Question 2(d)(i)
So: count the significant figures in the data the constant is calculated from, take the smallest, and give the constant to that number or one more.
Two sig figs isn't a safe default
In March 2025 /52 and November 2025 /51 and /53, a constant had to be given to 3 or 4 significant figures, so 2 lost the mark. Too many loses it as well. The November 2024 /52 report lists n given to one significant figure (too few) or five (too many). The June 2024 /51 report: “Some candidates gave k to too many significant figures.”
The final value
The last calculation in Question 2 asks for a value to a minimum of 2 significant figures, with the correct power of ten, in all four June 2026 schemes. (June 2026 /54 also allowed 1 s.f. if the uncertainty justified it.) Write the equation, the substitution and the answer with its unit.
Predict
Which value carries a sig-fig condition in the scheme?
Show the answer
The constant. June 2026 /51, /52, /53 and /54 all mark the constant from the gradient “given to 2 or 3 sf”. The gradient line in the same schemes checks read-offs and substitution, with no s.f. condition.
Predict · illustrative numbers
The raw data are to 2 s.f. Your calculator gives the constant as 0.38274. Which earns the mark?
Show the answer
0.38 or 0.383. Two significant figures matches the data, three is one more; both are inside the “2 or 3” range a scheme gives in this case. 0.4 has too few and 0.38274 too many, the two errors the November 2024 /52 report names.
Questions students ask
Do significant figures on the gradient matter in 9702 Paper 5?
Not for the gradient mark itself. No Paper 5 mark scheme from March 2021 to June 2026 sets a number of significant figures for the gradient or the y-intercept, so keep an extra figure there. The significant-figure mark comes on the constant you calculate from the gradient, and on the final value.
How many significant figures should the constant from the gradient have?
The same number as the least precise data it is calculated from, or one more. In June 2025 /51 E had two significant figures, so the constants R and Z had to be given to two or three. In November 2025 /51 one constant needed 3 or 4.
Should I use a log graph in Paper 5 Question 1?
Only when the power is unknown, as in June 2026 /51, where T depended on a to an unknown power q. When the relationship is already straight in known functions such as 1/r³, a log graph can lose the mark: 20 of the 42 schemes from March 2021 to June 2026 refuse one.