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Cambridge 2027 further mathematics: core methods and applications

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AI-assisted English content review; independent teacher approval pending.

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Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.

90 original questions across 15 selected topic areas. No account or payment required to practise.

Cambridge International AS & A Level Further Mathematics (9231), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.

5 questions in this topic

Exact English revisions checked in an AI-assisted academic review; independent teacher approval is still pending.

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Original written questions with indicative worked solutions. Conceptual mappings require teacher review; this is not a complete course or an official paper.

Subtopics in this topic (5)
  • Matrix addition, subtraction and multiplication
  • Common perpendicular to skew lines
  • Using supplied centres of mass of simple shapes
  • Normal test for a difference of population means
  • Single-sample sign and Wilcoxon signed-rank tests

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Question 1Let A=[[1,2],[-1,3]], B=[[2,0],[4,-1]], and C=[[1,0,2],[-1,3,1]]. (a) Calculate A+B and A-B. (b) Calculate AB and BA, showing row-column products, and explain whether they commute. (c) Calculate AC and state why CA is undefined.4 indicative marks
Calculate / Show / Explain4 indicative marksCalculator allowed

Let A=[[1,2],[-1,3]], B=[[2,0],[4,-1]], and C=[[1,0,2],[-1,3,1]]. (a) Calculate A+B and A-B. (b) Calculate AB and BA, showing row-column products, and explain whether they commute. (c) Calculate AC and state why CA is undefined.

Cambridge 2027: Direct conceptual match

Matrix addition, subtraction and multiplication: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. Add/subtract corresponding entries: A+B=[[1+2,2+0],[-1+4,3-1]]=[[3,2],[3,2]]; A-B=[[1-2,2-0],[-1-4,3-(-1)]]=[[-1,2],[-5,4]].
  2. AB=[[1*2+2*4,1*0+2*(-1)],[(-1)*2+3*4,(-1)*0+3*(-1)]]=[[10,-2],[10,-3]]. BA=[[2*1+0*(-1),2*2+0*3],[4*1+(-1)*(-1),4*2+(-1)*3]]=[[2,4],[5,5]]. Their (1,1) entries already differ, so AB!=BA.
  3. AC=[[1-2,0+6,2+2],[-1-3,0+9,-2+3]]=[[-1,6,4],[-4,9,1]]. A is 2 by 2 and C is 2 by 3, so AC is 2 by 3. CA would require 3=2 for its inner dimensions, which fails.

Self-marking points

View marking points
  1. Find A+B=[[3,2],[3,2]] and A-B=[[-1,2],[-5,4]].
  2. Find AB=[[10,-2],[10,-3]] and BA=[[2,4],[5,5]].
  3. Demonstrate noncommutativity.
  4. Find AC=[[-1,6,4],[-4,9,1]] and explain incompatible inner dimensions for CA.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 2Lines L and M are r=(0,0,0)+s(1,1,0) and r=(1,0,2)+t(1,0,1). (a) Show they are skew. (b) Let P on L and Q on M be endpoints of their common perpendicular. Determine s,t by scalar products. (c) Give P,Q, an equation of the common perpendicular, and its segment length. Verify both right angles.4 indicative marks
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Lines L and M are r=(0,0,0)+s(1,1,0) and r=(1,0,2)+t(1,0,1). (a) Show they are skew. (b) Let P on L and Q on M be endpoints of their common perpendicular. Determine s,t by scalar products. (c) Give P,Q, an equation of the common perpendicular, and its segment length. Verify both right angles.

Cambridge 2027: Direct conceptual match

Common perpendicular to skew lines: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. The directions are not scalar multiples. Intersection would require s=1+t, s=0 and 0=2+t; these give t=-2 but then s=-1, contradicting s=0. Hence the lines are skew.
  2. P=(s,s,0), Q=(1+t,0,2+t), so Q-P=(1+t-s,-s,2+t). Perpendicularity gives 1+t-2s=0 and 3+2t-s=0.
  3. Solving the first for t=2s-1 and substituting in the second gives 1+3s=0. Thus s=-1/3,t=-5/3; P=(-1/3,-1/3,0), Q=(-2/3,0,1/3).
  4. Q-P=(-1,1,1)/3, so the common perpendicular is r=(-1/3,-1/3,0)+lambda(-1,1,1). It reaches Q at lambda=1/3 and has segment length sqrt3/3. Dot products (-1,1,1).(1,1,0)=0 and (-1,1,1).(1,0,1)=0 verify both right angles.

Self-marking points

View marking points
  1. Show nonparallel directions and inconsistent intersection equations.
  2. Impose (Q-P).(1,1,0)=0 and (Q-P).(1,0,1)=0.
  3. Solve s=-1/3,t=-5/3 and give both endpoints.
  4. Give r=(-1/3,-1/3,0)+lambda(-1,1,1), segment length sqrt3/3 and verify both dot products are zero.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 3A uniform solid cone of radius 2 cm and height 6 cm is joined at its circular base to a uniform solid hemisphere of radius 2 cm made from the same material. Take the join as x=0, positive into the cone. You are given: cone volume pi*r^2*h/3 and centre h/4 from its base; hemisphere volume 2pi*r^3/3 and centre 3r/8 from its flat face. (a) Locate each centre and find the mass ratio. (b) Find the composite centre. (c) Explain which way the axis points when suspended freely from a point on the circular rim, using the centre-of-mass principle.4 indicative marks
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A uniform solid cone of radius 2 cm and height 6 cm is joined at its circular base to a uniform solid hemisphere of radius 2 cm made from the same material. Take the join as x=0, positive into the cone. You are given: cone volume pi*r^2*h/3 and centre h/4 from its base; hemisphere volume 2pi*r^3/3 and centre 3r/8 from its flat face. (a) Locate each centre and find the mass ratio. (b) Find the composite centre. (c) Explain which way the axis points when suspended freely from a point on the circular rim, using the centre-of-mass principle.

Cambridge 2027: Direct conceptual match

Using supplied centres of mass of simple shapes: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. The cone centre is h/4=6/4=3/2 cm into the cone. The hemisphere centre is 3r/8=6/8=3/4 cm on the opposite side, so its signed coordinate is -3/4.
  2. The volumes are pi*4*6/3=8pi and 2pi*8/3=16pi/3 cm^3. Equal density makes mass proportional to volume, hence cone:hemisphere=3:2.
  3. Represent the two masses as 3m and 2m at their supplied centres. Then 5m*x_bar=3m*(3/2)+2m*(-3/4)=3m. Thus x_bar=3/5 cm, on the cone side of the join. Symmetry puts it on the axis.
  4. At stable free suspension, the total weight acts through the composite centre and its line passes vertically below the suspension point. From a rim point the vector to the centre has axial component +3/5 and inward radial component 2 cm. This vector is vertical down in equilibrium, so the cone side of the axis has a downward component. This is a written model, not evidence of a practical suspension experiment.

Self-marking points

View marking points
  1. Use supplied centres x=3/2 and x=-3/4 with correct signs.
  2. Use volumes 8pi and 16pi/3 to obtain mass ratio 3:2.
  3. Take signed mass moments to obtain x_bar=3/5 cm.
  4. Explain the composite centre lies vertically below the rim suspension, with cone side of axis tilted down.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 4Independent random samples from two normal populations A and B have known population standard deviations 6 and 8. Sample sizes are 36 and 64; sample means are 52 and 49. (a) Choose and justify a test of H0:mu_A-mu_B=0 against H1:mu_A-mu_B>0 at 5%. (b) Calculate the standard error, test statistic and decision, using the upper 5% N(0,1) value 1.645. (c) Find a 95% two-sided confidence interval for the difference using 1.960 and explain its relation to the one-sided decision.4 indicative marks
Calculate / Show / Explain4 indicative marksCalculator allowed

Independent random samples from two normal populations A and B have known population standard deviations 6 and 8. Sample sizes are 36 and 64; sample means are 52 and 49. (a) Choose and justify a test of H0:mu_A-mu_B=0 against H1:mu_A-mu_B>0 at 5%. (b) Calculate the standard error, test statistic and decision, using the upper 5% N(0,1) value 1.645. (c) Find a 95% two-sided confidence interval for the difference using 1.960 and explain its relation to the one-sided decision.

Cambridge 2027: Direct conceptual match

Normal test for a difference of population means: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. Use a normal rather than pooled t test: the populations are normal, their variances are known, and the two samples are independent. Under H0, Xbar_A-Xbar_B is exactly N(0,626^{2}/36+828^{2}/64)=N(0,2).
  2. The standard error is sqrt2 and the observed difference is 3. Hence z=(3-0)/sqrt2=2.121, which exceeds 1.645. Reject H0 at 5%; there is evidence that A's population mean is greater than B's, not proof that the difference is exactly 3.
  3. The 95% two-sided interval is 3+/-1.960sqrt2=[0.228,5.772] to 3 decimals. It excludes zero here. The interval uses two tails of 2.5% each, while the requested one-sided test uses one tail of 5%; these decision rules are not generally equivalent even though they agree on these data.

Self-marking points

View marking points
  1. Choose a normal test because independent normal populations have known variances.
  2. Use SE=sqrt(36/36+64/64)=sqrt2.
  3. Obtain z=3/sqrt2=2.121 and reject at 5%.
  4. Obtain 3+/-1.960sqrt2, approximately [0.228,5.772], and distinguish tail conventions.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 5An independent random sample from a continuous population is 19,22,23,24,25,26,27,28. (a) Use an exact one-sided sign test at 5% for H0:median=20 against H1:median>20, stating its distribution and p-value. (b) Now assume the population is symmetric about its median. Use an exact one-sided Wilcoxon signed-rank test for the same hypotheses, showing ranks and obtaining the tail probability by enumerating subsets of ranks. (c) Explain why the two decisions can differ and which extra assumption the second test needs.4 indicative marks
Calculate / Show / Explain4 indicative marksCalculator allowed

An independent random sample from a continuous population is 19,22,23,24,25,26,27,28. (a) Use an exact one-sided sign test at 5% for H0:median=20 against H1:median>20, stating its distribution and p-value. (b) Now assume the population is symmetric about its median. Use an exact one-sided Wilcoxon signed-rank test for the same hypotheses, showing ranks and obtaining the tail probability by enumerating subsets of ranks. (c) Explain why the two decisions can differ and which extra assumption the second test needs.

Cambridge 2027: Direct conceptual match

Single-sample sign and Wilcoxon signed-rank tests: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. Differences from 20 are -1,2,3,4,5,6,7,8. There are no zeros. Under a continuous median-20 null, S, the positive count, is B(8,1/2). Observed S=7; the upper-tail probability is (C(8,7)+C(8,8))/2^8=9/256=0.03515625. Reject H0 at 5% in favour of a greater median.
  2. Under the additional symmetry assumption the signs are equally likely conditional on the absolute ranks. The absolute differences already have ranks 1,2,3,4,5,6,7,8. The sole negative has rank 1, so W-=1 and W+=35 (total 36).
  3. For a greater median, small W- is evidence. Of the 2^8 equally likely rank subsets, only {} and {1} have sum <=1. Thus p=2/256=0.0078125; reject at 5%.
  4. Both reject here, but the signed-rank test uses ranks of magnitudes as well as signs, so its p-value need not equal the sign test's and decisions can differ for other data or levels (here at 1% only signed-rank rejects). Symmetry is needed for the signed-rank calibration; the sign test does not require it. Neither is a test of a mean.

Self-marking points

View marking points
  1. Count 7 positives out of 8 and obtain sign p-value 9/256.
  2. Rank absolute differences 1 to 8, negative rank sum W-=1.
  3. Enumerate subsets {} and {1}, giving signed-rank p-value 2/256.
  4. Reject with both tests at 5%, explaining different statistics and symmetry requirement.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

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