November 2023 | Edexcel Past Paper I GCSE Maths Higher Paper 2 | Worked Solutions

EDEXCEL GCSE MATHS • HIGHER • PAPER 2H • CALCULATOR

November 2023 Paper 2 — Answers & Worked Solutions

MathsWealth answer guide for Pearson Edexcel GCSE Mathematics Higher Paper 2H.

80 marks • 1 hour 30 minutes • Calculator paper

How to use this page: Attempt each question before opening the solution. Show working clearly to earn method marks. For diagrams, graphs and constructions, use the original paper. These are independent worked explanations, not a replacement for the official question paper.

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Official Exam Materials

Open the original Pearson question paper and official mark scheme for the exact wording, diagrams and accepted answers.

Question Paper (PDF)Mark Scheme (PDF)Principal Examiner Feedback (PDF)

1MA1/2H · November 2023 · 80 marks · 1 hour 30 minutes

Worked Solutions

Topic classification, key working, final answer and mark-scheme focus for each numbered question.

Q1Expanding and factorising · Point: Expand the brackets, factorise the quadratic, and rearrange the formula as directed.

Expand the brackets, factorise the quadratic, and rearrange the formula as directed.

From f = 3g + 11, subtract 11 from both sides: f − 11 = 3g. Divide by 3 to get g = (f − 11)/3.

Answer: (a) 13y−1; (b) 3x(2x+5); (c) g=(f−11)/3

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Mark scheme focus: Show the first expansion or common factor to secure method marks.

Q2Percentage profit · Point: A trader buys items and sells them at the stated prices. Calculate the percentage profit.

A trader buys items and sells them at the stated prices. Calculate the percentage profit.

Income = 7.50 × 54 = £405. Outgoings total £100 + £120 + £80 = £300. Profit = £105. Percentage profit = 105 ÷ 300 × 100.

Answer: 35%

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Mark scheme focus: Percentage profit is calculated using profit ÷ cost/outgoings × 100.

Q3Compound interest · Point: £4500 increases by 3.4% per year for two years. Find the amount after the stated period.

£4500 increases by 3.4% per year for two years. Find the amount after the stated period.

Use a multiplier of 1.034 each year: 4500 × 1.034² = 4811.202…

Answer: £4811.20

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Mark scheme focus: Use compound multiplication, not simple interest added twice.

Q4Linear equation · Point: Solve the linear equation in the paper.

Solve the linear equation in the paper.

Starting from 5x − 14 = 52 − x, add x to both sides and add 14 to both sides: 6x = 66. Divide by 6 to obtain x = 11.

Answer: x = 11

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Mark scheme focus: Perform the same operation on both sides and show the isolation of x.

Q5Ratio and comparison · Point: Use the original ratio and the new ratio to determine the requested difference for Errol.

Use the original ratio and the new ratio to determine the requested difference for Errol.

The first ratio gives totals 90 : 120 : 60, with total 270. The new ratio has 2 + 5 + 3 = 10 parts, so one part is 27. Errol’s values are 2 × 30 = 60 and 3 × 27 = 81; difference = 21.

Answer: 21

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Mark scheme focus: Scale each ratio using its total, then compare the two relevant amounts.

Q6Bearings · Point: Use the bearings diagram to calculate the required bearing.

The bearing of port B from port A is 147°. Find the bearing of port A from port B.

Question 6: bearing schematic. The reverse bearing is 327°.

The reverse bearing differs by 180°. Calculate 147° + 180° = 327°. This is already between 000° and 360°, so the three-figure bearing is 327°.

Answer: 327°

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Mark scheme focus: Include the leading zero where needed and express the final bearing as three digits.

Q7Cylinder volume and rate · Point: A cylindrical container is filled at a constant rate. Calculate the time required.

A cylindrical container is filled at a constant rate. Calculate the time required.

Question 7: cylinder with radius 15 cm and height 43 cm, as stated in the official paper.

Volume = π × 15² × 43 = 9675π cm³ ≈ 30,394.9 cm³ = 30.3949 L. Divide by the fill rate 0.47 L/min.

Answer: Approximately 65 minutes

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Mark scheme focus: Convert cm³ to litres by dividing by 1000 before dividing by the rate in litres per minute.

Q8Bounds · Point: Write the bounds for a value rounded to one decimal place.

A number x rounds to 1.9 when written to 2 significant figures. Write the error interval for x.

A number that rounds to 1.9 to 1 decimal place is at least 1.85 but less than 1.95.

Answer: 1.85 ≤ x < 1.95

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Mark scheme focus: The lower bound is included; the upper bound is excluded.

Q9Expanding three brackets · Point: Expand the product of the three linear factors shown in the paper.

Expand the product of the three linear factors shown in the paper.

First multiply two brackets: (x + 7)(x − 2) = x² + 5x − 14. Then multiply by (x + 3): x³ + 8x² + x − 42.

Answer: x³ + 8x² + x − 42

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Mark scheme focus: Expand systematically and collect like terms only after all products have been written.

Q10Probability tree diagrams · Point: Complete the probability tree and find the probability of the required outcome.

Complete the probability tree and find the probability of the required outcome.

Question 10: probability tree using the stated probabilities and their complements.

Complete the tree using complementary probabilities: 1 − 0.13 = 0.87, 1 − 0.06 = 0.94, and 1 − 0.73 = 0.27. The route probabilities are 0.87 × 0.94 = 0.8178, 0.87 × 0.06 = 0.0522, 0.13 × 0.73 = 0.0949 and 0.13 × 0.27 = 0.0351. The required event is the first three routes, so 0.8178 + 0.0522 + 0.0949 = 0.9649 (equivalently, 1 − 0.0351).

Answer: 0.9649

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Mark scheme focus: Multiply along a path; add mutually exclusive paths. Use the probabilities shown in the completed tree.

Q11Ordering and classification · Point: Place the given values into the correct order/category.

Four graphs A, B, C and D show different types of proportionality. Match the graphs to y ∝ x², y ∝ x, y ∝ 1/x and y ∝ √x, in that order.

Question 11: characteristic proportionality graphs. Their qualitative shapes distinguish the four relationships.

Match each relationship to its characteristic graph: y ∝ x² is a curved graph through the origin (D); y ∝ x is a straight line through the origin (A); y ∝ 1/x is an inverse curve with two branches (B); and y ∝ √x is an increasing curve through the origin (C).

Answer: D, A, B, C

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Mark scheme focus: Check each item against the defining condition rather than relying on visual order.

Q12Coordinate geometry · Point: Find the length between the two given points using Pythagoras.

Find the length between the two given points using Pythagoras.

Question 12: midpoint coordinates M = (2,17) and N = (6,22), with MN shown.

The midpoint coordinates used in the scheme are M=(2,17) and N=(6,22). Distance = √[(6−2)²+(22−17)²] = √(16+25) = √41.

Answer: √41 ≈ 6.4

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Mark scheme focus: Subtract coordinates to find horizontal and vertical differences, then use Pythagoras.

Q13Recurring decimals proof · Point: Show that the recurring decimal in the question equals the stated fraction.

Show that the recurring decimal in the question equals the stated fraction.

Let x = 0.072372372… . Multiply to align the recurring digits: 10x = 0.723723…, and 10000x = 723.723723… . Subtract: 10000x − 10x = 723, so 9990x = 723. Therefore x = 723/9990 = 241/3330.

Answer: 241/3330

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Mark scheme focus: Use two equations with matching recurring digits; a decimal approximation is not a proof.

Q14Direct and inverse proportion · Point: Two variables are related by a combination of direct and inverse proportion. Find the requested value.

y is directly proportional to x², and y = 3 when x = 0.5. Also, x is inversely proportional to w, and x = 2 when w = 0.2. Find y when w = 2.

Since y ∝ x², let y = kx². Using y = 3 when x = 0.5 gives 3 = k(0.5)², so k = 12 and y = 12x². Since x ∝ 1/w, let x = j/w. Using x = 2 when w = 0.2 gives j = 2 × 0.2 = 0.4, so x = 0.4/w. When w = 2, x = 0.2. Therefore y = 12(0.2)² = 0.48.

Answer: 0.48

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Mark scheme focus: State the proportionality formula first; keep the constants consistent between the two equations.

Q15Histogram · Point: Complete the histogram and use it to estimate the required value.

The first histogram and table show race times for one group of students: use the histogram to find the missing frequency, then complete the histogram using the table. A second histogram shows times for 43 students in a different race. Estimate the median time for this second group.

Question 15: histogram layout. Bar heights must be set from the official table/diagram; this is a layout guide, not an exact data reconstruction.

For a histogram, frequency = frequency density × class width. The missing first frequency is 3; the two missing bar heights in the first histogram are 4.5 and 1.4. For the second histogram, there are 43 students, so the median is at the 22nd value. There are 12 students before the 16–20 second class, which contains 16 students (density 4 × width 4). The median is 22 − 12 = 10 students into this class, so the estimated time is 16 + (10/16) × 4 = 18.5 seconds.

Answer: Missing frequency 3; median estimate 18.5

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Mark scheme focus: Bar area represents frequency; the height alone does not represent frequency unless class widths are equal.

Q16Triangle area using sine rule · Point: Find the area of the triangle using the given sides and angles.

Find the area of the triangle using the given sides and angles.

Question 16: trapezium ABCD with AD ∥ BC and the given lengths and angles; the shaded triangle to calculate is BCD.

Use the sine rule to find BD: BD/sin(42°)=7.9/sin(112°), so BD≈10.9. Then area = 1/2 × BD × 15.3 × sin(42°).

Answer: 56.0 cm²

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Mark scheme focus: Find the missing side before applying the area formula ½ab sin C.

Q17Iteration and roots · Point: Show a root lies between 1 and 2, rearrange the equation for iteration, then calculate the root.

Show a root lies between 1 and 2, rearrange the equation for iteration, then calculate the root.

For f(x)=x³+2x−6, f(1)=−3 and f(2)=6, so the sign changes and a root lies between 1 and 2. Rearrange x³+2x=6 to x=∛(6−2x) and iterate as instructed.

Answer: 1.4496 (to 4 d.p.)

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Mark scheme focus: Use the required rearrangement for the iteration and do not round intermediate values too early.

Q18Vector proof · Point: Use vectors to prove the stated quadrilateral is a parallelogram.

Use vectors to prove the stated quadrilateral is a parallelogram.

Take A as the origin, with vectors AH = a, AE = b and DG = c. Since H is the midpoint of DA, AD = 2a. Since G is the midpoint of DC, DC = 2c, so AG = 2a + c. Also AB = 2b, and because F is the midpoint of BC, AF = (AB + AC)/2 = a + b + c. Hence EH = AH − AE = a − b and FG = AG − AF = (2a + c) − (a + b + c) = a − b. Therefore EH = FG: one pair of opposite sides is equal and parallel, so EFGH is a parallelogram.

Answer: Proof: opposite sides have equal vectors

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Mark scheme focus: Define each vector from the same origin consistently; equality of opposite side vectors proves the result.

Q19Composite functions · Point: Evaluate the composite function and rearrange the given function equation.

Evaluate the composite function and rearrange the given function equation.

Substitute x=−3 into the inner function first, then apply the outer function: 2(2×(−3)+3)²−1 = 2(−3)²−1 = 17. For the rearrangement, isolate the variable step by step.

Answer: (a) 17; (b) x = y/2 + 1/2

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Mark scheme focus: In a composite function, apply the inner function before the outer one.

Q20Algebraic fractions · Point: Simplify the algebraic expression by factorising and combining fractions.

Simplify the algebraic expression by factorising and combining fractions.

Factorise the numerator: (x+4)(2x−14)=2(x+4)(x−7). Use a common denominator, simplify, and combine the terms as in the mark scheme.

Answer: (6x + 37)/(6x − 42)

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Mark scheme focus: Factorise before cancelling; only cancel common factors, not terms in a sum.

Q21Ratio and algebra · Point: Use the given expressions for the numbers of gold items to find the total.

A Venn diagram shows medal counts for a collection: English medals are set A, gold medals are set B, and the regions are labelled 7x − 11 (A only), x (A ∩ B), 5x − 3 (B only), and 34 (outside both). Given P(English | gold) = 2/11, find the total number of medals.

A = English medals; B = gold medals. The number outside both sets is 34.

Among gold medals, the English gold medals are x and all gold medals total x + (5x − 3) = 6x − 3. Thus x/(6x − 3) = 2/11. Cross-multiplying gives 11x = 12x − 6, so x = 6. The total number of medals is (7x − 11) + x + (5x − 3) + 34 = 13x + 20 = 13(6) + 20 = 98.

Answer: 98

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Mark scheme focus: Keep the expressions symbolic until x is found, then substitute consistently.

MathsWealth revision note: These explanations are written for revision and may paraphrase the questions. Always check the exact wording, diagrams and any required units in the official Pearson paper. For questions based on graphs, constructions or a diagram, the original figure is essential.