November 2023 Edexcel GCSE Maths Higher Paper 1 — Answers & Worked Solutions | MathsWealth
EDEXCEL GCSE MATHS • HIGHER • PAPER 1H • NON-CALCULATOR
November 2023 Paper 1
— Answers & Worked Solutions
MathsWealth answer guide for Pearson Edexcel GCSE Mathematics Higher Paper 1H.
80 marks • 1 hour 30 minutes • Non-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.
Mark scheme focus: Multiply as integers, then place the decimal point using the two decimal places in the
factors.
Q2Indices · Point: Evaluate 5⁰ and 5⁻²; simplify the given powers of 2.
Evaluate 5⁰ and 5⁻². Then simplify (2⁵ × 2⁴) ÷ 2³ and write the result as a single power of 2.
Any non-zero number to the power 0 equals 1, so 5⁰ = 1. A negative index means reciprocal: 5⁻² = 1/5² = 1/25. For the final part, use index laws: (2⁵ × 2⁴) ÷ 2³ = 2⁵⁺⁴⁻³ = 2⁶ =
64.
Mark scheme focus: For HCF, multiply only the prime factors common to both numbers, using the smaller powers.
Q4Mean · Point: Five sticks have a mean length of
4.2 cm. One is 7 cm. Find the mean of the other four, then consider if that stick were 17 cm.
Five sticks have a mean length of 4.2 cm. One stick was measured as 7 cm. (a) Find the mean length of the other four sticks. (b) If that stick was actually 17 cm, explain how this changes the mean of the other four sticks and give the new mean.
The total length of all five sticks is 5 × 4.2 = 21 cm. (a) The other four total 21 − 7 = 14 cm, so their mean is 14 ÷ 4 = 3.5 cm. (b) If the known stick is 17 cm, the other four total 21 − 17 = 4 cm, giving a mean of 4 ÷ 4 = 1
cm. The mean of the other four is therefore lower.
Mark scheme focus: Convert a mean to a total before removing the known value.
Q5Geometric construction · Point: Construct a perpendicular (90°) at P on line AB using ruler and compasses.
Construct a perpendicular (90°) at P on line AB using ruler and compasses.
Question 5: perpendicular construction at P on line AB.
Construction arcs should be drawn by the student; this is a clean guide, not a substitute for construction marks.
Set the compass to a radius that reaches above and below the line. From two points equidistant from P, draw intersecting arcs above the line. Join P to the arc intersection.
Answer: A correctly constructed 90° line through P
Mark scheme focus: Leave
construction arcs visible; a protractor-only drawing is not the requested construction.
Q6Angles in an isosceles triangle · Point: Use BA = BD and the angle ratio shown in the diagram to find w.
Use BA = BD and the angle ratio shown in the diagram to find w.
Question 6: straight line ABC and isosceles triangle ABD,
with BA = BD. Angle labels are schematic; use the official figure for their precise placement.
Since BA = BD, triangle ABD is isosceles, so its base angles are equal. The ratio x:y = 2:1 and the angles on the straight line give 2x + y = 180°. There are 5 equal ratio parts, so each part is 180° ÷ 5 = 36°. Thus y = 36° and x = 72°. Since w and y lie on a straight line, w = 180° − 36° = 144°.
Mark
scheme focus: The ratio gives 36° per part; the required exterior angle is 180° − 36°.
Q7Forming and solving an equation · Point: Three shelves contain x, 3x + 1 and 2x − 5 books; there are 44 books altogether. Shelf B has mass 7500 g. Find the mass on shelf A.
Three shelves contain x, 3x + 1 and 2x − 5 books; there are 44
books altogether. Shelf B has mass 7500 g. Find the mass on shelf A.
x + (3x + 1) + (2x − 5) = 44, so 6x − 4 = 44 and x = 8. Shelf B has 3(8) + 1 = 25 books. If those 25 books have a total mass of 7500 g, each book has mass 7500 ÷ 25 = 300 g. Shelf A has 8 books, so its mass is 8 × 300 = 2400 g.
Mark scheme focus:</
b> Use the circumference formula to find the radius before using the area formula.
Q11Box plots and quartiles · Point: Complete a box plot for 80 days of traffic data and find the fraction of days with fewer than 1200 cars.
A box plot summarises the number of cars entering a village over 80 days. Complete the missing
quartiles and use the box plot to find the fraction of days with fewer than 1200 cars.
Question 11: reconstructed box plot using the official mark-scheme values: minimum 300, lower quartile 780, median 900, upper quartile 1200, maximum 1300.
From the box plot, the lower quartile is 780 and the upper quartile is 1200. The upper quartile marks the 75th percentile, so approximately three quarters of the 80 days had fewer than 1200 cars: (3/4) × 80 = 60 days. As a fraction of all days, this is 60/80 = 3/4.
Answer:
Lower quartile = 780; upper quartile = 1200; 60 days, i.e. 3/4 of the days
Mark scheme focus: Quartiles divide ordered data into four equal-sized groups.
Q12Perpendicular line · Point: Find the equation of the line perpendicular to the given line and passing through the
stated point.
Find the equation of the line perpendicular to 2y = 3x − 7 that passes through the point (6, −5).
The given line is 2y = 3x − 7, so y = (3/2)x − 7/2 and its gradient is 3/2. A perpendicular line has gradient −2/3. Substitute (6, −5) into y = −(2/3)x + c: −5 = −(2/3)(6) + c = −4 + c, so c = −1. Therefore the required equation is y = −(2/3)x − 1.
Mark scheme focus: M1 for the perpendicular gradient −2/3; M1 for substituting (6, −5) to find the intercept; A1 for y = −(2/3)x − 1 or an equivalent equation.
Q13Similar solids and volume scale factor · Point: Similar solids have corresponding heights in the ratio 2:5. The smaller solid has volume 12 cm³. Find the
larger volume.
Two similar solids have corresponding heights in the ratio 2:5. The smaller solid has volume 12 cm³. Find the larger volume.
The linear scale factor from the smaller solid to the larger is 5/2. Volume scales with the cube of the linear scale factor: (5/2)³ = 125/8. Therefore the larger volume is 12 × 125/8 = 187.5 cm³.
Mark scheme focus: M1 for correctly evaluating 272/3 as 9 or evaluating 23 as 8; A1 for 17.
Q15Distance-time graph ·
Point: Estimate the gradient of the curve at t = 3 using a tangent.
Estimate the gradient of the curve at t = 3 using a tangent.
Question 15: schematic distance–time curve with tangent position indicated. The original curve is required for a precise gradient estimate.
Draw a tangent at t = 3. Choose two well-separated points on the tangent and calculate rise ÷ run. The mark scheme accepts gradients in the range 22–28 with supporting tangent working.
Mark scheme focus: A tangent must be shown; a numerical answer alone may not earn marks.
Q16Recurrence relations · Point: Population follows Pₙ₊₁ = kPₙ. Given P₁ = 8 million and P₂ = 6 million, find P₃ and decide what
happens if k later increases by 0.3.
Population follows Pₙ₊₁ = kPₙ. Given P₁ = 8 million and P₂ = 6 million, find P₃ and decide what happens if k later increases by 0.3.
k = 6/8 = 0.75. P₃ = 6 × 0.75 = 4.5 million. If k later increases by 0.3, the new multiplier is 0.75 + 0.3 = 1.05. Since this is greater than 1, the population then grows by 5% per year.
Answer: P₃ = 4.5 million; after k increases to 1.05, the population increases year on year.
Mark scheme focus: A multiplier above 1 means growth; below 1 means decline.
Q17Factorising and inequalities · Point: Factorise 6x² − 5x − 4 and solve the associated inequality from the paper.
Factorise 6x² −
5x − 4, then solve the inequality 6x² − 5x − 4 < 0.
Factorise: 6x² − 5x − 4 = (2x + 1)(3x − 4). The roots are x = −1/2 and x = 4/3. The quadratic has a positive leading coefficient, so it is negative between its two roots. Hence −1/2 < x < 4/3.
Mark scheme focus: Match each side to the correct angle before substituting. Show the cosine-rule equation and the inverse cosine step.
Q21Circle
and simultaneous equations · Point: Sketch x² + y² = 169 and solve it simultaneously with 2y = 3x using the graph.
Sketch x² + y² = 169 and solve it simultaneously with 2y = 3x using the graph.
Question 21: circle x² + y² = 169 and line 2y = 3x on coordinate axes. The circle is centred at the origin with radius 13.
The circle has centre (0,0) and radius 13. The line is y = 1.5x. Read both intersection points from the graph; the solutions are equal in magnitude
with matching signs.
Answer: Approximately (7.2, 10.8) and (−7.2, −10.8)
1. The Eighth Wonder of the World Albert Einstein once famously called compound interest the "8th wonder of the world," stating, "He who understands it, earns it; he who doesn't, pays it." In the realm of GCSE Maths, Compound Interest is a highly tested crossover topic (usually appearing around Grade 4 to 5). But beyond the exam hall, understanding how compound interest works is the ultimate key to managing your real-world wealth—whether you are growing your savings or paying off a bank loan. Today, we will break down a real GCSE past paper question and reveal the most common mistakes students make. 2. The Core Formula Unlike simple interest, which only pays out on the original amount, compound interest pays interest on the interest you have already earned. The standard formula you need to know is: 3. Real GCSE Past Paper: Edexcel 2019 (Paper 3, Higher) Let’s apply this to a classic question that tricked many students. Question: Katy invests £200,000 in a saving...
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