M4.2 · Pythagoras and trigonometry
Explain — 5 marks
A roof frame is shaped as a right-angled triangle ABC, with the right angle at B. The horizontal beam AB has length 3.6 metres and the vertical support BC has length 2.7 metres. The sloping beam AC completes the triangle.
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(a) Calculate the length of the sloping beam AC. Give your answer in metres.
[2 marks]
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(b) Calculate the angle that the sloping beam AC makes with the horizontal beam AB. Explain why your method is appropriate for finding this angle.
[3 marks]
Show mark scheme
- (a) Method: correct substitution into Pythagoras' theorem (e.g., AC² = 3.6² + 2.7² or equivalent)
- (a) Answer: 4.5 metres (accept 4.50 or √20.25)
- (b) Method: correct use of trigonometry (e.g., tan⁻¹(2.7/3.6), sin⁻¹(2.7/4.5), or cos⁻¹(3.6/4.5))
- (b) Answer: 36.9° (accept 36.86° or 36.87° or 37°)
- (b) Explanation: identifies the sides used in relation to the angle (opposite and adjacent for tan, or opposite and hypotenuse for sin, or adjacent and hypotenuse for cos) and states this is why that trigonometric ratio is appropriate
M1.1 · Operations with integers, decimals, fractions
Calculate — 2 marks
A financial analyst is reviewing monthly profit and loss figures for a small business. In January, the business made a loss of £450.75. In February, the profit was £320.50. In March, there was another loss of £187.25. The analyst needs to calculate the overall financial position across these three months.
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(a) Calculate the net profit or loss for January and February combined. Give your answer as a positive or negative number.
[1 mark]
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(b) Calculate the overall profit or loss for all three months (January, February, and March combined).
[1 mark]
Show mark scheme
- (a) Correctly adds \(-450.75 + 320.50 = -130.25\) (or equivalent working showing subtraction of positive values with correct sign)
- (b) Correctly calculates \(-130.25 + (-187.25) = -317.50\) or \(-450.75 + 320.50 + (-187.25) = -317.50\) with correct final answer
M6.1 · Averages and spread
Explain — 5 marks
A café manager records the number of customers served each day over a week. The daily customer counts are: 24, 31, 28, 24, 35, 24, 28. The manager wants to understand the typical daily footfall and how consistent business is from day to day.
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(a) Calculate the mode, median, and mean number of customers for the week. Show your working.
[3 marks]
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(b) The manager claims that the mean is the best measure to use when planning staff levels. Explain why the median might actually be more useful in this situation.
[2 marks]
Show mark scheme
- (a) Mode identified as 24 (appears 3 times)
- (a) Median calculated as 28 (ordered data: 24, 24, 24, 28, 28, 31, 35; middle value is 28)
- (a) Mean calculated as 28.3 or \(28\frac{2}{7}\) (sum of 198 divided by 7)
- (b) Recognises that the mean is affected by the outlier value of 35, which is unusually high
- (b) Explains that the median (28) is closer to the most typical/common daily customer count and is not skewed by the one high value, making it more representative for practical staff planning
M1.2 · Factors, multiples, primes, HCF, LCM
Show — 4 marks
A museum has two antique clocks in its main hall. The first clock chimes every 24 minutes, and the second clock chimes every 36 minutes. Both clocks chime together at 10:00 am.
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(a) Find the HCF of 24 and 36.
[1 mark]
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(b) Show that the next time both clocks chime together is at 11:12 am.
[3 marks]
Show mark scheme
- (a) HCF = 12 (accept 2² × 3 from prime factorization)
- (b) Method to find LCM: either (24 × 36) ÷ 12, or prime factorization method, or listing multiples
- (b) LCM = 72 (minutes)
- (b) 72 minutes = 1 hour 12 minutes, hence 11:12 am
M1.2 · Factors, multiples, primes, HCF, LCM
Describe — 3 marks
A school organises two regular events: a Mathematics club that meets every 12 days and a Science club that meets every 18 days. Both clubs met on the same day this week. The school wants to understand the pattern of when these clubs coincide and needs to identify key numbers related to their schedules.
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(a) Describe the difference between the highest common factor (HCF) and the lowest common multiple (LCM) of 12 and 18. You should use the context of the clubs' meeting schedules in your explanation.
[2 marks]
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(b) The school also runs a coding club every 8 days. Describe how you would find when all three clubs (Mathematics, Science, and Coding) meet on the same day, and explain which mathematical concept from HCF or LCM you would use.
[1 mark]
Show mark scheme
- (a) Correctly identifies that HCF is the largest number that divides both 12 and 18 (HCF = 6), and explains this relates to the common factors of the meeting schedules
- (a) Correctly identifies that LCM is the smallest number that both 12 and 18 divide into (LCM = 36), and explains this relates to when both clubs meet together again (every 36 days)
- (b) Describes finding the LCM of 8, 12, and 18 (or equivalent method), and correctly identifies that LCM is the concept needed to find when all three clubs coincide on the same day
M1.4 · Estimation, rounding and bounds
Show — 2 marks
A financial analyst is reviewing quarterly revenue data for a retail company. The revenue for three consecutive quarters is recorded as £2.847 million, £3.154 million, and £2.896 million respectively. The analyst needs to round these figures to appropriate levels of precision for a board presentation and to establish realistic bounds for budget forecasting.
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(a) Round £2.847 million to 1 decimal place and to the nearest whole number. Show both answers.
[1 mark]
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(b) The total quarterly revenue is £8.897 million, rounded to 3 decimal places. Show that the lower bound of this total is £8.8965 million.
[1 mark]
Show mark scheme
- (a) Correctly rounds to 1 d.p.: £2.8 million (or 2.8) AND correctly rounds to nearest whole number: £3 million (or 3)
- (b) Identifies that the lower bound for a value rounded to 3 d.p. is the value minus half the place value of the last digit (0.0005), and calculates: 8.897 - 0.0005 = 8.8965
M4.5 · Vectors and transformations
Show — 4 marks
A surveyor is mapping a triangular plot of land. Point A is at the origin. Point B is located at position vector \(\begin{pmatrix} 8 \\ 6 \end{pmatrix}\) relative to A. Point C is located at position vector \(\begin{pmatrix} 2 \\ 10 \end{pmatrix}\) relative to A. The surveyor needs to verify geometric properties and then apply a transformation to create a scaled map.
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(a) Show that the vector from B to C is \(\begin{pmatrix} -6 \\ 4 \end{pmatrix}\).
[1 mark]
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(b) The triangle ABC undergoes an enlargement with centre at the origin and scale factor \(\frac{1}{2}\). Show that the image of point B after this enlargement has position vector \(\begin{pmatrix} 4 \\ 3 \end{pmatrix}\).
[1 mark]
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(c) After the enlargement, the triangle is translated by vector \(\begin{pmatrix} -1 \\ 2 \end{pmatrix}\). Show that the final position of the image of point B is \(\begin{pmatrix} 3 \\ 5 \end{pmatrix}\).
[2 marks]
Show mark scheme
- (a) Correctly calculates \(\vec{BC} = \vec{OC} - \vec{OB} = \begin{pmatrix} 2 \\ 10 \end{pmatrix} - \begin{pmatrix} 8 \\ 6 \end{pmatrix} = \begin{pmatrix} -6 \\ 4 \end{pmatrix}\)
- (b) Applies enlargement scale factor \(\frac{1}{2}\) to position vector of B: \(\frac{1}{2} \begin{pmatrix} 8 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}\)
- (c) Adds the translation vector to the enlarged position: \(\begin{pmatrix} 4 \\ 3 \end{pmatrix} + \begin{pmatrix} -1 \\ 2 \end{pmatrix}\)
- (c) Correctly evaluates to obtain \(\begin{pmatrix} 3 \\ 5 \end{pmatrix}\)
M3.1 · Ratio and proportion
Calculate — 2 marks
A shop manager is dividing a delivery of 120 notebooks between the stationery section and the office supplies section. The notebooks are to be shared in the ratio 3:5.
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(a) Calculate the total number of parts in the ratio 3:5.
[1 mark]
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(b) Calculate how many notebooks go to the stationery section.
[1 mark]
Show mark scheme
- (a) Correctly adds the ratio parts: 3 + 5 = 8
- (b) Correctly calculates the stationery share: \(\frac{3}{8} \times 120 = 45\) notebooks (or equivalent method such as \(120 \div 8 \times 3\))
M6.3 · Scatter graphs and correlation
Compare — 4 marks
A financial analyst collects data on 12 investment portfolios to investigate the relationship between the amount invested (in thousands of pounds) and the annual return (in thousands of pounds). The data collected is: Investment: 5, 8, 10, 12, 15, 18, 20, 22, 25, 28, 30, 35. Annual Return: 0.8, 1.2, 1.5, 1.8, 2.1, 2.9, 3.0, 3.2, 3.8, 4.1, 4.5, 5.2.
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(a) Calculate the correlation coefficient (to 2 decimal places) for this dataset and describe the type and strength of correlation shown.
[2 marks]
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(b) Compare the relationship shown in the data with the statement: 'A higher investment amount guarantees a higher annual return.' Explain why correlation does not prove causation and identify what other factors might affect annual return that are not accounted for in this analysis.
[2 marks]
Show mark scheme
- (a) Correlation coefficient calculated as 0.99 (or 0.98–1.00 range accepted)
- (a) Correctly identifies the correlation as positive and very strong
- (b) Correctly refutes the guarantee claim by explaining that strong correlation does not imply causation or certainty; identifies that the statement is too absolute
- (b) Identifies at least one other relevant factor affecting annual return (e.g. market conditions, investment type, economic climate, risk profile, management fees, inflation) that is not shown in the data
M3.3 · Direct and inverse proportion
Explain — 3 marks
A student investigates how the brightness of a light bulb changes with distance from a light source. She measures the light intensity at different distances from a lamp and records her results in a table.
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(a) The student finds that when she doubles the distance from the lamp, the light intensity becomes one quarter of its original value. Explain what type of proportion this relationship demonstrates.
[1 mark]
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(b) Explain why light intensity and distance follow this type of relationship rather than a direct proportion.
[2 marks]
Show mark scheme
- (a) This is an inverse proportion / inverse square law (1 mark)
- (b) Light spreads out in all directions from the source (1 mark)
- (b) The intensity is distributed over a larger surface area as distance increases / intensity is inversely proportional to the square of the distance (1 mark)
M1.4 · Estimation, rounding and bounds
Suggest — 3 marks
A furniture store is ordering new stock. The manager needs to estimate costs and quantities based on supplier information. The supplier provides measurements and prices that need careful rounding to make sensible business decisions.
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(a) A table costs £47.86. The manager rounds this price to the nearest pound to estimate the total cost of ordering 12 tables. Suggest what the manager's estimate would be.
[1 mark]
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(b) The length of a shelf is measured as 156.4 cm, rounded to 1 decimal place. Suggest the lower bound and upper bound for the actual length of the shelf.
[2 marks]
Show mark scheme
- (a) Rounds £47.86 to £48 and multiplies by 12 to give £576 (or states £48 × 12 = £576)
- (b) States lower bound as 156.35 cm (or ≥156.35)
- (b) States upper bound as 156.45 cm (or <156.45)
M2.2 · Solving equations and inequalities
Show — 3 marks
A student is investigating the relationship between the resistance of a wire and its length. They use the equation R = ρL/A, where R is resistance in ohms, ρ is resistivity, L is length in metres, and A is the cross-sectional area. For a particular wire, ρ = 1.7 × 10⁻⁸ Ω m and A = 2 × 10⁻⁶ m². The student needs to find the length of wire required to achieve a resistance of 8.5 Ω.
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(a) Show that the equation R = ρL/A can be rearranged to L = RA/ρ
[1 mark]
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(b) Using the rearranged equation and the values given in the context, show that the length of wire required is 1000 m.
[2 marks]
Show mark scheme
- (a) Multiply both sides by A and divide both sides by ρ to give L = RA/ρ (or equivalent algebraic steps shown)
- (b) Correct substitution of values: L = (8.5 × 2 × 10⁻⁶) / (1.7 × 10⁻⁸)
- (b) Correct final answer of 1000 m (or 1.0 × 10³ m) with working shown
GCSE Perfect