Pearson Edexcel International GCSE Mathematics A Paper 2F (Foundation Tier) - November 2023 Study Guide

Examination Overview and General Instructions

The Pearson Edexcel International GCSE in Mathematics A, Paper 2F (Foundation Tier), was administered on Friday, 10 November 2023. This paper holds a total of 100 marks and must be completed within a 2-hour duration. Candidates are required to use a calculator and specific geometric tools, including a ruler graduated in centimetres and millimetres, a protractor, a pair of compasses, a pen, an HB pencil, and an eraser. Tracing paper is also permitted. The instructions strictly mandate the use of black ink or ball-point pen for all written work. Candidates must provide sufficient working to be awarded marks; a correct answer without visible working may result in no credit. The paper comprises twenty-six questions that must be answered in the spaces provided. A critical instruction regarding the formulae page is that candidates must not write on it, as any content written there will result in no credit.

Fundamental Mathematics Formulae for Foundation Tier

The formulae sheet provided for the Foundation Tier includes essential geometric calculations. The area of a trapezium is calculated using the formula $\text{Area} = \frac{1}{2}(a + b)h$, where $a$ and $b$ represent the lengths of the parallel sides and $h$ represents the perpendicular height. For prisms, the volume is determined by the formula $\text{Volume of prism} = \text{area of cross section} \times \text{length}$. Specific formulae for cylinders include the volume, calculated as $\text{Volume of cylinder} = \text{π}r^2 h$, and the curved surface area, calculated as $\text{Curved surface area of cylinder} = 2\text{π}rh$, where $r$ is the radius and $h$ is the height.

Data Analysis: International Sugar Production

Question 1 provides a dataset regarding the weight of sugar produced by five specific countries in a single year. the weights, provided in tonnes, are as follows: Japan produced $72,900\,\text{tonnes}$, Barbados produced $15,745\,\text{tonnes}$, Kenya produced $592,668\,\text{tonnes}$, Gabon produced $23,787\,\text{tonnes}$, and Malaysia produced $28,149\,\text{tonnes}$. Candidates are tasked with identifying Kenya as the country with the greatest production weight. The number $28,149$ must be written out in words as "twenty-eight thousand, one hundred and forty-nine." In the number $23,787$, the value of the digit $8$ is specified as $80$ (or tens). Rounding the production weight for Barbados ($15,745\,\text{tonnes}$) to the nearest thousand yields a value of $16,000$.

Statistical Representation: Parcel Pictograms and Ratios

Question 2 utilizes a pictogram to describe the number of parcels posted by a company over four days (Monday through Friday). The key for the pictogram establishes that one symbol represents $16\,\text{parcels}$. On Tuesday, the pictogram displays one full symbol and a half-symbol, requiring calculation for the total. For Friday, candidates are informed that $24\,\text{parcels}$ were posted and must represent this on the pictogram using one and a half symbols ($16 + 8 = 24$). The question further requires calculating how many more parcels were posted on Wednesday than on Monday based on the symbols provided. Finally, a ratio must be established and simplified for the number of parcels posted on Monday compared to the number of parcels posted on Thursday.

Fractions, Decimals, and Percentages

Question 3 focuses on numerical conversions and comparisons. The decimal $0.03$ is converted to the fraction $\frac{3}{100}$. The decimal $0.9$ is converted to a percentage by multiplying by $100$, resulting in $90\%$. A set of decimals must be ordered from smallest to largest: $0.204$, $0.24$, $0.4$, $0.408$, and $0.48$. Additionally, a calculation involving decimals and fractions, specifically $0.93 + \frac{7}{10}$, must be computed to find the decimal sum of $1.63$.

Practical Calculations: Walking Distance and Unit Consistency

Question 4 describes Barney’s four walks on a Tuesday. The known lengths of the first three walks are $800\,\text{metres}$, $2\,\text{kilometres}$, and $1.7\,\text{kilometres}$. The final walk is expressed as $x\,\text{metres}$. To solve for $x$, all units must be consistent. Given the total length of the four walks is $6250\,\text{metres}$, the calculation follows: $800 + 2000 + 1700 + x = 6250$. This simplifies to $4500 + x = 6250$, leading to $x = 1750\,\text{metres}$.

Probability Scales and Counter Bags

Question 5 involves a bag containing $8\,\text{counters}$, where $6$ are orange and the remaining $2$ are purple. Delilah selects one counter at random. Candidates must mark the probability of selecting an orange counter on a scale from $0$ to $1$. The probability is calculated as $\frac{6}{8}$ or $0.75$. Furthermore, the probability of selecting a yellow counter must be marked. Since there are no yellow counters in the bag, the probability is $0$, representing an impossible event.

Geometry: Grid Drawings and Areas

Question 6 requires construction on a centimetre grid. Part (a) asks for the drawing of a right-angled triangle. Part (b) requires the drawing of a rectangle with an exact area of $20\,cm^2$. Possible dimensions for such a rectangle include $4\,cm \times 5\,cm$ or $2\,cm \times 10\,cm$.

Time Calculations and Bus Timetables

Question 7 covers time conversions and duration analysis. The time $4:30\,\text{pm}$ translates to $16:30$ in the 24-hour clock format. Using a bus timetable segment (Beetown at $08:45$, Corthill at $09:50$, and Pilton at $10:36$), candidates must determine the difference in travel times. The journey from Beetown to Corthill takes $65\,\text{minutes}$ ($08:45$ to $09:50$), while the journey from Corthill to Pilton takes $46\,\text{minutes}$ ($09:50$ to $10:36$). The difference is $65 - 46 = 19\,\text{minutes}$.

Angle Properties and Parallel Lines

Question 8 presents a diagram with straight lines $ABCD$ and $EFGH$, and parallel lines $KFBJ$ and $MGCL$. Angles provided include $\text{angle } ABJ = 125^∘$ and $\text{angle } BFG = 32^∘$. Candidates must find the value of $x$, which is given as $\text{angle } FGM$. Through parallel line properties (alternate or corresponding angles), $x$ is identified. To find the value of $y$ (angle $LCD$), candidates must apply geometric reasoning, such as angles on a straight line or interior angles between parallel lines, and must explicitly state the geometric reason for their answer.

Financial Proportions: Carrots and Potatoes

Question 9 details the cost of groceries in South African rand. The problem states that $3\,kg$ of carrots and $5\,kg$ of potatoes cost a total of $207\,\text{rand}$. Given that $2\,kg$ of carrots cost $48\,\text{rand}$, the price per kilogram of carrots is $24\,\text{rand}$. Therefore, $3\,kg$ of carrots cost $3 \times 24 = 72\,\text{rand}$. Subtracting this from the total gives the cost of the potatoes: $207 - 72 = 135\,\text{rand}$. The cost per kilogram of potatoes is then $135 \times \frac{1}{5} = 27\,\text{rand}$.

Number Machines and Algebraic Expressions

Question 10 introduces two number machines. The first involves an input which is multiplied by $y$ and then reduced by $17$ to produce an output. If an input of $7$ produces an output of $60$, the equation is $7y - 17 = 60$. Adding $17$ to both sides gives $7y = 77$, resulting in $y = 11$. The second machine takes an input $x$, adds $7$, and then divides the result by $5$. The algebraic expression for this output is written as $\frac{x + 7}{5}$.

Currency Conversion Graphs

Question 11 uses a conversion graph between Australian dollars (AUD) and euros (EUR). Using the graph, $35\,\text{Australian dollars}$ is converted to euros, and $20\,\text{euros}$ is converted back to Australian dollars. For a larger conversion, such as Lachlan changing $500\,\text{Australian dollars}$, candidates must use a known conversion factor from the graph (e.g., $50\,\text{AUD} \rightarrow \text{euros value}$) and multiply by $10$ to determine the total euros received.

Algebraic Manipulation, Expansion, and Factorization

Question 12 focuses on core algebraic skills. Part (a) requires expanding $x(x + 3)$, which results in $x^2 + 3x$. Part (b) asks to factorise $8p + 10$, which yields $2(4p + 5)$. Part (c) involves rearranging the formula $y = eh - f$ to make $e$ the subject. First, $f$ is added to both sides: $y + f = eh$. Then, dividing by $h$ results in $e = \frac{y + f}{h}$. Part (d) provides a constraint on a whole number $w$: $w > 7$ and $w \text{ ≤ } 10$. The possible whole number values for $w$ are $8$, $9$, and $10$.

Calculator Proficiency and Value for Money

Question 14 requires the use of a calculator to evaluate a complex numerical expression involving square roots and squares: $\sqrt{\frac{5.2 + 8.7}{14.5 - 2.5^2}}$. Candidates must write down all figures displayed on the calculator. Question 15 involves a comparative cost analysis for fudge bags. A small bag ($150\,g$) costs \text{}1.80, while a large bag ($400\,g$) costs \text{}5. To determine better value, candidates must calculate the cost per gram or grams per pound (e.g., small bag is $1.2\,\text{p per gram}$, large bag is $1.25\,\text{p per gram}$. Thus, the small bag is better value).

Geometry: Similar Triangles and Ratios

Question 18 involves triangles $ABC$ and $ECD$ where $ACD$ and $EBC$ are straight lines. Dimensions provided are $AB = 10\,cm$, $AC = 8\,cm$, $EB = 5\,cm$, and $CD = 14\,cm$. The length of $ED$ is labeled as $w\,cm$. By identifying the triangles as similar (as they share vertically opposite angles and alternate/corresponding properties if lines are parallel, though not explicitly stated), the scale factor is derived from known sides (e.g., $\frac{CD}{AC} = \frac{14}{8} = 1.75$). The value of $w$ is then found by multiplying the corresponding side by this scale factor: $w = 10 \times 1.75 = 17.5\,cm$. The answer must be given to one decimal place.

Advanced Algebraic Simplification and Exponents

Question 20 segments into four algebraic tasks. Part (a) requires solving an equation involving fractions: $\frac{2x - 5}{6} - \frac{2x - 5}{2}$. (Note: clear algebraic working is mandatory). Part (b) involves simplifying $\frac{h^{15}}{h^3}$, which follows exponent rules ($h^{15-3}$) to become $h^{12}$. Part (c) requires fully simplifying $(2g^3 k^5) \times (4g^2 k)$, resulting in $8g^5 k^6$. Part (d) requires finding $n$ given that $y^5 \times y^n \times y^7 = y^{12}$. Using exponent multiplication rules ($5 + n + 7 = 12$), it is found that $n + 12 = 12$, so $n = 0$.

Number Systems: Compound Interest and Mean Calculations

Question 22 requires showing that $3 \frac{3}{7} \times 2 \frac{2}{3} = 12/7$ (Note: transcription may vary, standard operation is conversion to improper fractions: $\frac{24}{7} \times \frac{8}{3} = \frac{64}{7} = 9 \frac{1}{7}$). Question 23 involves depreciation: a boat bought for $\$26,800$ loses $8\%$ value annually. After 3 years, the value is $\$26,800 \times (0.92)^3$. The result must be rounded to the nearest dollar. Question 24 addresses the mean: an 8-match mean of $6$ goals implies a total of $48$ goals. To reach a 10-match mean of $7$ ($70$ total goals), the team must score $70 - 48 = 22$ goals in the last two matches. If they score $k$ goals in each, $2k = 22$, so $k = 11$.

Coordinate Geometry and Hexagonal Area

Question 25 requires finding the equation of a line through $(0, -3)$ and $(2, 0)$. The gradient $m$ is $\frac{0 - (-3)}{2 - 0} = 1.5$, and the y-intercept $c$ is $-3$. The equation is $y = 1.5x - 3$. Question 26 involves a hexagon $ABCDEF$ with side lengths $AB = 25$, $BC = x + 2$, $CD = 8$, $EF = 7$, and $AF = x + 6$. The total area is $258\,cm^2$. To find $x$, the hexagon must be split into two rectangles. The area equation $(25(x+2)) + (7 \times 8)$ or similar combinations must be solved to find the value of $x$.