GCSE Mathematics Foundation Tier: Key Vocabulary & Formulas

Examination Structure, Administrative Rules, and Core Formulae

• General Certificate of Secondary Education (GCSE) Mathematics Unit M6 Paper 2 (With calculator) Foundation Tier [GMC62].

• Examination session parameters:

  • Time allocation: 1 hour ($10.45\,\text{am} - 11.45\,\text{am}$, Wednesday 3 June 2026).

  • Total score achievable: $50\,\text{marks}$ across 17 compulsory questions.

  • Allowed and required equipment: Electronic calculator, ruler, compasses, protractor, and HB pencil for all drawings, constructions, or sketches.

  • Permitted ink: Black ink only (gel pens strictly prohibited).

• Standard geometrical formulae provided on the reference sheet:

  • Area of a trapezium:     $\text{Area} = \frac{1}{2}(a + b)h$     where $a$ and $b$ denote the lengths of the two parallel sides, and $h$ represents the perpendicular distance (height) between them.     

    

  • Volume of a prism:     $\text{Volume} = \text{area of cross section} \times \text{length}$     

    

Distance Metric Conversions and Compass Bearings

• Conversion between imperial miles and metric kilometres (Larne to Ballymena):

  • Given distance: $20\,\text{miles}$.

  • Standard conversion equivalence: $5\,\text{miles} \approx 8\,\text{km}$ (or $1\,\text{mile} \approx 1.6\,\text{km}$).

  • Conversion calculation:     $20\,\text{miles} = 20 \times 1.6 = 32\,\text{km}$     $\text{Alternatively: } 20 \times \frac{8}{5} = 4 \times 8 = 32\,\text{km}$

  • Final converted distance: $32\,\text{km}$.

• Map scale factor calculation (Bangor to Coleraine):

  • Given representative scale: $1\,\text{cm} : 4\,\text{km}$.

  • Measured distance on map: $27.5\,\text{cm}$.

  • Actual real-world distance:     $\text{Actual distance} = 27.5 \times 4\,\text{km} = 110\,\text{km}$

• Reciprocal compass bearings:

  • Given direction: Comber is situated North East ($\text{NE}$) of Newry.

  • The reciprocal (opposite) direction of North East ($\text{NE}$, bearing $045^\circ$) is South West ($\text{SW}$, bearing $225^\circ$).

  • Therefore, Newry is South West of Comber.

Sample Space, Outcome Listings, and Independent Probabilities

• Two-match hockey series possible outcomes:

  • Each independent match allows three mutually exclusive results: Win ($W$), Draw ($D$), or Lose ($L$).

  • By the multiplication principle of combinatorics, total possible outcomes for two matches equal $3 \times 3 = 9$.

  • Systematic listing of all nine distinct outcomes:     $\{WW, WD, WL, DW, DD, DL, LW, LD, LL\}$

• Probability axiomatic boundaries:

  • Statement scenario: Ethan states $P(\text{rain}) = 0.7$, while Abbie asserts $P(\text{rain}) \approx 2$.

  • Abbie is incorrect because all classical probability values $P(E)$ must satisfy the fundamental inequality:     $0 \le P(E) \le 1$

  • A probability value cannot exceed $1$ (or $100\%$ certainty).

• Single selection of colored T-shirts:

  • Distribution of shirts in luggage:

  • Blue: $7$

  • Red: $5$

  • Green: $2$

  • Yellow: $1$

  • Total number of shirts:     $\text{Total} = 7 + 5 + 2 + 1 = 15$

  • Fractional probability of selecting a green T-shirt at random:     $P(\text{Green}) = \frac{2}{15}$

• Fair six-sided die throw:

  • Sample space: $S = \{1, 2, 3, 4, 5, 6\}$, where $n(S) = 6$.

  • Subset of outcomes strictly greater than 4: $\{5, 6\}$, where $n(\text{greater than } 4) = 2$.

  • Probability calculation:     $P(> 4) = \frac{2}{6} = \frac{1}{3}$

• Expected values and inverse sample size calculation:

  • Village morning traffic monitoring: $P(\text{driver speeding}) = 0.1$.

  • Observed number of speeding drivers: $12$.

  • Total number of drivers monitored ($N$):     $N = \frac{\text{Number of speeding drivers}}{P(\text{speeding})} = \frac{12}{0.1} = 120\,\text{drivers}$

• Coin tossing sample space and probability grid:

  • Two fair coins tossed independently, each with outcomes Heads ($H$) or Tails ($T$).

  • Grid of all joint possibilities:

  • $(H, H)$

  • $(H, T)$

  • $(T, H)$

  • $(T, T)$     

    

  • Probability of obtaining two tails simultaneously:     $P(TT) = \frac{1}{4} = 0.25$

Geometric Similarity, Linear Scale Factors, and Perimeter

• Enlargement of 2D polygons from Shape A to Shape B:

  • Corresponding side lengths:

  • Known dimension in Shape A: $6\,\text{cm}$.

  • Corresponding dimension in Shape B: $30\,\text{cm}$.

  • Linear scale factor ($k$):     $k = \frac{30\,\text{cm}}{6\,\text{cm}} = 5$

  • Finding unknown side length $x$ on Shape B:

  • Given corresponding base side on Shape A: $4\,\text{cm}$.

  • Dimension of $x$:       $x = 4 \times k = 4 \times 5 = 20\,\text{cm}$

  • Finding unknown side length $y$ on Shape B:

  • Total perimeter of enlarged Shape B: $110\,\text{cm}$.

  • Sum of known sides of Shape B: $30\,\text{cm}$, $25\,\text{cm}$, and $x = 20\,\text{cm}$.

  • Equation of perimeter:       $30 + 25 + 20 + y = 110$       $75 + y = 110$       $y = 110 - 75 = 35\,\text{cm}$

Linear Arithmetic Sequences and Tile Growth Models

• Tile arrangements utilizing black and white squares:   

  

  • Pattern 1: $5\,\text{black squares}$, $4\,\text{white squares}$, $9\,\text{total squares}$.

  • Pattern 2: $6\,\text{black squares}$, $6\,\text{white squares}$, $12\,\text{total squares}$.

  • Pattern 3: $7\,\text{black squares}$, $8\,\text{white squares}$, $15\,\text{total squares}$.

• Derivation of the $n$-th term formulae:

  • Number of black squares ($B_n$):     $B_n = 5 + (n - 1) \times 1 = n + 4$

  • Number of white squares ($W_n$):     $W_n = 4 + (n - 1) \times 2 = 2n + 2$

  • Total number of squares ($T_n$):     $T_n = B_n + W_n = (n + 4) + (2n + 2) = 3n + 6$

• Values for Pattern 5 ($n = 5$):

  • Black squares: $B_5 = 5 + 4 = 9$

  • White squares: $W_5 = 2(5) + 2 = 12$

  • Total squares: $T_5 = 3(5) + 6 = 21$

• White squares in the tenth pattern ($n = 10$):   $W_{10} = 2(10) + 2 = 22$

  • Correct value among choices ($13$, $40$, $22$, $31$): $22$.

Two-Way Contingency Tables and Categorical Probabilities

• Chess club membership breakdown:   

  

  • Year 8: $6\,\text{Girls}$, $11\,\text{Boys}$ (Subtotal = $17$)

  • Year 9: $9\,\text{Girls}$, $14\,\text{Boys}$ (Subtotal = $23$)

  • Year 10: $13\,\text{Girls}$, $7\,\text{Boys}$ (Subtotal = $20$)

  • Column sums:

  • Total Girls: $6 + 9 + 13 = 28$

  • Total Boys: $11 + 14 + 7 = 32$

  • Overall total club participants: $28 + 32 = 60$

• Probability determinations for a randomly chosen attendee:

  • Probability of selecting a Year 8 boy:     $P(\text{Year 8 Boy}) = \frac{11}{60}$

  • Probability of selecting a Year 9 student:     $P(\text{Year 9}) = \frac{23}{60}$

  • Probability of selecting a girl:     $P(\text{Girl}) = \frac{28}{60} = \frac{7}{15}$

  • Probability of selecting a student not in Year 8:     $P(\text{Not Year 8}) = 1 - P(\text{Year 8}) = 1 - \frac{17}{60} = \frac{43}{60}$     $\text{Alternatively: } \frac{23 + 20}{60} = \frac{43}{60}$

Financial Mathematics and Currency Conversions

• Parisian café pricing structure:

  • Coffee of the Day: €$4.00$

  • Café Espresso: €$5.00$

  • Café au Lait: €$4.50$

  • Café Mocha: €$7.00$

  • Toffee Latte: €$6.00$

• Gilbert family itemized order:

  • $1 \times \text{Coffee of the Day} = €4.00$

  • $1 \times \text{Café au Lait} = €4.50$

  • $2 \times \text{Toffee Latte} = 2 \times €6.00 = €12.00$

• Total cost in Euros (€):   $\text{Total (€)} = 4.00 + 4.50 + 12.00 = 20.50\,\text{€}$

• Conversion to British Pounds (£):

  • Official exchange rate: $£1 = €1.18$.

  • Division formula:     $\text{Cost in } £ = \frac{20.50}{1.18} = 17.37288...$

  • Rounding to two decimal currency places: $£17.37$.

Coordinate Transformations: Translations and Rotations

• Translation of polygon C to polygon D:   

  

  • Coordinate mapping analysis:

  • Vertex of C at $(2, 7)$ maps to vertex of D at $(-5, 6)$.

  • Vertex of C at $(5, 7)$ maps to vertex of D at $(-2, 6)$.

  • Vertex of C at $(5, 3)$ maps to vertex of D at $(-2, 2)$.

  • Vertex of C at $(2, 5)$ maps to vertex of D at $(-5, 4)$.

  • Transformation properties: Congruent, unchanged orientation, indicating a pure translation.

  • Vector calculation:     $\Delta x = -5 - 2 = -7$     $\Delta y = 6 - 7 = -1$

  • Full description: Translation by column vector $\begin{pmatrix} -7 \\ -1 \end{pmatrix}$.

• Anti-clockwise rotation about the origin:   

  

  • Transformation rule for a $90^\circ$ anti-clockwise rotation centered at the origin $(0, 0)$:     $(x, y) \rightarrow (-y, x)$

  • Applying rule to initial vertices:

  • Initial point $(1, 1) \rightarrow (-1, 1)$

  • Initial point $(4, 1) \rightarrow (-1, 4)$

  • Initial point $(4, 3) \rightarrow (-3, 4)$

  • The resulting rotated figure forms a congruent right-angled triangle in quadrant II with vertices at $(-1, 1)$, $(-1, 4)$, and $(-3, 4)$.

Discrete Probability Distributions with Symmetrical Unknowns

• Commuter travel methods for Jack: Car, Bus, Train, or Bicycle.

  • Given discrete probabilities:

  • $P(\text{Car}) = 30\%$

  • $P(\text{Bicycle}) = 6\%$

  • $P(\text{Bus}) = P(\text{Train})$

  • The sum of probabilities across all exhaustive, mutually exclusive categories equals $100\%$:     $P(\text{Car}) + P(\text{Bus}) + P(\text{Train}) + P(\text{Bicycle}) = 100\%$     $30\% + 2 \times P(\text{Bus}) + 6\% = 100\%$     $36\% + 2 \times P(\text{Bus}) = 100\%$     $2 \times P(\text{Bus}) = 64\%$     $P(\text{Bus}) = 32\%$     $P(\text{Train}) = 32\%$

  • Probability that Jack does not travel to work by bus:     $P(\text{Not Bus}) = 100\% - P(\text{Bus}) = 100\% - 32\% = 68\%$

Limits of Accuracy and Error Intervals

• Bag mass measured to the nearest kilogram:

  • Recorded measurement: $14\,\text{kg}$.

  • Measurement unit of precision: $1\,\text{kg}$.

  • Half-unit threshold: $\frac{1}{2} = 0.5\,\text{kg}$.

  • Error interval derivation:

  • Lower bound (minimum possible mass): $14 - 0.5 = 13.5\,\text{kg}$.

  • Upper bound (supremum mass): $14 + 0.5 = 14.5\,\text{kg}$.

  • Complete inequality specification:     $13.5\,\text{kg} \le \text{mass} < 14.5\,\text{kg}$

Geometric Constructions: Perpendicular Bisector

• Construction procedure using straightedge and pair of compasses:

  • Step 1: Place the compass point on one terminus of the line segment.

  • Step 2: Set compass width to an opening strictly greater than half the perceived length of the segment.

  • Step 3: Draw a continuous arc extending both above and below the line segment.

  • Step 4: Keeping the compass width fixed, transfer the compass point to the opposite endpoint of the segment.

  • Step 5: Draw a second arc intersecting the first arc above and below the segment at two distinct intersection points.

  • Step 6: Using a straightedge ruler, align and draw a straight pencil line through both arc intersection points.

  • Procedural rule: All construction arcs must remain clearly visible and un-erased.

Numerical Methods: Trial and Improvement

• Solving the non-linear equation $x^3 - x = 32$ correct to 1 decimal place:

  • Integer iteration:

  • Try $x = 3$:       $3^3 - 3 = 27 - 3 = 24 \quad (24 < 32, \text{ too small})$

  • Try $x = 4$:       $4^3 - 4 = 64 - 4 = 60 \quad (60 > 32, \text{ too big})$

  • Root lies in the interval $3 < x < 4$.

  • One decimal place iteration:

  • Try $x = 3.2$:       $3.2^3 - 3.2 = 32.768 - 3.2 = 29.568 \quad (29.568 < 32, \text{ too small})$

  • Try $x = 3.3$:       $3.3^3 - 3.3 = 35.937 - 3.3 = 32.637 \quad (32.637 > 32, \text{ too big})$

  • Root lies in the interval $3.2 < x < 3.3$.

  • Midpoint test for conclusive rounding:

  • Test $x = 3.25$:       $3.25^3 - 3.25 = 34.328125 - 3.25 = 31.078125 \quad (31.078125 < 32, \text{ too small})$

  • Because the value at $x = 3.25$ yields a result less than $32$, the true root must lie strictly between $3.25$ and $3.30$:       $3.25 < x < 3.3$

  • Every number in this interval rounds to $3.3$ to 1 decimal place.

  • Final root: $x = 3.3$.

Polygon Angles Meeting at a Common Vertex

• Geometry of regular polygons meeting at a point:   

  

  • Constituent shapes:

  • Shape P: Regular pentagon ($n = 5$ sides).

  • Shape T: Equilateral triangle ($n = 3$ sides).

  • Shape R: Rectangle ($n = 4$ sides).

• Interior angle calculations:

  • Interior angle of a regular pentagon ($P$):     $\text{Interior Angle} = \frac{(5 - 2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ$

  • Interior angle of an equilateral triangle ($T$):     $\text{Interior Angle} = \frac{180^\circ}{3} = 60^\circ$

  • Interior angle of a rectangle ($R$):     $\text{Interior Angle} = 90^\circ$

• Angle sum around a vertex:

  • The sum of all angles meeting at a single point on a plane equals $360^\circ$:     $x + 108^\circ + 60^\circ + 90^\circ = 360^\circ$     $x + 258^\circ = 360^\circ$     $x = 360^\circ - 258^\circ = 102^\circ$

  • Final angle: $x = 102^\circ$.

Experimental Probability and Relative Frequency

• Cumulative surveys of Year 13 students regarding part-time employment:   

  

  • Relative frequency formula:     $\text{Relative Frequency} = \frac{\text{Frequency of positive responses}}{\text{Total sample size}}$

• Table row computations:

  • Sample size $10$: Positive count = $6$, Relative frequency = $\frac{6}{10} = 0.6$

  • Sample size $40$: Positive count = $26$, Relative frequency = $\frac{26}{40} = 0.65$

  • Sample size $50$: Positive count = $36$, Relative frequency calculation:     $\text{Relative frequency} = \frac{36}{50} = 0.72$

  • Sample size $150$: Positive count = $111$, Relative frequency = $\frac{111}{150} = 0.74$

  • Sample size $200$: Relative frequency = $0.8$, Positive count calculation:     $\text{Number with a part-time job} = 200 \times 0.8 = 160$

  • Sample size $300$: Positive count = $231$, Relative frequency = $\frac{231}{300} = 0.77$

• Tabular completion values:

  • Missing relative frequency for $50$ students: $0.72$.

  • Missing student frequency for $200$ students: $160$.