2024 CCEA GCSE Mathematics 2210 Practice Paper with Answers

(a) \( \dfrac{5.0 \times 10^{-6}}{3.2 \times 10^{-8}} = 156.25 \), which rounds to 156 (3 s.f.). (b) \( 5.03 \times 10^{-6} \) rounded to 2 significant figures is \( 5.0 \times 10^{-6} \) m.

(a) M1: correct division set up \( 5.0\times10^{-6} \div 3.2\times10^{-8} \); A1: unrounded value 156.25 (or equivalent) seen; A1: 156 (3 s.f., ft from correct unrounded value). (b) A1: \( 5.0 \times 10^{-6} \) m correctly stated in standard form.

\( \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \); \( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \); so \( 5\sqrt{2} - 2\sqrt{2} = 3\sqrt{2} \).

M1: \( \sqrt{50} = 5\sqrt{2} \) correctly simplified; M1: \( \sqrt{8} = 2\sqrt{2} \) correctly simplified; A1: \( 3\sqrt{2} \).

Total parts = 3 + 4 + 5 = 12. Value of 1 part = \( 8400 \div 12 = £700 \). Amounts: \( 3 \times 700 = £2100 \); \( 4 \times 700 = £2800 \); \( 5 \times 700 = £3500 \).

M1: total parts = 12 found; M1: value of 1 part = £700 found; A1: two of the three amounts correct; A1: all three amounts correct (£2100, £2800, £3500).

This is an inverse proportion problem. Total worker-days needed = \( 5 \times 8 = 40 \). With 4 workers: \( 40 \div 4 = 10 \) days.

M1: total worker-days = 40 found (recognising inverse proportion); M1: 40 ÷ 4 set up; A1: 10 days.

Using the compound interest multiplier \( 1.025 \): total value \( = 3000 \times 1.025^3 = 3000 \times 1.076890625 = £3230.67 \)... recalculated precisely: \( 3000 \times 1.025^3 = 3230.859... \), which rounds to £3230.86.

M1: correct multiplier 1.025 used; M1: \( 3000 \times 1.025^3 \) calculated; A1: £3230.86 (accept values in the range £3230.85–£3230.86, depending on rounding method).

Upper bound of length = 8.45 cm; upper bound of width = 5.25 cm. Upper bound of area \( = 8.45 \times 5.25 = 44.3625 \) cm² (accept 44.4 cm² to 3 s.f.).

M1: upper bound length = 8.45 cm identified; M1: upper bound width = 5.25 cm identified; A1: 44.3625 cm² (accept 44.4 cm² to 3 s.f.).

Area of path \( = 12 \times 2 = 24 \) m². Area of one slab \( = 0.6 \times 0.4 = 0.24 \) m². Number of slabs \( = 24 \div 0.24 = 100 \).

M1: area of path = 24 m² found; M1: area of one slab = 0.24 m² found; A1: 100 slabs.

\( (2x+3)(x-5) = 2x^2 - 10x + 3x - 15 = 2x^2 - 7x - 15 \).

M1: at least 3 of the 4 terms correct from expansion; M1: correct four terms \( 2x^2 - 10x + 3x - 15 \); A1: simplified to \( 2x^2 - 7x - 15 \).

We need two numbers that multiply to give -14 and add to give -5: these are -7 and 2. So \( x^2 - 5x - 14 = (x-7)(x+2) = 0 \). Setting each factor to zero: \( x = 7 \) or \( x = -2 \).

M1: attempt to find factors of -14 summing to -5; M1: correct factorisation \( (x-7)(x+2) \); A1: x = 7; A1: x = -2; A1: both solutions clearly stated as the final answer.

Using \( x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \) with \( a=2, b=3, c=-5 \): \( x = \dfrac{-3 \pm \sqrt{9+40}}{4} = \dfrac{-3 \pm \sqrt{49}}{4} = \dfrac{-3 \pm 7}{4} \). So \( x = 1 \) or \( x = -2.5 \).

M1: correct substitution into the quadratic formula; M1: discriminant = 49 correctly calculated; A1: \( \sqrt{49} = 7 \); A1: x = 1; A1: x = -2.5.

Length = \( w+3 \). Area equation: \( w(w+3) = 40 \), so \( w^2 + 3w - 40 = 0 \). Factorising: \( (w+8)(w-5) = 0 \), giving \( w = 5 \) or \( w = -8 \). Since a width cannot be negative, \( w = -8 \) is rejected. So width = 5 cm, and length = \( 5+3 = 8 \) cm.

M1: correct equation formed \( w(w+3)=40 \) or \( w^2+3w-40=0 \); M1: correct factorisation \( (w+8)(w-5) \) or correct use of the quadratic formula; A1: w = 5, with w = -8 correctly rejected and a reason given; A1: length = 8 cm; A1: both dimensions clearly stated with units.

(a) \( 6x^2 - 24 = 6(x^2-4) = 6(x-2)(x+2) \), using the difference of two squares. (b) Setting \( 6(x-2)(x+2) = 0 \): since the factor 6 cannot be zero, either \( x-2=0 \) or \( x+2=0 \), giving \( x = 2 \) or \( x = -2 \).

(a) M1: 6 taken out as a common factor, giving \( 6(x^2-4) \); A1: fully factorised \( 6(x-2)(x+2) \) (difference of two squares). (b) M1: sets each factor to zero; A1: x = 2; A1: x = -2.

Multiplying both sides by \( (x-1) \): \( y(x-1) = 3x+2 \), so \( yx - y = 3x + 2 \). Collecting x terms on one side: \( yx - 3x = y + 2 \). Factorising: \( x(y-3) = y+2 \). Dividing: \( x = \dfrac{y+2}{y-3} \).

M1: multiplies both sides by (x-1) to clear the fraction; M1: expands to give \( yx - y = 3x + 2 \); M1: collects x terms on one side, \( yx - 3x = y+2 \); A1: factorises to \( x(y-3) = y+2 \); A1: \( x = \dfrac{y+2}{y-3} \).

From the second equation, \( x = y + 2 \). Substituting into the first equation: \( 3(y+2) + 2y = 16 \), so \( 3y + 6 + 2y = 16 \), giving \( 5y = 10 \), so \( y = 2 \). Then \( x = 2 + 2 = 4 \).

M1: rearranges one equation to make x or y the subject (e.g. x = y+2); M1: correct substitution into the other equation; M1: correct simplification to solve for one variable (5y=10); A1: y = 2; A1: x = 4.

Frequency density = frequency ÷ class width. For 0≤t<10 (width 10): \( 8 \div 10 = 0.8 \). For 10≤t<20 (width 10): \( 22 \div 10 = 2.2 \). For 20≤t<30 (width 10): \( 18 \div 10 = 1.8 \). For 30≤t<50 (width 20): \( 12 \div 20 = 0.6 \). A histogram would show frequency density on the vertical axis and time on the horizontal axis, with continuous bars (no gaps) drawn between the correct class boundaries at the heights calculated above.

M1: correct class widths identified (10, 10, 10, 20); M1: frequency density formula (frequency ÷ width) applied to at least two classes; A1: two frequency densities correct; A1: all four frequency densities correct (0.8, 2.2, 1.8, 0.6); A1: correct description of histogram construction (frequency density on the y-axis, continuous bars matching class boundaries, no gaps).

With n = 80: the median is the 40th value, which falls in the 10–15 kg class (cumulative frequency rises from 34 to 62 across this class of width 5 kg): \( \text{median} \approx 10 + \dfrac{40-34}{62-34}\times5 = 10+1.07 = 11.1 \) kg. The lower quartile is the 20th value, in the 5–10 kg class (cumulative frequency 10 to 34, width 5): \( Q_1 \approx 5 + \dfrac{20-10}{34-10}\times5 = 5+2.08 = 7.1 \) kg. The upper quartile is the 60th value, in the 10–15 kg class (cumulative frequency 34 to 62, width 5): \( Q_3 \approx 10 + \dfrac{60-34}{62-34}\times5 = 10+4.64 = 14.6 \) kg. Interquartile range \( = Q_3 - Q_1 \approx 14.6 - 7.1 = 7.6 \) kg (using unrounded intermediate values).

(a) M1: correct linear interpolation method shown/identified within the correct class; A1: median ≈ 11.1 kg (accept 10.9–11.2). (b) M1: correct method for Q1 (≈ 7.1 kg); M1: correct method for Q3 (≈ 14.6 kg); A1: IQR ≈ 7.6 kg (accept 7.3–7.8, error carried forward from Q1/Q3).

\( \Sigma(\text{goals} \times \text{frequency}) = (0\times3)+(1\times6)+(2\times7)+(3\times3)+(4\times1) = 0+6+14+9+4 = 33 \). Total number of matches = 20. \( \text{Mean} = 33 \div 20 = 1.65 \).

M1: correct products calculated (0, 6, 14, 9, 4); M1: sum of products = 33; M1: total frequency = 20 identified; A1: mean = 1.65.

\( P(\text{both red}) = \dfrac{5}{8}\times\dfrac{4}{7} = \dfrac{20}{56} \). \( P(\text{both blue}) = \dfrac{3}{8}\times\dfrac{2}{7} = \dfrac{6}{56} \). \( P(\text{same colour}) = \dfrac{20}{56}+\dfrac{6}{56} = \dfrac{26}{56} = \dfrac{13}{28} \).

M1: \( P(RR) = \dfrac{5}{8}\times\dfrac{4}{7} \) correctly calculated; M1: \( P(BB) = \dfrac{3}{8}\times\dfrac{2}{7} \) correctly calculated; M1: the two probabilities correctly added; A1: \( \dfrac{13}{28} \) (or equivalent, e.g. 0.464).

(a) By Pythagoras: \( \text{height} = \sqrt{6.5^2 - 2.5^2} = \sqrt{42.25-6.25} = \sqrt{36} = 6.0 \) m. (b) \( \cos\theta = \dfrac{2.5}{6.5} \), so \( \theta = \cos^{-1}(2.5/6.5) = 67.4° \) (1 d.p.).

(a) M1: correct Pythagoras set up \( 6.5^2 - 2.5^2 \); M1: = 36; A1: height = 6.0 m. (b) M1: correct trig ratio set up (e.g. \( \cos\theta = 2.5/6.5 \) or \( \tan\theta = 6/2.5 \)); M1: correct inverse trig operation used; A1: 67.4° (1 d.p.).

Since angle A = 35° and BC is opposite A while AB is adjacent to A: \( \tan(35°) = \dfrac{BC}{AB} \), so \( BC = 8 \times \tan(35°) = 8 \times 0.70021 = 5.6017 \), which rounds to 5.60 cm (3 s.f.).

M1: correct trigonometric ratio identified (\( \tan A = \text{opposite}/\text{adjacent} = BC/AB \)); M1: correct equation \( BC = 8\tan(35°) \) formed; A1: unrounded value ≈5.6017 seen; A1: 5.60 cm (3 s.f.).

Radius = 7 cm. (a) \( C = \pi d = \pi \times 14 = 43.98 \), rounding to 44.0 cm (3 s.f.). (b) \( A = \pi r^2 = \pi \times 7^2 = \pi \times 49 = 153.94 \), rounding to 154 cm² (3 s.f.).

(a) M1: \( C = \pi \times 14 \) set up; A1: 44.0 cm (3 s.f.). (b) M1: radius = 7 cm identified; M1: \( A = \pi \times 7^2 \) set up; A1: 154 cm² (3 s.f.).

\( \text{Area of sector} = \dfrac{\theta}{360}\times\pi r^2 = \dfrac{120}{360}\times\pi\times9^2 = \dfrac{1}{3}\times254.47 = 84.82 \), which rounds to 84.8 cm² (3 s.f.).

M1: correct fraction of circle identified (120/360 = 1/3); M1: full circle area calculated \( \pi \times 9^2 = 254.47 \); M1: correct formula applied (fraction × full area); A1: unrounded value ≈84.82; A1: 84.8 cm² (3 s.f.).

\( \text{Volume} = 25 \times 10 \times 2 = 500 \) m³. Converting to litres: \( 500 \times 1000 = 500\,000 \) litres.

M1: correct formula for volume of a cuboid identified (length × width × height); M1: \( 25\times10\times2 \) calculated; A1: 500 m³; M1: correct conversion factor used (×1000); A1: 500,000 litres.

\( 3x - 7 \leq 2x + 5 \); collecting x terms: \( 3x - 2x \leq 5 + 7 \), so \( x \leq 12 \). On a number line, this is represented with a closed (filled) circle at x = 12, since the inequality includes 12, with an arrow extending to the left, indicating all values less than or equal to 12.

M1: correctly collects x terms on one side (\( 3x - 2x \leq 5+7 \) or equivalent); A1: x ≤ 12; A1: correct description of number line representation (closed circle at 12, arrow pointing left/towards negative values).

Sum of interior angles of an n-sided polygon \( = (n-2)\times180° \); for an octagon, \( n=8 \): \( (8-2)\times180 = 1080° \). For a regular octagon, each interior angle \( = 1080 \div 8 = 135° \).

M1: sum of interior angles = \( (8-2)\times180 = 1080° \) (or exterior angle method: \( 360/8=45° \), interior \( =180-45 \)); A1: 135°.

(a) Under a 90° clockwise rotation about the origin, \( (x,y) \to (y,-x) \). So \( (1,1)\to(1,-1) \); \( (3,1)\to(1,-3) \); \( (1,4)\to(4,-1) \). (b) Under enlargement scale factor 2 centre the origin, \( (x,y)\to(2x,2y) \). So \( (1,1)\to(2,2) \); \( (3,1)\to(6,2) \); \( (1,4)\to(2,8) \). (c) Since triangle C was formed from triangle A by an enlargement of scale factor 2 centre the origin, the inverse transformation mapping C back onto A is an enlargement, scale factor \( \dfrac{1}{2} \), centre (0,0).

(a) M1: correct rotation rule applied \( (x,y)\to(y,-x) \); A1: two vertices correct; A1: all three vertices correct. (b) M1: correct enlargement rule \( (x,y)\to(2x,2y) \) applied; A1: all three vertices of C correct. (c) A1: enlargement identified with scale factor 1/2; A1: centre (0,0) stated.

(a) Gradient \( = \dfrac{13-5}{6-2} = \dfrac{8}{4} = 2 \). (b) Using \( y=mx+c \) with gradient 2 and the point (2,5): \( 5 = 2(2)+c \), so \( c=1 \); the equation is \( y=2x+1 \). (c) Parallel lines have the same gradient (2); using the point (0,-3), \( c=-3 \), giving \( y=2x-3 \).

(a) M1: correct gradient formula set up; A1: gradient = 2. (b) M1: substitutes a point and the gradient into y=mx+c to find c; A1: y = 2x + 1. (c) M1: recognises parallel lines share the same gradient; A1: y = 2x - 3.

Equating the expressions for y: \( x^2 - 2x = x + 4 \), so \( x^2 - 3x - 4 = 0 \). Factorising: \( (x-4)(x+1) = 0 \), giving \( x=4 \) or \( x=-1 \). When \( x=4 \): \( y = 4+4 = 8 \). When \( x=-1 \): \( y = -1+4 = 3 \).

M1: correctly equates the two expressions and rearranges to \( x^2-3x-4=0 \); M1: correct factorisation \( (x-4)(x+1) \); A1: x = 4, y = 8; A1: x = -1, y = 3.

The outcomes that sum to 9 are: (3,6), (4,5), (5,4), (6,3) — 4 outcomes. There are \( 6\times6=36 \) equally likely total outcomes. \( P(\text{total}=9) = \dfrac{4}{36} = \dfrac{1}{9} \).

M1: total number of equally likely outcomes = 36 identified; M1: 4 favourable outcomes correctly listed/identified; A1: probability = 1/9 (accept 4/36 unsimplified, or equivalent decimal 0.111).

Area scale factor \( = (\text{linear scale factor})^2 = \left(\dfrac{5}{3}\right)^2 = \dfrac{25}{9} \). Area of Q \( = 27 \times \dfrac{25}{9} = 75 \) cm².

M1: recognises area scale factor = (linear scale factor)²; M1: calculates \( (5/3)^2 = 25/9 \); M1: \( 27 \times 25/9 \) set up; A1: 75 cm².

Using inclusion-exclusion, the number studying French or Spanish (or both) \( = 18+15-7 = 26 \). Number studying neither \( = 30-26 = 4 \). \( P(\text{neither}) = \dfrac{4}{30} = \dfrac{2}{15} \).

M1: correct use of inclusion-exclusion \( (18+15-7) \); M1: = 26 correctly calculated; M1: \( 30-26=4 \) identified; A1: probability = 2/15 (or 4/30, accept equivalent decimal ≈0.133).

\( x^2+6x+5 = (x+3)^2 - 9 + 5 = (x+3)^2 - 4 \). Setting this to zero: \( (x+3)^2 = 4 \), so \( x+3 = \pm2 \), giving \( x=-1 \) or \( x=-5 \).

M1: correctly completes the square, \( (x+3)^2 - 9 \); M1: simplifies to \( (x+3)^2 - 4 = 0 \) or \( (x+3)^2 = 4 \); A1: x = -1; A1: x = -5.

\( 20t - 5t^2 = 15 \), so \( 5t^2 - 20t + 15 = 0 \), giving \( t^2 - 4t + 3 = 0 \). Factorising: \( (t-1)(t-3) = 0 \), so \( t=1 \) or \( t=3 \).

M1: correctly forms the equation \( 20t-5t^2=15 \); M1: rearranges and simplifies to \( t^2-4t+3=0 \); M1: correct factorisation \( (t-1)(t-3) \); A1: t = 1 s and t = 3 s both stated.

By Pythagoras: \( \text{distance} = \sqrt{12^2+9^2} = \sqrt{144+81} = \sqrt{225} = 15.0 \) km.

M1: correct Pythagoras set up \( 12^2+9^2 \); M1: = 225; A1: 15.0 km.

Since the tangent PT is perpendicular to the radius OT at T, triangle OTP is right-angled at T. By Pythagoras: \( PT = \sqrt{OP^2-OT^2} = \sqrt{13^2-5^2} = \sqrt{169-25} = \sqrt{144} = 12 \) cm.

M1: recognises triangle OTP is right-angled at T (tangent-radius property); M1: correct Pythagoras set up \( 13^2-5^2 \); A1: PT = 12 cm.

When \( y=0 \): \( 2x=12 \), so \( x=6 \), giving the point (6,0). When \( x=0 \): \( 3y=12 \), so \( y=4 \), giving the point (0,4).

M1: sets y = 0 and solves for x; A1: (6,0); A1: (0,4) with corresponding method for x = 0.

A suitable method would be to distribute a short questionnaire asking each pupil to record the average time (in minutes) spent on homework per week, alongside collecting their most recent test score from school records. To ensure the sample is representative of the whole school population, a stratified random sample should be used: pupils are selected from each year group (or class) in proportion to the size of that year group, with pupils within each stratum chosen using random sampling (e.g. random number generation), so that no particular year group or ability level is over- or under-represented.

1 mark: a suitable, valid method of collecting the required data described (e.g. questionnaire/survey recording time and score); 1 mark: stratified sampling identified as the method for selecting a representative sample; 1 mark: correct explanation that the sample is stratified by an appropriate variable (e.g. year group) in proportion to population size, with random selection within each stratum.

The lower bound of the volume, measured to the nearest cm³, is 215.5 cm³. The side length of the cube is \( \sqrt[3]{215.5} = 5.978 \), which rounds to 5.98 cm (3 s.f.).

M1: lower bound of volume identified as 215.5 cm³; M1: cube root of 215.5 calculated; A1: 5.98 cm (3 s.f.).

\( P(\text{same colour}) = P(RR)+P(BB)+P(GG)+P(YY) = 4\times\left(\dfrac{1}{4}\times\dfrac{1}{4}\right) = 4\times\dfrac{1}{16} = \dfrac{1}{4} \). (Equivalently: whatever colour is spun first, the probability the second spin matches it is always 1/4.)

M1: correct method (e.g. \( 4\times(1/4)^2 \), or recognising \( P(\text{match})=1/4 \) directly); A1: 1/4.

Numerator: \( x^2-9 = (x-3)(x+3) \). Denominator: \( x^2-x-6 = (x-3)(x+2) \). Cancelling the common factor \( (x-3) \): \( \dfrac{(x-3)(x+3)}{(x-3)(x+2)} = \dfrac{x+3}{x+2} \).

M1: both numerator and denominator correctly factorised; A1: correctly cancels the common factor to give \( \dfrac{x+3}{x+2} \).

(a) \( y = kx^2 \). Substituting \( x=4, y=48 \): \( 48 = k(4^2) = 16k \), so \( k=3 \); the formula is \( y=3x^2 \). (b) \( y = 3(6^2) = 3\times36 = 108 \).

(a) M1: correct proportional relationship set up (y=kx²) and substitution to find k; A1: \( y=3x^2 \). (b) M1: correct substitution x=6 into the formula; A1: y = 108.

(a) Area scale factor \( = (2/3)^2 = 4/9 \); new area \( = 18 \times 4/9 = 8 \) cm². (b) The enlarged shape is smaller than the original, because a scale factor between 0 and 1 (here, 2/3) produces a reduction rather than an increase in size.

(a) M1: area scale factor \( = (2/3)^2 = 4/9 \) identified; A1: 8 cm². (b) A1: smaller identified; A1: correct explanation that a scale factor between 0 and 1 produces a reduction.

Since \( P(\text{red}) = 0.35 \), the number of red marbles \( = 0.35\times20 = 7 \), so the number of green marbles \( = 20-7 = 13 \). \( P(\text{both green}) = \dfrac{13}{20}\times\dfrac{12}{19} = \dfrac{156}{380} = \dfrac{39}{95} \approx 0.411 \).

M1: number of green marbles = 13 correctly identified; M1: \( P(GG) = 13/20 \times 12/19 \) correctly set up; A1: 39/95 (or equivalent, e.g. 0.411, accept 156/380 unsimplified).

(a) \( \dfrac{dy}{dx} = 3x^2-6x-9 \). (b) Setting \( 3x^2-6x-9=0 \): dividing by 3, \( x^2-2x-3=0 \), so \( (x-3)(x+1)=0 \), giving \( x=3 \) or \( x=-1 \). When \( x=-1 \): \( y=(-1)^3-3(-1)^2-9(-1)+5=-1-3+9+5=10 \). When \( x=3 \): \( y=27-27-27+5=-22 \). Turning points: (-1,10) and (3,-22). (c) \( \dfrac{d^2y}{dx^2}=6x-6 \). At \( x=-1 \): \( 6(-1)-6=-12<0 \), so (-1,10) is a maximum. At \( x=3 \): \( 6(3)-6=12>0 \), so (3,-22) is a minimum.

(a) M1: correct differentiation of at least two terms; A1: \( 3x^2-6x-9 \). (b) M1: sets the derivative to zero and simplifies (e.g. \( x^2-2x-3=0 \)); M1: correct factorisation/solutions x=-1, x=3; A1: both coordinates correctly found ((-1,10) and (3,-22)). (c) M1: second derivative test (or valid alternative, e.g. sign change of gradient) correctly applied; A1: correctly identifies (-1,10) as a maximum and (3,-22) as a minimum.

(a) \( \dfrac{dy}{dx} = 2x-4 \); at \( x=1 \), gradient \( = 2(1)-4=-2 \). When \( x=1 \), \( y=1-4+7=4 \), so the point is (1,4). Equation of tangent: \( y-4=-2(x-1) \), so \( y=-2x+2+4=-2x+6 \). (b) The tangent crosses the x-axis when \( y=0 \): \( 0=-2x+6 \), so \( x=3 \), giving the point (3,0).

(a) M1: correct differentiation \( dy/dx=2x-4 \); M1: gradient at x=1 correctly found (-2); M1: y-coordinate at x=1 correctly found (y=4); A1: correct tangent equation y = -2x + 6. (b) M1: sets y=0 in the tangent equation; A1: (3,0), with correct working shown.

(a) The perpendicular from the centre to a chord bisects the chord, so half of AB = 8 cm. Using Pythagoras in the right-angled triangle formed by the radius, half-chord and perpendicular distance d: \( d = \sqrt{10^2-8^2} = \sqrt{100-64} = \sqrt{36} = 6.00 \) cm. (b) In the same right-angled triangle (hypotenuse 10 cm, opposite side to the half-angle = 8 cm): \( \sin(\theta/2) = \dfrac{8}{10} = 0.8 \), so \( \theta/2 = \sin^{-1}(0.8) = 53.13° \), giving angle AOB \( = 2\times53.13° = 106.3° \) (1 d.p.).

(a) M1: recognises the perpendicular from the centre bisects the chord (half-chord = 8 cm); M1: correct Pythagoras set up \( 10^2-8^2 \); A1: unrounded value 6; A1: d = 6.00 cm. (b) M1: correct trig ratio set up for the half-angle (e.g. \( \sin(\theta/2)=8/10 \)); M1: \( \theta/2 = 53.13° \) found; A1: angle AOB = 106.3° (1 d.p., doubling correctly applied).

The length of the space diagonal of a cuboid is given by \( d = \sqrt{l^2+w^2+h^2} = \sqrt{12^2+5^2+4^2} = \sqrt{144+25+16} = \sqrt{185} = 13.6 \) cm (3 s.f.).

M1: correct 3D Pythagoras formula identified \( \sqrt{l^2+w^2+h^2} \); M1: correct substitution \( 12^2+5^2+4^2=185 \); A1: \( \sqrt{185} \) evaluated; A1: 13.6 cm (3 s.f.).

Height to the base of the flagpole \( = 45\tan(32°) = 45\times0.62487 = 28.12 \) m. Height to the top of the flagpole \( = 45\tan(38°) = 45\times0.78129 = 35.16 \) m. Height of the flagpole itself is the difference: \( 35.16-28.12 = 7.04 \) m (3 s.f., using unrounded intermediate values).

M1: correct expression for height to the base of the flagpole (\( 45\tan32° \)); M1: correct expression for height to the top of the flagpole (\( 45\tan38° \)); M1: correctly subtracts the two heights to find the flagpole height alone; A1: 7.04 m (3 s.f., accept 7.03–7.04 depending on rounding).