2023 CCEA GCSE Mathematics 2210 Practice Paper with Answers

(a) As the number of hours revised increases, the test score also increases for 7 of the 8 students (roughly following score ≈ 7x + 31), so this is strong positive correlation.
(b) The point (2, 79) does not fit this trend — a student who revised for only 2 hours would be expected to score around 45%, not 79%. Possible reason: this student may already have known the material well from previous study, extra tuition, or a strong natural aptitude for the subject. Final answer: outlier (2, 79).

(a) [1] B1: correct description "(strong) positive correlation". (b) [1] B1: correct pair (2, 79) stated, with a sensible contextual reason for the anomaly (e.g. prior knowledge); accept any plausible reason. Reject an answer that just repeats "it's different" with no reason.

(a) By Pythagoras' theorem: cable² = 35² + 50² = 1225 + 2500 = 3725. Cable = \( \sqrt{3725} = 61.0327...\ \text{m} \approx 61.03 \) m (2 d.p.).
(b) Since only whole metres can be bought and the exact length (61.03 m) must be covered, round UP to 62 m.
(c) \( 61.0327... \div 1000 = 0.0610 \) km (3 s.f.).
Final answer (minimum metres to buy): 62 m.

(a) [3] M1: 35² + 50² (or equivalent use of Pythagoras); M1: √3725 evaluated; A1: 61.03 m (2 d.p., accept 61.0). (b) [1] B1 ft: correct round-UP of their (a) to next whole metre (62, or ft from their value) — must be rounded up, not to nearest. (c) [1] B1 ft: their (a) ÷ 1000 to 3 s.f., e.g. 0.0610 km.

Area of trapezium = \( \frac{1}{2}(a+b)h = \frac{1}{2}(8+14)(6) = \frac{1}{2}(22)(6) = 66 \) cm². Final answer: 66 cm².

[2] M1: correct substitution into \( \frac{1}{2}(a+b)h \); A1: 66 cm² (units required for A1).

(a) Students studying at least one language = 32 − 6 = 26. Using |F ∪ S| = |F| + |S| − |F ∩ S|: 26 = 18 + 15 − |F ∩ S| = 33 − |F ∩ S|, so |F ∩ S| = 33 − 26 = 7. Both languages: 7 students.
(b) French only = 18 − 7 = 11 students. P(French but not Spanish) = 11/32 (already in simplest form, since 11 is prime and does not divide 32). Final answer: 11/32.

(a) [2] M1: 32 − 6 = 26 (at least one language) and |F|+|S| = 33 seen; A1: both = 7. (b) [2] M1 ft: French only = 18 − their both; A1 ft: 11/32 (accept equivalent unsimplified fraction with correct numerator/denominator; ft their (a)).

Midpoint = \( \left( \frac{-3+9}{2}, \frac{7+(-1)}{2} \right) = \left( \frac{6}{2}, \frac{6}{2} \right) = (3, 3) \). Final answer: (3, 3).

[2] M1: correct method for at least one coordinate, e.g. \( \frac{-3+9}{2} \); A1: (3, 3), both coordinates correct.

(a) Gradient \( = \frac{20 - 0}{160 - 0} = \frac{20}{160} = 0.125 \) litres per km.
(b) Distance per litre \( = \frac{1}{0.125} = 8 \) km per litre (equivalently, 160 km ÷ 20 litres = 8 km per litre). Final answer: 8 km per litre.

(a) [2] M1: \( \frac{20}{160} \); A1: 0.125 litres/km (units required). (b) [1] B1 ft: 8 km per litre (ft as reciprocal of their (a), or independently as 160÷20).

(a) £68 represents 80% of the original price P: \( 0.8P = 68 \Rightarrow P = \frac{68}{0.8} = 85 \). Original price = £85.
(b) Final price after both discounts = \( 68 \times 0.85 = 57.80 \). Overall reduction from the original price = \( 85 - 57.80 = 27.20 \). Overall percentage discount \( = \frac{27.20}{85} \times 100\% = 32\% \) (equivalently, combined multiplier \( 0.8 \times 0.85 = 0.68 \), i.e. a 32% reduction, not 35%). The customer is NOT correct; the true combined discount is 32%.

(a) [3] M1: 0.8P = 68 (or equivalent, e.g. 68 ÷ 0.8); M1: correct rearrangement/method; A1: £85. (b) [2] M1: 68 × 0.85 (= 57.80) or combined multiplier 0.8 × 0.85 = 0.68; A1: correct conclusion that combined discount is 32%, not 35%, with the value 32% stated. No credit for stating 35% is correct.

\( \frac{x}{4} + 3 = 8 \Rightarrow \frac{x}{4} = 5 \Rightarrow x = 20 \). Final answer: x = 20.

[2] M1: \( \frac{x}{4} = 5 \) (subtracting 3 from both sides); A1: x = 20.

(a) \( 84 = 2^2 \times 3 \times 7 \); \( 126 = 2 \times 3^2 \times 7 \).
(b) LCM uses the highest power of each prime present: \( 2^2 \times 3^2 \times 7 = 4 \times 9 \times 7 = 252 \). Final answer: 252.

(a) [2] B1: \( 84 = 2^2 \times 3 \times 7 \); B1: \( 126 = 2 \times 3^2 \times 7 \). (b) [1] B1 ft: 252, using highest power of each prime from their (a).

(a) Area = length × width: \( x(x+5) = 84 \Rightarrow x^2 + 5x = 84 \Rightarrow x^2 + 5x - 84 = 0 \).
(b) Factorising: need two numbers multiplying to −84 and summing to 5: 12 and −7. \( (x+12)(x-7) = 0 \Rightarrow x = -12 \) or \( x = 7 \). Since \( x \) is a width, \( x > 0 \), so \( x = -12 \) is rejected. Final answer: x = 7 cm (length = 12 cm, check: 7 × 12 = 84 ✓).

(a) [1] B1: correct expansion/rearrangement shown to the given equation. (b) [2] M1: correct factorisation (x+12)(x-7) or use of the quadratic formula; A1: x = 7, with x = −12 explicitly rejected as not a valid width.

Common denominator is \( (x+2)(x-1) \): \( \frac{3(x-1)}{(x+2)(x-1)} + \frac{2(x+2)}{(x+2)(x-1)} = \frac{3(x-1) + 2(x+2)}{(x+2)(x-1)} \). Expanding the numerator: \( 3x - 3 + 2x + 4 = 5x + 1 \). Final answer: \( \frac{5x+1}{(x+2)(x-1)} \) (equivalently \( \frac{5x+1}{x^2+x-2} \)).

[3] M1: correct common denominator (x+2)(x-1); M1: correct numerator 3(x-1) + 2(x+2) formed; A1: simplified to \( \frac{5x+1}{(x+2)(x-1)} \) — no further common factor to cancel.

(a) Gradient of \( L_1 = \frac{10-4}{3-1} = \frac{6}{2} = 3 \). Using (1,4): \( 4 = 3(1) + c \Rightarrow c = 1 \). So \( L_1: y = 3x + 1 \).
(b) Parallel lines have equal gradient, so \( L_2 \) has gradient 3. Through (0,−2): \( c = -2 \), so \( L_2: y = 3x - 2 \). Testing (5,13): \( 3(5) - 2 = 15 - 2 = 13 \), which matches the y-coordinate, so (5, 13) DOES lie on \( L_2 \). Final answer: L2: y = 3x − 2; yes, (5,13) lies on L2.

(a) [2] M1: gradient = 3 found; A1: y = 3x + 1. (b) [2] M1 ft: y = 3x − 2 stated (gradient ft from (a)); A1: correct substitution showing 3(5) − 2 = 13 and correct conclusion that the point lies on the line.

(a) Pressure = Force ÷ Area = \( \frac{450}{0.36} = 1250 \) N/m² (Pa).
(b) \( 1250 \text{ Pa} \div 1000 = 1.25 \) kPa. Final answer: 1.25 kPa.

(a) [2] M1: 450 ÷ 0.36; A1: 1250 Pa (units required). (b) [1] B1 ft: their (a) ÷ 1000, e.g. 1.25 kPa.

Area of sector \( = \frac{\theta}{360} \times \pi r^2 = \frac{75}{360} \times \pi \times 8^2 = \frac{75}{360} \times \pi \times 64 = 41.887... \) m². Rounded to 3 s.f.: 41.9 m². Final answer: 41.9 m².

[3] M1: \( \frac{75}{360} \times \pi \times 8^2 \) — correct substitution; M1: correct method/evaluation; A1: 41.9 m² (3 s.f., units required).

(a) With n = 15 values in order, the lower quartile is the \( \frac{1}{4}(n+1) = 4\text{th} \) value = 3, and the upper quartile is the \( \frac{3}{4}(n+1) = 12\text{th} \) value = 9. IQR = 9 − 3 = 6 hours.
(b) The value 25 is a clear outlier compared with the rest of the data (which lie between 2 and 12), and this outlier pulls the mean (7) noticeably above the median (5). Because the median is not affected by extreme values, the median (5 hours) better represents a typical employee's overtime. Final answer: IQR = 6 hours; median is the more representative average.

(a) [2] M1: correct identification of Q1 = 3 and Q3 = 9 (or equivalent valid quartile method); A1: IQR = 6. (b) [2] B1: median selected; B1: correct reason referring to the outlier (25) distorting the mean / median unaffected by extreme values. No credit for "median" with no valid reason.

(a) The median is the \( \frac{60}{2} = 30\text{th} \) value. This lies in the 40–50 class (cumulative frequency rises from 22 to 40 across this class of 18 students). Median \( \approx 40 + \frac{30-22}{40-22} \times 10 = 40 + \frac{8}{18} \times 10 = 40 + 4.44... = 44.4 \) marks (1 d.p.).
(b) Number scoring below the pass mark (40) = 22 (cumulative frequency at 40). Number passing = 60 − 22 = 38. Percentage passing \( = \frac{38}{60} \times 100\% = 63.3\% \) (1 d.p.). Final answer: median ≈ 44.4 marks; 63.3% passed.

(a) [2] M1: correct interpolation set-up \( 40 + \frac{30-22}{18} \times 10 \); A1: 44.4 (1 d.p., accept 44.4–44.5 range from sensible graph reading). (b) [1] B1 ft: 38/60 × 100 = 63.3% (ft from their reading of cf at mark 40).

(a) A stratified sample divides the population into groups (strata) — here, Year 10 and Year 11 — and selects a sample from each group in proportion to the size of that group within the whole population, usually then chosen at random within each stratum.
(b) Total students = 180 + 150 = 330. Year 10 in sample \( = 55 \times \frac{180}{330} = 55 \times \frac{6}{11} = 30 \). Year 11 in sample \( = 55 \times \frac{150}{330} = 55 \times \frac{5}{11} = 25 \). Check: 30 + 25 = 55 ✓. Final answer: 30 Year 10 students, 25 Year 11 students.

(a) [1] B1: correct explanation referencing proportional representation of each stratum/group. (b) [3] M1: 330 total students identified; M1: correct method \( 55 \times \frac{180}{330} \) (or equivalent) for at least one group; A1: 30 and 25 both correct, summing to 55.

First find AC using Pythagoras in triangle ABC: \( AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 9^2} = \sqrt{144+81} = \sqrt{225} = 15 \) cm.
In triangle ACD, the right angle is at C, angle ADC = 32°, and AC = 15 cm is the side opposite angle ADC. CD is adjacent to angle ADC, so \( \tan(32°) = \frac{AC}{CD} \Rightarrow CD = \frac{15}{\tan(32°)} = \frac{15}{0.6249} = 24.005... \) cm. Rounded to 3 s.f.: 24.0 cm. Final answer: CD = 24.0 cm.

[4] M1: AC² = 12² + 9² (Pythagoras set up in triangle ABC); A1: AC = 15 cm; M1: correct trig ratio set up in triangle ACD, \( CD = \frac{15}{\tan 32°} \) (ft their AC); A1: CD = 24.0 cm (3 s.f., ft their AC).

(a) Base area \( = 3x \times x = 3x^2 \). The two long side walls each have area \( 3x \times (x-2) \), giving \( 2 \times 3x(x-2) = 6x^2 - 12x \). The two end walls each have area \( x \times (x-2) \), giving \( 2x(x-2) = 2x^2 - 4x \). Total: \( A = 3x^2 + (6x^2-12x) + (2x^2-4x) = 11x^2 - 16x \) cm², as required.
(b) At \( x = 6 \): base area \( = 3(6)(6) = 108 \) cm²; height \( = 6 - 2 = 4 \) cm. Volume = base area × height \( = 108 \times 4 = 432 \) cm³. Final answer: 432 cm³.

(a) [4] M1: base area 3x²; M1: two side walls 6x² − 12x; M1: two end walls 2x² − 4x; A1: correctly summed to 11x² − 16x (all terms shown, no errors). (b) [3] M1: base area at x=6 = 108 cm²; M1: height = 4 cm; A1: volume = 432 cm³ (units required).

Upper bound of distance = 340 + 5 = 345 km. Lower bound of time = 5.2 − 0.05 = 5.15 hours. To maximise speed (= distance ÷ time), use the largest possible distance with the smallest possible time: \( \text{speed}_{max} = \frac{345}{5.15} = 66.990... \) km/h. Rounded to 1 d.p.: 67.0 km/h. Final answer: 67.0 km/h.

[3] B1: upper bound of distance = 345 km; B1: lower bound of time = 5.15 hours; A1: 345 ÷ 5.15 = 67.0 km/h (1 d.p.) — no credit for using upper bound of time or lower bound of distance.

(a) Angle ABC = 90°, because the angle in a semicircle (subtended by the diameter AC) is always 90°.
(b) ABCD is a cyclic quadrilateral, and opposite angles of a cyclic quadrilateral sum to 180°. So angle ADC = 180° − angle ABC = 180° − 90° = 90°.
(c) In triangle ACD, the angles sum to 180°: angle CAD = 180° − angle ADC − angle ACD = 180° − 90° − 25° = 65°. Angle BAD = angle BAC + angle CAD = 40° + 65° = 105°. Final answer: angle BAD = 105°.

(a) [2] MA1: angle ABC = 90°; MA1: reason "angle in a semicircle is 90°". (b) [2] MA1 ft: angle ADC = 90° (ft their (a)); MA1: reason "opposite angles of a cyclic quadrilateral sum to 180°". (c) [2] M1: angle CAD = 180 − 90 − 25 = 65° found (ft their (b)); A1: angle BAD = 40 + 65 = 105°.

First take out the common factor x: \( 6x^3 - 5x^2y - 6xy^2 = x(6x^2 - 5xy - 6y^2) \). Now factorise the quadratic \( 6x^2 - 5xy - 6y^2 \): looking for factors of the form \( (3x + ay)(2x + by) \) where \( ab = -6 \) and \( 3b + 2a = -5 \); taking \( a = 2, b = -3 \) gives \( (3x+2y)(2x-3y) \). Check: \( 6x^2 - 9xy + 4xy - 6y^2 = 6x^2 - 5xy - 6y^2 \) ✓. Final answer: \( x(3x+2y)(2x-3y) \).

[3] B1: common factor x taken out correctly, x(6x² − 5xy − 6y²); M1: correct method to factorise the resulting quadratic (e.g. trial factors checked by expansion); A1: fully correct x(3x+2y)(2x−3y).

Multiply both sides by \( (x+1)(x-2) \): \( 4(x-2) + 3(x+1) = 2(x+1)(x-2) \).
LHS: \( 4x - 8 + 3x + 3 = 7x - 5 \).
RHS: \( 2(x^2 - x - 2) = 2x^2 - 2x - 4 \).
So \( 7x - 5 = 2x^2 - 2x - 4 \Rightarrow 0 = 2x^2 - 2x - 4 - 7x + 5 = 2x^2 - 9x + 1 \).
Using the quadratic formula with a=2, b=−9, c=1: \( x = \frac{9 \pm \sqrt{(-9)^2 - 4(2)(1)}}{2(2)} = \frac{9 \pm \sqrt{81-8}}{4} = \frac{9 \pm \sqrt{73}}{4} \).
\( \sqrt{73} = 8.544...\). \( x = \frac{9+8.544}{4} = 4.386 \approx 4.39 \), or \( x = \frac{9-8.544}{4} = 0.114 \approx 0.11 \). Neither value equals −1 or 2, so both are valid. Final answer: x = 4.39 or x = 0.11 (2 d.p.).

[7] M1: multiplying through by (x+1)(x−2); M1: correct LHS expansion 7x−5; M1: correct RHS expansion 2x²−2x−4; A1: correctly rearranged to 2x²−9x+1=0; M1: correct substitution into the quadratic formula; A1: x = 4.39; A1: x = 0.11 (both required, 2 d.p.).

(a) Frequency density = frequency ÷ class width. 20–30 (width 10): \( 9 \div 10 = 0.9 \). 50–70 (width 20): \( 12 \div 20 = 0.6 \).
(b) If frequency itself were plotted for classes of unequal width, wider classes would appear to have disproportionately tall bars even if the data is not more densely packed; frequency density (frequency ÷ width) accounts for the different widths so that the AREA of each bar, not its height, represents the frequency — giving a fair visual comparison.
(c) Cumulative frequencies: 9 (≤30), 24 (≤35), 48 (≤40), 78 (≤50), 90 (≤70). The median is the \( \frac{90}{2}=45\text{th} \) value, which falls in the 35–40 class (cumulative frequency rises from 24 to 48 across this class of 24 runners). Median \( \approx 35 + \frac{45-24}{48-24} \times 5 = 35 + \frac{21}{24} \times 5 = 35 + 4.375 = 39.375 \approx 39.4 \) minutes (1 d.p.). Final answer: densities 0.9 and 0.6; median ≈ 39.4 minutes.

(a) [2] B1: 0.9; B1: 0.6 (both with correct working frequency ÷ width). (b) [1] B1: correct explanation referencing area representing frequency / fair comparison across unequal widths. (c) [3] M1: correct cumulative frequencies built and 45th value located in 35–40 class; M1: correct interpolation set-up \( 35 + \frac{45-24}{24} \times 5 \); A1: 39.4 minutes (1 d.p., accept 39.3–39.5 range).

Numerator (difference of two squares): \( x^2 - 16 = (x-4)(x+4) \).
Denominator: \( 2x^2 + 9x + 4 = (2x+1)(x+4) \) — check: \( 2x^2 + 8x + x + 4 = 2x^2 + 9x + 4 \) ✓.
So \( \frac{(x-4)(x+4)}{(2x+1)(x+4)} = \frac{x-4}{2x+1} \), cancelling the common factor (x+4) (valid for x ≠ −4). Final answer: \( \frac{x-4}{2x+1} \).

[3] B1: numerator correctly factorised (x−4)(x+4); M1: denominator correctly factorised (2x+1)(x+4); A1: correctly cancelled to \( \frac{x-4}{2x+1} \).

(a) Total surface area of the cone = curved surface + base = \( \pi r l + \pi r^2 = \pi r(r+8) + \pi r^2 = \pi r^2 + 8\pi r + \pi r^2 = 2\pi r^2 + 8\pi r \).
Total surface area of the hemisphere = curved surface + flat circular base = \( 2\pi r^2 + \pi r^2 = 3\pi r^2 \).
Setting the two total surface areas equal: \( 2\pi r^2 + 8\pi r = 3\pi r^2 \). Dividing both sides by \( \pi r \) (valid since r > 0): \( 2r + 8 = 3r \Rightarrow r = 8 \), as required.
(b) Hemisphere TSA \( = 3\pi r^2 = 3\pi(8)^2 = 3\pi(64) = 192\pi \) cm² (exact). Numerically: \( 192\pi = 603.19 = 603 \) cm² (3 s.f.). Final answer: r = 8 cm; TSA = 192π cm² ≈ 603 cm².

(a) [4] M1: cone curved surface area πr(r+8); M1: cone total SA = 2πr² + 8πr; M1: hemisphere TSA = 3πr² stated; A1: equation solved correctly to r = 8 (division by πr shown/implied). (b) [2] B1: exact value 192π cm²; B1 ft: 603 cm² (3 s.f., ft their r).