ACSI Mock Paper D2.1 — Extra Practice on Q2, Q3, Q4

END OF YEAR EXAMINATION — SECONDARY 2

CAMBRIDGE INTERNATIONAL MATHEMATICS · Paper 2 · 1 hour

NAME: ______________________________ CLASS: ________________ MARKS: ______ / 50

INSTRUCTIONS

Answer all questions. You will need a graphic display calculator.
Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs.
Write your answer to each question in the space provided.
You should use a graphic display calculator where appropriate.
You may use tracing paper.
You must show all necessary workings clearly — marks are given for correct methods, including sketches, even if your answer is incorrect.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
For π, use your calculator value.

INFORMATION

The total mark for this paper is 50.
The number of marks for each question or part question is shown in brackets [ ].

List of Formulas

Area of triangle = ½ × base × height Volume of prism = area of cross-section × length Area of circle = πr² Volume of pyramid = ⅓ × base area × height Circumference of circle = 2πr Volume of cylinder = πr²h Curved surface area of cylinder = 2πrh Volume of cone = ⅓πr²h Curved surface area of cone = πrl Volume of sphere = ⁴⁄₃πr³ Surface area of sphere = 4πr² Arc length = (θ/360°) × 2πr Area of a sector = (θ/360°) × πr² For y = ax² + bx + c: x = −b/2a is the line of symmetry, and the roots are x = (−b ± √(b² − 4ac))/2a

Q1. (a) In triangle DEF, the angle at E is a right angle. G lies on DF and EG is perpendicular to DF.

Explain, giving geometric reasons, why triangle DEF is similar to triangle DEG.

[2]

(b) Triangle ABC is similar to triangle PQR, with the right angles at B and at Q.

(i) Find the value of x.

x = ______________________ cm [2]

(ii) The area of triangle PQR is 120 cm². Calculate the area of triangle ABC.

______________________ cm² [2]

Q2. (a) In triangle PQR, the angle at Q is a right angle. S lies on PR and QS is perpendicular to PR.

Explain, giving geometric reasons, why triangle PQR is similar to triangle PQS.

[2]

(b) Triangle LMN is similar to triangle XYZ, with the right angles at M and at Y.

(i) Find the value of x.

x = ______________________ cm [2]

(ii) The area of triangle LMN is 54 cm². Calculate the area of triangle XYZ.

______________________ cm² [2]

Q5. The distances travelled by 120 students are recorded in the table below.

Distance (x km)	0 < x ≤ 20	20 < x ≤ 40	40 < x ≤ 60	60 < x ≤ 80	80 < x ≤ 100
Frequency	16	24	34	30	16

(a) Calculate an estimate of the mean distance.

______________________ km [2]

(b) Give a reason why the mean distance is an estimate.

[1]

______________________ km [1]

Q6. The times taken by 200 runners to finish a race are recorded in the table below.

Time (t minutes)	0 < t ≤ 10	10 < t ≤ 20	20 < t ≤ 30	30 < t ≤ 40	40 < t ≤ 50
Frequency	30	50	60	40	20

(a) Calculate an estimate of the mean time.

______________________ minutes [2]

(b) Give a reason why the mean time is an estimate.

[1]

______________________ minutes [1]

Answer Key — ACSI Mock Paper D2.1

Total: 32 marks · 6 questions · more variations of D2 Q2, Q3 and Q4 (same skills and structure, fresh numbers). Q1/Q2 are similar triangles, Q3/Q4 square-based pyramids, Q5/Q6 grouped frequency tables. Method marks (M) are for a correct method even if the answer is wrong; accuracy marks (A) only for a correct answer.

Q1 (a) [M1 for the angle reasons, A1 for completing the argument]

Angle DGE = angle DEF = 90° (given: EG is perpendicular to DF, and the angle at E is a right angle). Angle EDG is common to both triangles (the same angle at D; that is angle GDE = angle FDE). Two pairs of equal angles → the triangles are similar (AA).

Q1 (b) (i) x = 24 [M1 for the ratio, A1]

Matching sides: PQ/AB = QR/BC → 10/5 = x/12 → x = 12 × 2 = 24. (Check: 5, 12, 13 and 10, 24, 26 are both right-angled triangles.)

Q1 (b) (ii) 30 cm² [M1 for the area scale factor, A1]

The linear scale factor from PQR to ABC is 1/2, so areas shrink by (1/2)² = 1/4. Area of ABC = 120 × 1/4 = 30 cm². (½ × 5 × 12 = 30, which checks.)

Q2 (a) [M1 for the angle reasons, A1 for completing the argument]

Angle PSQ = angle PQR = 90° (given). Angle QPS is common to both triangles (the same angle at P). Two pairs of equal angles → the triangles are similar (AA).

Q2 (b) (i) x = 8 [M1 for the ratio, A1]

Matching sides: XY/LM = YZ/MN → 6/9 = x/12 → x = 12 × 2/3 = 8. (Check: 9, 12, 15 and 6, 8, 10 are both right-angled triangles.)

Q2 (b) (ii) 24 cm² [M1 for the area scale factor, A1]

The linear scale factor from LMN to XYZ is 6/9 = 2/3, so areas shrink by (2/3)² = 4/9. Area of XYZ = 54 × 4/9 = 24 cm². (½ × 6 × 8 = 24, which checks.)

Q3 (a) 360 cm² [M1 for the slant height, M1 for the four triangles, A1]

Slant height = √(12² + 5²) = 13 cm. Four triangular faces: 4 × ½ × 10 × 13 = 260 cm². Square base: 10 × 10 = 100 cm². Total = 260 + 100 = 360 cm²

Q3 (b) 1350 cm³ [M1 for the area ratio, M1 for cubing, A1]

Area ratio B : A = 810 : 360 = 9/4, so the linear scale factor is 3/2. Volume ratio = (3/2)³ = 27/8. Volume of B = 400 × 27/8 = 1350 cm³

Q4 (a) 576 cm² [M1 for the slant height, M1 for the four triangles, A1]

Slant height = √(6² + 8²) = 10 cm. Four triangular faces: 4 × ½ × 16 × 10 = 320 cm². Square base: 16 × 16 = 256 cm². Total = 320 + 256 = 576 cm²

Q4 (b) 64 cm³ [M1 for the area ratio, M1 for cubing, A1]

Area ratio B : A = 144 : 576 = 1/4, so the linear scale factor is 1/2. Volume ratio = (1/2)³ = 1/8. Volume of B = 512 × 1/8 = 64 cm³

Q5 (a) mean ≈ 51.0 km [M1 for the midpoints × frequency, A1]

Midpoints 10, 30, 50, 70, 90: (10×16) + (30×24) + (50×34) + (70×30) + (90×16) = 160 + 720 + 1700 + 2100 + 1440 = 6120; ÷ 120 = 51.0 km

Q5 (b) [B1] — every value in a class is replaced by its midpoint, and the individual distances are not known, so the answer is an estimate

Q5 (c) IQR ≈ 39.0 km [B1]

Cumulative frequencies: 16, 40, 74, 104, 120. Q₁ is the 30th value → in 20 < x ≤ 40: 20 + 20 × (30 − 16)/24 = 31.666…. Q₃ is the 90th value → in 60 < x ≤ 80: 60 + 20 × (90 − 74)/30 = 70.666…. IQR = 70.667 − 31.667 = 39.0 km

Q6 (a) mean ≈ 23.5 minutes [M1 for the midpoints × frequency, A1]

Midpoints 5, 15, 25, 35, 45: (5×30) + (15×50) + (25×60) + (35×40) + (45×20) = 150 + 750 + 1500 + 1400 + 900 = 4700; ÷ 200 = 23.5 minutes

Q6 (b) [B1] — the midpoint of each class is used in place of the actual times, which are not known individually

Q6 (c) IQR ≈ 18.5 minutes [B1]

Cumulative frequencies: 30, 80, 140, 180, 200. Q₁ is the 50th value → in 10 < t ≤ 20: 10 + 10 × (50 − 30)/50 = 14.0. Q₃ is the 150th value → in 30 < t ≤ 40: 30 + 10 × (150 − 140)/40 = 32.5. IQR = 32.5 − 14.0 = 18.5 minutes