ACSI Mock Paper D2 — Mathematics Paper 2

Sec 2 End-of-Year Examination Practice — Cambridge International Mathematics
50 marks · 1 hour · Graphic display calculator allowed
Prepared by Miss Clarissa Ng
www.clartutors.com

END OF YEAR EXAMINATION — SECONDARY 2

CAMBRIDGE INTERNATIONAL MATHEMATICS · Paper 2 · 1 hour
NAME: ______________________________ CLASS: ________________ MARKS: ______ / 50

INSTRUCTIONS

INFORMATION

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
Questions 1 to 8 (50 marks) · ACSI 2023 Paper 2 skills · fresh questions

Q1. A map is drawn to a scale of 1 cm to 5 km.

(a) On the map, the length of a road is 3.5 cm. Find the actual length of the road in kilometres.

______________________ km    [1]

(b) A forest has an area of 225 km2. Find the area of the forest on the map.

______________________ cm2    [2]

Q2. (a) In triangle XYZ, the angle at Y is a right angle. W lies on XZ and YW is perpendicular to XZ.

XZYWNOT TOSCALE

Explain, giving geometric reasons, why triangle XYZ is similar to triangle XWY.

[2]

(b) Triangle ABC is similar to triangle PQR.

ABCPQR6.4 cm3.28.8 cmxNOT TOSCALE

(i) Find the value of x.

x = ______________________    [2]

(ii) The area of triangle PQR is 36 cm2. Calculate the area of triangle ABC.

______________________ cm2    [2]

Q3. The diagram shows a solid Pyramid A with a square base of sides 12 cm and height 8 cm.

8 cm12 cmPyramid A

(a) Find the total surface area of Pyramid A.

______________________ cm2    [3]

(b) A mathematically similar Pyramid B has a total surface area of 864 cm2. Find the volume of Pyramid B if the volume of Pyramid A is 576 cm3.

______________________ cm3    [3]

Q4. The distances travelled by 150 commuters are recorded in the table below.

Distance (x km)0 < x ≤ 1010 < x ≤ 2020 < x ≤ 3030 < x ≤ 4040 < x ≤ 5050 < x ≤ 70
Frequency82240362816

(a) Calculate an estimate of the mean distance.

______________________ km    [2]

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

[1]

(c) Calculate an estimate of the interquartile range of the distances.

______________________ km    [1]

Q5. The diagram shows a pair of axes.

xy

(a) On the diagram, sketch the graph of y = 0.5x2 + x − 4.    [2]

(b) Write down the coordinates of the turning point.

( ______ , ______ )    [1]

(c) Write down the coordinates of the axes intercepts.

( ______ , ______ )    ( ______ , ______ )    ( ______ , ______ )    [3]

(d) State the equation of the line of symmetry.

______________________    [1]

(e) The equation 0.5x2 + x − 4 = k has only one solution. State the value of k.

k = ______________________    [1]

(f) A straight line has equation y = mx + c, where m and c are integers and m ≠ 0, c ≠ 0. Given that 0.5x2 + x − 4 = mx + c has no solutions, find the equation of the line.

y = ______________________    [2]

Q6. A car travels 420 km from Town A to Town B. The average speed is x km/h.

(a) Write down an expression, in terms of x, for the time taken for this journey.

______________________ hours    [1]

(b) On the return journey from Town B to Town A, the car travels faster by 20 km/h. Write down an expression, in terms of x, for the time taken for the return journey.

______________________ hours    [1]

(c) The return journey took 30 minutes less. Write down an equation in x and show that it simplifies to x2 + 20x − 16 800 = 0.

[3]

(d) Solve the equation x2 + 20x − 16 800 = 0. Give your answers correct to the nearest whole number.

x = ______________________ or x = ______________________    [2]

(e) The car left Town A at 07 15. Use your answer to part (d) to find the time it arrived at Town B.

______________________    [2]

Q7. The diagram shows the cross-section of a tunnel. This cross-section is made up of a triangle and a sector of radius 10 m and angle 300°. The tunnel is 10 m wide at the base.

300°h10 m10 mNOT TOSCALE

(a) The height of the triangle is h. Show that h = 8.66, correct to 3 significant figures.    [2]

(b) Calculate the area of the cross-section.

______________________ m2    [3]

(c) The tunnel has a length of 1.2 km. Calculate the volume of earth that was removed to make the tunnel. Give your answer in cubic metres.

______________________ m3    [2]

Q8. The diagram shows a solid cone of radius r cm and height 2r cm.

2rrNOT TOSCALE

The total surface area of the cone is 824 cm2. Calculate the value of r.

r = ______________________    [5]

End of Paper 2. Check your work — give every non-exact answer to 3 significant figures (1 decimal place for angles) and show all working.

Answer Key — ACSI Mock Paper D2

Total: 50 marks · 8 questions · modelled on the 2023 ACSI paper (Paper 2 skills, fresh numbers). Method marks (M) are awarded for a correct method even if the final answer is wrong; accuracy marks (A) only for a correct answer.
Q1 (a) 17.5 km  [A1]
Scale 1 cm : 5 km, so 3.5 cm × 5 = 17.5 km
Q1 (b) 9 cm2  [M1 for squaring the scale factor, A1]
For area the scale factor is squared: 1 cm2 : 52 = 25 km2. So the map area is 225 ÷ 25 = 9 cm2
Q2 (a)  [M1 for the angle reasons, A1 for completing the argument]
Angle XWY = angle XYZ = 90° (given: the right angles marked at W and at Y). Angle YXW is common to both triangles (the same angle at X; equivalently angle WXY = angle YXZ). Two pairs of equal angles → the triangles are similar (AA).
Q2 (b) (i) x = 4.4  [M1 for the ratio, A1]
Matching sides: AC/PR = CB/RQ → 6.4/8.8 = 3.2/x → x = 8.8 × 3.2 ÷ 6.4 = 4.4
Q2 (b) (ii) 19.0 cm2  [M1 for the area scale factor, A1]
The linear scale factor from PQR to ABC is 6.4/8.8 = 8/11, so areas shrink by (8/11)2. Area of ABC = 36 × (8/11)2 = 36 × 0.5290… = 19.04… = 19.0 cm2 (3 s.f.)
Q3 (a) 384 cm2  [M1 for the slant height, M1 for the four triangles, A1]
Slant height = √(82 + 62) = 10 cm. Four triangular faces: 4 × ½ × 12 × 10 = 240 cm2. Square base: 12 × 12 = 144 cm2. Total = 240 + 144 = 384 cm2
Q3 (b) 1944 cm3  [M1 for the area ratio, M1 for cubing, A1]
Area ratio B : A = 864 : 384 = 9/4, so the linear scale factor is 3/2. Volume ratio = (3/2)3 = 27/8. Volume of B = 576 × 27/8 = 1944 cm3
Q4 (a) mean ≈ 32.3 km  [M1 for the midpoints × frequency, A1]
Midpoints 5, 15, 25, 35, 45, 60: (5×8) + (15×22) + (25×40) + (35×36) + (45×28) + (60×16) = 40 + 330 + 1000 + 1260 + 1260 + 960 = 4850; ÷ 150 = 32.333… = 32.3 km (3 s.f.)
Q4 (b)  [B1] — each distance is not known exactly: every value in a class is replaced by its midpoint, so the answer is an estimate
Q4 (c) IQR ≈ 20.4 km  [B1]
Cumulative frequencies: 8, 30, 70, 106, 134, 150. Q1 is the 37.5th value → in 20 < x ≤ 30: 20 + 10 × (37.5 − 30)/40 = 21.875. Q3 is the 112.5th value → in 40 < x ≤ 50: 40 + 10 × (112.5 − 106)/28 = 42.321. IQR = 42.321 − 21.875 = 20.446… = 20.4 km
Q5 (a)  [M1 for a parabola opening upwards, A1 for the correct shape and intercepts]
An upward parabola crossing the x-axis at −4 and 2, crossing the y-axis at −4, with its lowest point at (−1, −4.5)
Q5 (b) turning point (−1, −4.5)  [A1]
x = −b/2a = −1/(2 × 0.5) = −1; y = 0.5(−1)2 + (−1) − 4 = 0.5 − 5 = −4.5
Q5 (c) intercepts (−4, 0), (0, −4), (2, 0)  [A1 for each]
0.5x2 + x − 4 = 0 → x2 + 2x − 8 = 0 → (x + 4)(x − 2) = 0 → x = −4 or 2; at x = 0, y = −4
Q5 (d) x = −1  [A1] — the line of symmetry passes through the turning point
Q5 (e) k = −4.5  [A1] — one solution means the line y = k touches the curve at the turning point
Q5 (f) any line lying wholly below the curve, e.g. y = x − 12  [M1 for a line with integer m and c, A1 for checking it misses the curve]
Substitute: 0.5x2 + x − 4 = x − 12 → 0.5x2 + 8 = 0 → x2 = −16, which has no real solutions. The line must pass below the turning point (−4.5) and its intercept must not be 0 — any such line is accepted, but the gradient must not be 0.
Q6 (a) 420/x hours  [B1]  —  (b) 420/(x + 20) hours  [B1]
Q6 (c)  [M1 for the difference of the two times = 0.5, M1 for multiplying out, A1 for the printed form]
420/x − 420/(x + 20) = 0.5 → 420(x + 20) − 420x = 0.5x(x + 20) → 8400 = 0.5x2 + 10x → 16 800 = x2 + 20x → x2 + 20x − 16 800 = 0
Q6 (d) x = 120 or x = −140  [M1 for the formula, A1]
x = (−20 ± √(202 + 4 × 16 800))/2 = (−20 ± √67 600)/2 = (−20 ± 260)/2. The speed must be positive, so x = 120 km/h (the other root, −140, is rejected)
Q6 (e) arrival 10 45  [M1 for the journey time, A1]
Time from Town A to Town B = 420 ÷ 120 = 3.5 h = 3 h 30 min. 07 15 + 3 h 30 min = 10 45
Q7 (a)  [M1 for the right-angled triangle, A1 for the value to 3 s.f.]
The triangle has two sides of 10 m (both radii) and a base of 10 m, so each half of the base is 5 m. h = √(102 − 52) = √75 = 8.6602… = 8.66 m (3 s.f.). (The triangle is in fact equilateral, since all three sides are 10 m.)
Q7 (b) 305 m2  [M1 for the triangle, M1 for the sector, A1]
Triangle = ½ × 10 × 8.6603 = 43.301 m2. Sector = (300/360) × π × 102 = 261.799 m2. Total = 43.301 + 261.799 = 305.10 = 305 m2 (3 s.f.)
Q7 (c) 3.66 × 105 m3  [M1 for the length in metres, A1]
1.2 km = 1200 m, so volume = 305.10 × 1200 = 366 120 m3 = 3.66 × 105 m3 (3 s.f.)
Q8   r = 9.00 cm  [M1 for the slant height, M1 for the surface-area formula, M1 for substituting, M1 for rearranging, A1]
Slant height l = √(r2 + (2r)2) = r√5. Total surface area = πr2 + πrl = πr2(1 + √5) = 824, so r2 = 824/(π(1 + √5)) = 81.05… and r = 9.003… cm = 9.00 cm (3 s.f.)
Miss Clarissa Ng | www.clartutors.com