ACSI Mock Paper A2 — 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
Questions 1 to 9 (50 marks)

Q1. A map is drawn to a scale of 1 : 150 000.

(a) The distance on the map between two towns is 8.4 cm. Calculate the actual distance, in kilometres, between the two towns.    [2]

(b) A reservoir covers an area of 2.25 km². Find the area, in square centimetres, covered by the reservoir on the map.    [2]

(c) On a second map, the same reservoir covers an area 9 times that on the first map. Find the scale of the second map, giving your answer in the form 1 : n.    [1]

Q2. The table shows the heights of 60 boys studying at a school.

Height, h (cm)140 < h ≤ 150150 < h ≤ 160160 < h ≤ 170170 < h ≤ 180
Number of boys820x12

(a) Show that x = 20.    [1]

(b) Hence, estimate the mean height of the boys.    [4]

Q3. A small pyramid-shaped paperweight has a height of 6 cm and a volume of 240 cm³. A larger paperweight is geometrically similar to it and has a height of 9 cm.

(a) Find the volume of the larger paperweight.    [3]

(b) The base area of the smaller paperweight is 60 cm². Find the base area of the larger paperweight.    [2]

Q4. The figure shows two triangles, ABC and ACE. Triangle ABC is right-angled at B, with AB = 9 cm and BC = 12 cm. CE = 8 cm and AE = 17 cm.

A B C E 9 cm 12 cm 8 cm 17 cm 15 cm ?

(a) Show that AC = 15 cm.    [2]

(b) Show that triangle ACE is right-angled.    [2]

(c) Hence, find the area of triangle ACE.    [1]

Q5. The graph of y = −2x2 + 4x + 6 crosses the y-axis at P and the x-axis at Q and R, where Q is to the left of R.

(a) Write down the coordinates of P.    [1]

(b) Find the coordinates of Q and of R.    [2]

(c) Find the coordinates of the turning point of the graph, and state whether it is a maximum or a minimum point.    [2]

Q6. In Diagram 1, OAB is a sector of a circle, centre O, with radius 15 cm and ∠AOB = 144°. The sector is folded by joining OA and OB to form the cone shown in Diagram 2.

O A B 15 cm 144° Diagram 1 r 15 cm Diagram 2

(a) Find the arc length of AB, leaving your answer in terms of π.    [2]

(b) Show that the base radius r of the cone is 6 cm.    [2]

(c) Find the height of the cone.    [2]

Q7. 15 students took part in a puzzle-solving competition. Their times, in seconds, are shown in the stem-and-leaf diagram below.

42  5  6  9
51  3  3  7
60  2  4  8
73  5
81

Key: 4 | 2 represents 42 seconds

(a) Write down the mode.    [1]

(b) Find the median.    [1]

(c) Find the range.    [1]

(d) Find the interquartile range.    [1]

(e) The slowest 40% of the students do not qualify for the next round. Find the time of the slowest student who qualifies.    [2]

Q8. A van travels 480 km from Town A to Town B at an average speed of x km/h.

(a) Write down an expression, in terms of x, for the time taken, in hours, for the journey from Town A to Town B.    [1]

(b) On the return journey, the van's average speed is 8 km/h less. Write down an expression, in terms of x, for the time taken, in hours, for the return journey.    [1]

(c) The return journey takes 1 hour 30 minutes longer than the journey to Town B. Form an equation and show that it reduces to x2 − 8x − 2560 = 0.    [3]

(d) Solve x2 − 8x − 2560 = 0, giving your answers correct to 2 decimal places.    [2]

Q9. A closed cylindrical container has a base radius of 7 cm and a height of 12 cm.

(a) Find the volume of the container, correct to 3 significant figures.    [2]

(b) A larger container is geometrically similar to it and has a volume 8 times as large. Find the height of the larger container.    [2]

(c) Find the total surface area of the larger container, correct to 3 significant figures.    [2]

End of Paper 2. Check your work — non-exact answers should be given correct to 3 significant figures, and all working should be shown.

Answer Key — ACSI Mock Paper A2

Total: 50 marks · 9 questions. Method marks (M) are awarded for a correct method even if the final answer is wrong; accuracy marks (A) only for a correct answer. This is a calculator paper, so non-exact answers are given to 3 s.f.
Q1 (a) 12.6 km  [M1 for multiplying by the scale, A1]
8.4 × 150 000 = 1 260 000 cm = 12 600 m = 12.6 km
Q1 (b) 1 cm²  [M1 for converting to cm² and dividing by the square of the scale, A1]
2.25 km² = 2.25 × 1010 cm². Area on map = 2.25 × 1010 ÷ (150 000)² = 1 cm²
Q1 (c) 1 : 50 000  [A1]
Area ratio 9 → length ratio √9 = 3, so the scale is 150 000 ÷ 3 = 1 : 50 000
Q2 (a) x = 20  [A1]
Total number of boys is 60, so x = 60 − 8 − 20 − 12 = 20
Q2 (b) 161 cm  [M1 for midpoints, M1 for multiplying, M1 for the total, A1]
Midpoints: 145, 155, 165, 175.
(8 × 145) + (20 × 155) + (20 × 165) + (12 × 175) = 1160 + 3100 + 3300 + 2100 = 9660
Estimated mean = 9660 ÷ 60 = 161 cm
Q3 (a) 810 cm³  [M1 for the length ratio, M1 for cubing it, A1]
Length ratio = 9 ÷ 6 = 1.5, so volume ratio = 1.5³ = 3.375. Volume = 240 × 3.375 = 810 cm³
Q3 (b) 135 cm²  [M1 for squaring the ratio, A1]
Area ratio = 1.5² = 2.25, so base area = 60 × 2.25 = 135 cm²
Q4 (a) AC = 15 cm  [M1 for Pythagoras, A1]
AC² = AB² + BC² = 9² + 12² = 81 + 144 = 225 → AC = 15 cm
Q4 (b) Right-angled at C  [M1 for squaring the sides, A1]
AC² + CE² = 15² + 8² = 225 + 64 = 289 and AE² = 17² = 289. Since AC² + CE² = AE², triangle ACE is right-angled at C
Q4 (c) 60 cm²  [A1]
Area = ½ × 15 × 8 = 60 cm²
Q5 (a) P(0, 6)  [A1]
At the y-axis, x = 0 → y = 6, so P is (0, 6)
Q5 (b) Q(−1, 0) and R(3, 0)  [M1 for solving, A1 for both]
−2x² + 4x + 6 = 0 ÷ (−2) → x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0 → x = 3 or x = −1, so Q(−1, 0) and R(3, 0)
Q5 (c) (1, 8), maximum  [M1 for x = 1, A1 for y = 8 and maximum]
Line of symmetry x = −b ÷ 2a = −4 ÷ (−4) = 1, so y = −2(1) + 4 + 6 = 8. Turning point = (1, 8), and it is a maximum because the coefficient of x² is negative.
Q6 (a) 12π cm  [M1 for the arc length formula, A1]
Arc AB = (144 ÷ 360) × 2π × 15 = 0.4 × 30π = 12π cm
Q6 (b) r = 6 cm  [M1 for arc = circumference of base, A1]
The arc becomes the circumference of the base: 2πr = 12π → r = 6 cm
Q6 (c) 13.7 cm  [M1 for Pythagoras in the cross-section, A1]
The slant edge is the radius of the sector, 15 cm, so height² = 15² − 6² = 225 − 36 = 189 → height = 13.7 cm (3 s.f.)
Q7 (a) 53 s  [A1] — it appears twice, more than any other time.
Q7 (b) 57 s  [A1] — the 8th of the 15 values.
Q7 (c) 39 s  [A1] — 81 − 42 = 39.
Q7 (d) 19 s  [A1] — lower quartile = 4th value = 49 s, upper quartile = 12th value = 68 s, so IQR = 68 − 49 = 19 s.
Q7 (e) 60 s  [M1 for 40% of 15 = 6 students excluded, A1]
40% of 15 = 6 students do not qualify, so the 6 slowest (81, 75, 73, 68, 64, 62) are out. The slowest student who qualifies took 60 s.
Q8 (a) 480x hours  [A1]
Q8 (b) 480x − 8 hours  [A1]
Q8 (c) x² − 8x − 2560 = 0  [M1 for the equation, M1 for multiplying by x(x − 8), A1]
480/(x − 8) − 480/x = 1.5 → multiply every term by x(x − 8):
480x − 480(x − 8) = 1.5x(x − 8) → 3840 = 1.5x² − 12x → ÷ 1.5: x² − 8x − 2560 = 0
Q8 (d) x = 54.75  [M1 for the quadratic formula, A1]
x = [8 ± √(64 + 10240)] ÷ 2 = (8 ± √10304) ÷ 2 → x = 54.7543… or x = −46.75… The speed must be positive, so x = 54.75 (2 d.p.)
Q9 (a) 1850 cm³  [M1 for V = πr²h, A1]
V = π × 7² × 12 = 1847.25… = 1850 cm³ (3 s.f.)
Q9 (b) 24 cm  [M1 for the volume ratio → length ratio, A1]
Volume ratio = 8, so length ratio = ∛8 = 2 → height = 12 × 2 = 24 cm
Q9 (c) 3340 cm²  [M1 for the larger radius, M1 for the surface area formula, A1]
Larger radius = 7 × 2 = 14 cm. Total surface area = 2πr² + 2πrh = 2π(14²) + 2π(14)(24) = 392π + 672π = 1064π = 3342.65… = 3340 cm² (3 s.f.)
Miss Clarissa Ng | www.clartutors.com