ACSI Mock Paper C2 — Mathematics Paper 2

Sec 2 End-of-Year Examination Practice — Cambridge International Mathematics
50 marks · 1 hour · 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 2024 Paper 2

Q1. A map is drawn to a scale of 1 : 200 000.

(a) The distance between two towns on the map is 12.5 cm. Find the actual distance between the towns, in kilometres. [1]

(b) The actual area of a lake is 1.01 km2. Find the area of the lake on the map, in cm2. [2]

(c) A second map, Map B, is drawn to a different scale. The area of the same lake on Map B is 4 times its area on the first map. Find the scale of Map B in the form 1 : n. [2]

Q2. The heights of 80 girls are measured and grouped as shown.

Height (cm)145155165175
Frequency12273x5

(a) Given that the total frequency is 80, show that 3x = 36 and find the value of x. [1]

(b) Calculate an estimate of the mean height of the girls. [2]

Q3. Two water tumblers are geometrically similar. The larger tumbler has a base radius of 7 cm and a height of 28 cm. The smaller tumbler has a height of 20 cm.

(a) Find the base radius of the smaller tumbler. [1]

(b) The larger tumbler holds 350 ml of water. Find the capacity of the smaller tumbler, correct to the nearest millilitre. [2]

Q4. In the diagram, triangle ABC is right-angled at B with AB = 12 cm and BC = 16 cm. Triangle CDE is congruent to triangle ABC and is right-angled at D. The two triangles meet at C, and angle ACE = 90°.

ABCDE12 cm16 cm16 cm12 cm

(a) Find the length of AC. [2]

(b) Show that AE = 20√2 cm. [2]

Q5. The graph of y = −23x2 + 32x + 1 is to be drawn for −1 ≤ x ≤ 4.

(a) Complete the table of values, giving each value correct to 1 decimal place. [2]

x−101234
y

(b) Draw the graph of y = −23x2 + 32x + 1 for −1 ≤ x ≤ 4. [1]

(c) Use your graph to find the values of x for which y = −2, giving your answers correct to 2 decimal places. [2]

(d) A suitable straight line is inserted on the graph to solve −23x2 + 2x = −2.

(i) State the equation of the line to be inserted. [1]

(ii) Hence, solve −23x2 + 2x = −2, giving your answers correct to 2 decimal places. [2]

Q6. A sector of a circle of radius 12 cm and angle 120° is cut out and folded to form a cone, with the two straight edges joined.

(a) Find the length of the arc of the sector, in terms of π. [2]

(b) Find the base radius of the cone. [1]

(c) Find the height of the cone, and hence its volume. [3]

(d) The cone is melted down and recast into a sphere. Find the radius of the sphere, correct to 3 significant figures. [2]

Q7. 15 students took part in a reaction test. Their times, in seconds, were:

88   91   83   90   69   68   77   70   62   88   99   81   76   72   74

(a) Draw an ordered stem-and-leaf diagram for these times. [3]

(b) Write down the mode. [1]

(c) Find the median. [1]

(d) Find the range. [1]

(e) Find the interquartile range. [2]

(f) The slowest 40% of the students do not qualify for the next round. Find the number of students who qualify, and the slowest time that qualifies. [2]

Q8. Yuto drives a distance of 508 km from Town P to Town Q at an average speed of x km/h. On the return journey he drives 10 km/h faster, and the return journey takes 1.25 hours less.

(a) Write down, in terms of x, the time taken for the journey from P to Q, in hours. [1]

(b) Write down, in terms of x, the time taken for the return journey, in hours. [1]

(c) Show that x2 + 10x − 4064 = 0. [3]

(d) Solve x2 + 10x − 4064 = 0, giving your answers correct to 2 decimal places. [2]

(e) Yuto arrives at Town Q at 4.12 pm. Find the time he left Town P. [2]

End of Paper 2. Check your work — make sure every answer is rounded as asked and that all working is shown.

Answer Key — ACSI Mock Paper C2

Total: 50 marks · 8 questions · the 2024 ACS paper (calculator). 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) 25 km  [A1]
12.5 × 200 000 = 2 500 000 cm = 25 000 m = 25 km
Q1 (b) 0.2525 cm2  [M1 for the area scale, A1]
1.01 km2 = 1.01 × 1010 cm2; the area scale factor is (200 000)2 = 4 × 1010, so the map area = 1.01 × 1010 ÷ 4 × 1010 = 0.2525 cm2
Q1 (c) 1 : 100 000  [M1 for the length scale factor, A1]
Area on Map B = 4 × 0.2525 = 1.01 cm2, so the area scale factor from Map B to the first map is 4, and the length scale factor is √4 = 2. Map B’s scale is therefore 200 000 ÷ 2 = 1 : 100 000
Q2 (a) x = 12  [A1]
12 + 27 + 3x + 5 = 44 + 3x = 80 → 3x = 36 → x = 12
Q2 (b) mean = 159.25 cm  [M1 for the midpoints, A1]
Using midpoints 145, 155, 165, 175 with frequencies 12, 27, 36, 5: Σfx = 145(12) + 155(27) + 165(36) + 175(5) = 1740 + 4185 + 5940 + 875 = 12 740; mean = 12 740 ÷ 80 = 159.25 cm
Q3 (a) 5 cm  [A1]
Length scale factor = 20 ÷ 28 = 5/7, so the radius = 7 × 5/7 = 5 cm
Q3 (b) 128 ml  [M1 for the volume ratio, A1]
Volume scale factor = (5/7)3 = 125/343, so the capacity = 350 × 125/343 = 127.55… ≈ 128 ml
Q4 (a) AC = 20 cm  [M1 for Pythagoras, A1]
AC2 = 122 + 162 = 144 + 256 = 400 → AC = 20 cm
Q4 (b) shown  [M1 for CE = 20, A1 for the calculation]
The triangles are congruent, so CE = AC = 20 cm. Angle ACE = 90°, so by Pythagoras AE2 = AC2 + CE2 = 400 + 400 = 800 → AE = √800 = √(400 × 2) = 20√2 cm
Q5 (a) −1.2, 1, 1.8, 1.3, −0.5, −3.7  [M1 for two correct, A1 for all]
Substituting: x = −1 → −0.667 − 1.5 + 1 = −1.1667 ≈ −1.2; x = 0 → 1; x = 1 → −0.667 + 1.5 + 1 = 1.8333 ≈ 1.8; x = 2 → −2.667 + 3 + 1 = 1.3333 ≈ 1.3; x = 3 → −6 + 4.5 + 1 = −0.5; x = 4 → −10.667 + 6 + 1 = −3.6667 ≈ −3.7
Q5 (b) smooth curve through the six points  [A1]
Q5 (c) x ≈ 3.53 or x ≈ −1.28  [M1 for reading, A1]
−(2/3)x2 + (3/2)x + 1 = −2 → ×6: −4x2 + 9x + 6 = −12 → 4x2 − 9x − 18 = 0 → x = (9 ± √(81 + 288)) ÷ 8 = (9 ± √369) ÷ 8 → x = 3.53 or −1.28
Q5 (d)(i) y = −½x − 1  [A1]
Rearranging the curve: −(2/3)x2 = −2x − 2, so y = (−2x − 2) + (3/2)x + 1 = −½x − 1. The line is y = −½x − 1
Q5 (d)(ii) x ≈ 3.79 or x ≈ −0.79  [M1 for the quadratic, A1]
−½x − 1 = −(2/3)x2 + (3/2)x + 1 → ×6: −3x − 6 = −4x2 + 9x + 6 → 4x2 − 12x − 12 = 0 → x2 − 3x − 3 = 0 → x = (3 ± √21) ÷ 2 → x = 3.79 or −0.79
Q6 (a) 8π cm  [M1 for the arc formula, A1]
Arc = (120/360) × 2π(12) = 8π cm
Q6 (b) 4 cm  [A1]
The arc becomes the base circumference: 2πr = 8π → r = 4 cm
Q6 (c) height 8√2 cm ≈ 11.31 cm, volume ≈ 189.6 cm3  [M1 for Pythagoras, M1 for the volume, A1]
h2 = 122 − 42 = 144 − 16 = 128 → h = 8√2 ≈ 11.31 cm. Volume = ⅓π(42)(8√2) = (128√2/3)π ≈ 189.6 cm3
Q6 (d) 3.56 cm  [M1 for equating the volumes, A1]
(4/3)πR3 = (128√2/3)π → R3 = 32√2 = 45.255 → R = 3.56 cm (3 s.f.)
Q7 (a) stem-and-leaf  [M1 for the ordered list, M1 for the stems and leaves, A1 for the key]
Ordered: 62, 68, 69, 70, 72, 74, 76, 77, 81, 83, 88, 88, 90, 91, 99
6 | 2 8 9    7 | 0 2 4 6 7    8 | 1 3 8 8    9 | 0 1 9    Key: 6 | 2 means 62 seconds
Q7 (b) 88  [A1]  —  (c) 77  [A1]  —  (d) 37  [A1]
(b) 88 appears twice. (c) The 8th of the 15 values is 77. (d) 99 − 62 = 37
Q7 (e) 18  [M1 for the quartiles, A1]
Lower quartile = the 4th value = 70; upper quartile = the 12th value = 88; interquartile range = 88 − 70 = 18
Q7 (f) 9 students qualify; the slowest qualifying time is 81 seconds  [M1 for 60% of 15, A1]
60% of 15 = 9 students. The 9 fastest times are the first 9 in order, and the 9th value is 81 seconds
Q8 (a) 508 ÷ x hours  [A1]
Q8 (b) 508 ÷ (x + 10) hours  [A1]
Q8 (c) shown  [M1 for the equation, M1 for clearing the fractions, A1]
508/x − 508/(x + 10) = 1.25 → 508(x + 10) − 508x = 1.25x(x + 10) → 5080 = 1.25x2 + 12.5x → ×8: 40 640 = 10x2 + 100x → ÷10: x2 + 10x − 4064 = 0
Q8 (d) x = 58.95 or −68.95  [M1 for the formula, A1]
x = (−10 ± √(100 + 16 256)) ÷ 2 = (−10 ± √16 356) ÷ 2 = (−10 ± 127.89) ÷ 2 → x = 58.95 or −68.95 (x > 0, so x = 58.95)
Q8 (e) 7.35 am  [M1 for the time, A1]
Time = 508 ÷ 58.945 = 8.6182 h = 8 h 37 min. He arrived at 16:12, so he left at 16:12 − 8 h 37 min = 07:35 (7.35 am)
Miss Clarissa Ng | www.clartutors.com