ACSI Mock Paper C1 — Mathematics Paper 1

Sec 2 End-of-Year Examination Practice — Cambridge International Mathematics
50 marks · 1 hour · No calculator
Prepared by Miss Clarissa Ng
www.clartutors.com

END OF YEAR EXAMINATION — SECONDARY 2

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

INSTRUCTIONS

INFORMATION

Questions 1 to 16 (50 marks) · ACSI 2024 Paper 1

Q1. The two circles below are each divided into 8 equal sections, and each section is numbered. Some sections are shaded.

(a) Shade one more section on the first circle so that the overall shape has exactly one line of symmetry. [1]

1234567812345678

(b) Shade two more sections on the second circle so that the overall shape has rotational symmetry of order 2. [1]

1234567812345678

Q2. Solve the simultaneous equations

x = 2y + 13     and     2x + y = 6

x = ______________________     y = ______________________    [3]

Q3. Given that p = 8 × 107 and q = 2 × 106, evaluate the following, giving your answers in standard form.

(a) p ÷ q    [2]

(b) 1q2    [3]

Q4. (a) Simplify 15x−3, leaving your answer in positive index form. [1]

(b) Evaluate 81¾. [2]

Q5. 5 × ∛125 = 5n. Find the value of n. [2]

Q6. Simplify √50 − 2√8. [2]

Q7. Rationalise the denominator of 213 − √2, giving your answer in its simplest form. [3]

Q8. Expand and simplify (a − 2)(a + 2)(a2 + 4). [3]

Q9. Factorise completely 8a2b3 − 18b. [3]

Q10. (i) Factorise 2x2 − 5x − 3. [2]

(ii) Hence, simplify 5x − 3 − 72x2 − 5x − 3 as a single fraction in its simplest form. [2]

Q11. Solve the equation (x − 3)(x − 4) = 6.

x = ______________________ or x = ______________________    [4]

Q12. The table below shows the number of erasers that a group of students have.

Number of erasers0123
Frequency28x1

(i) Write down the largest possible value of x if the mode is 1.    [1]

(ii) Write down the value of x if the median is 1.5.    [1]

Q13. In the diagram, triangle XYZ is similar to triangle XUF. Given that XY = 8 cm, YF = 4 cm, XU = 6 cm and UZ = x cm, find the value of x.

XYZFU8 cm4 cm6 cmx cm

x = ______________________ cm    [2]

Q14. Given that 3ab + 8cd − 8cb − 3ad = 0 and b ≠ d, find the value of ca. [3]

Q15. The diagram shows the graph of y = x(x − 6). The graph passes through the origin O and the point A on the x-axis.

1234567−1−2−3−4−5−6−7−8−9xyAO

(i) Find the coordinates of A.    [2]

(ii) Write down the equation of the line of symmetry of the graph.    [1]

(iii) Find the smallest value of y.    [2]

Q16. Given that √((z − yy)) = 1x, express y in terms of x and z. [4]

End of Paper 1. Check your work — make sure every answer is in its simplest form and that all working is shown.

Answer Key — ACSI Mock Paper C1

Total: 50 marks · 16 questions · the 2024 ACS paper. 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) Shade inner section 1 — the shape then has one line of symmetry, the vertical one [1]
1234567812345678
Q1 (b) Shade inner section 5 and outer section 8 — a rotation of 180° then maps the shape onto itself, so the order is 2 [1]
1234567812345678
Q2   x = 5, y = −4  [M1 for substituting, A1 for y, A1 for x]
x = 2y + 13 → 2(2y + 13) + y = 6 → 5y = −20 → y = −4, so x = 2(−4) + 13 = 5
Q3 (a) 4 × 101 (= 40)  [M1 for the division, A1 for standard form]
p ÷ q = (8 × 107) ÷ (2 × 106) = 4 × 101
Q3 (b) 2.5 × 10−13  [M1 for squaring q, M1 for the reciprocal, A1]
q2 = (2 × 106)2 = 4 × 1012, so 1 ÷ q2 = 0.25 × 10−12 = 2.5 × 10−13
Q4 (a) x3 ÷ 5  [A1]
1 ÷ (5x−3) = (1/5)x3 = x3/5
Q4 (b) 27  [M1 for the fourth root, A1]
81¾ = (81¼)3 = 33 = 27
Q5   n = 2  [M1 for ∛125 = 5, A1]
∛125 = 5 (because 53 = 125), so 5 × ∛125 = 5 × 5 = 52 → n = 2
Q6   √2  [M1 for simplifying each surd, A1]
√50 = 5√2 and 2√8 = 2(2√2) = 4√2, so 5√2 − 4√2 = √2
Q7   9 + 3√2  [M1 for the conjugate, M1 for the denominator, A1]
21/(3 − √2) × (3 + √2)/(3 + √2) = 21(3 + √2)/(9 − 2) = 21(3 + √2)/7 = 3(3 + √2) = 9 + 3√2
Q8   a4 − 16  [M1 for (a − 2)(a + 2) = a2 − 4, M1 for the second product, A1]
(a − 2)(a + 2) = a2 − 4, then (a2 − 4)(a2 + 4) = a4 − 16
Q9   2b(2ab − 3)(2ab + 3)  [M1 for 2b, M1 for the difference of two squares, A1]
8a2b3 − 18b = 2b(4a2b2 − 9) = 2b(2ab − 3)(2ab + 3)
Q10 (i) (2x + 1)(x − 3)  [M1 for the factors, A1]
Q10 (ii) 2(5x − 1) ÷ ((2x + 1)(x − 3))  [M1 for the common denominator, A1]
5/(x − 3) − 7/((2x + 1)(x − 3)) = [5(2x + 1) − 7]/((2x + 1)(x − 3)) = (10x − 2)/((2x + 1)(x − 3)) = 2(5x − 1)/((2x + 1)(x − 3))
Q11   x = 1 or x = 6  [M1 for expanding, M1 for the quadratic, M1 for the factors, A1]
x2 − 7x + 12 = 6 → x2 − 7x + 6 = 0 → (x − 1)(x − 6) = 0 → x = 1 or 6
Q12 (i) 7  [A1]  —  (ii) 9  [A1]
(i) For 1 to be the mode it must have the highest frequency, so x < 8; the largest integer is 7. (ii) The total is 2 + 8 + x + 1 = 11 + x; for a median of 1.5 the two middle values must be 1 and 2, which needs x = 9 (20 values, so the 10th and 11th are 1 and 2)
Q13   x = 6 cm  [M1 for the ratio, A1]
XF = 8 − 4 = 4 cm. Similar triangles give XU/XZ = XF/XY, so 6/(6 + x) = 4/8 = ½ → 12 = 6 + x → x = 6 cm
Q14   c/a = 3/8  [M1 for grouping, M1 for factorising, A1]
3ab − 3ad + 8cd − 8cb = 3a(b − d) − 8c(b − d) = (b − d)(3a − 8c) = 0. Since b ≠ d, 3a = 8c, so c/a = 3/8
Q15 (i) A = (6, 0)  [M1, A1]  —  (ii) x = 3  [A1]  —  (iii) smallest y = −9  [M1, A1]
(i) y = 0 at x = 0 (the origin) and at x − 6 = 0 → A = (6, 0). (ii) The line of symmetry is midway between the roots: x = 3. (iii) At x = 3, y = 3(3 − 6) = −9
Q16   y = x2z ÷ (1 + x2)  [M1 for squaring, M1 for clearing the fraction, M1 for collecting y, A1]
(√((z − y)/y))2 = (1/x)2 → (z − y)/y = 1/x2 → x2(z − y) = y → x2z = y + x2y = y(1 + x2) → y = x2z/(1 + x2)
Miss Clarissa Ng | www.clartutors.com