ACSI Mock Paper A1 — 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)

Q1. The two circles below are each divided into 8 equal sections. Some sections are shaded.

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

12345678

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

12345678

Q2. Solve the simultaneous equations.

3x + 2y = 16

x = y + 2    [3]

x = ______________    y = ______________

Q3. Given that p = 4 × 107 and q = 8 × 103, evaluate the following, giving your answers in standard form.

(a) p ÷ q    [2]

(b) 1/p    [3]

Q4. (a) Simplify 12x-3, leaving your answer in positive index. [1]

(b) Evaluate 272/3.    [2]

Q5. Find the value of n.

2n × √64 = 29    [2]

n = ______________

Q6. Simplify √75 − 2√12.    [2]

Q7. Rationalise the denominator of 184 − √7.    [3]

Q8. Expand and simplify.

(3p − q)(3p + q)(9p2 + 1)    [3]

Q9. Factorise completely.

20p4q − 45p2q3    [3]

Q10. (i) Factorise 3x2 − 10x − 8.    [2]

(ii) Hence simplify 4x − 4 − 63x2 − 10x − 8.    [2]

Q11. Solve the equation.

(x − 2)(x − 5) = 10    [4]

x = ______________  or  x = ______________

Q12. The table below shows the number of pets owned by a group of students.

Number of pets0123
Frequency37x6

(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 PQR is similar to triangle PST. Given that PQ = 10 cm, QS = 5 cm, PR = 12 cm and RT = x cm, find the value of x.    [2]

P Q R S T 10 cm 5 cm 12 cm x cm

x = ______________

Q14. Given that 4ab + 9cd − 9cb − 4ad = 0 and b ≠ d, find the value of ac.    [3]

Q15. The diagram shows the graph of y = x(x − 4). The graph passes through the origin O and the point A.

x y O A (2, −4) Graph of y = x(x − 4)

(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 √(w − v)v = 1y, express v in terms of w and y.    [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 A1

Total: 50 marks · 16 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.
Q1 (a) Shade section 7 — the shape then has 4 lines of symmetry [1]
12345678Added section 7
4 lines of symmetry
Q1 (b) Shade sections 5 and 6 — the shape then has rotational symmetry of order 2 [1]
12345678Added sections 5 and 6
rotates onto itself every 180°
Q2   x = 4, y = 2  [M1 for substituting, A1 for y = 2, A1 for x = 4]
x = y + 2 → 3(y + 2) + 2y = 16 → 5y + 6 = 16 → 5y = 10 → y = 2, so x = 2 + 2 = 4
Q3 (a) 5 × 103  [A1 for 5, A1 for 103]
4 × 107 ÷ 8 × 103 = 0.5 × 104 = 5 × 103
Q3 (b) 2.5 × 10-8  [M1 for 1 ÷ 4, A1 for 10-7, A1 for standard form]
1 ÷ (4 × 107) = 0.25 × 10-7 = 2.5 × 10-8
Q4 (a) x³2  [A1]
1 ÷ (2x-3) = x3 ÷ 2 = x³/2
Q4 (b) 9  [M1 for cube root, A1]
272/3 = (271/3)2 = 32 = 9
Q5   n = 6  [M1 for √64 = 23, A1]
√64 = 8 = 23, so 2n × 23 = 29 → n + 3 = 9 → n = 6
Q6   √3  [M1 for simplifying either surd, A1]
√75 = 5√3 and 2√12 = 2(2√3) = 4√3, so 5√3 − 4√3 = √3
Q7   8 + 2√7  [M1 for the conjugate, A1 for the denominator, A1 for the answer]
18/(4 − √7) × (4 + √7)/(4 + √7) = 18(4 + √7)/(16 − 7) = 18(4 + √7)/9 = 2(4 + √7) = 8 + 2√7
Q8   81p4 + 9p2 − 9p2q2 − q2  [M1 for (3p − q)(3p + q), A1 for the expansion, A1 for all four terms correct]
(3p − q)(3p + q) = 9p2 − q2; then (9p2 − q2)(9p2 + 1) = 81p4 + 9p2 − 9p2q2 − q2
Q9   5p2q(2p − 3q)(2p + 3q)  [M1 for the common factor, A1 for the difference of two squares, A1 for both brackets]
20p4q − 45p2q3 = 5p2q(4p2 − 9q2) = 5p2q(2p − 3q)(2p + 3q)
Q10 (i) (3x + 2)(x − 4)  [M1 for a correct pair of brackets, A1]
3x2 − 10x − 8 = 3x2 + 2x − 12x − 8 = x(3x + 2) − 4(3x + 2) = (3x + 2)(x − 4)
Q10 (ii) 2(6x + 1)(3x + 2)(x − 4)  [M1 for the common denominator, A1]
4/(x − 4) − 6/[(3x + 2)(x − 4)] = [4(3x + 2) − 6] / [(3x + 2)(x − 4)] = (12x + 2)/[(3x + 2)(x − 4)] = 2(6x + 1)/[(3x + 2)(x − 4)]
Q11   x = 0 or x = 7  [M1 for expanding, A1 for x² − 7x = 0, A1 for factorising, A1 for both roots]
(x − 2)(x − 5) = 10 → x2 − 7x + 10 = 10 → x2 − 7x = 0 → x(x − 7) = 0 → x = 0 or x = 7
Q12 (i) 6  [A1]  — for the mode to be 1, the frequency of 1 must be the largest, so x < 7; the largest whole-number value is 6.
Q12 (ii) 4  [A1]  — with x = 4 the total frequency is 20, so the median is the mean of the 10th and 11th values, which are 1 and 2 → median = 1.5. (x = 3 + 7 − 6)
Q13   x = 6  [M1 for the correct ratio, A1]
PS = 10 + 5 = 15 cm. Since the triangles are similar, PS/PQ = PT/PR → 15/10 = (12 + x)/12 → 1.5 × 12 = 12 + x → 18 = 12 + x → x = 6
Q14   a/c = 9/4  [M1 for grouping, A1 for the common factor, A1 for the answer]
4ab − 4ad + 9cd − 9cb = 4a(b − d) − 9c(b − d) = (b − d)(4a − 9c) = 0. Since b ≠ d, b − d ≠ 0, so 4a = 9c → a/c = 9/4
Q15 (i) A(4, 0)  [M1 for setting y = 0, A1]
y = 0 → x(x − 4) = 0 → x = 0 (the origin O) or x = 4, so A is (4, 0)
Q15 (ii) x = 2  [A1]  — midway between the two roots 0 and 4, shown dashed on the graph.
Q15 (iii) −4  [M1 for substituting x = 2, A1]
Smallest y is at x = 2: y = 2(2 − 4) = 2 × (−2) = −4
Q16   v = wy²y² + 1  [M1 for squaring, M1 for rearranging, A1 for factorising, A1 for the answer]
√((w − v)/v) = 1/y → (w − v)/v = 1/y² → y²(w − v) = v → wy² − vy² = v → wy² = v(1 + y²) → v = wy²/(y² + 1)
Miss Clarissa Ng | www.clartutors.com