ACSI Mock Paper D1 — 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 13 (50 marks) · ACSI 2023 Paper 1 skills · fresh questions

Q1. (a) For the diagram below, write down the order of rotational symmetry. [1]

(b) Shade in two more squares so that the diagram below has rotational symmetry of order 2 and no lines of symmetry. [1]

(c) Complete the following statements with a word from the list. [2]

rectangle    square    parallelogram    kite    rhombus
(i) A ____________________ has exactly one line of symmetry.(ii) A ____________________ has rotational symmetry of order 2 but no lines of symmetry.

Q2. Solve the simultaneous equations

3x + 4y = 23     and     7x − 2y = 31

x = ______________________     y = ______________________    [3]

Q3. 15 bean plants are measured. The height of each plant in centimetres, correct to 1 decimal place, is listed below.

12.613.912.214.514.4
13.213.715.114.513.9
12.814.515.313.414.1

(a) Complete the ordered stem and leaf diagram to show this information. One value has been entered for you. [3]

StemLeaf
12
13
14
153

Key: ____________  |  ____________   represents   ____________________

(b) Work out the range.

____________________ cm    [2]

(c) State the mode.

____________________ cm    [1]

(d) State the median.

____________________ cm    [1]

(e) State the upper quartile.

____________________ cm    [1]

Q4. A gardener records the following about the flowers he grows.

A the type of flower  ·  B the height of the plant  ·  C the number of seeds in a pod  ·  D the colour of the flower.

(a) Which one of A, B, C or D represents discrete data? [1]

(b) Which one of A, B, C or D represents continuous data? [1]

Q5. Expand and simplify (3x − 2)(x + 1)(x − 5). [3]

Q6. (a) Simplify √75 − √12. [2]

(b) Rationalise the denominator and simplify 2 + √54 − √5. [3]

Q7. (a) Work out 4.2 × 1017 + 4.2 × 1018. Give your answer in standard form. [2]

(b) In this calculation, the three numbers are written in standard form.

(1.5 × 104) × (1.5 × 10n) = 2.25 × 10−3

n is an integer. Find the value of n.

n = ______________________    [1]

Q8. Rearrange the formula M = (x − c)2 + d to write x in terms of c, d and M.

x = ______________________    [3]

Q9. Simplify 9p2 − 163px + 6p + 4x + 8. [3]

Q10. Write as a single fraction in its simplest form.

36 − 2x  +  x − 4x − 3    [3]

Q11. Solve the equation   3 − 4x + 2 = 52x − 3.

x = ______________________ or x = ______________________    [4]

Q12. The areas of the two rectangles below are equal. Find the value of x. Show all your working.

3x + 22x4xx + 3NOT TOSCALE

x = ______________________    [3]

Q13. (a) Simplify y0 × (ab)−4a6. [2]

(b) Evaluate ( 2549 )−½. [2]

(c) Find the value of x when 27x+1 = 32x+5.

x = ______________________    [2]

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 D1

Total: 50 marks · 13 questions · modelled on the 2023 ACSI paper (Paper 1 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) 8  [A1] — eight blades, each 45° apart, map onto each other eight times in a full turn
(b) Shade the bottom-right cell and the third cell of the bottom row  [A1] — those are the 180° images of the two shaded cells, so the shape now maps onto itself under a half-turn (order 2). No mirror line works: the shaded cells have no partner across any of the four lines.
(c) (i) kite  (ii) parallelogram  [A1, A1]
Q2   x = 5, y = 2  [M1 for substituting/rearranging, A1 for x, A1 for y]
From 7x − 2y = 31, y = (7x − 31)/2. Substitute into the first: 3x + 2(7x − 31) = 23 → 3x + 14x − 62 = 23 → 17x = 85 → x = 5, so y = (35 − 31)/2 = 2
Q3 (a) Stem-and-leaf  [B1 for the ordered leaves, M1 for the key, A1 for all values correct]
12 | 2 6 8    13 | 2 4 7 9 9    14 | 1 4 5 5 5    15 | 1 3  —  key: 12 | 2 represents 12.2 cm
Q3 (b) Range = 3.1 cm  [M1 for 15.3 − 12.2, A1]
Q3 (c) Mode = 14.5 cm  [B1] — it appears three times
Q3 (d) Median = 13.9 cm  [B1] — 15 values, so the 8th in order
Q3 (e) Upper quartile = 14.5 cm  [B1] — the 12th of 15 values (¾(15 + 1) = 12)
Q4 (a) C — the number of seeds in a pod (counted)  [A1]  —  (b) B — the height of the plant (measured)  [A1]
Discrete data can only take particular values (a count); continuous data is measured and can take any value in a range. A and D are not numerical at all.
Q5   3x3 − 14x2 − 7x + 10  [M1 for one product, M1 for the second product, A1]
(x + 1)(x − 5) = x2 − 4x − 5, then (3x − 2)(x2 − 4x − 5) = 3x3 − 12x2 − 15x − 2x2 + 8x + 10 = 3x3 − 14x2 − 7x + 10
Q6 (a) 3√3  [M1 for simplifying the surds, A1]
√75 = 5√3 and √12 = 2√3, so 5√3 − 2√3 = 3√3
Q6 (b) (13 + 6√5) ÷ 11  [M1 for the conjugate, M1 for the denominator, A1]
(2 + √5)/(4 − √5) × (4 + √5)/(4 + √5) = (8 + 2√5 + 4√5 + 5)/(16 − 5) = (13 + 6√5)/11 — not further simplified since 11 does not divide 13 or 6
Q7 (a) 4.62 × 1018  [M1 for factorising out 1017, A1 for standard form]
4.2 × 1017 + 4.2 × 1018 = 4.2 × 1017(1 + 10) = 46.2 × 1017 = 4.62 × 1018
Q7 (b) n = −7  [A1]
1.5 × 1.5 = 2.25 is already correct, so the indices must satisfy 4 + n = −3 → n = −7
Q8   x = c + √(M − d)  [M1 for isolating the square, M1 for square-rooting, A1]
M − d = (x − c)2 → x − c = √(M − d) → x = c + √(M − d)
Q9   (3p − 4) ÷ (x + 2)  [M1 for factorising the numerator, M1 for grouping the denominator, A1]
9p2 − 16 = (3p − 4)(3p + 4); the denominator groups as 3p(x + 2) + 4(x + 2) = (x + 2)(3p + 4). The (3p + 4) cancels: (3p − 4)/(x + 2)
Q10   (2x − 11) ÷ (2(x − 3))  [M1 for the common denominator, M1 for combining, A1]
6 − 2x = −2(x − 3), so 3/(6 − 2x) = −3/(2(x − 3)). Then (−3/2 + x − 4)/(x − 3) = (x − 11/2)/(x − 3) = (2x − 11)/(2(x − 3))
Q11   x = 8/3 or x = −1  [M1 for the common denominator, M1 for cross-multiplying, M1 for the quadratic, A1]
3 − 4/(x + 2) = (3x + 6 − 4)/(x + 2) = (3x + 2)/(x + 2). Cross-multiplying: (3x + 2)(2x − 3) = 5(x + 2) → 6x2 − 5x − 6 = 5x + 10 → 6x2 − 10x − 16 = 0 → 3x2 − 5x − 8 = 0 → (3x − 8)(x + 1) = 0 → x = 8/3 or x = −1
Q12   x = 4  [M1 for equating the areas, M1 for the quadratic, A1]
2x(3x + 2) = 4x(x + 3) → 6x2 + 4x = 4x2 + 12x → 2x2 − 8x = 0 → 2x(x − 4) = 0. x ≠ 0 because it is a length, so x = 4
Q13 (a) 1 ÷ (a10b4)  [M1 for handling the indices, A1]
y0 = 1 and (ab)−4 = a−4b−4, so the fraction is a−4−6b−4 = a−10b−4 = 1/(a10b4)
Q13 (b) 7/5  [M1 for the reciprocal, A1]
(25/49)−½ = (49/25)½ = √49 ÷ √25 = 7/5 = 1.4
Q13 (c) x = 2  [M1 for 27 = 33, A1]
27 = 33, so 33(x+1) = 32x+5 → 3x + 3 = 2x + 5 → x = 2
Miss Clarissa Ng | www.clartutors.com