Maths: Full Practice Paper (Additional, Level 3, Set 3) - Worksheets, Questions and Revision

22 original exam-style questions - 6 pages of questions with a full mark scheme - free printable PDF.

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13+ · Mathematics

CE.M23 Maths: Full Practice Paper (Additional, Level 3, Set 3)

ISEB COMMON ENTRANCE CE.M23 · Calculator allowed · about 60 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Time allowed: 60 minutes. Answer ALL 22 questions. Write your answers in the spaces provided. You ARE allowed to use a calculator. You must show all your working; marks are available for a correct method as well as for the final answer. The number of marks for each question, or part-question, is shown in brackets. This is a third practice paper covering stretch and scholarship-track topics from the Additional (Level 3) syllabus: index laws, expanding pairs of brackets, factorising, solving simultaneous equations by substitution, Pythagoras' theorem, right-angled trigonometry, compound measures, and harder probability.

Write 2³ * 2⁴ as a single power of 2. (1 mark)

(Total for Question 1 is 1 mark)

Work out the value of 9⁰ + 5². (1 mark)

(Total for Question 2 is 1 mark)

Simplify p¹¹ / p⁵, giving your answer as a single power of p. (1 mark)

(Total for Question 3 is 1 mark)

Simplify (z⁴)² * z³, giving your answer as a single power of z. (2 marks)

(Total for Question 4 is 2 marks)

Expand and simplify (x + 6)(x - 2). (2 marks)

(Total for Question 5 is 2 marks)

Expand and simplify (4x - 3)(2x + 7). (3 marks)

(Total for Question 6 is 3 marks)

Factorise x² + 3x - 18. (2 marks)

(Total for Question 7 is 2 marks)

Factorise x² - 36. (2 marks)

(Total for Question 8 is 2 marks)

Solve the simultaneous equations using substitution. y = 4x - 3 and x + y = 17 (3 marks)

(Total for Question 9 is 3 marks)

Solve the simultaneous equations using substitution. y = 2x + 7 and 4x + y = 25 (4 marks)

(Total for Question 10 is 4 marks)

A right-angled triangle has two shorter sides of length 8 cm and 15 cm. Calculate the length of the hypotenuse. (3 marks)

(Total for Question 11 is 3 marks)

A right-angled triangle has a hypotenuse of length 20 cm and one shorter side of length 12 cm. Calculate the length of the other shorter side. (3 marks)

(Total for Question 12 is 3 marks)

An engineer installs a zip-wire that runs in a straight line from the top of a launch platform down to a landing point on level ground, 24 m from the base of the platform. The zip-wire is 25 m long. Calculate the height of the platform. (4 marks)

(Total for Question 13 is 4 marks)

In a right-angled triangle, one of the angles is 48 degrees. The side adjacent to this angle has length 11 cm. Calculate the length of the side opposite this angle, correct to 1 decimal place. (3 marks)

(Total for Question 14 is 3 marks)

In a right-angled triangle, the side opposite an angle x has length 9 cm and the side adjacent to x has length 6 cm. Calculate the size of angle x, correct to the nearest degree. (3 marks)

(Total for Question 15 is 3 marks)

Nia jogs 10.8 km in 54 minutes at a constant speed. Work out her average speed in km/h. (3 marks)

(Total for Question 16 is 3 marks)

A metal bar has a mass of 712.8 g and a volume of 88 cm³. Work out the density of the metal, in g/cm³. (3 marks)

(Total for Question 17 is 3 marks)

A jar contains 7 red beads, 5 blue beads and 3 green beads. A bead is picked at random from the jar. Work out the probability that the bead is blue. Give your answer as a fraction in its simplest form. (2 marks)

(Total for Question 18 is 2 marks)

A box contains 6 orange sweets and 8 lemon sweets. Two sweets are taken from the box at random, one after the other, without replacement. Work out the probability that both sweets are the same flavour. (4 marks)

(Total for Question 19 is 4 marks)

From the top of a lighthouse, 45 m above sea level, Ravi observes a small boat out at sea. The angle of depression from Ravi to the boat is 18 degrees.

(a)Calculate the horizontal distance from the base of the lighthouse to the boat, correct to 1 decimal place. (3 marks)(3)

(b)The boat then sails directly towards the base of the lighthouse until the angle of depression becomes 33 degrees. Using your answer to part (a), calculate how far the boat has sailed towards the lighthouse, correct to 1 decimal place. (2 marks)(2)

(Total for Question 20 is 5 marks)

A cafe sells coffees and teas. Four coffees and two teas cost 15.60 pounds in total. One coffee costs 0.90 pounds more than one tea.

(a)Using t for the cost of one tea, in pounds, write down an expression for the cost of one coffee, in pounds. (1 mark)(1)

(b)Form and solve a pair of simultaneous equations to find the cost of one tea and the cost of one coffee. (4 marks)(4)

(Total for Question 21 is 5 marks)

The three sides of a right-angled triangle have lengths x cm, (x + 3) cm and (x + 6) cm, where (x + 6) cm is the hypotenuse.

(a)Use Pythagoras' theorem to form an equation in x. (1 mark)(1)

(b)Expand and simplify your equation to show that it can be written as x² - 6x - 27 = 0. (2 marks)(2)

(c)Factorise your equation and hence find the value of x, giving a reason for rejecting the other solution. (2 marks)(2)

(Total for Question 22 is 5 marks)

Mark scheme · CE.M23 Maths: Full Practice Paper (Additional, Level 3, Set 3)

Question 1

B1 2⁷ oe (accept 128) cao
Answer: 2⁷

Question 2

B1 26 cao
Answer: 26

Question 3

B1 p⁶ cao
Answer: p⁶

Question 4

M1 (z⁴)² = z⁸ seen
A1 z¹¹ cao
Answer: z¹¹

Question 5

M1 x² - 2x + 6x - 12, or three correct terms of the expansion, oe
A1 x² + 4x - 12 cao
Answer: x² + 4x - 12

Question 6

M1 8x² + 28x - 6x - 21, or a correct partial expansion, oe
M1 collects the two middle terms: 28x - 6x = 22x
A1 8x² + 22x - 21 cao
Answer: 8x² + 22x - 21

Question 7

M1 identifies a factor pair of -18 that sums to 3 (6 and -3)
A1 (x + 6)(x - 3) cao
Answer: (x + 6)(x - 3)

Question 8

M1 recognises the difference of two squares, x² - 6²
A1 (x - 6)(x + 6) cao
Answer: (x - 6)(x + 6)

Question 9

M1 substitutes y = 4x - 3 into the second equation: x + (4x-3) = 17
M1 simplifies to 5x - 3 = 17 and solves to find x = 4
A1 x = 4, y = 13 cao
Answer: x = 4, y = 13

Question 10

M1 substitutes y = 2x + 7 into the second equation: 4x + (2x+7) = 25
M1 collects terms to give 6x + 7 = 25
A1 x = 3
A1 y = 13 (follow through 2 x their x + 7)
Answer: x = 3, y = 13

Question 11

M1 8² + 15² seen
M1 64 + 225 = 289
A1 √289 = 17 cm cao
Answer: 17 cm

Question 12

M1 20² - 12² seen
M1 400 - 144 = 256
A1 √256 = 16 cm cao
Answer: 16 cm

Question 13

M1 recognises Pythagoras applies with the zip-wire as the hypotenuse
M1 25² - 24² seen
M1 625 - 576 = 49
A1 √49 = 7 m cao
Answer: 7 m

Question 14

M1 tan(48) = opposite / 11 oe
M1 opposite = 11 * tan(48)
A1 12.2 cm (1 d.p.) cao
Answer: 12.2 cm

Question 15

M1 tan(x) = 9 / 6 oe
M1 x = arctan(1.5)
A1 56 degrees (nearest degree) cao
Answer: 56 degrees

Question 16

M1 converts 54 minutes to 0.9 hours (or 9/10 hour), oe
M1 speed = 10.8 / 0.9
A1 12 km/h cao
Answer: 12 km/h

Question 17

M1 quotes density = mass / volume
M1 712.8 / 88
A1 8.1 g/cm³ cao
Answer: 8.1 g/cm³

Question 18

M1 total number of beads = 15
A1 1/3 cao
Answer: 1/3

Question 19

M1 P(both orange) = 6/14 * 5/13
M1 P(both lemon) = 8/14 * 7/13
M1 adds the two probabilities: 15/91 + 28/91 oe
A1 43/91 cao
Answer: 43/91

Question 20

(a) M1 identifies the angle at the boat, between the sea and the line of sight, as 18 degrees (alternate angles), with tan(18) = 45 / d oe
(a) M1 d = 45 / tan(18)
(a) A1 138.5 m (1 d.p.) cao
(a) Answer: 138.5 m
(b) M1 finds the new distance 45 / tan(33), and subtracts it from their part (a) distance
(b) A1 69.2 m (1 d.p.) cao (follow through their part (a))
(b) Answer: 69.2 m

Question 21

(a) B1 t + 0.9 oe
(a) Answer: t + 0.9
(b) M1 forms 4(t+0.9) + 2t = 15.6, using their part (a)
(b) M1 expands and simplifies to 6t + 3.6 = 15.6
(b) A1 t = 2.00 (tea costs 2.00 pounds)
(b) A1 coffee costs 2.90 pounds (follow through their t + 0.9)
(b) Answer: Tea = 2.00 pounds, coffee = 2.90 pounds

Question 22

(a) B1 x² + (x+3)² = (x+6)² oe
(a) Answer: x² + (x+3)² = (x+6)²
(b) M1 expands both brackets correctly: x² + x² + 6x + 9 = x² + 12x + 36
(b) A1 correctly rearranges to x² - 6x - 27 = 0
(b) Answer: x² - 6x - 27 = 0
(c) M1 factorises to (x-9)(x+3) = 0
(c) A1 x = 9, rejecting x = -3 since a length cannot be negative
(c) Answer: x = 9

Question 22

5 marks

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