Maths: Full Practice Paper (Additional, Level 3, Set 2) - 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.M17 Maths: Full Practice Paper (Additional, Level 3, Set 2)

ISEB COMMON ENTRANCE CE.M17 · 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 second 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 3⁴ * 3² as a single power of 3. (1 mark)

(Total for Question 1 is 1 mark)

Work out the value of 6⁰ + 3³. (1 mark)

(Total for Question 2 is 1 mark)

Simplify n⁹ / n⁴, giving your answer as a single power of n. (1 mark)

(Total for Question 3 is 1 mark)

Simplify (y²)³ * y⁴, giving your answer as a single power of y. (2 marks)

(Total for Question 4 is 2 marks)

Expand and simplify (x + 4)(x - 6). (2 marks)

(Total for Question 5 is 2 marks)

Expand and simplify (3x - 2)(2x + 5). (3 marks)

(Total for Question 6 is 3 marks)

Factorise x² + 2x - 15. (2 marks)

(Total for Question 7 is 2 marks)

Factorise x² - 16. (2 marks)

(Total for Question 8 is 2 marks)

Solve the simultaneous equations using substitution. y = 3x - 2 and x + y = 14 (3 marks)

(Total for Question 9 is 3 marks)

Solve the simultaneous equations using substitution. y = 2x + 5 and 3x + y = 20 (4 marks)

(Total for Question 10 is 4 marks)

A right-angled triangle has two shorter sides of length 9 cm and 12 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 25 cm and one shorter side of length 7 cm. Calculate the length of the other shorter side. (3 marks)

(Total for Question 12 is 3 marks)

A builder rests a straight metal ramp from the edge of a raised loading platform down to a point on level ground. The ramp is 13 m long, and the point on the ground is 12 m from the base of the platform. 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 52 degrees. The side adjacent to this angle has length 9 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 8 cm and the side adjacent to x has length 5 cm. Calculate the size of angle x, correct to the nearest degree. (3 marks)

(Total for Question 15 is 3 marks)

Tom drives 133 km in 1 hour 45 minutes at a constant speed. Work out his average speed in km/h. (3 marks)

(Total for Question 16 is 3 marks)

A metal sphere has a mass of 316.8 g and a volume of 24 cm³. Work out the density of the metal, in g/cm³. (3 marks)

(Total for Question 17 is 3 marks)

A drawer contains 6 black socks, 4 grey socks and 2 navy socks. A sock is picked at random from the drawer. Work out the probability that the sock is grey. Give your answer as a fraction in its simplest form. (2 marks)

(Total for Question 18 is 2 marks)

A tin contains 5 mint sweets and 7 toffee sweets. Two sweets are taken from the tin 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 vertical cliff, 84 m above sea level, Priya observes a boat out at sea. The angle of depression from Priya to the boat is 22 degrees.

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

(b)The boat then sails directly towards the base of the cliff until the angle of depression becomes 40 degrees. Using your answer to part (a), calculate how far the boat has sailed towards the cliff, correct to 1 decimal place. (2 marks)(2)

(Total for Question 20 is 5 marks)

A stationery shop sells notebooks and pens. Five notebooks and three pens cost 21.50 pounds in total. One notebook costs 1.50 pounds more than one pen.

(a)Using p for the cost of one pen, in pounds, write down an expression for the cost of one notebook, in pounds. (1 mark)(1)

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

(Total for Question 21 is 5 marks)

The three sides of a right-angled triangle have lengths x cm, (x + 7) cm and (x + 9) cm, where (x + 9) 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 - 4x - 32 = 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.M17 Maths: Full Practice Paper (Additional, Level 3, Set 2)

Question 1

B1 3⁶ oe (accept 729) cao
Answer: 3⁶

Question 2

B1 28 cao
Answer: 28

Question 3

B1 n⁵ cao
Answer: n⁵

Question 4

M1 (y²)³ = y⁶ seen
A1 y¹⁰ cao
Answer: y¹⁰

Question 5

M1 x² - 6x + 4x - 24, or three correct terms of the expansion, oe
A1 x² - 2x - 24 cao
Answer: x² - 2x - 24

Question 6

M1 6x² + 15x - 4x - 10, or a correct partial expansion, oe
M1 collects the two middle terms: 15x - 4x = 11x
A1 6x² + 11x - 10 cao
Answer: 6x² + 11x - 10

Question 7

M1 identifies a factor pair of -15 that sums to 2 (5 and -3)
A1 (x + 5)(x - 3) cao
Answer: (x + 5)(x - 3)

Question 8

M1 recognises the difference of two squares, x² - 4²
A1 (x - 4)(x + 4) cao
Answer: (x - 4)(x + 4)

Question 9

M1 substitutes y = 3x - 2 into the second equation: x + (3x-2) = 14
M1 simplifies to 4x - 2 = 14 and solves to find x = 4
A1 x = 4, y = 10 cao
Answer: x = 4, y = 10

Question 10

M1 substitutes y = 2x + 5 into the second equation: 3x + (2x+5) = 20
M1 collects terms to give 5x + 5 = 20
A1 x = 3
A1 y = 11 (follow through 2 x their x + 5)
Answer: x = 3, y = 11

Question 11

M1 9² + 12² seen
M1 81 + 144 = 225
A1 √225 = 15 cm cao
Answer: 15 cm

Question 12

M1 25² - 7² seen
M1 625 - 49 = 576
A1 √576 = 24 cm cao
Answer: 24 cm

Question 13

M1 recognises Pythagoras applies with the ramp as the hypotenuse
M1 13² - 12² seen
M1 169 - 144 = 25
A1 √25 = 5 m cao
Answer: 5 m

Question 14

M1 tan(52) = opposite / 9 oe
M1 opposite = 9 * tan(52)
A1 11.5 cm (1 d.p.) cao
Answer: 11.5 cm

Question 15

M1 tan(x) = 8 / 5 oe
M1 x = arctan(1.6)
A1 58 degrees (nearest degree) cao
Answer: 58 degrees

Question 16

M1 converts 1 hour 45 minutes to 1.75 hours (or 7/4 hour), oe
M1 speed = 133 / 1.75
A1 76 km/h cao
Answer: 76 km/h

Question 17

M1 quotes density = mass / volume
M1 316.8 / 24
A1 13.2 g/cm³ cao
Answer: 13.2 g/cm³

Question 18

M1 total number of socks = 12
A1 1/3 cao
Answer: 1/3

Question 19

M1 P(both mint) = 5/12 * 4/11
M1 P(both toffee) = 7/12 * 6/11
M1 adds the two probabilities: 20/132 + 42/132 oe
A1 31/66 cao
Answer: 31/66

Question 20

(a) M1 identifies the angle at the boat, between the sea and the line of sight, as 22 degrees (alternate angles), with tan(22) = 84 / d oe
(a) M1 d = 84 / tan(22)
(a) A1 207.9 m (1 d.p.) cao
(a) Answer: 207.9 m
(b) M1 finds the new distance 84 / tan(40), and subtracts it from their part (a) distance
(b) A1 107.8 m (1 d.p.) cao (follow through their part (a))
(b) Answer: 107.8 m

Question 21

(a) B1 p + 1.5 oe
(a) Answer: p + 1.5
(b) M1 forms 5(p+1.5) + 3p = 21.5, using their part (a)
(b) M1 expands and simplifies to 8p + 7.5 = 21.5
(b) A1 p = 1.75 (pen costs 1.75 pounds)
(b) A1 notebook costs 3.25 pounds (follow through their p + 1.5)
(b) Answer: Pen = 1.75 pounds, notebook = 3.25 pounds

Question 22

(a) B1 x² + (x+7)² = (x+9)² oe
(a) Answer: x² + (x+7)² = (x+9)²
(b) M1 expands both brackets correctly: x² + x² + 14x + 49 = x² + 18x + 81
(b) A1 correctly rearranges to x² - 4x - 32 = 0
(b) Answer: x² - 4x - 32 = 0
(c) M1 factorises to (x-8)(x+4) = 0
(c) A1 x = 8, rejecting x = -4 since a length cannot be negative
(c) Answer: x = 8

Question 22

5 marks

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