Maths: Full Practice Paper (Additional, Level 3, Set 5) - 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.M33 Maths: Full Practice Paper (Additional, Level 3, Set 5)

ISEB COMMON ENTRANCE CE.M33 · 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 fifth practice paper covering stretch and scholarship-track topics from the Additional (Level 3) syllabus: index laws (including fractional and negative indices), harder quadratic factorisation, solving simultaneous equations that require rearranging first, Pythagoras' theorem, right-angled trigonometry, compound measures, and harder probability without replacement.

Write 5⁴ * 5⁵ as a single power of 5. (1 mark)

(Total for Question 1 is 1 mark)

Simplify t¹³ / t⁶, giving your answer as a single power of t. (1 mark)

(Total for Question 2 is 1 mark)

Work out the value of 16^1/2 + 3^-2, giving your answer as a fraction. (2 marks)

(Total for Question 3 is 2 marks)

Simplify (2a³)² * a⁴, giving your answer as simply as possible. (2 marks)

(Total for Question 4 is 2 marks)

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

(Total for Question 5 is 2 marks)

Expand and simplify (6x - 5)(2x + 3). (3 marks)

(Total for Question 6 is 3 marks)

Factorise x² - 2x - 24. (2 marks)

(Total for Question 7 is 2 marks)

Factorise 3x² - 5x - 2. (3 marks)

(Total for Question 8 is 3 marks)

Solve the simultaneous equations using substitution. y = 4x - 5 and x + y = 15 (3 marks)

(Total for Question 9 is 3 marks)

Solve the simultaneous equations using substitution. 2y = 3x + 1 and x + y = 8 (4 marks)

(Total for Question 10 is 4 marks)

A right-angled triangle has two shorter sides of length 20 cm and 21 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 37 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)

A kite is flying above a level field. The kite string runs in a straight line from Priya's hand, at ground level, to the kite. The string is taut and 85 m long. The point on the ground directly below the kite is 84 m from Priya. Calculate the height of the kite above the ground. (4 marks)

(Total for Question 13 is 4 marks)

In a right-angled triangle, one of the angles is 27 degrees. The side adjacent to this angle has length 18 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 13 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)

Chloe cycles 15.6 km in 40 minutes at a constant speed. Work out her average speed in km/h. (3 marks)

(Total for Question 16 is 3 marks)

A metal ingot has a mass of 891 g and a volume of 55 cm³. Work out the density of the metal, in g/cm³. (3 marks)

(Total for Question 17 is 3 marks)

A box contains 9 red discs, 15 blue discs and 6 green discs. A disc is picked at random from the box. Work out the probability that the disc is green. Give your answer as a fraction in its simplest form. (2 marks)

(Total for Question 18 is 2 marks)

A pencil case contains 6 red pens and 9 blue pens. Two pens are taken from the pencil case at random, one after the other, without replacement. Work out the probability that the two pens are the same colour. (4 marks)

(Total for Question 19 is 4 marks)

From a point P on level ground, Marcus measures the angle of elevation to the top of a wind turbine as 18 degrees. Point P is 130 m from the base of the turbine.

(a)Calculate the height of the turbine, correct to 1 decimal place. (3 marks)(3)

(b)Marcus then walks towards the turbine until he reaches a point Q, where the angle of elevation to the top of the turbine has increased to 31 degrees. Using your answer to part (a), calculate the distance PQ, correct to 1 decimal place. (2 marks)(2)

(Total for Question 20 is 5 marks)

An ice cream van sells cones and lollies. Four cones and three lollies cost 18.00 pounds in total. One lolly costs 1.00 pound less than one cone.

(a)Using c for the cost of one cone, in pounds, write down an expression for the cost of one lolly, in pounds. (1 mark)(1)

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

(Total for Question 21 is 5 marks)

The three sides of a right-angled triangle have lengths x cm, (x + 17) cm and (x + 18) cm, where (x + 18) 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 - 2x - 35 = 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.M33 Maths: Full Practice Paper (Additional, Level 3, Set 5)

Question 1

B1 5⁹ oe cao
Answer: 5⁹

Question 2

B1 t⁷ oe cao
Answer: t⁷

Question 3

M1 16^1/2 = 4 and 3^-2 = 1/9 both identified
A1 37/9 oe (accept 4 1/9) cao
Answer: 37/9

Question 4

M1 (2a³)² = 4a⁶ seen
A1 4a¹⁰ cao
Answer: 4a¹⁰

Question 5

M1 x² - 4x + 7x - 28, or three correct terms of the expansion, oe
A1 x² + 3x - 28 cao
Answer: x² + 3x - 28

Question 6

M1 12x² + 18x - 10x - 15, or a correct partial expansion, oe
M1 collects the two middle terms: 18x - 10x = 8x
A1 12x² + 8x - 15 cao
Answer: 12x² + 8x - 15

Question 7

M1 identifies a factor pair of -24 that sums to -2 (-6 and 4)
A1 (x - 6)(x + 4) cao
Answer: (x - 6)(x + 4)

Question 8

M1 recognises that ac = -6 (3 x -2) and identifies two numbers multiplying to -6 and summing to -5 (-6 and 1)
M1 splits the middle term and factorises by grouping: 3x² - 6x + x - 2 = 3x(x-2) + 1(x-2)
A1 (3x + 1)(x - 2) cao
Answer: (3x + 1)(x - 2)

Question 9

M1 substitutes y = 4x - 5 into the second equation: x + (4x-5) = 15
M1 simplifies to 5x - 5 = 15 and solves to find x = 4
A1 x = 4, y = 11 cao
Answer: x = 4, y = 11

Question 10

M1 rearranges 2y = 3x + 1 to make y the subject: y = (3x+1)/2, oe
M1 substitutes into x + y = 8 and clears the fraction: 2x + 3x + 1 = 16, oe
A1 x = 3
A1 y = 5 (follow through their x)
Answer: x = 3, y = 5

Question 11

M1 20² + 21² seen
M1 400 + 441 = 841
A1 √841 = 29 cm cao
Answer: 29 cm

Question 12

M1 37² - 12² seen
M1 1369 - 144 = 1225
A1 √1225 = 35 cm cao
Answer: 35 cm

Question 13

M1 recognises Pythagoras applies with the string as the hypotenuse
M1 85² - 84² seen
M1 7225 - 7056 = 169
A1 √169 = 13 m cao
Answer: 13 m

Question 14

M1 tan(27) = opposite / 18 oe
M1 opposite = 18 * tan(27)
A1 9.2 cm (1 d.p.) cao
Answer: 9.2 cm

Question 15

M1 tan(x) = 13 / 6 oe
M1 x = arctan(2.1667)
A1 65 degrees (nearest degree) cao
Answer: 65 degrees

Question 16

M1 converts 40 minutes to 2/3 hour (or 0.6 recurring hours), oe
M1 speed = 15.6 / (2/3)
A1 23.4 km/h cao
Answer: 23.4 km/h

Question 17

M1 quotes density = mass / volume
M1 891 / 55
A1 16.2 g/cm³ cao
Answer: 16.2 g/cm³

Question 18

M1 total number of discs = 30
A1 1/5 cao
Answer: 1/5

Question 19

M1 P(red then red) = 6/15 * 5/14
M1 P(blue then blue) = 9/15 * 8/14
M1 adds the two probabilities: 30/210 + 72/210 oe
A1 17/35 cao
Answer: 17/35

Question 20

(a) M1 tan(18) = height / 130 oe
(a) M1 height = 130 * tan(18)
(a) A1 42.2 m (1 d.p.) cao
(a) Answer: 42.2 m
(b) M1 finds the new horizontal distance using their part (a) height and tan(31), then subtracts it from 130 m
(b) A1 59.7 m (1 d.p.) cao (follow through their part (a))
(b) Answer: 59.7 m

Question 21

(a) B1 c - 1 oe
(a) Answer: c - 1
(b) M1 forms 4c + 3(c-1) = 18, using their part (a)
(b) M1 expands and simplifies to 7c - 3 = 18
(b) A1 c = 3.00 (cone costs 3.00 pounds)
(b) A1 lolly costs 2.00 pounds (follow through their c - 1)
(b) Answer: Cone = 3.00 pounds, lolly = 2.00 pounds

Question 22

(a) B1 x² + (x+17)² = (x+18)² oe
(a) Answer: x² + (x+17)² = (x+18)²
(b) M1 expands both brackets correctly: x² + x² + 34x + 289 = x² + 36x + 324
(b) A1 correctly rearranges to x² - 2x - 35 = 0
(b) Answer: x² - 2x - 35 = 0
(c) M1 factorises to (x-7)(x+5) = 0
(c) A1 x = 7, rejecting x = -5 since a length cannot be negative
(c) Answer: x = 7

Question 22

5 marks

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