Visuteach provides 11 plus maths practice papers that help prepare for GL Assessment (formerly known as NFER Nelson), Essex CSSE (Consortium for Selective Schools in Essex), Northern Ireland PPTC & AQE CEA (Common Entrance Assessment) and for the North London Independent Girls’ Schools’ Consortium Group 2 tests. Our practice papers are available for immediate download on our Purchase page.

Visuteach also provides interactive online maths tests in our Maths Membership sites. We have a GL Assessment/PPTC maths membership site and a North London Independent Girls’ Schools’ Consortium (NLIGSC) Group 1 membership site.

The GL/PPTC maths membership site consists of 7 GL Assessment style maths tests (each consisting of 50 questions). These 7 tests contain the questions used in our 5 paper-based GL Assessment tests, and there are 2 tests that are only available in our membership site and which are not available in paper format.

The North London Independent Girls’ Schools’ Consortium (NLIGSC) Group 1 maths membership site contains online answers and explanations to the Consortium’s past Group 1 maths papers from 2008, 2009, 2010, 20111, 2012 and 2013. The NLIGSC Group 1 membership site also contains one Visuteach 11+ online Group 1 style maths test.

Visuteach does not sell CEM 11 plus papers. However, the maths in our papers and online tests can provide useful preparation for CEM tests since they cover similar Key Stage 2 topics.

Our maths practice papers and online tests contain explanations (not just answers). For example our GL Assessment style maths practice paper 1 contains 19 pages of explanations for the 50 questions in the test.

To open free sample papers for these tests, click on the blue links in the Sample Paper column below. To download the free sample papers, right-click on the blue links and choose ‘Save Target As’ from the pop-up menu.

Visuteach 11 Plus Maths Papers | Format | Number of questions | Time Allowed | Sample Paper |

GL Assessment Style | Multiple-choice | 50 | 50 mins | GL Sample Test |

Essex CSSE | Standard | 20 | 35 mins | |

PPTC | Multiple-choice | 45 | 45 mins | PPTC Sample Test |

AQE CEA (combined maths & English) | Standard | 32 Maths, 26 English | 1 hr | |

North London Independent Girls’ Schools’ Consortium | Standard | Anything between 39-42 | 1 hr 15 mins | NLIGSC Sample Test |

Visuteach provides two online interactive maths membership sites/packages – a GL Assessment/PPTC maths membership and a North London Independent Girls’ Schools’ Consortium (NLIGSC) Group 1 membership site/package.

To view demos, please click on the blue links in the tables below. This will open new windows or tabs containing demos of our membership site software.

GL/PPTC Maths Membership Demo 1 Untimed GL Assessment style 10 question untimed test |
GL/PPTC 11+ Maths Demo 1 Untimed |

GL/PPTC Maths Membership Demo 2 Untimed GL Assessment style 10 question untimed test |
GL/PPTC 11+ Maths Demo 2 Untimed |

GL/PPTC Maths Membership Demo 1 Timed GL Assessment style 10 minute, 10 question timed test |
GL/PPTC 11+ Maths Demo 1 Timed |

GL/PPTC Maths Membership Demo 2 Timed GL Assessment style 10 minute, 10 question timed test |
GL/PPTC 11+ Maths Demo 2 Timed |

North London Independent Girls’ Schools’ Consortium Group 1 Maths Membership Demo (Sample answers to some questions from Group 1 2008 maths paper) Membership site contains links to Group 1 maths papers from 2008-2013 together with answers and explanations for the more difficult questions. The membership site also contains a Visuteach Group 1 style maths test where the questions and answers are displayed in software rather than on paper. |
NLIGSC Group 1 Demo |

For more information on our maths membership sites, please click here

Visuteach provides five types of 11+ maths practice papers. We provide GL Assessment style tests (often formerly referred to as NFER Nelson style tests), Essex CSSE (Consortium for Selective Schools in Essex) style tests, Northern Ireland PPTC transfer style tests (which are also GL Assessment style papers), Northern Ireland AQE CEA combined maths and English practice papers and common entrance style maths practice papers for Group 2 schools of the North London Independent Girls’ Schools’ Consortium.

Visuteach’s GL Assessment style 11 plus maths practice papers are multiple-choice tests and are similar to the GL Assessment/NFER sample papers. Our questions cover the topics that are tested for in these papers. Our GL Assessment style tests are 50 minute tests and contain 50 questions.

Visuteach’s Essex CSSE style maths practice papers are standard-format papers similar in style to the CSSE maths papers. They contain 20 questions which are to be completed in 35 minutes. Our papers contain explanations (not just answers).

Visuteach’s PPTC transfer test maths practice papers are an adaptation of our GL Assessment style 11 plus maths practice papers. They contain 45 multiple-choice questions instead of the 50 in our GL papers, and the question format is slightly different in order to match the format of PPTC questions. The answer sheet is also different in order to conform to the format of the PPTC exam. If you are taking the PPTC exam, then you should buy our PPTC practice papers and not our general GL Assessment papers. More information on our PPTC style tests is available on our PPTC Transfer Test page.

Our AQE papers are standard format papers, based on the type of questions found in the AQE CEA (Common Entrance Assessment) sample papers for the AQE Northern Ireland transfer test. The AQE papers contain 58 questions covering maths and English. There are 32 maths questions and 26 English questions. More information on our AQE papers can be found on our AQE CEA Transfer Test page.

Visuteach’s North London Independent Girls’ Schools’ Consortium Group 2 style 11 plus maths tests are in standard format (i.e. they are not multiple-choice tests). These tests are 1 hour 15 minute tests and contain 40 questions. More information on our North London Independent Girls’ Schools’ Consortium tests is available on our 11 Plus Independent Schools page.

Jack was x years old 4 years ago. How old will he be 6 years from now?

Choose one of the following answers:

x+2

x+4

x+6

x+10

x+8

Answer: x+10

Jack was x years old 4 years ago, so he is now x+4 years old. In 6 years’ time he will be 6 years older i.e. he will be x+4+6 = x+10 years old

A, B and C are the angles of an isosceles triangle. Angle A measures 70°. Angle B is greater than 60°. What is the size of angle C?

Choose one of the following answers:

70°

50°

60°

55°

40°

Answer: 40°

An isosceles triangle has at least 2 equal sides and 2 equal angles. We also know that the angles inside a triangle add up to 180°.

Angle A is 70°, angle B is greater than 60° and therefore angle C is less than 50°. We know that two of these angles must be equal in order for this triangle to be isosceles and the only way that this can be is if angle B = angle A i.e. angle B = 70°. Therefore angle C = 40°

The table shows some values in a conversion table of metres to feet and feet to metres. From the table you can see that 5 m = 16.4 feet and that 5 feet = 1.52 m. What figure should replace the question mark?

Choose one of the following answers:

25.40 ft

26.25 ft

23.80 ft

25.80 ft

24.90 ft

Answer: 26.25 ft

The ? represents the value of 8 metres in feet. In the table we are given the values of 5 m, 10 m and 13 m in feet, and we can use this information to find the value of 8 m in feet. We do this by seeing that 8 = 13 – 5 and therefore the value of 8 m in feet will be the value of 13 m in feet minus the value of 5 m in feet = 42.65 – 16.40 = 26.25 feet (i.e. 26.25 ft)

What is the perimeter of this shape?

Choose one of the following answers:

2x+2y+2z

w+2x+2z

y+w+x+2z

2w+y+2z

y+w+x+z

Answer: 2x+2y+2z

the perimeter of this shape is the sum of the lengths of the horizontal segments and the vertical segments

the sum of the lengths of the horizontal segments = AJ+BC+DE+FG+HI

BC+DE+FG+HI is the same as the length of AJ (i.e. z), so

the sum of the lengths of the horizontal segments = AJ+(BC+DE+FG+HI) = z+z = 2z

the sum of the lengths of the vertical segments = JI+AB+CD+FE+GH

AB+GH is the same length as JI (i.e. y), so

the sum of the lengths of the vertical segments = JI+AB+CD+FE+GH = JI + (AB+GH) +CD+FE = y+y+CD+FE = 2y+CD+FE

CD is of length x and CD is the same length as FE, so we have

the sum of the lengths of the vertical segments = 2y+CD+FE = 2y+x+x = 2y+2x

so the perimeter of the shape = sum of the lengths of the horizontal segments + sum of the lengths of the vertical segments = 2z+2y+2x = 2x+2y+2z

Which of these shapes have rotational symmetry?

Choose one of the following answers:

A and B

B and E

C and D

A and C

C and E

Answer: A and C

shape A has rotational symmetry of order 2 (i.e. it fits onto itself twice as it is turned through 360°). This can be seen in (i) and (ii) above. We have put a blob on the shape in order to more easily see what happens when we rotate the shape. We start from position (i) and end up at position (ii) after rotating the shape 180° clockwise. From position (ii) we end up back at position (i) by rotating the shape a further 180° clockwise. So there are 2 positions in which the shape looks the same when being rotated through 360° and therefore shape A has rotational symmetry of order 2.

shape C has rotational symmetry of order 4 (i.e. it fits onto itself four times as it is turned through 360°). This can be seen in (iii), (iv), (v) and (vi) above. We have put a blob on the shape in order to more easily see what happens when we rotate the shape. We start from position (iii) and end up at position (iv) after rotating the shape 90° clockwise. From position (iv) we end up at position (v) after rotating the shape 90° clockwise. From position (v) we end up at position (vi) after rotating the shape 90° clockwise. From position (vi) we end up back at position (iii) after rotating the shape a further 90° clockwise. So there are 4 positions in which the shape looks the same when being rotated through 360° and therefore shape C has rotational symmetry of order 4.

If 5x – 6 = 7x + 4, what is x?

Choose one of the following answers:

3

-3

-5

5

Answer: -5

5x-6 = 7x+4

to solve for x you need to isolate all of the x terms on one side of the equation. Performing the same operation on both sides of an equation leaves the equation unchanged (i.e. the equality still holds true). We use this fact to help us isolate x terms. First we subtract 5x from both sides of the equation i.e.

5x – 6 – 5x = 7x + 4 – 5x which gives us -6 = 2x + 4

then we subtract 4 from both sides of the equation and we get

- 6 – 4 = 2x + 4 – 4 i.e. -10 = 2x

then we divide both sides of the equation by 2 and we get

-5 = x i.e. x = -5

A radio controlled car needs to be guided along the white squares from point A to point B, avoiding the houses on the way. The car can only move FORWARD, TURN RIGHT 90° and TURN LEFT 90°.

Which instructions should you use to guide the car?

Choose one of A, B, C or D

A) FORWARD 2, TURN RIGHT 90°,

FORWARD 1, TURN LEFT 90°,

FORWARD 4, TURN LEFT 90°,

FORWARD 4, TURN RIGHT 90°,

FORWARD 2

B) FORWARD 2, TURN LEFT 90°,

FORWARD 1, TURN RIGHT 90°,

FORWARD 4, TURN LEFT 90°,

FORWARD 1, TURN RIGHT 90°,

FORWARD 2

C) FORWARD 2, TURN RIGHT 90°,

FORWARD 1, TURN LEFT 90°,

FORWARD 3, TURN LEFT 90°,

FORWARD 3, TURN RIGHT 90°,

FORWARD 3

D) FORWARD 2, TURN RIGHT 90°,

FORWARD 1, TURN LEFT 90°,

FORWARD 4, TURN LEFT 90°,

FORWARD 3, TURN RIGHT 90°,

FORWARD 2

Answer: D

The correct answer is D. The car moves forward 2, so it moves from position A to position 1. The car is then facing towards the east as shown by the arrow in the square at position 1. The car then turns right 90° (i.e. it turns 90° in a clockwise direction), so it is now facing south. It then moves forward 1 and ends up in position 2. The car then turns left 90° (i.e. it turns 90° in an anti-clockwise direction), so it is now facing east. The car then moves forward 4 so it ends up at position 3. The car then turns left 90° (i.e. it turns 90° in an anti-clockwise direction), so it is now facing north. The car then moves forward 3 so it ends up at position 4. The car then turns right 90° (i.e. it turns 90° in a clockwise direction), so it is now facing east. The car then moves forward 2 so it ends up at position B.

Answer A is incorrect, because the car does not end up at position B, as shown in the diagram below:

Answer B is incorrect, because the car crashes into the house at position X, as shown in the following diagram:

Answer C is incorrect, because the car crashes into the house at position X, as shown in the diagram below:

The diagram shows exterior angles for 5 regular polygons. Which one of these angles is the smallest?

Choose one of A, B, C, D or E

Answer: E

There are a number of ways of working this question out. Two ways are as follows :

a) for a regular polygon, all of its sides are equal in length and all of its interior angles are equal. Each exterior angle of the polygon has a corresponding interior angle and the exterior angle = 180° – interior angle. Therefore the larger the interior angle of the polygon is, then the smaller the exterior angle is. By looking at the polygons in the question, we can see that the more sides the polygon has, the greater is each interior angle and therefore the smaller is each exterior angle. Therefore the polygon with the greatest number of sides has the smallest exterior angle. The octagon has the most sides and therefore the answer is E (i.e. angle E)

b) the sum of the exterior angles of any polygon is 360°. A polygon with n sides has n interior angles and n corresponding exterior angles. For a regular polygon, all of its interior angles are equal to each other and also its exterior angles are equal to each other. For a regular polygon with n sides, each of its n exterior angles is equal and the sum of its n exterior angles is 360°, so each exterior angle will be equal to 360°/n. Therefore the larger the value of n, the smaller the value of the exterior angle (e.g. if n = 3 then each exterior angle will be 360/3 = 120° and if n = 6 then each exterior angle will be 360/6 = 60°). So the regular polygon with the greatest number of sides will be the one that has the smallest exterior angle. The octagon has the most sides and therefore the answer is E (i.e. angle E)

How many of these shapes contain interior reflex angles?

Choose one of the following answers:

1

2

3

4

5

Answer: 2

a reflex angle is an angle between 180° and 360°. The interior reflex angles of the shapes in the question are shown below

so there are two shapes which contain interior reflex angles i.e. shapes D and E. So the answer is 2

A milkman delivers milk to certain housing estates. The street layout of the housing estates is shown in the diagram above. The milkman wants to avoid visiting the same street more than once, but there are no restrictions on him passing over the same street corners. On which housing estate is this possible?

Choose one of A, B, C, D or E

Answer: E

You can solve this problem by trying to trace a path through the streets of each housing estate and finding the housing estate that allows the milkman to visit each street once only. However, there is a quicker, easier way to work the problem out.

We will first define a vertex as being a point at which paths (i.e. streets in this example) meet. The degree of the vertex is the number of paths that meet at the vertex. The degree of a vertex is odd if there are an odd number of paths meeting at the vertex, and is even if there are an even number of paths meeting at the vertex.

Looking at the housing estates in the question, we can draw rings around the vertices of each housing estate. Doing this we get the following diagram

The key facts you need to know to solve this problem are :

If there are two or less vertices of odd degree then there is a path around the housing estate which allows the milkman to visit each street once only.

If there are more than two vertices of odd degree then there is no path around the housing estate which allows the milkman to visit each street once only.

So the solution can be found by eliminating any housing estate that has more than 2 vertices of odd degree, and looking for the housing estate that has 2 or less vertices of odd degree.

A is incorrect because housing estate A has 4 vertices of degree 3 (i.e. vertices 1, 2, 4 and 5) and 1 vertex of degree 2 (i.e. vertex 3). Therefore it has 4 vertices of odd degree and 1 vertex of even degree. It has more than two vertices of odd degree and is therefore not the correct answer.

B is incorrect because housing estate B has 4 vertices of degree 3 (i.e. vertices 1, 2, 4 and 5) and 1 vertex of degree 2 (i.e. vertex 3). Therefore it has 4 vertices of odd degree and 1 vertex of even degree. It has more than two vertices of odd degree and is therefore not the correct answer.

C is incorrect because housing estate C has 4 vertices of degree 3 (i.e. vertices 1,2,4 and 6) and 2 vertices of degree 2 (i.e. vertices 3 and 5). Therefore it has 4 vertices of odd degree and 2 vertices of even degree. It has more than two vertices of odd degree and is therefore not the correct answer.

D is incorrect because housing estate D has 4 vertices of degree 3 (i.e. vertices 1, 3, 4 and 7) and 3 vertices of degree 4 (i.e. vertices 2, 5 and 6). Therefore it has 4 vertices of odd degree and 3 vertices of even degree. It has more than two vertices of odd degree and is therefore not the correct answer.

E is correct because housing estate E has 2 vertices of degree 3 (i.e. vertices 3 and 4), 2 vertices of degree 2 (i.e. vertices 2 and 5) and 1 vertex of degree 4 (i.e. vertex 1). Therefore it has 2 vertices of odd degree and 3 vertices of even degree. It has two or less vertices of odd degree and is therefore the correct answer.

In summary, to solve these problems quickly, all you need to do is to eliminate any housing estate that has more than 2 vertices of odd degree. The solution will be the housing estate that has 2 or less vertices of odd degree.

Note that a path (which covers every street without visiting a street more than once) around housing estate E can be found by starting at a vertex of odd degree. For example, we can start at vertex 3 (which has an odd degree of 3), and one correct path is:

vertex 3 to vertex 2, vertex 2 to vertex 1, vertex 1 to vertex 3, vertex 3 to vertex 4, vertex 4 to vertex 1, vertex 1 to vertex 5 and vertex 5 to vertex 4.