For information on all of our membership packages and their prices, please click on **Membership Packages and Prices**

**Step 1 : Join (register an account)**

Click on the Join menu item shown in the image below and you will be taken to the Join page.

Enter a username and password along with your email address in the fields shown below, then click on the Register button which will automatically take you to the Login page.

**Step 2: Log in**

Enter your username and password in the fields shown below, then click on the Login button which will automatically take you to the Membership Welcome Start Page which lists any current memberships you may have and allows you to buy new memberships.

**Step 3: Purchase a membership**

Scroll down the Membership Welcome Start Page until you see the table listing the membership packages and their prices. To purchase a package, click on the Buy Now button next to that package. Note that you must pay for each package separately i.e. you must complete the purchase of one package before you can purchase another package. The memberships are non-recurring which means that you will not be automatically rebilled when a membership expires. You can choose to renew a membership, it will not be done for you automatically.

**Step 4: Log in to the system**

Click on the Login menu item shown below and you are taken to the Login page.

After you have purchased a membership package, you will already be logged in and all you need to do is to click on the Welcome Page link at the bottom of the Login page as shown below. Clicking on the Welcome Page link takes you to the Membership Welcome Start Page.

Note that if you chose a usename of andrew, for example, the system will indicate that andrew is already logged in by showing the username in capitalised form (i.e. as ANDREW) by default

**Step 5: Access your membership package(s)**

Near the top of the Membership Welcome Start Page you will see a list of your active membership packages. To access your package(s), click on the Welcome Page link next to the package name as shown below.

Note that the expiry date of your package is shown next to its name and the expiry date is in the format yyyy-mm-dd. Therefore the expiry date shown below is 4/11/2015 i.e. 4th November 2015.

If you have any questions about the join process, please send us an email at visuteach@hotmail.com

Visuteach provides 11 plus maths practice papers for immediate download as well as interactive online maths membership packages.

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.

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’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 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.

Visuteach also provides interactive online maths tests in our Maths Membership sites. Our membership sites are online only i.e. questions, answers and explanations are displayed online rather than in paper format and cannot be printed out. Payments to join a membership site are one-off payments (i.e. are non-recurring), so no money will be automatically charged to your card when your membership expires.

To join a membership package you need to register (i.e. join) on the **JOIN** page, log in, then pay, then go to **LOG IN** page and then click on the Welcome Page link at the bottom of that page to access the membership package.

Our membership packages work on all devices – PCs, laptops, tablets and Android, Apple and Windows phones. All questions and tests can be tried and retaken as often as wanted during the membership period.

Our maths membership sites are shown below:

PACKAGE |
COST |

GL Assessment-PPTC Maths Membership (1 month, non-recurring)
500 GL Assessment style maths questions. Answers and explanations provided. Questions presented as 10 fifty question fifty minute tests as well as 50 ten question ten minute tests. The package includes all of the questions in our 5 paper-based GL Assessment tests as well as questions in 5 unique 50 question tests exclusive to the online membership package. |
£12.50 |

GL Assessment-PPTC Maths Membership (3 months, non-recurring)
500 GL Assessment style maths questions. Answers and explanations provided. Questions presented as 10 fifty question fifty minute tests as well as 50 ten question ten minute tests. The package includes all of the questions in our 5 paper-based GL Assessment tests as well as questions in 5 unique 50 question tests exclusive to the online membership package. |
£25.00 |

Essex CSSE Maths Membership Package 1 (expires on 27 Sep 2016)
Essex CSSE Maths Tests 1 to 5. Similar in style to the Essex CSSE 2015 maths sample paper. Five 24 question 60 minute maths tests. Answers and explanations provided. |
£20.00 |

North London Independent Girls’ Schools’ Consortium Group 1 Maths Membership (3 months non-recurring)
Answers and explanations to 2008, 2009, 2010, 2011, 2012, 2013, 2014 and 2015 Group 1 maths papers |
£35.00 |

North London Independent Girls’ Schools’ Consortium Group 2 Maths Membership (3 months non-recurring)
Answers and explanations to 2008, 2009, 2010, 2011, 2012, 2013, 2014 and 2015 Group 2 maths papers |
£35.00 |

North London Independent Girls’ Schools’ Consortium Group 1 & Group 2 Maths Membership (3 months non-recurring)
Answers and explanations to 2008, 2009, 2010, 2011, 2012, 2013, 2014 and 2015 Group 1 and Group 2 maths papers |
£60.00 |

St Paul’s Girls’ School Maths Membership (3 months non-recurring)
Answers and explanations to two St Paul’s Girls’ School maths sample papers |
£25.00 |

Maths Pack 1 Membership (3 months non-recurring)
Answers and explanations to two Merchant Taylors’ School maths sample papers and to one St Albans School maths sample paper |
£20.00 |

Habs Boys’ Maths Pack 1 Membership (3 months, non-recurring)
Answers and explanations for Haberdashers’ Aske’s Boys’ School maths papers for years 2009, 2010, 2011, 2013, 2014 and 2015 |
£30.00 |

Our independent school maths membership packages provide links to all of the independent school question papers which can be printed out and downloaded. Some of the question papers can be seen on the schools’ websites linked to below:

North London Independent Girls’ Schools’ Consortium Group 1 Past 11+ Maths Papers

North London Independent Girls’ Schools’ Consortium Group 2 Past 11+ Maths Papers

St Paul’s Girls’ School Sample 11+ Maths Papers

Merchant Taylors’ School Sample 11+ Maths Papers

St Albans School Sample 11+ Maths Paper

Haberdashers’ Aske’s Boys’ School 11+ Entrance Exam Papers

Demos for the GL/PPTC 11+ maths, Essex CSSE 11+ maths and North London Independent Girls’ Schools’ Consortium Group 1 11+ maths membership packages are available below. Our other independent school maths membership packages display answers and explanations in the same way as the North London Independent Girls’ Schools’ Consortium Group 1 11+ maths membership 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 |

The demo below has a maths test of 5 out of the full 24 questions and should therefore be completed in 12.5 minutes rather than 60 minutes. The maths tests in our membership packages are 24 question tests.

Print out the blank answer sheet and write your answers on it. When you have completed the test, check your answers against the completed answer sheet.

Click on the links in the table below.

Essex Maths | Blank Answer Sheet |
Demo Test |
Completed Answer Sheet |
Answers & Explanations for Demo Test |

For more information, please click on our our Essex CSSE 11+ page Essex CSSE 11+.

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 2008-2015 together with answers and explanations for the more difficult questions. |
NLIGSC Group 1 Demo |

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

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.