Eleven Plus Maths

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Why buy Visuteach's maths papers?

Our maths papers and interactive tests contain detailed explanations (not just answers). This helps you understand how to solve the problems.

For example our GL Assessment style maths paper 1 contains 19 pages of explanations for the 50 questions in the test.

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More eleven plus maths & non verbal reasoning tests

Essex 11 plus maths tests & more Essex 11 plus English tests

Northern Ireland PPTC English transfer tests & AQE transfer tests

Visuteach provides three types of sample 11 plus maths papers. We provide GL Assessment style tests (often formerly referred to as NFER Nelson style tests), PPTC transfer style tests (which are also GL Assessment style papers) and common entrance style maths tests for the North London Independent Girls' Schools' Consortium.

Visuteach's GL Assessment style 11 plus maths tests 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 PPTC transfer test maths papers are an adaptation of our GL Assessment style 11 plus maths papers. They contain 45 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 papers and not our general GL Assessment papers. More information on our PPTC style tests is available on our PPTC Transfer Test page.

Visuteach's North London Independent Girls' Schools' Consortium 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.

Open or download our 11 plus maths sample paper by clicking on the link below. The sample maths practice paper contains 10 questions and answers rather than the 50 questions available in the full product. You can open the sample test paper by clicking on the link below or you can download it by right-clicking on the link below and choosing the 'Save Target As' menu option 

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Some sample questions from our practice papers are given below. You can also try our interactive tests. There is an untimed test, with answers and explanations immediately after each question, as well as a timed test, with answers and explanations at the end of the test. If you purchase the interactive tests, the software is displayed in full-screen mode rather than the smaller version shown on this page.

Our 11 plus practice papers and interactive tests can be purchased and downloaded immediately on our 11 Plus Exam Products page.

Use the scrollbar on the right of each test to scroll down and view the answers and explanations

11 Plus Maths Exam - 10 Question Untimed Test

11 Plus Maths Exam - 10 Question Timed Test

11 Plus Maths Exam Question 1

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

11 Plus Maths Exam Question 2

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°

11 Plus Maths Exam Question 3

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)

11 Plus Maths Exam Question 4

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

11 Plus Maths Exam Question 5

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.

11 Plus Maths Exam Question 6

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

11 Plus Maths Exam Question 7

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

11 Plus Maths Exam Question 8

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)

11 Plus Maths Exam Question 9

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

11 Plus Maths Exam Question 10

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.