Maths Learning Resources
Free for parents and fun for children! Discover a library of primary school maths worksheets, games, tips for parents, and more! Perfect for learning remotely, preparing for the 11 Plus exam, and making progress in maths.
Learn how to Solve 11+ Exam Maths Questions
ry to learn how to quickly recognise the type of question and then apply the correct problem-solving method. Practising this will help your child to complete the test successfully.
Counting Problem
Modified from City of London School for Girls 11+ Maths Entrance Examination 2011 #30
Counting Problem
In the following pattern, you can only move from the letter you are currently at to one of the two letters in the row below immediately to the left or the right.
One route from A to Q is ACEHLQ.
- How many possible routes are there from A to Q?
- How many possible routes are there from A to S?
Answer and solutions
Answer:
a) 5. b) 10
Solution:
We can solve this question by listing all the possible routes, but it’s easy to repeat or omit some routes. We can use the “Marking Method” to help us find the answer quickly and accurately.
As shown in the figure below, the number besides each letter indicates the number of routes from A to this letter. Each number equals to the sum of the two adjacent numbers above.
From the figure we can easily find that there are 5 possible routes from A to Q, and 10 possible routes from A to S.
Nets
Modified from Sevenoaks School 11+ Maths Entrance Examination 2015 #17
Nets
The diagram shows the net of a cube. Which edge meets the edge P when the net is folded to form a cube?
Answer and solutions
Answer: C
Solution:
How can you find which two edges will meet when the net is folded? It is difficult to rely solely on spatial imagination. There is a simple way to solve this question.
We can start by observing a folded cube.
In the cube above, the line segment between vertex A and vertex G is a body diagonal. A body diagonal of a cube is a line connecting two vertices that are not on the same face. Each vertex of the cube is only connected to only one other by a body diagonal. On a net of this cube, the line between A and G will pass through two faces, and it must be the diagonal of the rectangle formed by the two square faces. The figure below shows one possible case (only a part of the net is shown).
Here is a useful rule: starting from a point on the net and moving along the diagonal of a double-grid rectangle to another point, you will reach the other end point of the body diagonal. If you move diagonally across another double-grid rectangle from where you end up, you will return to the end point of the body diagonal you started at, but you will be at a different point on the net. So, the point you started at and the point you end up at must meet at the same vertex of the folded cube. For example:
For this question, we mark the two endpoints of edge P as M and N.
Move from M twice: M-M1-M2, M and M2 will meet at the same vertex of the cube by the rule above.
Move from N four times: N-N1-N2-N3-N4, N, N2 and N4 will meet at the same vertex of the cube by the rule above.
So, MN and M2N4 will meet to form an edge of the cube. That is to say, edge C meets edge P when the net is folded to form a cube.
Indefinite Equations Word Problem
Modified from Whitgift School Croydon 11+ Math Exam Syllabus Sample Question #20
Indefinite Equations Word Problem 1
Pip wants to reserve some boats. Each small boat seats 3 people and each big boat seats 8 people. What is the minimum number of boats he needs to reserve to seat 68 of his friends and himself, without any empty seats?
Answer and solutions
Answer:
13 (7 small boats and 6 big boats).
Solution:
Let x be the number of small boats, y be the number of big boats.
3x+8y =68+1=69
Normally, students will list the possible numbers of x and y until they have found all the answers. This would take a long time. This problem can be solved more quickly by considering the relationship between multiples of the same number.
We can find that 69 is a multiple of 3, 3x is also a multiple of 3, so 8y must be a multiple of 3 as well. Then y must be a multiple of 3.
When y=3×0=0, 3x+8×0=69 , x=23.
When y=3×1=3, 3x+8×3=69 , x=15.
When y=3×2=6, 3x+8×6=69 , x=7.
If the value of y continues to increase in multiples of 3, the value of x will become less than0. Therefore, there are only three pairs of solutions:
7+6=13<15+3<23+0, so Pip needs to reserve at least 13 boats.
Modified from Bancroft's School 11+ Maths Entrance Examination 2018 #31
Indefinite Equations Word Problem 2
Bud bought several chocolate cakes and strawberry cakes for her family. A chocolate cake costs ￡10, and a strawberry cake costs ￡9. She paid ￡46 in total. How many chocolate cakes and strawberry cakes did she buy?
Answer and solutions
Answer:
Bud bought 1 chocolate cake and 4 strawberry cakes.
Solution:
Let x=the number of chocolate cakes, y=the number of strawberry cakes.
10x+9y=46
This problem can be solved quickly by analysing the last digit.
The last digit of 10x is 0, and the last digit of 46 is 6, so the last digit of 9y should be 6. When y=4, 9×4=36, so the last digit of 9y is 6. When y=4, 4y=36, 10x+36=46, x=1.
If the value of y continues to increase, the value of x will be less than 0.
Therefore, there is only one pair of solutions.: 
In other words, Bud bought 1 chocolate cake and 4 strawberry cakes.
Remainder Problem
Modified from PMC February 2016 #20
Remainder Problem 1
John has less than 100 marbles in a bag.
If he counts them in groups of 3, he will get a remainder of 2.
If he counts them in groups of 4, he will get a remainder of 2.
If he counts them in groups of 5, he will get a remainder of 2.
How many marbles does John have?
Answer and solutions
Answer: 62
Solution:
Through observation, we can see that these remainders are all the same. Therefore, if we subtract 2 from the total number of marbles, the result will be a multiple of 3, 4 and 5.
The lowest common multiple of 3, 4 and 5 is 60. 60+2=62.
After 60, 120 is the next smallest common multiple of 3, 4 and 5. But this would mean John has 120+2=122 marbles, and 122>100. Therefore, John must have 62 marbles.
Remainder Problem 2
Pip thinks of a number.
When the number is divided by 5, the remainder is 4.
When the number is divided by 6, the remainder is 5.
When the number is divided by 9, the remainder is 8.
Pip’s number is less than 100. What is the number?
Answer and solutions
Answer: 89
Solution:
Through observation, we can see that the remainder of each division is 1 less than the divisor. Therefore, as long as Pip’s number is increased by 1, it can be divided by each of these divisors with no remainder.
Let the number be x. x+1 should be a common multiple of 5, 6 and 9.
The lowest common multiple of 5, 6 and 9 is 90. When x+1=90, x=89. 89 meets all the requirements of this question.
After 90, 180 is the next smallest common multiple of 5, 6 and 9. When x+1=180, x=179>100. Therefore, Pip’s number can only be 89.
Number Puzzles Questions
Modified from 2019 Junior Mathematical Challenge #22
Number Puzzles Question
In the calculation below, all of the digits are replaced with letters. The same letters represent the same digits. None of A, B or C is zero. Work out what digit each of the letters stands for in this calculation:
ABC+ABC+ABC=BBB.
Answer and solutions
Answer:
A=1, B=4, C=8.
Solution:
Through observation, we can find that BBB is a multiple of 111. We can use this to find the following:
BBB=B×111=B×37×3
ABC×3=BBB=B×37×3
ABC =B×37
Therefore, the product of B×37 should be a three-digit figure.
37×1 is a two-digit figure, so B≠1
37×2 is a two-digit figure, so B≠2
37×3 is a three-digit figure, 37×3=111 . It doesn’t meet the requirement of ABC×3=BBB, so B≠3
37×4 is a three-digit figure, 37×4 =148. It meets the requirement of ABC×3=BBB.
If the value of B continues to increase, the values of A, B, and C won’t meet the requirement of ABC×3=BBB . The only solution to this problem is A=1, B=4, C=8.
Number Machine
Modified from City of London School for Girls 11+ Maths Entrance Examination 2015 #15
Number Machine
A number machine produces the input and output table below.
What is the rule for this number machine?
Answer and solutions
Answer:
Divide by 3 and then add 6.
Solution:
We can compare the variation of the differences between different inputs and between the corresponding outputs.
+3 in the input results in +1 in the output, +6 in the input results in +2 in the output, etc. So, we can try to divide the input numbers by 3. After that we can find the difference between the new number and the output is 6.
Therefore, the rule is “divide by 3 and then add 6”.
Logical Deduction
Modified from Bancroft’s School 11+ Maths Entrance Examination 2018 #28
Logical Deduction 1
There are 5 thief suspects. Only one of them is the real thief, and only three of them tell the truth. Can you help the police find the thief according to their statements below?
Suspect A: “Suspect D is the thief.”
Suspect B: “I am not the thief!”
Suspect C: “I’m quite sure Suspect E is not the thief.”
Suspect D: “Suspect A is telling a lie.”
Suspect E: “Suspect B is telling the truth.”
Answer and solutions
Answer:
E is the thief.
Solution:
Finding contradictions between different statements can help us to solve problems like this one. We can find that suspect A’s statement and suspect D’s statement contradict each other (they cannot both be true), which means one out of suspect A and suspect D must be lying, and the other is telling the truth.
What’s more, suspect B’s statement and Suspect E’s statement are equivalent, which means suspects B and E are either both telling lies or both telling the truth.
According to the given information, only three of them tell the truth, which means only two of them tell lies.
If both suspects B and E were telling lies, there would be at least 3 suspects lying in total (as either suspect A or suspect D is lying), which would contradict the fact that only two of them lie. Therefore, both suspects B and E are telling the truth.
Now there are three people telling the truth (B, E and either A or D), which means that suspect C is lying, and suspect E is the thief.
Logical Deduction 2
Jessica, Ethan, and Ben are comparing how many sweets they have with each other.
Jessica: “I have 13 sweets, 3 less than Ethan, and 1 more than Ben”
Ethan: “I do not have the lowest amount, Ben has 4 more or less than me, and Jessica has 11 sweets”
Ben: “I have less than Jessica, Jessica has 10 sweets, and Ethan has 2 more than Jessica”
If for each person’s three statements one is incorrect and the other two are correct, how many sweets does the person with the least number of sweets have?
Pip’s number is less than 100. What is the number?
Answer and solutions
Answer: 9
Solution:
When there are so many known conditions, tabulation can help us understand the problem.
From the table above, it is easy to find that the three red statements are contradictory and the two blue statements are contradictory.
We also find that there are two contradictory statements between Jessica and Ben, so each of them is correct about one and incorrect about the other of these two statements (as only one red statement can be correct and only one blue statement can be correct, but each person is only incorrect about one of their statements). That is to say, Jessica has either 13 or 10 sweets, and so “Jessica has 11 sweets” is incorrect.
Therefore, the other statements are correct.
From the correct statements, we can find B<J (from Ben). Then from Ethan’s statements, we know that Ethan has 4 more or less than Ben, but also that Ethan does not have the least, so Ethan must have 4 more than Ben. Finally, from Jessica’s statements, we know that Jessica has 1 more than Ben, so must have just 3 less than Ethan, so Jessica’s second statement J=E-3 is correct. “Jessica has 13 sweets” must then be incorrect, so Jessica must have 10 sweets. Ben has the least number of sweets. He has 9 sweets.