3. It’s all in the mind
Being able to find missing information in a problem is an important mathematical skill to have.
Think of a number and add 1 to it.
Multiply by 3.
Subtract 4.
Add the number you first used.
Add nine.
Divide by 4.
Minus the number you first thought of. 2
Teacher notes
Everyone should end up with the number 2 if they have followed all the instructions carefully and correctly. Students should be encouraged to try again with a different number. As they are working through different possibilities, you may wish to get them to assign an algebraic notation to the number they started with. Some students may already have done this and might be working towards an algebraic solution. Encourage students to think about what steps have resulted in everyone arriving at the same solution.
What number did you end up with?
Did everyone end up with the same number? Why?

4. Spookily accurate Teacher notes
With this activity, students will be asked to go through the same process as on the previous slide. The great thing about this activity is that it will generate a new set of instructions for your students to follow each time it is used. For those students who have already grasped the concept of using algebraic notation to work out the correct number, encourage them to write an algebraic formula for each step of the instruction. You may also at this point want to ask them to write their own ‘magic’ formula.

5. As clever as it seems
We can explain how this ‘trick’ works every time by using an algebraic expression to mirror the instructions given.
Using x for the missing number, an expression can be formed.
3(x + 1) – 4 + x + 9 4
– x = ?
3x + 3 – 4 + x + 9 4
– x = ?
4x + 8
– x = ? 4
Teacher notes
If you wish, you could add the text from slide 3 to this slide so that the students can see how the instructions and the algebraic notation go together:
Think of a number and add 1 to it.
Multiply by 3.
Subtract 4.
Add the number you first used.
Add nine.
Divide by 4.
Minus the number you first thought of.
x + 2 – x = ?
? = 2

6. Card trick
From a normal pack of playing cards, choose one card with a value between 1 and 9.
Add 6 to it.
Multiply it by 2.
Add 4.
Divide this by 2.
Subtract the number you started with.
Teacher notes
This will provide the students with further practice of using algebraic notation to help ‘crack’ a set of instructions and it should help increase their confidence working with algebra to solve a problem. This is an important part of the new GCSE qualification and if students feel comfortable analysing and breaking down simple problems like this it will help improve their confidence.
The value of the initial card is x.
The instructions for the algebraic expression would be: (2(x + 6) + 4) ÷ 2– x = ?
(2x ) ÷ 2 – x = ?
(2x + 16) ÷ 2 – x = ?
x + 8 – x = ?
? = 8
At this stage, encourage students to split into groups and create their own set of ‘mystery number’ instructions that they can use and try out on other students.
What number are you left with?
Write your own algebraic expression that would solve the example above.

8. Shapes and angles
Lisa has drawn some diagrams of four different shapes and has used algebra to note the lengths of their sides.
Although her algebra is good, her diagrams are poor.
Teacher notes
The shape with sides of length a, a, 2a and 2a is a kite. The shape with sides of length y, y, y and y is a rhombus. The shape with sides of length 2x, 2x, 5x and 5x is a parallelogram and the shape with sides of length 4p, 2p, 2p and 5p is a trapezium.
What shapes could these be if none of the angles are right angles? The diagrams are not to scale.

10. Consecutive numbers Teacher notes
Students should notice that the sum of the first and last numbers and the sum of the middle two numbers is always equal, no matter what set of consecutive numbers is chosen. Get the students to test this theory with large numbers and small numbers alike.
Ask the students to think about what is happening and why it is happening. Is there any pattern that they can decipher? Can they possibly use algebra to explain what is happening?

11. An algebraic solution?
When dealing with this investigation, it is possible to develop an algebraic solution.
Let x represent the first number.
This means that x + 1, x + 2 and x + 3 are the next 3 consecutive numbers.
Adding together the first and last number in the set gives: x + x + 3 = 2x + 3.
Adding together the middle two numbers in the set gives: x x + 2 = 2x + 3
Teacher notes
Whatever the value of the first number, this expression will always give these totals.
Substituting x = 4 into the expression, 2x + 3 = 11
Substituting x = 7 into the expression, 2x + 3 = 17
The sum of the first and last numbers will always equal the sum of the middle two numbers.

13. Four is the magic number
There are lots of different number relationships that can be explained algebraically.
Choose 3 consecutive odd numbers.
Multiply the first and the last two numbers together.
Square the middle number.
What is the difference between the two values that you have?
Teacher notes
The following slide gives an algebraic expression of what is happening in this example. Again, you should encourage students to test the theory for lots of different examples so that they can check to see if there are exceptions to the rule.
Try this for several sets of numbers.
Can you use algebra to explain what is happening?

14. Easily explained?
This number relationship can be solved if we use algebra to explain what is happening.
Let n be the first odd number.
n + 2 and n + 4 are the consecutive odd numbers.
Multiplying the first and last numbers gives:
Squaring the middle number gives:
Teacher notes
By this stage, pupils should be quite comfortable with the idea of algebraic notation. With that in mind, you may wish to get your pupils to make up their own rules and check them. Can they uncover any number relationships that they hadn’t been aware of previously? If so, what are they? Encourage them to test their classmates’ work to see if they can spot any exceptions to the rules or spot any flaws in their conclusions.
n(n +4)
= n² + 4n
(n +2)²
= n² + 4n + 4
The difference between the two expressions is always 4.

16. Hundred square This image shows a small part of a hundred square.
Multiply the opposite corners of the rectangle.
Add together the middle 2 numbers.
What do you notice?
Extend the pattern with the next row. Repeat the instructions.
Teacher notes
4 × 16 = 64; 6 × 14 = 84; = 20
84 – 64 = 20.
6 × 24 = 144; 26 × 4 = 104; = 40
144 – 104 = 40.
By changing the numbers to letters where x = 4, x + 1 = 5, etc, you can see that the difference between the opposite corners of the rectangle is equal to the value of the square at the middle of the bottom row and the one above it added together.
What do you notice?
Can you explain what is happening?

17. Hundred square investigations
Teacher notes
This slide contains a very open-ended activity that can be used for very general number investigations or for very specific number relationships and seeing how they operate together. Use the paintbrush tools to highlight any numbers that you want the students to focus their investigations on. Use the annotation tools to highlight factors, prime numbers, etc. if you so wish.

18. Magic square
In a magic square, each row, column and diagonal adds up to the same total.
Complete this magic square by finding the values of a, b, c and d.
Can you create your own algebraic magic square?
Teacher notes
As the text says, a magic square has rows, columns and diagonals that add up to the same total. In this magic square, the total can be found by adding the three numbers from the bottom row: = 21.
We can then use this information to work out the value of a, c and d.
a = 21; a = 21 – (11 + 6) = 21 – 17 = 4.
d = 21; d = 21 – (10 + 6) = 21 – 16 = 5.
c = 21; c = 21 – (11 + 3) = 21 – 14 = 7.
b + c + d = 21; b = 21 – (c + d) = 21 – 12 = 9.
Encourage the students to substitute the values back in to the magic square and check all the rows, columns and diagonals. This should help them ensure they haven’t made any mistakes anywhere.

19. More magic squares Teacher notes
Again, this activity provides some real open-ended practice for students to get to grips with the idea of magic squares. Magic squares will quite often crop up in exam papers as test of the students’ ability to problem solve given a specific amount of information. Students that are able to fully comprehend the process behind the magic squares should have a head start in their problem-solving skills.

21. Patio problems The diagrams below show measurements
for patios in 2 different houses.
x + 3 2x
2x + 3
2x + 1
x + 2
3x + 4 4
The perimeter of the trapezium shaped patio is equal to 24 m.
Teacher notes
The trapezium patio perimeter: (2x + 1) + (x + 2) + (3x + 4) + (2x + 1) = 24; 8x + 8 = 24; 8x = 16; x = 2 m
The perimeter of the other patio is: (2x + 3) + (x + 3) + (x + 3) + (2x) + (2x + 3)
Substituting x = 2 into this gives: (2 × 2 + 3) + (2 + 3) + (2 + 3) + (2 × 2) + (2 × 2 + 3) = = 28 m
This question is a real test of the students’ ability to apply the correct mathematical method in order to find the correct answer. Students should know that the area of a trapezium is equal to: ½(a + b)h.
As a result, the area of the trapezium is: ½ × (4 + 10) × 4 = 28 m². At £2.50 per m², the cost is: £2.50 × 28 = £70.
Working out the area of the second patio will be far more difficult for the students. The easiest way of doing this is to calculate the area of the complete rectangle, then subtract the area of the triangle.
The area of the complete rectangle is: (2x + 3) × (2x + 3) = 7 × 7 = 49 m².
The area of the small triangle missing from the rectangle is: ½bh = ½ × x × 3 = ½ × 2 × 3 = 3 m².
The total area of the patio is = 49 m² – 3 m² = 46 m².
Cost of decking this area = £2.50 × 46 = £115.
What does the perimeter of the other patio equal?
Decking costs £2.50 per m². How much will it cost to deck each patio? Show your working.

22. More perimeter questions
Teacher notes
The correct answers are as follows:
Trapezium: Length of sides = x + 2, x, x, 6. Perimeter: 26 m. x = 6 m.
Rectangle: Length of sides = x – 3, x – 3, x + 2, x + 2. Perimeter = 46 m. x = 12 m.
Parallelogram: Length of sides = x – 1, x – 1, 2x, 2x. Perimeter = 28 m. x = 5 m.
5 sided polygon: Length of sides = x – 2, x – 1, x + 5, x + 1, x. Perimeter = 26 m. x = 4.6 m.
4 sided polygon: Length of sides = x + 5, 2x, 2x, x + 3. Perimeter = 50 m. x = 7.
With all of these questions, the option is there for you to add any necessary annotations to the image so that the students can practice calculating the areas of the patios. If decking is sold at the same price as on the last slide, which patio is most expensive? By how much?

23. Perimeter and area Teacher notes
This open-ended activity deals with different sized rectangular shapes and could be a good extension activity for the students in this context. Hide the grid, and any three numbers, including the perimeter or area. Use your knowledge of algebra to find the missing numbers, and then reveal them to check your answers.
You could encourage students to create their own shape for a patio by drawing it on cm grid paper. Ask them to calculate the perimeter of the shape, the area of the shape and how much it will cost to deck the patio.
As an extension, you could also limit the amount of money that you wish to spend and say that you want to maximize the amount of space that you have in your patio. What shape and what dimensions would work best?

24. An impossible task? Here is another diagram for a patio of a house.
In this problem there appear to be 2 possible perimeters for this trapezium shaped patio.
3x + 3
x – 6
2x + 3
Teacher note
This is a good example of the practicalities of solving this type of problem.
3x x – 6 + 3x x + 3 = 75, 9x + 3 = 75, 9x = 72, x = 8
3x x – 6 + 3x x + 3 = 30, 9x + 3 = 30 , 9x = 27, x = 3
If x = 3, the side of the trapezium that is of length x – 6 would equal –3. Having a negative length is impossible. Therefore the perimeter of the shape can’t equal 30 feet.
The possible perimeters have been identified as being 75 feet and 30 feet. Why is one of these impossible?

26. Prove yourself useful
Sally has drawn a quadrilateral, shape ABCD, and labelled the sides of the quadrilateral using algebra.
(3x – 10) cm D C A B
5 cm
5 cm
Teacher notes
By working out the value of x, we can work out the length of side AB and the length of side CD. From the diagram, we already know that AB = CD and AC = BD.
AB = CD
3x – 10 = 2x – 5
x – 10 = –5
x = 5
Substituting x = 5 into the length AB, we can tell that the length of AB = 3 × 5 – 10 = 15 – 10 = 5 cm
Substituting x = 5 into the length CD, we can tell that the length of CD = 2 × 5 – 5 = 10 – 5 = 5 cm.
This means that all 4 sides of the quadrilateral have a length of 5 cm. We know that a square has 4 equally sized lengths, so therefore quadrilateral ABCD could be a square. We don’t know for sure that it is a square as we don’t know the size of the angles of the quadrilateral. It could also be a rhombus.
(2x – 5) cm
Show that quadrilateral ABCD could be a square.