maths gcse coursework tasks

Mr Barton's Rich Tasks

On this page I have collected together my favourite rich tasks that I have used over the last 11 years, in an accessible, easy to use format. I also invite teachers to share their ideas for interesting probing questions and lines of inquiry for students to investigate.

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For me, a rich task is one that both engages and challenges students with a wide level of mathematical ability. They need to be “low barrier, high ceiling”, by which I mean students need to have found success/made progress with the task within the first 30 seconds, but there is still enough meat left to keep them thinking 30 minutes (or even 3 lessons) later.

I feel activities like these are crucial for students’ mathematical development. They allow them to be creative, and work together in meaningful and positive ways. When developing our Scheme of Work (you can read my series of blog posts about it), we decided to include a compulsory rich task for all students each topic unit, and many of those can be found below.

The key to a good rich task are the questions that accompany it. This is where effective differentiation happens. All students begin the task in exactly the same way, but once an initial stage has been reached, students (individual or in groups) are free to pursue different investigations, probing questions and lines of inquiry. These can be provided by the teacher, or even by the students themselves.

The strength of the rich task lies in these questions. So here is my plan: I am going to share as many of my favourite rich tasks as possible, and hopefully teachers from around the world are going to provide the questions. These can be lines of inquiry, investigations, prompts, hypotheses, extensions, simplifications, modifications, whatever you like. Crucially, you do not need to know the answer yourself. Just throw it out there! There will be space for these in the Comments section at the bottom of each TES Resource page, and I will always get the ball rolling with a few questions of my own.

Please join in. Please spread the word. Please just share even one question. And then the tasks will keep getting better, and better, and better. 

The Rich Tasks keyboard_arrow_up Back to Top

Task 1 - Positive Differences Brief Description: Students build simple number pyramids by taking the positive difference of pairs of numbers Potential Skills Involved: Arithmetic, Fractions, Decimals, Writing Expressions, Proof

Task 2 - The Factors and Multiples Game Brief Description: Students play a strategic game on a 1-100 number grid, crossing off factors and multiples Potential Skills Involved: Arithmetic, Factors, Multiples, Primes, Proof

Task 3 - Choose 3 Numbers Brief Description: Students try to guess each other's starting numbers by working backwards from the sums of pairs of numbers Potential Skills Involved: Arithmetic, Writing expressions, Solving Equations

Task 4 - Will they meet? Brief Description: Can you help Romeo and Juliet get back together in my first ever romantic maths activity? Potential Skills Involved: Enlargement, Vectors, Similar Shapes, Rotation

Task 5 - Number Shacks Brief Description: Can you figure out how the numbers of these shacks are formed and use this to predict answers and spot patterns? Potential Skills Involved: Arithmetic, Writing Expressions

Task 6 - Averaging it out Brief Description: What happens when we continually take the mean of sets of numbers? Potential Skills Involved: Averages, ICT

Task 7 - Fraction Arrangement Brief Description: Can you order different digits to produce the biggest and smallest possible answers for these fraction problems? Potential Skills Involved: Operations with fractions

Task 8 - Diffy Brief Description: The first lesson our new bunch of Year 7s experience, and one of my all time favourites Potential Skills Involved: Arithmetic, Writing Expressions, Proof

Task 9 - Simultaneous Equations Staircase Brief Description: Why does everyone get the same answer to these simultaneous equation problems? Potential Skills Involved: Simultaneous Equations, Proof

Task 10 - How many angles? Brief Description: Using a geoboard, how many angles between 10 and 180 can you make? Potential Skills Involved: Angle Facts, Circle Theorems

Task 11 - Number Reverse Brief Description: What happens when we reverse the digits of numbers and perform operations on them? Potential Skills Involved: Arithmetic, Writing expressions, Proof

Task 12 - Multiplication Reduction Brief Description: Follow the rule to reduce a number in size using multiplication. Does anything interesting happen? Potential Skills Involved: Arithmetic, Writing Expressions

Task 13 - How many quadrilaterals? Brief Description: Using a geoboard, how many different quadrilaterals can you make? Potential Skills Involved: Properties of shapes, Angle facts

Task 14 - 1089 Brief Description: Why is the number 1089 so special? Potential Skills Involved: Arithmetic, Writing expressions, Proof

Task 15 - Square Co-ordinates Brief Description : What do the co-ordinates of the corners of squares have in common? Potential Skills Involved: Co-ordinates, Properties of shapes, Vectors, Proof

Task 16 - Polar Bears Brief Description : Can you figure out how to get the totals in this dice game? Potential Skills Involved: Arithmetic

Task 17 - Pascal's Triangle Brief Description : What maths can you discover hiding in Pascal's triangle? Potential Skills Involved: Sequences

Task 18 - Entrapment Brief Description : A fun strategy game using all of the transformations Potential Skills Involved: Reflection, Rotation, Translation, Enlargement

Task 19 - Fire Hydrants Brief Description : Where is the optimum position to place these fire hydrants to maximise their coverage? Potential Skills Involved: Geometrical Reasoning

Task 20 - Diagonals of Rectangles Brief Description : How many squares does the diagonal of a rectangle pass through? Potential Skills Involved: Arithmetic, Sequences, Factors, Multiples, Primes

Task 21 - T-totals Brief Description : How can you work out the T-number in this classic piece of maths coursework? Potential Skills Involved: Arithmetic, Writing Expressions, Proof

Task 22 - Number Snakes Brief Description : What is the longest number snake you can make using these simple rules? Potential Skills Involved: Arithmetic, Properties of Numbers, Writing Expressions

Task 23 - Summing Consecutive Numbers Brief Description : Which numbers can be made using the sums of consecutive numbers? Potential Skills Involved: Arithmetic, Writing Expressions

Task 24 - NIM Brief Description : The wonderful strategy game using piles of counters Potential Skills Involved: Strategy, Factors, Multiples, Primes

Task 25 - Function Machines Brief Description : Why do these function machines seem to give the same difference? Potential Skills Involved: Arithmetic, Order of Operations, Writing Expressions, Expanding Brackets

Task 26 - Leap Frog Brief Description : If you leap over this set of 3 points enough times, what do you notice? Potential Skills Involved: Co-ordinates, Construction, Vectors

Task 27 - Solving Linear Equations Brief Description : By arranging sets of digits, what types of solutions can you generate to these simple linear equation problems? Potential Skills Involved: Solving linear equations

Task 28 - Decimal Arithmetic Brief Description : By arranging sets of digits, can you make the biggest and smallest decimal totals possible? Potential Skills Involved: Arithmetic, Decimals, Place Value

Task 29 - 24 Cubes Brief Description : What different 3D objects can you make with 24 cubes and what do you notice about their properties? Potential Skills Involved: Surface Area, Volume, Similarity

Task 30 - Tilted Squares Brief Description : How many squares with area 1-20 can you create? Potential Skills Involved: Area, Pythagoras

Frequently Asked Questions keyboard_arrow_up Back to Top

Are you saying we should be doing this every lesson? No. Definitely not. I am acutely aware of the need for students to gain practise in key mathematical skills. But I do strongly believe that regular lessons like this are just as important for a student's mathematical development and to increase their enjoyment in the subject. They should not be seen as one-offs. Both students and teachers should value them as highly as any type of lesson. What do I do if a child doesn't engage with a particular question or line of inquiry? The simple answer is that I give them another one! I have to make a judgment call as to whether the student has genuinely tried and not just given up too easily. But if, for whatever reason, a probing question or line of inquiry hasn't resonated with a student, then I will set them off on something else. Indeed, the beauty of having lots of questions up your sleeve is that you are far more likely to find ones that engage your students than if you just have one line of investigation that the whole class is following. Would you do this type of lesson for an observation? Yes, I definitely would. 100%. Sure, such lessons are a little bit on the risky side as you don't know what is going to happen. But they are also incredibly flexible. Imagine you had meticulously planned a lesson with a 40 slide PowerPoint and 5 beautifully prepared, differentiated worksheets. And then you find that the students don't understand even the basics. Or, they understand far more than you anticipated. You might be in a bit of trouble. But with a lesson rammed full of probing questions, you can just try them out on another line of inquiry. Or, better still, get them to come up with their own.

© 2003 Andrew Martin Last updated:27 th September 2003 http://www.mymathsteacher.com/GCSEMaths/

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Number Grid Coursework

Maths Investigation 1

Higher Tier Task - Number Grid

Section 1: 2x2 Box on Width 10 Grid

1) Introduction

I was given a number grid, like the one in Fig 1.1. The task was to, in the 2x2 box, find the product of top-left (TL) and bottom-right (BR) numbers, and the product of the top-right (TR) and bottom-left (BL) numbers and then to calculate the difference of these two products.

This calculation rule was to be followed throughout the investigation. Having done this, I found the difference of the two products to be 10  and I wondered what would happen if the box was placed in other locations on the grid.

To discover this, I will calculate the difference of the two products for 5 different random locations of the box within the grid. As it would be impractical (and impossible if the grid extended to more than 10 rows) to do all possible calculations, 5 should be enough to display any patterns that may lie therein.

3) Data Collection

Here are the results of the 5 calculations for 2x2 Box  on Width 10 Grid :

4) Data Analysis

From the table, it is very easy to see that on all tested locations of the box, the difference of the two products was 10.

5) Generalisation

Using this apparently constant number, it can be assumed that for all possible locations of the 2x2 box on the width 10 grid, that the difference is always 10. Therefore, the following equation should be satisfied with any real value of a , where:

a  is the top-left number in the box;

( a  + 1) is the top-right number in the box, because it is always “1 more” than a ;

( a + 10) is the bottom-left number in the box, because it is always “10 more” than a ;

( a + 11) is the bottom-right number in the box, because it is always “11 more” than a .

( a  + 1)( a  + 10) - a ( a  + 11) = 10

The basic algebraic labelling of the box is:

This means that I predict that with a 2x2 box on a width 10 grid, the difference of the two products will always be 10.

My formula works as shown with the following, previously unused values:

1) Where a  = 5

2) Where a  = 64

7) Justification

The formula can be proven to work with the following algebra:

The formula works because the “10” term is only produced in the expansion of the two brackets on the left of the minus sign, and not from the more simply factorised “ a ( a  + 11)” term. The a 2  term is present on both sides of the minus sign, as is the 11 a  term therefore, they cancel each other out to leave the number 10.

8) Conclusion

After this justification, it can now be said that for every 2x2 box on a Width 10 Grid, the difference of the two products will always be 10.

9) Extension

Having done this, I saw that my formula would only work for 2x2 boxes on a Width 10 grid. To improve the usefulness of my formula, I wondered what would happen to the difference of the two products if I varied the width of the grid on which the 2x2 box was placed.

Section 2: 2x2 Box on Width “ z ” Grid

Throughout this section, the variable z  will be used to represent the width of the grid i.e. the number of columns on the grid. The variable a  will continue to be used for the top-left number in the box i.e. the location of the box upon the grid.

Varying values of z will be tested to give different widths of the grid. Grid sizes to be used for data collection will range from 11 to 15. On these grids, from 5 different random positions of the 2x2 box, the differences of the two products will be calculated. Again, I believe 5 calculations are enough to display any patterns.

Fig 2.1 to Fig 2.5 are the grids used for the varying values of z. An example of the 2x2 box has been highlighted on each one.

(a) Here are the results of the 5 calculations for 2x2 Box  on Width 11 Grid (Fig 2.1):

(b) Here are the results of the 5 calculations for 2x2 Box  on Width 12 Grid (Fig 2.2):

(c) Here are the results of the 5 calculations for 2x2 Box  on Width 13 Grid (Fig 2.3):

(d) Here are the results of the 5 calculations for 2x2 Box  on Width 14 Grid (Fig 2.4):

(e) Here are the results of the 5 calculations for 2x2 Box  on Width 15 Grid (Fig 2.5):

This is a preview of the whole essay

From these results, it is possible to take the calculated difference of the two products, and plot this against the width of the grid ( z ):

From the tables (a)-(e), it is possible to see that with a 2x2 box, the difference of the two products always equals z : the width of the grid. When plotted on a graph (Fig 2.6), the relationship is clearly visible as a perfect positive correlation. A graph is a good choice to show this relationship however, a straight line drawn between these would not be correct because with this particular problem, the values to be inputted into equations must be natural numbers i.e. Integers > 0 .

Using this apparent relationship, it can be assumed that, when a 2x2 box is placed anywhere on the grid, the difference of the two products will be z  for all possible widths. Therefore, the following equation should be satisfied with any real value of a  and any real value of z  where:

( a + z ) is the bottom-left number in the box, because it is always “the grid width ( z )” more than a ;

( a + z  + 1) is the bottom-right number in the box, because it is always “the grid width ( z ) plus 1” more than a .

( a  + 1)( a  + z ) - a ( a  + z  + 1) = z

This means that I predict that with a 2x2 box on a width z  grid, the difference of the two products will always be z , the width of the grid.

1) Where a  = 9, and z = 16

2) Where a  = 72 and z = 17  

The formula works because the “ z ” term is only produced in the expansion of the two brackets on the left of the minus sign, and not from the more simply factorised “ a ( a  + z  + 1)” term. The a 2  term is present on both sides of the minus sign, as are the az  and a  terms. Therefore, they cancel each other out to leave z .

After this justification, it can now be said that for every 2x2 box on a Width z  Grid, the difference of the two products will always be z .

Having done this, I saw that my formula would only work for 2x2 boxes on a Width z  grid. To improve the usefulness of my formula, I wondered what would happen to the difference of the two products if I varied the length of the box i.e. made it a 3x3 or 4x4.

Section 3: “ p  x p ” Box on Width 10 Grid

Throughout this section, the variable p  will be used to represent the length of the square box. The variable a  will continue to be used for the top-left number in the box i.e. the location of the box upon the grid.

Varying values of p will be tested to give different lengths of sides for the boxes. The lengths will range from 3 to 7. With these boxes, in 5 different random locations on the width 10 grid, the differences of the two products will be calculated. Again, I believe 5 calculations are enough to display any patterns.

Fig 3.1 is the width 10 grid, with 3x3 to 7x7 example boxes.

a) Here are the results of the 5 calculations for 3x3 Box  on Width 10  Grid:

b) Here are the results of the 5 calculations for 4x4 Box  on Width 10  Grid:

c) Here are the results of the 5 calculations for 5x5 Box  on Width 10  Grid:

d) Here are the results of the 5 calculations for 6x6 Box  on Width 10  Grid:

e) Here are the results of the 5 calculations for 7x7 Box  on Width 10  Grid:

From tables (a)-(e), it is possible to see that all the differences of the products tested are multiples of 10. It also seems that there is a pattern, where the difference equals “the length of the box minus 1” squared multiplied by 10.

Using this apparent rule, it can be assumed that for all possible locations of a box of side p  on a width 10 grid, the difference of the two products is always “the length of the box minus 1” squared, multiplied by 10. Therefore the following equation should be satisfied with any real value of a , and any real value of p  where:

( a  + [ p  – 1]) is the top-right number in the box, because it is always “[ p  – 1] more” than a ;

( a + 10[p – 1]) is the bottom-left number in the box, because it is always “10 multiplied by [p – 1]” more than a ;

( a  + 10[ p  – 1] + [ p  – 1]) is the bottom-right number in the box, because it is always “[p – 1] plus 10 multiplied by [ p  – 1]” more than a .

For simplicity, d  = [ p  – 1]:

( a  + d )(a + 10 d ) - a ( a  + 10 d + d ) = 10 d 2

          a + 10[p - 1]       a + 10[p - 1] + [ p - 1]

This means that I predict that with a p  x p  box on a width 10 grid, the difference of the two products will always be 10[ p  – 1] 2 ,  “the length of the box minus 1” squared, multiplied by 10.

1) Where a  = 12, and p =   8 : I predict 10 x [ p  – 1] 2  = 10 x 49 = 490

2) Where a  = 2, and p =   9 : I predict 10 x [ p  – 1] 2  = 10 x 64 = 640

The formula can be proven to work with the following algebra (where d = [ p  – 1]):

The formula works because the “10 d 2 ” term is only produced in the expansion of the two brackets on the left of the minus sign, and not from the more simply factorised “ a ( a  + 10 d + d )” term. The a 2  term is present on both sides of the minus sign, as is the 11 ad term. Therefore, they cancel each other out to leave “10 d 2 ” .

After this justification, it can now be said that for every possible p x p box on a Width 10 Grid, the difference of the two products will always be 10( p  – 1) 2 .

Having done this, I saw that my formula would only work for square boxes on a width 10 grid. To improve the usefulness of my formula, I wondered what would happen to the difference of the two products if I varied the length of the box and the width of the box i.e. made it a 2x3, 2x4, 3x5, 3x2, 4x7, 4x9 etc.

Section 4: “ p  x q ” Box on Width 10 Grid

Throughout this section, the variable q  will be used to represent the length of the box extending down   the grid and the variable p  will be used to represent the length of the box extending across the grid. The variable a  will continue to be used for the top-left number in the box i.e. the location of the box upon the grid.

Random varying values and combinations of p  and q     will be tested to give different lengths and widths of sides for the boxes. The lengths will range from 2 to 5. With these boxes, in 5 different random locations on the width 10 grid, the differences of the two products will be calculated. Again, I believe 5 calculations are enough to display any patterns.

Referring back to Fig 3.1, it shows examples of the varying box sizes. The only difference is, the boxes may now be made into rectangles.

a) Here are the results of the 5 calculations for 2x4 Box  on Width 10  Grid:

b) Here are the results of the 5 calculations for 3x6 Box  on Width 10  Grid:

c) Here are the results of the 5 calculations for 5x3 Box  on Width 10  Grid:

From ‘Section 3’, I proved the formula, “Difference = 10( p  – 1) 2 ”. If ( p – 1) is the variable representing the side of the square, then the area calculated from that is ( p  – 1) 2 , or ( p  – 1)( p  – 1). Because each of these brackets represents the side of the square, when it becomes a rectangle, it is safe to assume that the area will be:

(p – 1)(q – 1)

Bearing this in mind whilst analysing this data helped me realise that the difference of the two products, say in a 5x3 box, was 4x2x10 = 80, and in a 3x6 box, was 2x5x10 = 100. The relationship was such that the difference was the “length of the box minus 1”, times the “width of the box minus 1” times 10. I noticed the “10” from, again, all the differences being multiples of 10.

Using this apparent relationship, it can be assumed that for all possible locations of a box of one side p , and the other q , on a width 10 grid, the difference of the two products is always “the length of the box minus 1” times the “width of the box minus 1” multiplied by 10. Therefore the following equation should be satisfied with any real value of a , any real value of p , and any real value of q  where:

( a + 10[ q  – 1]) is the bottom-left number in the box, because it is always “10 multiplied by [ q  – 1]” more than a ;

( a  + 10[ q  – 1] + [ p  – 1]) is the bottom-right number in the box, because it is always “10 times [ q  – 1]” plus      “[p – 1]” more than a .

For simplicity, d  = [ p  – 1], and e  = [ q  – 1]:

( a  + d )(a + 10 e ) - a ( a  + 10 e + d ) = 10 de

   a + 10[ q  – 1]                 a + 10[ q  - 1] + [ p - 1]

This means that I predict that with a p  x q  box on a width 10 grid, the difference of the two products will always be 10( p  – 1)( q  – 1),  “the length of the box minus 1”  multiplied by “the width of the box minus 1”, multiplied by 10.

1) Where a  = 12,   p =   4, and q = 5 : I predict 10 x [ p  – 1][ q  – 1] = 10 x 3 x 4 = 120

2) Where a  = 16, p =  6 and q  = 7 : I predict 10 x [ p  – 1][ q  – 1] = 10 x 5 x 6 = 300

The formula can be proven to work with the following algebra (where d = [ p  – 1], and e  = [ q  – 1]):

The formula works because the “10 de ” term is only produced in the expansion of the two brackets on the left of the minus sign, and not from the more simply factorised “ a ( a  + 10 e + d )” term. The “ a 2 ” term is present on both sides of the minus sign, as is the “10 ae ”   term and the “ ad ” term. Therefore, they cancel each other out to leave “10 de ”.

After this justification, it can now be said that for every possible p x q box on a Width 10 Grid, the difference of the two products will always be 10( p  – 1)( q  – 1).

Having done this, I saw that my formula would only work for boxes on a width 10 grid. To improve the usefulness of my formula, I wondered what would happen to the difference of the two products if I varied the length of the box, the width of the box and  the grid width.

Section 5: “ p  x q ” Box on Width z Grid

Throughout this section, the variable q  will be continue to be used to represent the length of the box extending down   the grid and the variable p  will continue to be used to represent the length of the box extending across the grid. The variable z  will continue to be used for the width of the grid and the variable a  will continue to be used for the top-left number in the box i.e. the location of the box upon the grid.

Varying values and combinations of p  and q     will be tested to give different lengths and widths of sides for the boxes. Simultaneously, varying values of z will be tested to give different widths of grids. With these boxes, in 5 different random locations on the width z  grid, the differences of the two products will be calculated. Again, I believe 5 calculations are enough to display any patterns.

Referring back to Figs 2.1 to 2.5, these will continue to be used as examples of grid widths 11 to 15. Fig 3.1 shows examples of the varying box sizes, except that they can be rectangular.

a) Here are the results of the 5 calculations for 2x4 Box  on Width 11  Grid:

b) Here are the results of the 5 calculations for 3x6 Box  on Width 13  Grid:

c) Here are the results of the 5 calculations for 5x3 Box  on Width 14  Grid:

As found in ‘Section 4’, (p – 1)(q – 1 ) is the formula for an area relating to the box. When the grid width was 10, as in the last section, the formula for the difference of the two products was 10 (p – 1)(q – 1) . Being 10, I guessed that the ‘10’ referred to the grid width.

As it happened, I noticed that all the differences tested in this section were multiples of their respective grid widths i.e. 112 is a multiple of 14 and 33 is a multiple of 11. I also noticed that the (p – 1)(q – 1 ) was still correct. The “length of the box minus 1”, times the “width of the box minus 1”, times the “grid width” equals the difference.

Using this apparent relationship, it can be assumed that for all possible locations of a box of one side p , and the other q , on a width z  grid, the difference of the two products is always “the length of the box minus 1” times the “width of the box minus 1” times the “grid width”. Therefore the following equation should be satisfied with any real value of a , any real value of p  , any real value of q , and any real value of z  where:

( a + z [ q  – 1]) is the bottom-left number in the box, because it is always “ z  multiplied by [ q  – 1]” more than a ;

( a  + z [ q  – 1] + [ p  – 1]) is the bottom-right number in the box, because it is “ z  multiplied by [ q  – 1]” plus             “[ p  – 1]” more than a .

( a  + d )(a + ze ) - a ( a  + ze   + d ) = zde

    a + z [ q  – 1]                   a + z [ q  - 1] + [ p - 1]

This means that I predict that with a p  x q  box on a width z  grid, the difference of the two products will always be z ( p  – 1)( q  – 1),  “the length of the box minus 1”  multiplied by “the width of the box minus 1”, multiplied by the “grid width”.

1) Where a  = 12,   p =   4, q = 5, and z  = 17 : I predict z  x [ p  – 1][ q  – 1] = 17 x 3 x 4 = 204

2) Where a  = 16, p =  6, q  = 7, and z  = 23 : I predict z  x [ p  – 1][ q  – 1] = 23 x 5 x 6 = 690

After this justification, it can now be said that for every possible p x q box on a Width z  Grid, the difference of the two products will always be z ( p  – 1)( q  – 1).

Investigation Conclusion and Evaluation

From the simple study of a 2x2 square box on a width 10 grid, I have been able to progress all the way to the formula to find the difference of the two products on any rectangular box on any width grid. The 2x2 box given in the project brief was useful because, now, I can see that it gives the number “1” in ( p  – 1)( q  – 1). This is, therefore, very useful for spotting patterns i.e. the differences with different grid widths. However, when advancing into more complex areas of the project, like the last section where three different variables were altered, spotting patterns became more difficult and knowledge from the prior sections was required to find the formula.

Overall, the best way of presenting the results was in tables because there was an obvious pattern of values in the “Difference” column, because they were all in one line. A graph or diagram would not have been as suitable because the patterns would not have been as apparent. However, a graph was useful in proving that the grid width has a bearing on the difference with a 2x2 box. It showed a perfect positive correlation which meant that the formula was visibly true for all real grid width values. (N.B. “Real”, in this investigation, meant a number that could be used within the problem itself – as it happens, these were natural numbers.)

As a result of this investigation, I can now present a formula which has been proven to work for all real values inputted:

z ( p  – 1)( q  – 1)

Document Details

• Word Count 5993
• Page Count 13
• Subject Maths

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Year 9 Maths Skills for GCSE Science

Subject: Physics

Age range: 14-16

Resource type: Lesson (complete)

Last updated

22 August 2024

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