GCSE: TTotal
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 Level: GCSE
 Questions: 75

Ttotal coursework
5 star(s)is (n19) as it has been decreased by 1. The cell to the right of (n18) is (n17) as it is 1 more than (n18). When these 5 terms are added together I get: (n) + (n9) + (n17) + (n18) + (n19) = 5n  63 The calculation above shows that the sum of the 5 terms within the Tshape is 5n  63, therefore I can make a proper formula: T = 5n  63 where T is the Ttotal and n is the Tnumber Tnumber (n) Ttotal (T) Ttotal using formula (5n63) 20 37 (5x20)= 100 10063 = 37 26 67 (5x26)= 130 13063 = 67 50 187 (5x50)= 250 25063 = 187 80 337
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TTOTALS
4 star(s)* When the TNumber is even, the TTotal is even. I will now find a rule which links the Tnumber with the TTotal: n+(n8)+(n16)+ (n18)+(n17) =5n56 When n=36 =(5x36)56=124 Testing: 19+20+21+28+36=124 As you can see my rule has worked. TTotals  Any sized Grid I will now find the general rule for any sized grid, which links the TNumber with the TTotal. n+(nG)+(n2G) +(n2G1)+(n2G+1) = 5n7G When n=65, and G=10 =(5x65)(7x10)= 255 44+45+46+55+65= 225 As you can see my rule has worked. Translation: If I translate the T 3 Vectors right, it will become: 22+23+24+33+43= 145 25+26+27+36+46= 160 TNumber=43 TNumber=46 TTotal= 145 TTotal=160 * The TTotal has increased by 15.
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Ttotals. I am going to investigate the relationship between the ttotal, T, and the tnumber, n. The tnumber is always the number at the bottom of the tshape when it is orientated upright.
The ttotal, T, can therefore be written in terms of the tnumber, n, as T= 5n  63. Using similar reasoning we can express T in terms of n in an 8×8 and 10×10 grid. Tshapes in an 8×8 grid 1 2 3 4 5 6 9 10 11 12 13 14 17 18 19 20 21 22 25 26 27 28 29 30 33 34 35 36 37 38 41 42 43 44 45 46 n T T=5n56 18 34 5(18)56=37 21 49 5(21)56=52 42 154 5(42)56=67 45 169 5(45)56=172 The ttotal increases by 120 when the tnumber is increased by 24 and by 15 when the tnumber increases by 3.
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Objectives Investigate the relationship between the ttotals and tnumbers. To translate the tshape to different parts of the grid.
(x*5), where x is the number of times you are moving to the right, so if I translate to the right once it would be new Ttotal = current Ttotal + (1x5), Ttotal = Ttotal + 5, if I translate 5 times to the right it would be new Ttotal = current Ttotal + (5*5), Ttotal +25. Proof At T18, if I translated twice to the right, Formula: 'Current Ttotal + (x*5)' x is the number of times you translate to a certain direction, in this case the right.
 Word count: 7061

Relationship between Ttotal and Tnumber I am going to investigate Ttotals and Tnumbers on a 9x9 grid.
So far I have noticed that when the T number increases by one, the T total increases by 5. I predict that if I move the T shape across once more I will get a T total of 52. I am going to do 2 more T shapes to confirm my prediction. T total = 4 + 5 + 6 + 14 + 23 = 52 T number = 23 T total = 5 + 6 + 7 + 15 + 24 = 57 T number = 24 I have tested my prediction and have found that my prediction was right.
 Word count: 986

TShapes Coursework
2 6 10 16 3 9 11 20 4 12 12 24 5 15 13 28 6 18 14 32 4) Data Analysis From the table, it is possible to see a couple of useful patterns: 1) The Sum of the Wing is always 3 times the Middle Number; 2) The Sum of the Tail is always 8 more than the Middle Number; 5) Generalisation It can be assumed that for all possible locations of the 3x1 "T" on the width 8 grid, these patterns will be true.
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Mathematics Coursework  T Shapes
17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 Here is a table showing the results so far for the different grid sizes, tnumbers, ttotals and the vector.
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T shapes. I then looked at more of these TShapes from the grid in sequence and then by tabulating these results I could then work out a formula.
T Number times 5 TTotal 100 37 105 42 110 47 115 52 120 57 125 62 From these results I can now predict that relationship formula is that 5 times the TNumber  63. To help prove that this is true I will now rather than calling it the TNumber I will call it 'n'. This then means that: Formula= 5n  63 n19 n20 n21 n9 n If now use the Formula in the Tshape and use 'n' it is now shown that there are 5 'n' in the Tshape.
 Word count: 1852

Investigate the relationship between the Ttotal and the Tnumber in the 9 by 9 grid.
In this way we have a Tshape where the bottom (T) number is 'n' and from this we can work out the others. The number above 'n' will be 9 less that 'n' itself because of the size of this particular grid in which there are nine numbers in each row. The number above that one will be 'n18', because it will again be 9 less than the last number as the properties of the number square dictate. The two either side of this will have to be 'n19' and 'n17', because of the fact that the number square goes up in units of one.
 Word count: 1853

T Total and T Number Coursework
N=25 30 35 5 5 The formula for this is Tn535 These are the formulas to work out the ttotals on the grid sizes that they are written next to. The Different Formula for each Grid After investigating on the five different grid sizes I have found five different formulas. I shall use these different formulas to find a general formula for the Ttotal of any grid size. So far the formulas that I have found are; 9x9 Grid; T=5n63 8x8 Grid; T=5n56 7x7 Grid; T=5n49 6x6 Grid; T=5n42 5x5 Grid: T=5n35 Now that I have found the individual formulas I shall check that they are correct using a different T number of the particular grid in question.
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In this section there is an investigation between the ttotal and the tnumber.
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 We all ready know the answer to the one in red. To work out the one in green all we have to do is work out the difference in the tnumber and in this case it is 54.
 Word count: 4144

For my investigation, I will be investigating if there is a relationship between ttotal and tnumber. I will first try to find a relationship between Tnumber and TTotal on a 9x9 grid then change the variables such as grid size.
56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 First I put the T shape onto my 9x9 grid and translated it right by 1 space each time. As shown above I started on 20 and finished on 25 I then constructed the tale below. TNumber (T) TTotal (N) Difference 20 1+2+3+11+20=37  21 2+3+4+12+21=42 5 22 3+4+5+13+22=47 5 23 4+5+6+14+23=52 5 24 5+6+7+15+24=57 5 25 6+7+8+16+25=62 5 The table above shows the difference between the consecutive TTotals as the TNumber increases by one.
 Word count: 2692

For my investigation, I will be investigating if there is a relationship between ttotal and tnumber. I will first try to find a relationship between Tnumber and TTotal on a 9x9 grid
56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 First I put the T shape onto my 9x9 grid and translated it right by 1 space each time. As shown above I started on 20 and finished on 25 I then constructed the tale below. TNumber (T) TTotal (N) Difference 20 1+2+3+11+20=37  21 2+3+4+12+21=42 5 22 3+4+5+13+22=47 5 23 4+5+6+14+23=52 5 24 5+6+7+15+24=57 5 25 6+7+8+16+25=62 5 The table above shows the difference between the consecutive TTotals as the TNumber increases by one.
 Word count: 2692

Tshapes. In this project we have found out many ways in which to solve the problem we have with the tshape being in various different positions with different sizes of grids.
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 We all ready know the answer to the one in red. To work out the one in green all we have to do is work out the difference in the tnumber and in this case it is 54.
 Word count: 4144

Translate the Tshape to different positions on the grid Investigate the relationship between the Ttotal and the Tnumber
is 12. * The difference between the differences along the bottom row ( 4, 5, 6 ) is also 12. T = (n  19) + (n  18) + (n  17) + (n  9) + n = 5n  63 T total = 5n  63 T = (n  19) + (n  18) + (n  17) + (n  9) + n = 5n  63 T total = 5n  63 I can use these equations to find the ttotals just from knowing the Tnumber. I predict that if the Tnumber is 50 then (5x50)
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TTotal. I will take steps to find formulae for changing the position of the T in many ways using methods such as translation and rotation.
then I will work on 180 degrees * Rotation will be my final part of my investigation * All the way through my work I will be including explanations and diagrams * As well as using explanations of what I am doing, I will explain why I am doing it and why I get the answers I do * I will be stating all the variables and when I add a new variable I will clearly state what it is.
 Word count: 4873

The investiagtion betwwen the relationship of the Tnumber and Ttotal
55 5 26 60 5 27 65 5 28 70 5 From these results I worked out that the Nth term was 5Tn  70. However this only applied a 10 by 10 grid and so if I wanted a formula that applied to any grid then I would have to make all the parts to my formula dependant upon the grid size. For this I needed some new lettering. I needed letters that were dependant on the grid size so I used M to represent the increase in value in one movement down the grid, and R to represent the increase in value in one movement right in the grid.
 Word count: 1988

TShapes Coursework
This is an example TShape that I will be using to show you how to find the TTotal and TNumber, which I took from the above 9x9 grid. 1 2 3 11 20 TTotal: Found, by adding up all the numbers within the Tshape. The sum of these numbers equals the Ttotal. Therefore, for example the Tshape above, the Ttotal would be 1 + 2 + 3 + 11 + 20, so therefore the Ttotal is 37. TNumber: The number 20, which is at the bottom of the above, is referred to as the Tnumber, or Tn.
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T totals. In this investigation I aim to find out relationships between grid sizes and T shapes within the relative grids, and state and explain all generalizations I can find
61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 From these Extra T Shapes we can plot a table of results. TNumber (x) TTotal (t) 20 37 26 67 49 182 50 187 52 197 80 337 From this table the first major generalization can be made, The larger the TNumber the larger the TTotal The table proves this, as the TNumbers are arranged in order (smallest first)
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Ttotal Investigation
The first part of my formula is therefore 5T. It is 5T because that is how many numbers inside the T. I multiplied the difference between the Ttotal which is 5 by the Tno. Underneath is my working out: 22 x 5 =110 23 x 5=115 24 x 5=120 I then took away the Ttotal numbers so that I could find the difference between the Tno and the numbers in each of the 4 squares inside the T. 110 40 =70 115 45 =70 120 50 = 70 I found out that the difference is 70.
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T totals  translations and rotations
The number directly above this is 1 place back in the grid so it is N11= N2. The two remaining numbers in the T shape are N2+9 and N29. Thus the T total is: N+ (N1) + (N2) + (N2+9) + (N29) = 5N63 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56
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Ttotals, Main objective of this project of Ttotals coursework is to find an interrelationship between the Ttotal and the Tnumber.
may be too small for the process of either horizontal or vertical movement of a particular tshape to be restricted. This booklet describes methods and calculations used in effort to complete the main objectives of the GCSE Mathematics Coursework 20072008 (Ttotals). Contents: Introduction. Ttotal, Tnumber.............................................................3 Methods.........................................................................................5 Evaluation of Results.........................................................................8 Introduction. Ttotal, Tnumber. This coursework is about trying to find a connection between the ttotals and tnumber according to the tshape Here is an example of a Tshape drawn on a 9 x 9 number grid 1 2 3 4 5 6 7 8 9 10 11 12 13 14
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In this investigation Im going to find out relationships between the grid sizes and T shapes within the relative grids, and state an explanation to generalize the finding using the TNumber
of 30, and the Ttotal (t) adds up to 87 (11+12+13+21+30). With the second T shape with a T number of 31, the Ttotal adds up to 92, by looking at the two results a trend can be seen therefore suggesting the larger the T number the larger the total. By looking at the TShapes we can plot a table of results. TNumber (n) TTotal (t) 30 87 31 92 32 97 33 102 34 107 By looking at my table of results a pattern can be seen between the TNumber and the TTotal, there's also a relationship between the TNumber and the TTotal because a trend occurs as you move it over different parts of the grid and it gives a ratio of 1:5.
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TTotals (A*) Firstly I have chosen to look at the 9 by 9 grid. I will be taking five tnumbers in a row and investigating the ttotals for them. Once I have completed all five, I will then look for a formula to link those five
The ttotal is 32+23+15+13+14 which will give us 97. Tnumber: 32 Ttotal: 97 Number 4: 14 15 16 24 33 The tnumber in this case will be 33. The ttotal is 33+24+14+15+16 which will give us 102. Tnumber: 33 Ttotal: 102 Number 5: 15 16 17 25 34 The tnumber in this case will be 34. The ttotal is 34+25+15+16+17 which will give us 107. Tnumber: 34 Ttotal: 107 Formula: After investigating the tnumbers from 30 to 34 and comparing them with their ttotals, I have noticed that every time I increase the tnumber by one the ttotal goes up by five.
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I am going to investigate the relationship between the TTotals and Tnumbers when the Tshape is translated in different sizes of grids
it shows us 37 and when skipping one place you get 47 the same thing is done to 42 and skipping a place will give you 52. Which will give me the following pattern: T + 5 N + 1 ratio 1:5 Now we got this information so we can find the formula. I have found a formula which is 5N  63 = T I have worked this formula by TNumber subtracting The T total which would look like this: (N11)
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