• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  • Level: GCSE
  • Subject: Maths
  • Word count: 2735

T-Total Investigation

Extracts from this document...

Introduction

Daniel Hughes 4C        T-Total Investigation        28/04/07

T-Total Investigation

        I am going to look at and investigate the relationship between the T-total, R-number and grid size of a T-shape on a numbered grid.

  1. Investigate the relationship between the T-total and the T-number drawn on a 9 by 9 number grid.

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

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

        There is a possibility of 196 different translations of T-shape on this grid.  I am going to investigate on this grid the T-shape shown above.

        Here is a table with T-totals from the grid above:

T-number

T-total

20

(20+11+1+2+3) 37

21

(21+12+2+3+4) 42

22

(22+13+3+4+5) 47

23

(23+14+4+5+6) 52

24

(24+15+5+6+7) 57

        To find a relationship between the T-total and T-number I used the difference method.

T-TotalT-numberDifference

20                37                5        

21                42                5

22                47                5

23                52                5

24                57                

T = T-total

        The reason why the difference increases by 5 each time, is because there is 5 numbers in the T.  As the T moves across the each number increases by one.  Five times each one equals 5.

T = 5n – X

   = (5 x 20) - X = 100 – X = 37

   = 63

So, 5n – 63 is the formula to work out the t-total for this grid.

Eg.        Formula = 5n – 63

                  = (5x24) - 63

                  = 120 – 63

                  = 57

24 + 15 + 5 + 6 + 7 = 57

        Formula = 5n – 63

                  = (5x50) – 63

                  = 250 – 63

                  = 187

50 + 41 + 31 + 32 + 33 = 187

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

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

        The T though can be translated to different positions.

Eg.

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

        In different positions though the formula does not work.

Eg.                = 5n – 63

                = (5x4) – 63

                = 20 – 63

                = -43

...read more.

Middle

                  = 285 + 7

                  = 292

57 + 58 + 59 + 50 + 68 = 292

        I predict to work out the reverse of the sideways T in the opposite direction, you have to change the formula to 7 from –7.

Eg.        Formula = 5n – 7

                  = (5x12) – 7

                  = 60 – 7

                  = 53

12 + 11 + 10 + 1 + 19 = 53

         Formula = 5n –7

                  = (5x16) –7

                  = 80 – 7

                  = 73

16 + 15 + 14 + 5 + 23 = 73

        Formula = 5n – 7

                  = (5x68) – 7

                  = 340 – 7

                  = 333

68 + 67 + 66 + 57 + 75 = 333

  1. Uses grids of different sizes.  Translate the T– shape to different positions.  Investigate relationships between the T-total, the T-numbers and the grid size.

Here is a grid 3 by 3:

1

2

3

4

5

6

7

8

9

        I am going to concentrate on the original translation of the T.  The T-total of 8 in this grid is 19.  Adding another column (to make it 3 by 4) will give me two T’s to do the difference method with.

1

2

3

4

5

6

7

8

9

10

11

12

T-numberT-total        Difference

10                22                5n

11                27                5n

        5n = (5x10)

             = 50 –22 = 28

                        = 5n – 28

Eg.        Formula = 5n – 28

                  = (5x11) – 28

                  = 55 – 28

                  = 27

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

T-numberT-total        Difference

12                25                5n

13                30                5n

14                35                5n

        The difference between the T-totals is always 5n.

        5n – x = (5x12) – x

                   = 60 – x

              25 = 60 – x

                x = 35

                   = 5n – 35

Eg.        Formula = 5n –35

                  = (5x14) – 35

                  = 70 – 35

                  = 35

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

T-numberT-totalDifference

14                28                5n

15                33                5n

16                38                5n

17                43                5n

        It is 5n again.

5n – X = (14x5) –X

        = 70 – X

       28 = 70 – X

        X = 42

        = 5n – 42

Eg. Formula = 5n – 42

                 = (5x17) – 42

                 = 85 – 42

                 = 43

I noticed a pattern in the formulas.

3x4        5n – 28        7

3x5        5n – 35        7

3x6        5n - 42         7

        The difference between the number you take away increases by 7 every time.

...read more.

Conclusion

22 – 4 = 18x10 = 180

13 – 4 = 9x10 = 90

        What I was trying to do was to develop a formula from the first formula with 3 rows and T-number 4.  For example the number 90, + 33 = 123, which is the number taken away for the formula for T-number 13.  180 added to 33 is 213, which is the number taken away from 15n, for the T-number 22.  This has to be expressed into a formula, so here is.

15n – (10(n - 4) + 33)

        Formula = 15n – (10(n - 4) + 33)

                  = (15x7) – (10(7 – 4) + 33)

                  = 105 – (10x3 + 33)

                  = 105 (30 + 33)

                  = 105 – 63

                  = 42

7 + 8 + 9 + 6 _+ 12 = 42

        Formula = 15n – (10(n - 4) + 33)

                  = (15x13) – (10(13-4) + 33)

                  = 195 – (10x9 + 33)

                  = 195 – 123

                  = 72

13 + 14 + 15 + 12 + 18 = 72

        Formula = 15n – (10(n - 4) + 33)

                  = (15x22) – (10(22-4) + 33)

                  = 330 – (10x18 + 33)

                  = 330 – (180 + 33)

                  = 330 – 213

                  = 117

22 + 23 + 24 + 21 + 17 = 117

         As the formula doesn’t change when a column is added, this is the general formula for this sideways T.

Eg.

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

        Formula = 15n – (10(n – 4) + 33)

                  = (15x7) – (10(7 - 4) + 33)

                  = 105 – (10x3 + 33)

                  = 105 – (30 + 33)

                  = 105 – 63

                  = 42

7 + 8 + 9 + 3 + 15 = 42

        Formula = 15n – (10(n – 4) + 33)

                  = (15x16) – (10(16 - 4) + 33)

                  = 240 – (10x12 + 33)

                  = 240 –153

                  = 87

16 + 17 + 18 + 12 + 24 = 87

        Formula = 15n – (10(n – 4) + 33)

                  = (15x27) – (10(27 - 4) + 33)

                  = 405 – (10x23 + 33)

                  = 405 – ( 230 + 33)

                  = 405 – 263

                  = 142

27 + 28 + 29 + 23 + 35 = 142

        I therefore predict if you reverse the T round the other way that the formula will be, 15n + (n + 4)-10 + 33).

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

        Formula = 15n + (n + 4)-10 + 33)

                  = (15x9) + (9 + 4)-10 + 33)

                  = 135 + (13x-10 + 33)

                  = 135 + (-130 + 33)

                  = 135 – 97

                  = 38

9 + 8 + 7 + 1 + 13 = 38

        Formula = 15n + (n + 4)-10 + 33)

                  = (15x30) + (30 + 4)-10 + 33)

                  = 450 + (34x-10 + 33)

                  = 450 + (-340 + 33)

                  = 450 – 373

                  = 72

30 + 29 + 28 + 22 + 34 = 143

        Formula = 15n + (n + 4)-10 + 33)

                  = (15x22) + (22 + 4)-10 + 33)

                  = 330 + (26x-10 + 33)

                  = 330 + (-260 + 33)

                  = 330 – 227

                  = 103

22 + 21 +14 + 20 + 26 = 103

...read more.

This student written piece of work is one of many that can be found in our GCSE T-Total section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE T-Total essays

  1. T-Total Investigation

    T=5v-18 We can even say that; T=5v-2g As 9 is the grid size, and all numbers in our relationship grid were related around 9, so we can concluded that it is related to the grid size. We can test this on different positions on the 9x9 grid.

  2. Objectives Investigate the relationship between ...

    17 24 25 26 33 34 35 SUM method: 15+16+17+25+34=107 Algebraic Formula (5n-63): 5x34-63 = 107 * T34 90� Rotation 25 26 27 34 35 36 43 44 45 27+36+45+35+34=177 T-shape T-total Increment T34 107 T34 (90�) 177 +70 We also have an increment of '+70', therefore we know that,

  1. T-Totals (A*) Firstly I have chosen to look at the 9 by 9 grid. ...

    Checking: I will now be checking if the formula I have come up with was correct, I will do this by choosing a random number on the 10 by 10 grid and substituting the t-number into the equation and coming up with the t-total without adding up the rest of the numbers.

  2. Maths Coursework T-Totals

    This will also work in reverse, as we can find any T-Number's value at any time by using v-g. Translations Vertical Again, we shall use our standard gird size and position to establish our basic starting point; 1 2 3 4 5 6 7 8 9 10 11 12 13

  1. T-Totals. Firstly I am going to do a table of 5 x 5 and ...

    + 3 + 4 + 8 + 13 = 30 I will continue to place the T-shape in different location and once I have enough data I will put it into a table. The T-number is 14 3 + 4 + 5 + 9 + 14 = 35 T-total =

  2. T-Total.I aim to find out relationships between grid sizes and T shapes within the ...

    + ( 2 x 5 ) 25 (55 - 30) 8 30 t = (5 x 8) + ( 2 x 5 ) N/a From this we can see that 25 is the "magic" number for vertical translations by +1 on a grid width of 5, from this I can see a link with the "magic"

  1. The aim of this investigation is to find a relationship between the T-total and ...

    I am going to multiply the first T-number in the table by 5, which gives me 50. I am now going to subtract 50 by 22, this gives me 28. The sum I have come up with by doing these calculations is: Ttot = 5Tn - 28 Test: I am

  2. Mathematics Coursework - T-totals

    Then I found out that five (that is the number of numbers in a T-shape) times the T-number minus the sum of each of the other number in the T-shape subtracted from the T-number gave you the T-total. From this I could derive a formula.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work