# T-Totals. I am going to investigate the relationship between the T-total and the Tnumber. I will use a 9 x 9 grid. I will draw different Tshapes onto the grid.

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Introduction

MATHS COURSEWORK

T-TOTALS

BY EMMET MURPHY 11K

Aim: I am going to investigate the relationship between the T-total and the T–number. I will use a 9 x 9 grid. I will draw different T–shapes onto the grid. Then I will find the T–total and the T–number. I will put the answers into a table of results. I will try to find the relationship between these numbers.

Method: To find the T-total I had to add up all the numbers in the T- shape. For example:

1+2+3+11+20 = 37

2+3+4+12+21 = 42

3+4+5+13+22 = 47

4+5+6+14+23 = 52

5+6+7+15+24 = 57

6+7+8+16+25 = 62

7+8+9+17+26 = 67

10+11+12+20+29 = 82

11+12+13+21+30 = 87

12+13+14+22+31 = 92

13+14+15+23+32 = 97

14+15+16+24+33 = 102

15+16+17+25+34 = 107

16+17+18+26+35 = 112

17+18+19+27+36 = 117

18+19+20+28+37 = 122

Table of Results

T-number | T-total | Difference |

20 | 37 | 5 |

21 | 42 | |

22 | 47 | 5 |

23 | 52 | |

24 | 57 | 5 |

25 | 62 | |

26 | 67 | 5 |

29 | 82 | |

30 | 87 | 5 |

31 | 92 | |

32 | 97 | 5 |

33 | 102 | |

34 | 107 | 5 |

35 | 112 | |

36 | 117 | 5 |

37 | 122 |

The relationship between the T-number and the T-total is:

Middle

5n – 63.

I will substitute 3 terms in for n:

The first term:

(5 x 20) – 63

100 – 63

= 37

The seventh term:

(5 x 26) – 63

130 – 63

= 67

The fifteenth term:

(5 x 36) – 63

180 – 63

= 117

10x10 grid:

1+2+3+12+22 = 40

2+3+4+13+23 = 45

3+4+5+14+24 = 50

4+5+6+15+25 = 55

5+6+7+16+26 = 60

6+7+8+17+27 = 65

7+8+9+18+28 = 70

8+9+10+19+29 = 75

Table of Results:

T-number | T-total | Difference |

22 | 40 | 5 |

23 | 45 | |

24 | 50 | 5 |

25 | 55 | |

26 | 60 | 5 |

27 | 65 | |

28 | 70 | 5 |

29 | 75 |

Analysis of Results: In order to find the formula for the T-shape on a 10x10 grid I took the first T-number, 22 and multiplied it by 5(the difference). This totalled 110. To get the T-total I subtracted 70 from 110 which equalled 40. When put into a formula it looks like this:

5n – 70

I will substitute three numbers in for n:

The first term:

(5 x 22) – 70

110 – 70

= 40

The seventh term:

(5 x 28) – 70

140 – 70

= 70

The eighth term:

(5 x 29) – 70

145 – 70

= 75

Step 2:

Now I will investigate the relationship between the T-number and the T- total using a 10x10 grid and rotating the T- shape 90o clockwise.

Method:

Conclusion

(5 x 2) + 70

10 +70

= 80

The seventh term:

(5 x 8) + 70

40 + 70

= 110

The fifteenth term:

(5 x 16) + 70

80 + 70

= 150

Conclusion: Using the 10x10 grid I have found a pattern in the formulae.

Upright T: 5n – 70

Rotated 90o: 5n + 7

Rotated 180o: 5n + 70

I noticed that all the formulae started with 5n. However for the T-shape rotated 180 o the function sign was opposite to the upright T-shape i.e. 5n – 70 and 5n + 70. Based on this I predict that a T-shape rotated 270o will have the opposite function sign to the T-shape rotated 90o ie. 5n – 7

To test my prediction, I will look at three T-shapes:

1+11+21+12+13 = 58

2+12+22+13+14 = 63

3+13+23+14+15 = 68

(5 x 13) – 7 = 58

(5 x 14) – 7 = 63

(5 x 15) – 7 = 68

From these results I can see that my hypothesis correct.

To conclude my investigation I have shown that the difference is always 5 for a T-total in any grid size. Also when the T-shape is rotated 180o (reflected) the function signs are opposite.

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

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