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• Level: GCSE
• Subject: Maths
• Word count: 3047

# GCSE Maths Bad Tomato Investigation

Extracts from this document...

Introduction

## GCSE Maths Bad Tomato Investigation

A problem has emerged involving bad tomatoes. I am going to investigate it and find some solutions to the problem. At first there is a regular tray of good tomatoes. Then suddenly, one goes bad. An hour later, all the ones that the bad tomato is touching go bad. This continues until the whole tray goes bad. The problem gets worse when the trays are stacked on top of each other, but I’ll look at that later.

I’m going to start by investigating a normal square box of tomatoes. It would look like this to start with:

Then let’s say tomato 1 goes bad:

After one hour, tomatoes 1, 2 and 5 will be bad:

Then after another hour tomatoes 3, 6 and 9 will go bad. This will continue until the whole tray of tomatoes is bad:

After looking at this process, a pattern became obvious. This was that the quickest legal route to the furthest tomato away from the original bad tomato would be the number of hours for the whole tray to go bad. Basically, if tomato X goes bad, then the quickest legal route to tomato Y, being the furthest tomato away from tomato X would be the number of hours for the whole tray to go bad. Let’s call this the XY rule. This 4 tomato by 4 tomato tray would take 6 hours to go bad:

The order of tomatoes going bad is illustrated in this table:

Hour (H)              No. of bad toms after that hour (N)               Tomatoes which are bad

0                                        1                                                                                     1

1                                        3                                                                                 1, 2, 5.

2                                        5                                                                           1, 2, 5, 3, 6, 9.

3                                        9                                                        1, 2, 5, 3, 6, 9, 4, 7, 10, 13.

4                                        12                                       1, 2, 5, 3, 6, 9, 4, 7, 10, 13, 8, 11, 14.

5                                        15                          1, 2, 5, 3, 6, 9, 4, 7, 10, 13, 8, 11, 14, 12, 15.

6                                              16                        1, 2, 5, 3, 6, 9, 4, 7, 10, 13, 8, 11, 14, 12, 15, 16.

Middle

9x9                                                          8

11x11                                                     10

There is another obvious pattern in this table, it’s that the time taken for the whole tray to go bad is always size L minus 1. This is because going along diagonally from the bad tomato in the middle, it will always be half of (L - 1) no. of toms away:

Along with the two tomatoes on route from middle to corner, the two below them need to be taken to make a staircase type route from middle to corner. This is a quickest route from bad tom to furthest tom i.e. XY rule, which proves that it takes 4 hours for this tray to go bad with the first bad tom in middle. Therefore, an expression can be drawn to show how long it takes for trays to go bad when a bad tomato is in the middle of an odd numbered L x L tray:   L - 1 = T  .

Now the bad tomato will be on the edge of a square tray. This is more difficult to work out because the tomato could be in the middle of the edge, near the corner but still on the edge or not in the middle or the corner and be in no real place that can be described. So as not to complicate this problem any further I’ll look at a tomato on the edge of a 4x4 tray:

Here, if say, tomato 5 goes bad, it takes 5 hours for the tray to go bad.

Conclusion

Now if the tomatoes were in a box, stacked in a 3-D form, they would effectively have a Z-axis, aswell as an X and Y to contend with, and infect. If a tomato went bad in the corner of a set of 3 3x3 trays stacked on top of each other, it would take 6 hours for the set to go bad:

Hour (H)             No. of tomatoes going bad at hour H (B)   Total No. of bad toms (N)

0                                                  1                                                           1

1                                                  3                                                           4

2                                                  6                                                          10

3                                                  7                                                          17

4                                                  6                                                          23

5                                                  3                                                          26

6                                                  1                                                           1

Now I’ll make a table showing how long it takes other cubed sets of trays to go bad:

Size (LxLxL)              Time taken for all toms to go bad in hours (T)

2                                                      3

3                                                      6

4                                                      9

5                                                     12

6                                                     15

From this the obvious formula for the amount of time for a cube of tomatoes, or stacked set of trays to go bad is 3L-3 = T . It would be very difficult to try to find any other formulas for cubes of fixed size as specificity of the tomato going bad becomes very complex in 3-D.

The last thing which I will look at is an infinitely big 3 dimensional cubed tray with a the first tomato going bad in the middle. It’s growth rate pattern will probably be very similar to a 2-D tray, the 2-D pattern was 4H. Here is a table showing howmany tomatoes were going bad each hour, it’s on the next page:

:

Hour (H)              No. of toms going bad at hour H (B)

1                                                6

2                                               18

3                                               38

4                                               66

5                                              102

The pattern here is in fact very similar to the 2-D infinite square starting at middle one, it’s 4N2+2 = B   . The squared part of the formula is to add an extra dimension to it. Here Concludeth my Investigation.

David Langer 10SM

This student written piece of work is one of many that can be found in our GCSE Bad Tomatoes section.

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