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  • Level: GCSE
  • Subject: Maths
  • Word count: 2600

Mathematics Coursework : The Phone Box Problem

Extracts from this document...

Introduction

Mathematics Coursework : The Phone Box Problem

Introduction: A woman only has 10 and 20 pence coins to use in a phone box. I have to investigate the number of different ways she could use 10 and 20 pence coins in a phone box. I am also investigating a man who only has 10 and 50 pence coins. I will start off with the woman.
10 pence call
                                             10=10p =1 way
20 pence call
20=10p+10p
                                              20=20p =2 ways
30 pence call
30=10p+10p+10p
                                          30=20p+10p =3 ways
30=10p+20p
40 pence call
40=10p+10p+10p+10p
40=20p+10p+10p
                                      40=10p+20p+10p =5 ways
40=10p+10p+20p
40=20p+20p
50 pence call
50=10p+10p+10p+10p+10p
50=20p+10p+10p+10p
50=10p+20p+10p+10p
                                 50=10p+10p+20p+10p =8ways
50=10p+10p+10p+20p
50=20p+20p+10p
50=20p+10p+20p
50=10p+20p+20p

Pascal´s triangle theory
                                       This should help me to work out long and write out long equations like the ones I am doing

E.g. 50 5 10p =1
                           1 20p and 3 10p=4 =8 ways
2 20p and 1 10p=3

To get this I used the calculator button which has a big c you put the total number of units at the top then you put the number of 20 pence coins at the bottom. You use the biggest one at the bottom I use 20 pence because it is relevant.

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Middle


Investigation 2
In this investigation I am investigating a man who only has 10 and 50 pence coins to use in a phone box. I am going to use Pascal´s triangle to help me in this investigation as well.
 
10 pence call
10=1 10p=
This would equal 1 different combinations

20 pence call
20=2 10p=
This would equal 1 different combinations

30 pence call
30=3 10p=
This would equal 1 different combinations

40 pence call
40=4 10p=
This would equal 1 different combinations

Results
 From these results so far it looks like they are going up by the number 1 so I predict that for a 50 pence call there should be 1 combination. I will now test out this theory by working out how many combinations there are for a 50 pence call.
50 pence call
50=5 10p=1
50=1 50p=1
This would equal 2 different combinations
This is wrong from my prediction so my prediction is wrong. I will now test 60-90 pence calls then look for any pattern.
60 pence call
60=6 10p=1
60=1 50p and 1 10p=2
This would equal 3 different combinations

70 pence call
70=7 10p=1
70=1 50p and 2 10p=3
This would equal 4 different combinations
80 pence call
80=8 10p=1
80=1 50p and 3 10p=4
This would equal 5 different combinations

80 pence call
80=8 10p=1
80=1 50p and 3 10p=4
This would equal 5 different combinations
90 pence call
90=9 10p=1
90=1 50p and 4 10p=5
This would equal 6 different combinations
Results

 
       

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Conclusion


80p=20p+30p+30p =1
There are 4 different combinations
90 pence call
90p=30p+30p+30p =1
90p=30p+20p+20p+20p =1
90p=20p+30p+20p+20p =1
90p=20p+20p+30p+20p =1
90p=20p+20p+20p+30p =1
There are 4 different combinations
Results
    From these if you look my theory only comes in after the 5 term (the 50 pence call). If my theory is correct a 100 pence call should have 7 different combinations.

I will now test my theory for how many combinations there are in a 100 pence call
100 pence call
100p=20p+20p+20p+20p+20p =1
100p=30p+30p+20p+20p =1
100p=20p+30p+20p+30p =1
100p=30p+20p+30p+20p =1
100p=20p+20p+30p+30p =1
100p=30p+20p+20p+30p =1
100p+20p+30p+30p+20p =1
There are 7 different combinations

My prediction was right and there were 7 different combinations for a 100 pence call so my formula is right for two variables:
Tn=(Tn-V1/10)+(Tn-V2/10)
Tn= term number
V= variable
/= Divide
I think that I can adapt my how far you have to go forward until you use the formula I think it would be:
Formula 4
H=V1/10+V2/10
H= how far you have to go back until you can use a formula
V= variable
/= Divide
E.g. 20 and 30 pence
H=V1/10+V2/10
H=20/10+30/10
H=2+3
H=5
Conclusion
          If you use formula 3 and formula 4 together you can work out how far forward you can use your formula and what is the next term.

...read more.

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