# Mathematics Coursework : The Phone Box Problem

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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.

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

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.

This student written piece of work is one of many that can be found in our GCSE Pay Phone Problem section.

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