The woman is going to make a phone call costing any multiple of 10p. I am going to investigate the number of different ways she could put the 10p and 20p coins into the payphone.
Introduction
A pay phone will take only 10p, 20p, 50p, and £1 coins. A woman has plenty of 10p and 20p coins. She has no other coins. She can put the coins into the pay phone in any order.
To make a call costing 50p, she could put in the coins in any order;
20p, 20p, 10p
or 10p, 20p, 20p
or 10p, 10p, 20p, 10p
There are more ways of making 50p with only 10p and 20p coins.
. The woman is going to make a phone call costing any multiple of 10p. I am going to investigate the number of different ways she could put the 10p and 20p coins into the payphone.
A man also wants to use the pay phone. He has plenty of 10p and 50p coins. He has no other coins. He wishes to make a telephone call costing any multiple of 10p.
2. I am going to investigate the number of different ways he has of entering the 10p and 50p coins into the telephone.
3. I will then investigate the more general cases leading into special cases.
There can be made 8 different combinations using only 10p and 20p coins to make a call costing 50p.
These are;
. 10,10,10,10,10,10
2. 10,10,10,20
3. 20,10,10,10
4. 10,20,10,10
5. 10,10,20,10
6. 10,20,20
7. 20,10,20
8. 20,20,10
. The woman is going to make a phone call costing any multiple of 10p. Investigate the number of different ways she could put the 10p and 20p coins into the pay phone.
Multiple of 10p Combinations for 10p, 20p Total no of combinations (Fibonacci sequence) Algebra/Terms
0p 10 1 T1
20p 10,1020 2 T2
30p 10,10,1010,2020,10 3 T3 = T2+T1
40p 10,10,10,10 20,2010,10,2010,20,1020,10,10 5 T4=T3+T2
50p 10,10,10,10,1020,20,1020,10,2010,20,2010,10,10,2020,10,10,1010,20,10,1010,10,20,10 8 T5=T4+T3
Fibonacci sequence
This sequence in which each term is the sum of the two preceding terms is very well known. It is called the Fibonacci sequence after the Italian mathematician who investigated it. The nth term Un, is very complicated and you probably don´t need to know how to find before your A-levels.
Prediction
Looking at the table above, we know that;
30p = T3 = T2+T1
40p = T4 = T3 +T2
50p = T5 = T4+T3
So, 60p´s prediction will be
60p = T6 = T5+T4
Explanation
As every "total no of combinations" of putting the coins into the pay phone are found by adding the 2 previous terms, the result of Tx can be solved by this formula;
Tx = T(x-1)+T(x-2)
Prediction & Explanation
So, for 60p we know it is the equivalent of T6. Let us put this into the formula, to find the "number of ways".
Tx = T(x-1)+T(x-2)
T6 = T(6-1)+T(6-2)
T6 = T5+T4
T6 = 8+5
T6 = 13
Testing
The prediction will be tested as to see whether it will work out.
Multiple of 10p Combinations for 10p, 20p Total no of combinations Algebra/Terms
60p 10,10,10,10,10,1020,20,2010,10,10,10,2020,10,10,10,1010,20,10,10,1010,10,20,10,1010,10,10,20,1010,10,20,2020,10,10,2020,20,10,1010,20,20,1010,20,10,2020,10,2010, 13 T6 = T5+T4
The pattern of the different number of ways a multiple of 10p can be used, using 10p and 20p coins only, is recorded below;
0p- 1
20p- 2
30p- 3
40p- 5
50p- 8
60p- 13
It can also be shown in a dot pattern
0p- .
20p- ..
30p-...
40p- .....
50p- ........
60p- .............
I am now going to use some "imaginary coins". Starting of with 10p and 30p, I will investigate the number of different ways you could put the 10p and 30p coins into the pay phone. The phone call will cost any multiple of 10p.
Multiple of 10p Combinations for 10, 30p Total no of combinations Algebra/Terms
0p 10 1 T1
20p 10,10 1 T2
30p 10,10,10,30 2 T3
40p 10,10,10,1010,3030,10 3 T4 = T3+T1
50p 10,10,10,10,1010,10,3030,10,1010,30,10 4 T5 = T4+T2
60p 10,10,10,10,10,1030,3010,10,10,3030,10,10,1010,30,10,1010,10,30,10 6 T6 = T5+T3
Prediction
Looking at the table at the table above, we know that;
40p = T4 = T3+T1
50p = T5 = T4+T2
60p = T6 = T5 +T3
So, 70p´s prediction will be
70p = T7 =T6+T4
Explanation
As every "total no of combinations" of putting the coins into the pay phone are found by adding the previous term together with the third previous term, the result of Tx can be solved by this formula;
Tx = T(x-1)+T(x-3)
Prediction & Explanation
So for 70p, we know it is the equivalent of T7. Let us put this into the formula, to find the "number of ways".
Tx = T(x-1)+T(x-3)
T7 = T(7-1)+T(7-3)
T7 = T6+T4
T7 = 6+3
T7 = 9
Testing
The prediction will be tested as to see whether it will work out.
Multiple of 10p Combinations for 10p, 30p Total no of combinations Algebra/Terms
70p 10,10,10,10,10,10,1030,30,1030,10,3010,30,3010,10,10,10,3030,10,10,10,1010,30,10,10,1010,10,30,10,1010,10,10,30,10 9 T7 = T6+T4
The pattern of the different number of ways a multiple of 10p can be used, using 10p and 30p coins only, is recorded below;
0p- 1
20p- 1
30p- 2
40p- 3
50p- 4
60p- 6
70p- 9
It can also be shown in a dot pattern
0p- .
20p- .
30p- ..
40p- ...
50p- ....
60p- ......
70p- .........
The ...
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Multiple of 10p Combinations for 10p, 30p Total no of combinations Algebra/Terms
70p 10,10,10,10,10,10,1030,30,1030,10,3010,30,3010,10,10,10,3030,10,10,10,1010,30,10,10,1010,10,30,10,1010,10,10,30,10 9 T7 = T6+T4
The pattern of the different number of ways a multiple of 10p can be used, using 10p and 30p coins only, is recorded below;
0p- 1
20p- 1
30p- 2
40p- 3
50p- 4
60p- 6
70p- 9
It can also be shown in a dot pattern
0p- .
20p- .
30p- ..
40p- ...
50p- ....
60p- ......
70p- .........
The next "imaginary coin" I will be investigating is 40p together with 10p. The number of different ways you could put the two coins into the pay phone is shown in the table on the next page. The phone call will cost any multiple of 10p.
Multiple of 10p Combinations for 10p,40p Total no of combinations Algebra/Terms
0p 10 1 T1
20p 10,10 1 T2
30p 10,10,10 1 T3
40p 10,10,10,1040 2 T4
50p 10,10,10,10,1010,4040,10 3 T5 = T4+T1
60p 10,10,10,10,10,1010,10,4040,10,1010,40,10 4 T6 = T5+T2
70p 10,10,10,10,10,10,10,10,10,10,4040,10,10,1010,40,10,1010,10,40,10 5 T7 = T6+T3
80p 10,10,10,10,10,10,10,1040,4010,10,10,10,4040,10,10,10,1010,40,10,10,1010,10,40,10,1010,10,10,40,10 7 T8 = T7+T4
90p 10,10,10,10,10,10,10,10,1010,40,4040,10,4040,40,1010,10,10,10,10,4040,10,10,10,10,1010,40,10,10,10,1010,10,40,10,10,1010,10,10,40,10,1010,10,10,10,40,10 10 T9 = T8+T5
Prediction
Looking at the table above, we know that;
50p = T5 = T4+T1
60p = T6 = T5+T2
70p = T7 = T6+T3
80p = T8 = T7+T4
90p = T9 = T8+T5
So, £1´s prediction will be
£1 = T10 = T9+T6
Explanation
As every "total no of combinations" of putting the coins into the pay phone are found by adding the previous term together with the forth previous term, the result of Tx can be solved by this formula;
Tx = T(x-1)+T(x-4)
Prediction & Explanation
So for £1, we know that it is the equivalent of T10. Let us put this into the formula, to find the "number of ways"
Tx = T(x-1)+T(x-4)
T10 = T(10-1)+T(10-4)
T10 = T9+T6
T10 = 10+4
T10 = 14
Testing
The prediction will be tested as to see whether it will be solved.
Multiple of 10p Combinations for 10p, 40p Total no of combinations Algebra/Terms
£1 10,10,10,10,10,10,10,10,10,1010,10,40,4040,10,10,4040,40,10,1010,40,40,1010,40,10,4040,10,40,1010,10,10,10,10,10,4040,10,10,10,10,10,1010,40,10,10,10,10,1010,10,40,10,10,10,1010,10,10,40,10,10,1010,10,10,10,40,10,1010,10,10,10,10,40,10 14 T10 = T9+T6
The pattern of the different number of ways a multiple of 10p can be used, using 10p and 40p coins only is recorded below;
0p- 1
20p- 1
30p- 1
40p- 2
50p- 3
60p- 4
70p- 5
80p- 7
90p- 10
£1- 14
It can also be shown in a dot pattern
0p- .
20p- .
30p- .
40p- ..
50p-...
60p- ....
70p- .....
80p- .......
90p- ..........
£1- ..............
2. A man wants to use the pay phone. He has plenty of 10p and 50p coins. He has no other coins. He wishes to make a telephone call costing any multiple of 10p.
Investigate the number of different ways he has of entering the 10p and 50p coins into the telephone.
Having done and tested the formulae for:
0p, 20p Tx = T(x-1)+T(x-2)
0p, 30p Tx = T(x-1)+T(x-3)
0p, 40p Tx = T(x-1)+T(x-4)
I will now start predicting the formula for 10p, 50p.
Prediction
When comparing the three formulae, it is noticeable that the second part of the equation on the right-hand side is decreasing with one number each time.
Eg
Tx = T(x-1)+T(x-2)
Tx = T(x-1)+T(x-3)
Tx = T(x-1)+T(x-4)
So a possible prediction for the formula of 10p, 50p would be;
Tx = T(x-1)+T(x-5)
Testing
For testing the formula to see whether the prediction is right, some stages of work done for the previous three formulae have to be worked.
Multiple of 10p Combinations for 10p, 50p Total no of combinations Algebra/Terms
0p 10 1 T1
20p 10,10 1 T2
30p 10,10,10 1 T3
40p 10,10,10,10 1 T4
50p 10,10,10,10,1050 2 T5
60p 10,10,10,10,10,1010,5050,10 3 T6 = T5+T1
70p 10,10,10,10,10,10,1010,10,5050,10,1010,50,10 4 T7 = T6+T2
80p 10,10,10,10,10,10,10,1010,10,10,5050,10,10,1010,50,10,1010,10,50,10 5 T8 = T7+T3
90p 10,10,10,10,10,10,10,10,1010,10,10,10,5050,10,10,10,1010,50,10,10,1010,10,50,10,1010,10,10,50,10 6 T9 = T8+T4
As we already know the "total no of combinations" you can put the 10p and 50p into a payphone for a call costing 90p, the predicted formula will be tested for this;
Tx = T(x-1)+T(x-5)
T9 = T(9-1)+T(9-5)
T9 = T8+T4
T9 = 5+1
T9 = 6
3) Investigating the more general cases
General Rule for (10p, np)
In the previous pages, I worked out the combinations for; (10p, 20p), (10p, 30p), (10p, 40p) and (10p, 50p) and found the formulae, which are shown below:
Coins for combinations Formulae
0p, 20p10p, 30p10p, 40p10p, 50p Tx = T(x-1)+T(x-2)Tx = T(x-1)+T(x-3)Tx = T(x-1)+T(x-4)Tx = T(x-1)+T(x-5)
From the above table it is clear that there is relationship between the coins for combination and the relevant formulae. The last numbers of the formulae changes to 2, 3, 4 and 5 when we use 20p, 30p, 40p and 50p respectively. This also shows us that when 20 is divided by 10 it equals 2 and 30 divided by 10 equals 3 and so on.
As the 10p is constant throughout the tables, the first part of the formulae (Tx-1) is also remaining constant and 10 divided by 10 equals 1!
From these formulae, it is possible to find the General Rule for (10p, np) where;
n = any multiple of 10 and
np >10p
So General Rule for (10p, np) would be;
Tx = T(x-1)+T(x-n/10)
Testing
In the following example, I am going to test the total no of combinations for 50p made by 10p and 20p coins. T5 = 50p
(10p, np) Tx = T(x-1)+T(x-n/10)
(10p, 20p) T5 = T(5-1)+T(5-20/10)
T5 = T4+T3 (see next page)
T5 = 5+3
T5 = 8 (see page 2)
In the next example, I am going to test the total no of combinations for 60p made by 10p and 40p coins. T6 = 60p.
(10p, 40p) T6 = T(6-1)+T(6-40/10)
T6 = T5+T2
T6 = 3+1
T6 = 4 (see page 6)
Finding the General Rule for (20p, np)
Investigating the total no of combinations you could put 20p, 30p coins into the pay phone.
Phone call costing Combinations for 20p, 30p Total no of combinations Algebra/Terms
20p 20 1 T1
30p 30 1 T2
40p 20,20 1 T3
50p 20,3030,20 2 T4 = T2+T1
60p 20,20,2030,30 2 T5 = T3+T2
70p 20,20,3030,20,2020,30,20 3 T6 =T4+T3
80p 20,20,20,2030,30,2020,30,3030,20,30 4 T7 = T5+T4
90p 30,30,3020,20,20,3030,20,20,2020,30,20,2020,20,30,20 5 T8 = T6+T5
Explanation
As every "total no of combinations" of putting the coins into the pay phone are found by adding the second previous term with the third precious term, the result of Tx can be solved by this formula;
Tx = T(x-2)+T(x-3)
(Note the connection between the 20p and 30p coins with the numbers 2 and 3 respectively in the formula)
Investigating the total no of combinations you could put 20p, 40p coins into the pay phone.
Phone call costing Combinations for 20p, 40p Total no of combinations Algebra/Terms
20p 20 1 T1
30p - - T2
40p 20,2040 2 T3
50p - - T4
60p 20,20,2040,2020,40 3 T5 = T3+T1
70p - - T6 = T4+T2
80p 20,20,20,2040,4020,20,4040,20,2020,40,20 5 T7 = T5+T3
90p - - T8 = T6+T4
£1 20,20,20,20,2020,20,20,4040,20,20,2020,40,20,2020,20,40,2040,40,2020,40,4040,20,40 8 T9 = T7+T5
Explanation
As every "total no of combinations" of putting the coins into the pay phone are found by adding the second previous term with the forth previous term, the result of Tx can be solved by this formula;
Tx = T(x-2)+T(x-4)
(Note the connection between the 20p and 40p coins with the numbers 2 and 4 respectively in the formula)
Predicting the formula for (20p, 50p)
As we know the formulae for (20p, 30p) and (20p, 40p) it is possible to find the formula for (20p, 50p) by predicting.
Coins for combinations Formulae
20p, 30p Tx = T(x-2)+T(x-3)
20p, 40p Tx = T(x-2)+T(x-4)
The connections between the second coins with the second part of the equations are very clear.
Eg 30p - (x-3)
40p - (x-4)
and 20p stays constant - (x-2)
Having compared the formulae for (20p, 30p) and (20p, 40p), the prediction for (20p, 50p) formula will be;
Tx = T(x-2)+T(x-5)
Testing
A short testing will be done to see whether the formula is suitable.
I am going to find the two terms, which will make up the 7th term (80p) in the sequence of (20p, 50p) using the formula above and then test it by making a table.
Tx = T(x-2)+T(x-5)
T7 = T(7-2)+T(7-5)
T7 = T5+T2
Phone call costing Combinations for (20p, 50p) Total no of combinations Algebra/Terms
20p 20 1 T1
30p - - T2
40p 20,20 1 T3
50p 50 1 T4
60p 20,20,20 1 T5
70p 20,5050,20 2 T6 = T4+T1
80p 20,20,20,20 1 T7 = T5+T2
The formula seems to be right!
I can now go on from this, using the table above, to find the total no of combinations for T7;
T7 = T5+T2
T7 = 1+0
T7 = 1
Having worked out the formulae for (20p, 30p), (20p, 40p) and (20p, 50p), I will now start predicting the formulae for (20p, np).
Coins for combinations Formulae
20p, 30p Tx = T(x-2)+T(x-3)
20p, 40p Tx = T(x-2)+T(x-4)
20p, 50p Tx = T(x-2)+T(x-5)
From the above table it is clear that there is relationship between the coins for combination and the relevant formulae. The last numbers of the formulae changes to 3, 4 and 5 when we use 30p, 40p, and 50p respectively. This also shows us that when 30 is divided by 10 it equals 3 and 40 divided by 10 equals 4 and so on.
As the 20p is constant throughout the tables, the first part of the formulae (Tx-2) is also remaining constant and, 20 divided by 10 equals 2!
From these formulae, it is possible to find the General Rule for (20p, np) where;
n = any multiple of 10 and
np >20p
So General Rule for (20p, np) would be;
Tx = T(x-2)+T(x-n/10)
Testing
In the following example, I am going to test the total no of combinations for 60p made by 20p and 30p coins. T4 = 60p
(20p, np) Tx = T(x-2)+T(x-n/10)
(20p, 30p) T4 = T(4-2)+T(4-30/10)
T4 = T2+T1
T4 = 1+1
T4 = 2 (see page 11)
In the next example, I am going to test the total no of combinations for 80p made by 20p and 50p coins. T7 = 80p.
(20p, 50p) T7 = T(7-2)+T(7-50/10)
T7 = T5+T2
T7 = 1+0
T7 = 1 (see page 13)
Predicting the General Rule for (30p,np)
When comparing the General Rule for (10p,np) and (20p,np), it is possible to predict the General Rule for (30p,np)
General Rule for General Rule
(10p,np) Tx = T(x-1)+T(x-n/10)
(20p,np) Tx = T(x-2)+T(x-n/10)
When observing the table above, the two formulae have only one difference. The General Rule for (10p,np) has got (x-1) whereas the General Rule for (20p,np) has got (x-2) instead.
So, when predicting the General Rule for (30p,np), the formula would be;
Tx = T(x-3)+T(x-n/10)
Testing
To test the General Rule for (30p,np), I am going to make some tables to get results from.
Phone call costing Combinations for 30p, 40p Total no of combinations Algebra/Terms
30p 30 1 T1
40p 40 1 T2
50p - - T3
60p 30,30 1 T4
70p 30,4040,30 2 T5 = T2+T1
80p 40,40 1 T6 = T3+T2
90p 30,30,30 1 T7 = T4+T3
I am going to test the General Rule for (30p,np) by taking the 7th term and putting it in the formula ;
Tx = T(x-3)+T(x-n/10)
T7 = T(7-3)+T(7-40/10)
T7 = T(7-3)+T(7-4)
T7 = T4+T3
It seems to be working!
Predicting the Overall Rule for any term.
Having got the General rules for;
(10p,np) Tx = T(x-1)+T(x-n/10)
(20p,np) Tx = T(x-2)+T(x-n/10)
(30p,np) Tx = T(x-3)+T(x-n/10)
The overall Rule would not be much different from these formulae.
As only the first part of the formulae changes for each "coins for combination", the Overall Rule would also need to have changed this part.
The connection between the first coin and the first part of the formulae is obvious.
Eg
(10p,np) Tx = T(x-1)+T(x-n/10)
(20p,np) Tx = T(x-2)+T(x-n/10)
(30p,np) Tx = T(x-3)+T(x-n/10)
This possible tells us that
0/10 = 1
which is a link between the coins and formula.
Also,
20/10 = 2
which is shown as a link
and at last,
30/10 = 3
which is also shown as link between the coins and formula.
If the first coin = m, and
second coin = n
the Overall Rule would possible look like this;
Tx = T(x-m/10)+T(x-n/10)
Testing
I am going to test the Overall Rule by putting in terms from previous tables and then look back to see if it is right.
I am going to start off with the 10p and 20p coins to make a phone call costing 60p. T6 = 60p
(mp,np) = (10p,20p)
Tx = T(x-m/10)+T(x-n/10)
T6 = T(6-10/10)+T(6-20/10)
T6 = T(6-1)+T(6-2)
T6 = T5+T4 (see page 3)
Now I am using 10p and 30p coins to make a phone call costing 40p. T4 = 40p
(mp,np) = (10p,30p)
Tx = T(x-m/10)+T(x-n/10)
T4 = T(4-10/10)+T(4-30/10)
T4 = T(4-1)+T(4-3)
T4 = T3+T1 (see page 4)
Lastly, I am going to use 20p and 40p coins to make a phone call costing 60p.
T5 = 60p
(mp,np) = (20p,40p)
Tx = T(x-m/10)+T(x-n/10)
T5 = T(5-20/10)+T(5-40/10)
T5 = T(5-2)+T(5-4)
T5 = T3+T1 (see page 12)
Finding the General Rules for same coins
Having done the Overall Rule for any term, I am going to investigate the General Rule for same coins.
I am going to start off with 10p and 10p.
Phone call costing Combinations for10p,10p Total no of combinations Algebra/Terms
0p 10 1 T1 = T(1-1)
20p 10,10 1 T2 = T1
30p 10,10,10, 1 T3 = T2
40p 10,10,10,10 1 T4 = T3
50p 10,10,10,10,10 1 T5 = T4
60p 10,10,10,10,10,10, 1 T6 = T5
70p 10,10,10,10,10,10,10, 1 T7 = T6
As all the combinations of the different phone calls are same, a possible General Formula for (10p,10p) would be
Tx = T(x-1)
This is not the only possible formula for (10p, 10p), but I have decided to use this one, as it will be easier to predict the Overall Rule for the same coins afterwards.
Now, I am going to investigate the General Rule for (20p, 20p)
Phone call costing Combinations for20p,20p Total no of combinations Algebra/Terms
20p 20 1 T1
30p - - T2
40p 20,20 1 T3 = T1
50p - - T4 = T2
60p 20,20,20 1 T5 = T3
70p - - T6 = T4
80p 20,20,20,20 1 T7 = T5
As any term is equal to the term, which is the second previous to it, the General Formulae for (20p, 20p) would be;
Tx = T(x-2)
Investigating the General Rule for (30p,30p)
Phone call costing Combinations for 30p, 30p Total no of combinations Algebra/Terms
30p 30 1 T1
40p - - T2
50p - - T3
60p 30,30 1 T4 = T1
70p - - T5 = T2
80p - - T6 = T3
90p 30,30,30 1 T7 = T4
As any term is equal to the term, which is the third previous to it, the General Formulae for (30p, 30p) would be;
Tx = T(x-3)
Finding the Overall Rule for same coins
Having got the General Rules for;
(10p, 10p) T(x-1)
(20p, 20p) T(x-2)
(30p, 30p) T(x-3)
The Overall Rule would be easy to predict.
As only the numbers of the formulae are changing it has a link with the coins.
Eg
0/10 = 1,
20/10 = 2,
30/10 = 3 and so on.
Let the same coins be (np,np).
So the Overall Rule would be;
Tx = T(x-n/10)
Testing
I am going to test the Overall Rule by putting in terms from previous tables and then look back to see if it is right.
I am going to start off with the 10p and 10p coins to make a phone call costing 60p. T6 = 60p
(np,np) = (10p,10p)
Tx = T(x-n/10)
T6 = T(6-10/10)
T6 = T5 (see page 17)
Next I am going to test 30p and 30p coins to make a phone call costing 90p.
T7 = 90p
(np,np) = (30p ,30p)
Tx = T(x-n/10)
T7 = T(7-30/10)
T7 = T4 (see page 17)
Connecting Pascal´s Triangle with Fibonacci sequence (page 2)
In Pascal´s Triangle, each number is the sum of the two numbers immediately above it, left and right.
For example, 2 is the sum of 1 and 1, and 4 is the sum of 3 and 1. Furthermore, the sum of each row equals a power of 2.
For example, the sum of the 3rd row is 4 = 2x2 the sum of the 4th row is
8 = 2x2x2.
The Pascal´s Triangle above shows the connection with the Fibonacci sequence (see page 2). The total no of combinations 1, 2, 3, 5, 8, 13,21 etc. are equal to the Fibonacci sequence and the sum of the rows drawn in the Pascal´s Triangle.
Conclusion
When working through this coursework, I have been able to produce simple, general and overall formulae. In every case, the simple formulae lead me to the general formulae leading to the overall formulae. These formulae helped me to tackle this problem the best way by predicting, explaining and testing throughout the work. I have been able to break the task into smaller tasks; some simple and some special cases and I have also extended my work by researching about Pascal´s triangle to connect this with the problem. I also extended my investigation by going further and researching about the same coins and found an overall rule for this as well. Most of my results have been tabulated in order to make the important data clearly expressed.
By doing this coursework, I have learnt how important it is to understand the basic work before moving towards the more complicated cases. As this is my first coursework, at the beginning it seemed to be difficult to start it off. After having done a couple of cases it suddenly became easier to solve the problems. But when approaching the final stages of work it took some time to think it through as the previous stages of work had to be compared to come to a decision. For example, when predicting the Overall rule for any term, I had to compare the general rules for (10p,np), (20p,np) and (30p,np).
From this coursework I have learnt how to set out my reasons for my conclusions.
I believe that this piece of work has given me an idea of how to approach another coursework, next time when given.