It is not necessary to use algebra to solve this problem. A little time looking at possibilities by substituting values for the number of adults and looking at possible combinations of pensioners and children will generate the solutions and give an insight into what is happening. This is really worth looking at, especially for the first part of the problem. Some solutions to the first part of the question were sent in from Jonathan, Garvin, Isabelle, Carmen and Kimberley from Waverley Christian College in Melbourne, Australia as well as several other anonymous entries. I have used these to form the basis of the first section below. After this I have included the thoughts of Jack and Robbie of Waverley Christian College, who begin to examine the algebra and finally a solution from Andrei of school 205 Bucharest.

10 adults would use all the £100 but note there would only be 10 in the audience.

With 9 adults you would have £90 and have 91 places left. 91 pupils would raise £9.10 so you would need some pensioners:
9 adults 90 pupils and 1 pensioner gives £99.50 (too little)
9 adults 89 pupils and 2 pensioners gives £99.90 (too little)
9 adults 88 pupils and 3 pensioners gives £100.30 (too much).
So it is not possible to have 9 adults.

Start with 8 adults and share the rest of the 100 tickets between the pensioners and the pupils. If you have too little money replace a pensioner by a pupil and if too much vice versa. Every time you replace a pensioner by a pupil you lose 50p and gain 10p so the amount drops by 40p, similarly replacing a pupil by a pensioner means the addition of 40p, so you have to be adrift by a multiple of 40p for this to work.

By taking your first estimate - it is clear that the difference between where you are and where you want to be is a multiple of 40p and therefore it is possible to keep replacing pensioners by pupils (or the other way around) until you reach the total of exactly £100.

There are 8 adults = 80 pounds
There are 27 pensioners = 13.50 pounds
There are 65 children = 6.50 pounds

******

And Jack and Robbie add:

Let 'a' be the adult 's ticket cost.
Let 'p' be the pensioner's ticket cost.
Let 'c' be child's ticket cost.

8a + 37p +55c = 109

8a + 32p + 60c = 102

8a + 28p + 65 = 100.5

8a + 26p + 65c = 99.5

So, a correct answer is....

8a + 27p + 65c = 100

To get exactly one hundred pounds from one hundred tickets, the cinema must admit eight adults, twenty-seven pensioners and sixty-five children.

****

The algebraic solution sent by Andrei is given below.

Use a to represent the number of adults, p the number of pensioners and c the number of children.

For the first situation there are 100 persons:

a + p + c = 100

And they took £100:

10a + 0.5p + 0.1c = 100

I observe that there is a maximum of 9 adults, and substituting _AMP_quot;a_AMP_quot; with numbers from 1 to 9, I obtain two equations with two unknowns.

a = 0

a = 1

a = 2

p +c = 98
5p + c = 800
4p = 702
p > 100 so there is no solution

a = 3

p +c = 97
5p + c = 700
4p = 603
p > 100 so there is no solution

a = 4

p +c = 96
5p + c = 600
4p = 504
p > 100 so there is no solution

a = 5

p +c = 95
5p + c = 500
4p = 405
p > 100 so there is no solution

a = 6

p +c = 94
5p + c = 400
4p = 306
p = 76.5 this is not an integer

a = 7

p +c = 93
5p + c = 300
4p = 207
p this is not an integer

a = 8

p +c = 92
5p + c = 200
4p = 108
p = 27
c = 65
SOLUTION

a = 9

p +c = 90
5p + c = 100
4p = 900
p = 225 > 100 so there is no solution

So, there is only one solution: a = 8, p = 27 and c = 65.

For the second possibility, I have:

a + p + c = 100
10a + p + 0.5c = 100

Again, a must be smaller than 10, so I analyze each possibility:

a = 0

p + c = 100
10p + c = 1000
9p = 900
p = 100
c = 0
SOLUTION

a = 1

p + c = 99
10p + c = 900
9p = 801
p = 89
c = 10
SOLUTION

a = 2

p + c = 98
10p + c = 800
9p = 702
p = 78
c = 20
SOLUTION

a = 3

p + c = 97
10p + c = 700
9p = 603
p = 67
c = 30
SOLUTION

a = 4

p + c = 96
10p + c = 600
9p = 504
p = 56
c = 40
SOLUTION

a = 5

p + c = 95
10p + c = 500
9p = 405 ? p = 45 ? c = 50
SOLUTION

a = 6

p + c = 94
10p + c = 400
9p = 306
p = 34
c = 60
SOLUTION

a = 7

p + c = 93
10p + c = 300
9p = 207
p = 23
c = 70
SOLUTION

a = 8

p + c = 92
10p + c = 200
9p = 108
p = 12
c = 80
SOLUTION

a = 9

p + c = 91
10p + c = 100
9p = 9
p = 1
c = 90
SOLUTION

Here there are 10 solutions.