Cinema problem

We had loads of really great solutions to this problem, so many that unfortunately we can't mention you all by name, but thanks to anyone who submitted a solution.

 

Thank you to Jaclyn from Brighton College for her solution to this problem.

     Solution: 

     3 adults, 47 pensioners, 50 children.

     Steps:

     1. Children can only go up in 10s.

     2. I tried 10 children but it didn't add up to 100.

     2. I tried 20 children but it didn't work.

     2. I tried 30 children with 3 adults and it didn't work.

     2. I tried 40 children and ... it didn't' work.

     3. Finally, I tried 50 children and it worked perfectly with 3 adults and 47 pensioners!

We have a couple of questions about this solution. Firstly, why does the number of children have to be a multiple of 10? Secondly, what do you think Jaclyn meant by 'it didn't work'?

Jaclyn also gave these solutions for the later ticket prices in the problem.

These give us one solution for each set of ticket prices, but are there any more solutions for these?

Here's a solution from Ellie from Hardwick Middle School - she used trial and error, but made in a very logical way.

My solution was that there were $3$ adults, $47$ pensioners and $50$ children. I mainly used trial and error, which is where you pick numbers from an educated guess and see if it works. If it doesn't you then change your answers. I knew that there would be very few adults because their prices were too high at £3.50 and if $28$ adults visited that would take the price to £98 meaning there would not be enough money remaining to have $100$ people in the cinema. So I started with the adults at two because that made a whole number( £6). I then had $50$ pensioners and $48$ children. That equalled $100$ people but £97.80 which was too little. To take the price up I changed $2$ adults to $3$ and $50$ pensioners to $49$. This equalled £100.30 though, which was too much. I changed the $49$ pensioners to $48$ pensioners and $48$ children to $49$. It equalled £100.15 which was £0.15 too much but I had realized that the difference between pensioners and children was £0.15. Therefore I changed $48$ pensioners to $47$ and $49$ children to $50$ which saved £0.15 and then the number of people in the cinema was 100 and ticket sales equalled £100 meaning I had solved the problem.

Abhishek from William Law CofE sent us this solution, which uses the idea of lowest common multiples:

From the question we know that everyone in the cinema hall is not a pensioner. If everyone is a child, the total cost will be too low because $100 \times 0.85 = 85$. If everyone is an adult, the total cost will be too high because $100 \times 3.50 = 350$. On an average, everybody's ticket costs £1 so if there are only pensioners and children, the total cost will be too low. If there are only pensioners and adults, the total cost will be too high. If there are only adults and children, we can find a solution but the answer is not a whole number. Therefore, there are adults, children and pensioners in the cinema hall. For every adult, the price is £2.50 more than the average price. For every child, the price is 15p less than the average price. The LCM of $250$p and $15$p is $750$p, i.e. £7.50. £2.50 $\times3$ is £7.50 and $15$p $\times 50$ is £7.50. Therefore, for every $50$ children, there are $3$ adults.

We can conclude that there are $3$ adults, $47$ pensioners and $50$ children.

9MAT1A from Diocesan Girls School in New Zealand sent us this solution, which takes a similar approach using simultaneous equations:

Initially we have two pairs of simultaneous equations and $3$ variables. If $A$

is the number of adults, $C$ the number of children and $P$ the number of

pensioners then we find:

$$3.5A+0.85C+P=100$$

$$A+C+P=100$$

We realised that the number of pensioners would be equal to the amount the

pensioners paid so really there are only two variables:

$$3.5A+0.85C=100-p=A+C$$

so $$2.5A=0.15C$$

We decided to make the coefficients whole numbers, since $A, C$ and $P$ are all

discrete variables.

$$250A=15C$$

so $$50A=3C$$

We can see then that one solution is $A=3$ and $C=50$. This will force $P = 47$.

All other whole number solutions will be in the ratio $3:50$ adults to children and this means other solutions have $C$ greater than $100$ which cannot be true as there are only $100$ seats in the cinema.

Niharika from Rugby School also investigated other combinations of ticket prices:

$l$ is number of adults

$m$ is number of pensioners

$n$ is number of children

Using prices in pence, if adults pay $400$p, pensioners pay $100$p, children pay $50$p

$$400l+100m+50n=10000$$

(they pay £100 i.e. 10 000p in total)

$$l+m+n=100$$

Multiplying by $400$,

$$400l+400m+400n=40000$$

Subtract this from the first equation:

$$300m+350n=30000$$

so $$m=\frac{600-7n}{6}$$

so $$m=100-\frac{7n}{6}$$

So $n$ is a multiple of $6, 0 \le n \le 100$. Values of $l$ and $m$ can be found for any of these values of $n$.

If adults pay $500$p, pensioners pay $250$p, children pay $50$p

$$500l+250m+50n=10000$$

so $$10l+5m+n=200$$

and $$l+m+n=100$$

Multiply the last equation by $10$ and subtract the first equation:

$$5m+9n=80$$

so $$m=40-\frac{9n}{5}$$

So there are solutions if $n$ is a multiple of $5$ and $0\le n\le 100$.

If we want to find a set of prices so that $25$ adults and $75$ children pay £100 in total

An adult pays $a$p

A child pays $c$p

so $$25a+75c=10000$$

so $$a+3c=400$$

If we assume that $a \ge c$ we get that $a \ge 100$ and $c\le 100$.

If we want to find a set of prices so that there are $40$ adults, $40$ children and $20$ pensioners,

A pensioner pays $p$ p

then $$40a+40c+20p=10000$$

so $$2a+2c+p=500$$

Some examples of when there are many solutions or just one solution:

Going back to earlier situations where children paid $85$p, pensioners paid $100$p and adults paid $350$p, we had

$$m=100-\frac{53n}{50}$$

$n$ must be a multiple of $50$. If $n>50$, $\frac{53n}{50}>100$ so $m<0$, which is impossible. So we only have one possible value of $n$, and one possible solution.

If children paid $50$p, pensioners paid $100$p and adults paid $400$p, we had

$$m=100-\frac{7n}{6}$$

Then we have solutions as long as $n$ is a multiple of $6$, so there are many solutions.

Some more examples:

If adults pay £3.00, pensioners pay £2.00 and children pay £1.50 there are many solutions.

If adults pay £3.50, pensioners pay £1.00 and children pay £0.65 there is only one solution.

Well done!