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Thank you very much to everyone who submitted solutions to this problem about seating in the cinema. Next time you are watching a film at the cinema, you can try and estimate the number of adults, children and pensioners and then work out the total takings for the screening. How many screenings of this film per day? How many per week? How many cinemas do you think show this film per week? How much is spent in total going to see this film every week (assume all cinemas charge similar prices)? Oh yes, and remember to enjoy the film!

This problem had two parts; the first part was divided into four questions and these provided the building blocks to help you complete the final challenge. The techniques and insights from the first four questions hint at ways to tackle the final challenge. This is common in many mathematical problems; techniques from simpler problems can be applied and perhaps combined to solve more complex questions.

Lina from the Whitby Maths Club submitted a great solution, with clear explanations for Questions A and B:

Question A:The minimum takings when $100$ tickets are sold would be £$50$ - all the tickets would be sold to pensioners.

The second lowest takings would be £$59.50$. That is what the cinema would take if $1$ adult and $99$ pensioners bought the tickets. The next lowest takings would be £$69$ which is what the cinema would take from $2$ adults and $98$ pensioners. And so on, and so on, until the takings reach £$1000$ when $100$ adults buy tickets!

The amount of adults and pensioners that the tickets were sold to on the day the cinema took £$449$ is $42$ adults and $58$ pensioners.

Question B:

The minimum takings would be £$10$, when all the audience were children. The next smallest amount would be £$19.90$ ($1$ adult and $99$ children). Next would be £$29.80$ ($2$ adults and $98$ children). And so on, and so on, and so on...

The number of adults and children there would be if the cinema got £$208$ would be $20$ adults and $80$ children.

Well done also to Max from St. Peter's C.E.V.C Primary, Millija from Charters School, Paul and Elliott from Wilson's School, and Savan from Canon Lane Middle School. These students were among several who submitted correct solutions to Questions A and B.

Joy, Cedlidh, Courtney and Ellie, from The Corsham School used algebra to tackle the questions. Here is their answer to the second part of Question A: The cinema takes £$449$. How many adults and how many pensioners will be there?

We found using algebra and creating two equations helped us to find the answers.Adults and pensioners need to add to $100$ so $a + p = 100$.

Also each adult pays £$10$ so the total money taken from the adults will be $10 \times a$ ($10$a), and each pensioner pays $50$pence so the total money from the pensioners will be $0.50\times p$ ($0.5p$).

The total money needs to add to £$449$ so $10a + 0.5p = 449$

Putting the two equations together we have:

\begin{align}

a+p & =100 \\

10a +0.5 p &=449

\end{align}

By doubling the second equation and then subtracting the first equation we get:

$19a = 798$ so $a = 42$ meaning there must be $42$ adults in the cinema.

In total there are $100$ seats so if there are $42$ adults there must be $58$ pensioners!

Max from St. Peter's C.E.V.C Primary submitted a lovely solution to all of the questions. Here is his answer for Questions C and D:

Question C:

The minimum possible takings were £$10.00$

The next minimum was £$10.40$

To find your next minimum you need to add $40$ pence on the amount of money that you already had.

For the takings to be £$20$, there must have been $25$ Pensioners and $75$ Children.

Question D:

For the takings to be exactly £$60$ I found seven ways....

$6$ Adults and $0$
Pensioners

$5$ Adults and $20$
Pensioners...

...and you keep adding
$20$ Pensioners and taking away one Adult until there are no more
adults.

When children were
included I found $427$ combinations.

As before, if you have
$6$ Adults, you have $0$ Pensioners, and if you have $5$ Adults you
have $20$ Pensioners.

But now you can convert
Pensioners to Children....

You have $20$ Pensioners
when you have $5$ Adults, but you can change Pensioners into
children:

$1$ Pensioner equals $5$
Children

So if you take away one
Pensioner from the $20$ Pensioners you have to add $5$ children on,
so you can have $5$ Adults and $5$ Children and $19$
Pensioners.

You evenually have to
change all $20$ Pensioners so that would give you $21$ different
combinations that include $5$ Adults...

...but then you have to think about
the solutions with $4$ Adults, and $3$ Adults, and so on down to
$0$ Adults.

Before children were
introduced we established that

when there are $4$ Adults
there will be $40$ Pensioners,

for $3$ Adults there will
be $60$ Pensioners,

for $2$ Adults $80$
Pensioners,

for $1$ Adult $100$
Pensioners

and for $0$ Adults $120$
Pensioners.

When we include children,
in each category there is one
more solution than the number of pensioners, so that means
that the number of solutions is

$21 + 41 + 61 + 81 + 101
+ 121 = 427$

A special mention also goes to Elliott for his solution to Questions C and D.

Questions A to D provided a "warm-up" for the final challenge. We received several correct solutions to this final challenge, some using trial and improvement, with others using algebra.

It is not necessary to use algebra to solve this final challenge. Building on the insights from the earlier questions, 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 . 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 pupil by a pensioner, and if too much, replace a
pensioner by a pupil.

Every time you replace a pensioner by a pupil you lose $50\text{p}$
and gain $10\text{p}$ so the amount drops by $40\text{p}$.
Similarly, replacing a pupil by a pensioner means the addition of
$40\text{p}$, so you have to be adrift by a multiple of
$40\text{p}$ to hit the target.

After taking your first estimate, it is easy to work out if the difference between where you are and where you want to be is a multiple of $40\text{p}$ and if 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 $=
£80$

There are $27$ pensioners $= £13.50$

There are $65$ children $= £6.50$

**************

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.

So, a correct answer is....

\begin{equation}8a + 27p + 65c = 100\end{equation}

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:

\begin{equation}a+p+c=100\end{equation}

And they took $£100$:

\begin{equation}10a + 0.5p + 0.1c = 100\end{equation}

I observe that there is a maximum of $9$ adults, and substituting $a$ with numbers from $0$ to $9$, I obtain two equations with two unknowns.

For

\begin{align}p +c &=
100\\5p + c & = 1000\\4p &= 900\\p &= 225 >
100\end{align}

so there is no solution

\begin{align}p +c &=
99\\5p + c &= 900\\4p &= 801\\p &>
100\end{align}

so there is no solution

\begin{align}

p +c& = 98\\

5p + c &= 800\\

4p &= 702\\

p &> 100

\end{align} so there is no solution

\begin{align}

p +c &= 97\\

5p + c &= 700\\

4p &= 603\\

p &> 100

\end{align} so there is no solution

\begin{align}

p +c &= 96\\

5p + c& = 600\\

4p &= 504\\

p &> 100

\end{align} so there is no solution

\begin{align}

p +c &= 95\\

5p + c &= 500\\

4p &= 405\\

p &> 100

\end{align} so there is no solution

\begin{align}

p +c &= 94\\

5p + c & = 400\\

4p & = 306\\

p &= 76.5

\end{align} this is not an integer, so there is no solution

\begin{align}

p +c &= 93\\

5p + c &= 300\\

4p & = 207

\end{align}

$p$ is not an integer, so there is no solution

\begin{align}

p +c &= 92\\

5p + c &= 200\\

4p& = 108\\

p &= 27\\

c& = 65

\end{align}

SOLUTION

\begin{align}

p +c &= 91\\

5p + c & = 100\\

4p & = 9

\end{align}

$p$ is not an integer, so there is no solution

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

Other students submitting correct solutions to this final challenge included: Max, from St. Peter's C.E.V.C Primary, a year nine class from The Grange, Christchurch, Nick, Chris, Bill, Emma from Edgworth Primary School, Savan from Cannon Lane Middle School, Rajeev from Fair Field Junior School, and Elliott from Wilson's School.

Alex from Stoke-on-Trent Sixth Form College submitted this very thorough solution. Note that this is more complex than you would normally be expected to do at Stage Three.

Well done to everyone!