# Cinema Surprises In Going to the Cinema, you are invited to fill 100 seats at the cinema and take exactly £100. What if we change the prices?

Can there be 100 people and takings of exactly £100 if the prices are:

Pensioners £1.00
Children £0.50

What if the prices are:

Pensioners £2.50
Children £0.50

Here are some questions you might like to consider:

• How many solutions are there for each set of prices?
• If I can find one solution, can I use it to help me find all the other solutions?
• Can you find alternative sets of prices that offer many solutions? What about exactly one solution?
• If a children's film has an audience of 3 children for every adult (no pensioners), how could the prices be set to take exactly £100 when all the seats are sold?
• What about a family film where adults, children and pensioners come along in the ratio 2:2:1?

### I'm presuming for the first

I'm presuming for the first one there has to be a mixture of ages, or can you just say there would be 100 pensioners paying £1 each? ### It's a solution!

Hi Rebecca, 100 pensioners paying £1 is a valid solution but perhaps not a very interesting one! Looking for other solutions where there are a mixture of ages sounds like a good idea...

### cinema

or you could just say that there are 200 children

### Too many children!

But there are only 100 seats in the cinema so they wouldn't all get in.

### Cinema Surprises

One solution!
30 pensioners - £30
60 children - £30
Total audience 100 – total amount £100
But I can’t see how to get there quickly using a formula. And that would be so helpful...

### Re: Cinema Surprises

That's a good solution - well done!

I think there are some other solutions as well. Finding some of these might help you to work out what connects all of these. What other solutions can anyone find?

### Cinema Surprises

65 pensioners - £65
30 children - £15

£15 + £65 + £20 = £100

### math

That is very well put. Can you show us the answer in a formula?

### Cinema Surprises

86 pensioners - £86
12 children - £6

£8 + £86 + £6 = £100

### Cinema Surprises

93 Pensioners
6 Children

86 Pensioners
12 Children

79 Pensioners
18 Children

72 Pensioners
24 Children

30 Pensioners
60 Children

Pattern:
Pensioners: -7
Children: +6

Formula:
a = 1 to 14 Adults
p = 100 - 7a Pensioners
c = 6a Children

### Cinema Surprises

Can you explain why the pattern works?

### Children's Film

For a children's film, with 3 children per adult and no pensioners, we could have these prices:

Children Price: 75p

Children Price: 50p

Children Price: 25p

The formula I found was:

$n = 1, 2$ or $3$
Adult Price $= 1 + 0.75 \times n$
Children Price $= 1 - 0.25 \times n$

### Children's Film

How did you find that formula? While it does work, you might be able to find another formula that can give you many more solutions.

### Cinema Surprises (a set of prices that has only one solution)

We found a set of prices with only one solution:

£3-pensioners
£0.5-children

We first used trial and error to find out what numbers could work, then we proved it by using a formula.

$7a+3p+0.5c = a+p+c$
$6a+2p = 0.5c$

So:
$7a+3p+6a+2p = 100$
$13a+5p = 100$
$p = 20-\frac{13a}{5}$

$a$ cannot be more than 10 because if we input it in the equation $p = 20-\frac{13a}{5}$, it would give us $p=-6$ and we can't have negative pensioners. So the number that would work would have to be 5 Then we know that the number of adults would be 5 and the number of pensioners would be 7.

So, eventually the answer for the number of people would be: 5 adults, 7 pensioners, and 88 children, and the earnings would be: £35 from adults, £21 from pensioners, and £44 from children.

### MATHS-BISSPUXI

We found a set of prices that only has one solution:

Pensioners £5
Children £0.50

$a+p+c=100$
$6a+5p+0.5c=100$
$12a+10p+c=200$
$c=200-12a-10p$

Substituting:
$a+p+200-12a-10p=100$
$-11a-9p=-100$
$11a+9p=100$
$11a=100-9p$
$a=\frac{100-9p}{11}$

For $p=5$, we find that $a=5$. If we increase $p$, the next $100-9p$ that is divisible by 11 is negative, so we only have one solution. This is 5 adults, 5 pensioners and 90 children.

### Cinema Surprises

Q1
Adult tickets are £5, Pensioners tickets £2.5 and Children tickets £0.5

1 6 93
2 12 86
3 18 78

4 24 72
5 30 65
6 36 68
7 42 51
8 48 44
9 54 37
10 60 30
11 66 23
12 72 16
13 78 9
14 84 2

Formula: A = Number of Adults
6A = Number of Children
100 - 7A = Number of Pensioners

Therefore, we have 14 solutions for this set of prices.

Q2
Adult tickets are £5, Pensioners tickets £2.5 and Children tickets £0.5

$A+C+P = 100$
$5A+0.5C+2.5P = 100 = A+C+P$
$4A+1.5P = 0.5C$

Substituting:
$5A+4A+1.5P+2.5P = 100$
$9A = 100-4P$
$A = (100-4P)÷9$
$4P = 100-9A$
$P = (100-9A)÷4 = 25-(9A÷4)$

Therefore, for 0 adults, we have 25 pensioners. This leaves 75 children, which gives us $0\times 5 +25\times 2.5 + 75\times 0.5 = 100$.
For 4 adults, we get 16 pensioners and 80 children. This gives us $4\times 5 +16\times 2.5 + 80\times 0.5 = 100$.
For 8 adults, we get 7 pensioners and 85 children. This gives us $8\times 5 +7\times 2.5 + 85\times 0.5 = 100$.
For 12 adults, we get -2 pensioners, which we can't have, so there are no more solutions. Therefore, there are 3 solutions for this set of prices.

Q3
Prices with only one solution: Adult tickets are £7, Pensioners tickets £3 and Children tickets £0.5.

$A+C+P = 100$
$7A+3P+0.5C = 100 = A+C+P$
$6A+2P = 0.5C$

Substituting:
$7A+3P+6A+2P = 100$
$13A+5P = 100$
$P = (100-13A)÷5 = 20-(13A÷5)$

If $A = 5$, then $P = 20-(13\times 5÷7) = 7$

Therefore, there is only one combination of five Adults, seven Pensioners and 88 Children if Adult tickets are £7, Pensioners tickets £3 and Children tickets £0.5.

### Cinema

The ticket price must average $1 / seat. Since pensioners pay the desired rate, arrange adults to children ratios to average$1/ seat i. e. 1 adult to six children ($7 / 7 seats) . Now the cinema can be filled by any no. of groups of 7 adults and children from 1 to 14 and make up the numbers to 100 with pensioners. If the prices change use the same logic i.e. the average ticket price must equal$1.