1. Pathways
  2. Freedom and Constraints
  3. A Trip to the Cinema
  4. Cinema Surprises

Cinema Surprises

strip of cinematic film
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:

Adults £4.00
Pensioners £1.00
Children £0.50

What if the prices are:

Adults £5.00
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?

Comments

Submitted by Rebecca on Fri, 10/02/2015 - 16:14

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?

ajk44's picture

Submitted by ajk44 on Mon, 10/05/2015 - 09:15

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...

Submitted by michael on Tue, 11/10/2015 - 21:46

or you could just say that there are 200 children

Submitted by cfg21 on Thu, 11/12/2015 - 13:53

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

Submitted by Suzanne on Mon, 11/30/2015 - 09:43

One solution!
10 adults - £40
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...

Submitted by dp419 on Mon, 11/30/2015 - 10:03

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?

Submitted by Anonymous on Wed, 05/18/2016 - 20:16

5 adults - £20
65 pensioners - £65
30 children - £15

£15 + £65 + £20 = £100

Submitted by Lucas svyrydenko on Mon, 02/13/2017 - 15:17

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

Submitted by Asm Hotmail on Wed, 05/18/2016 - 20:39

2 adults - £8
86 pensioners - £86
12 children - £6

£8 + £86 + £6 = £100

Submitted by Asm Hotmail on Wed, 05/18/2016 - 20:55

1 Adult
93 Pensioners
6 Children

2 Adults
86 Pensioners
12 Children

3 Adults
79 Pensioners
18 Children

4 Adults
72 Pensioners
24 Children

10 Adults
30 Pensioners
60 Children

Pattern:
Adults: +1
Pensioners: -7
Children: +6

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

Submitted by cfg21 on Thu, 05/19/2016 - 11:02

Can you explain why the pattern works?

Submitted by Asm Hotmail on Wed, 05/18/2016 - 21:20

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

Adult Price: £1.75
Children Price: 75p

Adult Price: £2.50
Children Price: 50p

Adult Price: £3.25
Children Price: 25p

The formula I found was:

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

Submitted by cfg21 on Thu, 07/28/2016 - 15:07

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.

Submitted by Qiyi Choong , C... on Wed, 06/15/2016 - 07:36

We found a set of prices with only one solution:

£7-adult
£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.

Submitted by Lydia&Kelly on Wed, 06/15/2016 - 15:41

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

Adults £6
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.

Submitted by Yurim Bae, Yuen... on Wed, 06/15/2016 - 23:43

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

Adult Children Pensioners
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.

Submitted by Terrence Thompson on Tue, 11/29/2016 - 04:27

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.