As the auditor for my local theme park, I noticed that on Saturday there were 4,296 children and 2,143 adults and the takings were £98,718.
However, on Sunday, there were 5,146 children and 2,807 adults and the takings were £122,570.
How much were the children's tickets and adult's tickets?
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Hint
This is quite a tricky puzzle, and knowledge of algebra would certainly help.
Answer
The children tickets were £14, and the adult tickets were £18.
Reasoning
There are a number of methods for solving this problem, including:
Using a spreadsheet.
Using a computer program.
Using the intersection of lines on a graph.
Using an online equation solver.
Solving simultaneous equations using algebra.
Solving simultaneous equations using inverse matrices.
Here is my solution using simultaneous equations and algebra.
First construct two algebraic equations, where 'c' is the number of children, and 'a' is the number of adults:
[1] 4296c + 2143a = 98718
[2] 5146c + 2807a = 122570
To make the number in front of 'c' the same on both, we multiply [1] by 5146 and [2] by 4296 to give:
[3] 22107216c + 11027878a = 508002828
[4] 22107216c + 12058872a = 526560720
Now we can do [4] – [3] to give:
1030994a = 18557892
Divide throughout by 1030994 so that:
a = 18
Substituting a = 18 in [1] will give:
4296c + 2143 x 18 = 98718
4296c + 38574 = 98718
4296c = 60144
c = 14
Double-Checking
c = 14 and a = 18
4,296 x 14 + 2,143 x 18 = 98,718
and
5,146 x 14 + 2,807 x 18 = 122,570
