[X] Privacy Policy + T&C We use cookies to personalise our content and ads, and for traffic analysis. Information about your use of our site is shared with our advertising and analytics providers. You also agree to our T&C.

As the auditor for my local theme park, I noticed that on Saturday there were 4296 children and 2143 adults and the takings were £98,718.

However, on Sunday, there were 5146 children and 2807 adults and the takings were £122,570.

How much were the adults tickets and children's tickets?

This is easily solved using a common method for solving simultaneous equations.

First construct two algebraic equations, where C is the number of children, and A is the number of adults:

4296C + 2143A = 98718              [1]
5146C + 2807A = 122570             [2]

To make the number in front of C the same on both we multiple [1] by 5146 and [2] by 4296 to give:

22107216C + 11027878A = 508002828  [3]
22107216C + 12058872A = 526560720  [4]

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

As required. QED.

Back to the puzzles...

Top 10 Illusions

 Shadow Illusion Level Balance? Same Length Red Lines? Spinning Dancer Purple Nurple